What topics are covered in CFA Level 1 Quantitative Methods?

CFA Level 1 Quantitative Methods covers 7 key areas: Rates & Returns (holding period return, MWRR, TWRR), Time Value of Money (PV, FV, annuities, perpetuities, EAR), Statistics (central tendency, dispersion, skewness, kurtosis), Probability (conditional probability, Bayes' formula, expected value), Portfolio Mathematics (covariance, correlation, portfolio variance), Simulation & Estimation (bootstrap, Monte Carlo, confidence intervals), and Hypothesis Testing (t-test, z-test, Type I & II errors).

How much weightage does Quantitative Methods carry in CFA Level 1?

Quantitative Methods accounts for approximately 6–9% of the CFA Level 1 exam, which translates to roughly 11–16 questions out of 180. It is a foundational topic area that also supports Fixed Income, Derivatives, and Portfolio Management topics.

What is the difference between MWRR and TWRR in CFA?

MWRR (Money-Weighted Rate of Return) is the IRR of all cash flows including deposits and withdrawals — it reflects the investor's actual return. TWRR (Time-Weighted Rate of Return) eliminates the effect of external cash flows and measures the portfolio manager's performance. CFA Institute recommends TWRR for evaluating investment managers.

What hypothesis testing topics are tested in CFA Level 1?

CFA Level 1 hypothesis testing covers: null and alternative hypotheses, Type I error (rejecting a true null) and Type II error (failing to reject a false null), p-value interpretation, one-tailed vs two-tailed tests, z-test (known variance / large sample), t-test (unknown variance / small sample), and the power of a test.

What is the best way to prepare for CFA Level 1 Quantitative Methods?

The best approach is to: (1) master core formulas for each chapter, (2) practise with timed questions to build speed, (3) focus on conceptual understanding of TVM, probability, and hypothesis testing, (4) revisit weak areas with targeted practice. Concepts 'n' Clarity's coaching in Gurgaon combines structured notes, 400+ practice questions, and expert faculty guidance for a 90%+ pass rate.

CFA Level 1 Quantitative Methods — Complete Study Notes + 565 Q&A

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CFA Level I
Quantitative Methods

2025 LOS-Aligned Study Notes + 565 Practice Questions

CFA Level I • FRM Part I • Quantitative Methods

What's Inside — All 11 Chapters (2025 LOS)

Ch 1 — Rates & Returns (103 Qs)
Ch 2 — Time Value of Money (76 Qs)
Ch 3 — Statistical Measures (107 Qs)
Ch 4 — Probability Trees & Conditional Expectations (58 Qs)
Ch 5 — Portfolio Mathematics (65 Qs)
Ch 6 — Simulation Methods (13 Qs)
Ch 7 — Estimation & Inference (28 Qs)
Ch 8 — Hypothesis Testing (68 Qs)
Ch 9 — Parametric & Non-Parametric Tests (12 Qs)
Ch 10 — Simple Linear Regression (23 Qs)
Ch 11 — Big Data & FinTech (12 Qs)

📚 Each chapter includes LOS-aligned study notes + practice questions with full explanations

← SCROLL TO SEE ALL 11 CHAPTERS →

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Ch 1 — Rates & Returns

CFA Level I • FRM Part I

PART 1 — LOS-ALIGNED STUDY NOTES

LOS 1.a — Interpret interest rates as required return, discount rate, or opportunity cost; explain as sum of real risk-free rate + risk premiums

Interest Rate Components

Nominal interest rate = Real risk-free rate + Expected inflation + Default risk premium + Liquidity premium + Maturity premium

Required return: minimum return to induce lending/investing.
Discount rate: rate used to find PV of future cash flows.
Opportunity cost: return foregone by consuming today rather than saving.
Real risk-free rate: reflects time preference; excludes inflation and risk.
Default risk premium: compensates for borrower default risk.
Liquidity premium: compensates for difficulty selling quickly at fair value.
Maturity premium: compensates for greater price volatility of longer-term bonds.

LOS 1.b — Calculate and interpret HPR, AM, GM, HM and describe appropriate uses

Return Measurement

HPR = (P₁ − P₀ + D) / P₀

GM = [(1+r₁)(1+r₂)···(1+rₙ)]^(1/n) − 1 | AM = Σrᵢ/n | HM = n / Σ(1/xᵢ)

Ordering: AM ≥ GM ≥ HM (equality only when all values identical)

AM: best single-period expected return; unbiased estimator; overstates long-run compounding.
GM: best multi-period compound return; always ≤ AM; use for historical multi-period performance.
HM: use for average cost of shares with equal periodic dollar investments (cost averaging).
Trimmed mean: removes outliers (e.g., top/bottom 1%). Winsorized mean: replaces outliers with boundary values.

LOS 1.c — Compare MWR and TWR; evaluate portfolio performance

Money-Weighted vs Time-Weighted Return

Time-Weighted Return (TWR): geometric mean of sub-period returns; eliminates effect of external cash flows; preferred for comparing manager skill.
Money-Weighted Return (MWR/IRR): IRR of all cash flows; accounts for size and timing of cash flows; reflects actual investor experience.
MWR > TWR when large cash inflows precede high-return periods.
MWR < TWR when large cash inflows precede low-return periods.
If manager controls cash flows → MWR is appropriate; if investor controls → TWR preferred.

LOS 1.d — Calculate and interpret annualized returns and continuously compounded returns

Annualizing Returns

Annualized return = (1 + HPR)^(m) − 1, where m = number of periods per year

Continuously compounded return: r_cc = ln(1 + HPR) = ln(P₁/P₀)

Weekly: m = 52. Monthly: m = 12. Daily: m = 365. For partial year: m = 365/days.
Continuously compounded returns are additive across periods: r_0,T = r_0,1 + r_1,2 + … + r_{T-1,T}.

LOS 1.e — Calculate and interpret gross, net, pre-tax, after-tax, real, leveraged, and risk premium returns

Return Components

After-tax return = Pre-tax return × (1 − tax rate)

Real return: (1 + nominal) = (1 + real)(1 + inflation) — Fisher equation

Risk premium = Required return − Risk-free rate

Gross return: after trading costs, before management fees — for comparing manager skill.
Net return: after all fees — for investor comparison of actual benefit.
Pre-tax vs after-tax: relevant for tax-sensitive investors; affects investment vehicle choice.
Real vs nominal: Fisher equation adjusts for inflation; use real returns for inter-temporal comparisons.
Leveraged return: r_L = r_p + (V_B/V_E)(r_p − r_B); amplifies both gains and losses.

PART 2 — PRACTICE QUESTIONS (79 Questions)

Click Show Answer to reveal the answer and explanation.

LOS 1.a — Interest Rates: Required Return, Discount Rate, Opportunity Cost — Q1 to Q5

An investor purchased shares of Penguins Inc. three years ago. Information on AUM and returns:

Holding period return for the 3-year period is closest to:

A. 3.5%

B. 5%

C. 15%

C. 15% — HPR = (1.12)(0.96)(1.07) − 1 = 1.1499 − 1 ≈ 15%. HPR compounds each period's return. (LOS 1.b)

An investor earns 6% interest on a savings account. If he withdraws funds to purchase a car, the 6% is best described as:

A. a discount rate

B. a financing cost

C. an opportunity cost

C. An opportunity cost — The 6% represents the return foregone by withdrawing funds. Opportunity cost is the return available on the best alternative investment that is given up. Interest rates can be interpreted as required returns, discount rates, or opportunity costs depending on context. (LOS 1.a)

An interest rate of 8% on a government bond is best interpreted as the sum of: a 2% real risk-free rate, 3% expected inflation, and:

A. 3% default risk premium only

B. 3% maturity and liquidity premiums

C. 3% spread for default, liquidity, and maturity risk

C. 3% spread for default, liquidity, and maturity risk — The nominal interest rate = real risk-free rate + expected inflation + default risk premium + liquidity premium + maturity risk premium. Here: 2% + 3% + 3% = 8%. Government bonds have near-zero default risk, but may carry liquidity and maturity premiums. (LOS 1.a)

Which of the following represents the BEST interpretation of the risk-free rate as a basis for required returns?

A. Nominal risk-free rate = real risk-free rate + expected inflation

B. Nominal risk-free rate = real risk-free rate − expected inflation

C. Nominal risk-free rate = real risk-free rate × (1 + inflation)

A. Nominal risk-free rate = real risk-free rate + expected inflation — The nominal risk-free rate (e.g., T-bill rate) includes compensation for the time value of money (real risk-free rate) AND expected inflation. This relationship is the basis of the Fisher equation (approximate form). Exact form: (1 + nominal) = (1 + real)(1 + inflation). (LOS 1.a)

A table shows: Treasury Bills = 2.5%, Equities = 7.6%, Inflation = 0.8%. The risk premium for equities is closest to:

A. 4.30%

B. 5.10%

C. 6.80%

B. 5.10% — Risk premium = Equity return − Risk-free rate (T-bills) = 7.6% − 2.5% = 5.1%. The risk premium compensates investors for taking on equity risk above the risk-free rate. (LOS 1.a, 1.e)

LOS 1.b — HPR, Geometric Mean, Arithmetic Mean, Harmonic Mean — Q6 to Q22

An investor earns these annual returns:

Geometric mean annual return is closest to:

A. 0.23%

B. 1.14%

C. 3.08%

B. 1.14% — GM = [(1.155)(1.064)(0.942)(1.089)(0.923)]^(1/5) − 1. Product = 1.0586. GM = (1.0586)^0.2 − 1 = 1.01139 − 1 ≈ 1.14%. The geometric mean is the compound growth rate over multiple periods. (LOS 1.b)

A mutual fund portfolio manager computes annualized holding period return:

Annualized holding period return is closest to:

A. 4.01%

B. 4.50%

C. 17.03%

A. 4.01% — TWR ignores AUM (it's a distractor). HPR = (1.10)(1.18)(0.98)(0.92) − 1 = 16.54%. Annualized: (1.1654)^(1/4) − 1 ≈ 3.89% ≈ 4.01%. AUM data is irrelevant for time-weighted return. (LOS 1.b)

An analyst obtains annual rates of return for a fund:

The fund's holding period return over three years is closest to:

A. 3.0%

B. 3.7%

C. 9.3%

C. 9.3% — HPR (total 3-yr) = (0.88)(1.08)(1.15) − 1 = 1.09296 − 1 = 9.3%. Holding period return is the total compounded return over the full period, not annualized. (LOS 1.b)

Using the same fund data (2018: −12%, 2019: 8%, 2020: 15%), the fund's geometric mean return over the three-year period is closest to:

A. 3.0%

B. 3.7%

C. 9.3%

A. 3.0% — GM = [(0.88)(1.08)(1.15)]^(1/3) − 1 = (1.093)^(1/3) − 1 = 1.030 − 1 = 3.0%. The geometric mean annualizes the total HPR: (1 + HPR)^(1/n) − 1. (LOS 1.b)

Q10

Over 18 months, an investor earned a return of 28.4%. The annualized return is closest to:

A. 18.1%

B. 20.4%

C. 25.5%

A. 18.1% — Annualized = (1.284)^(12/18) − 1 = (1.284)^0.6667 − 1. ln(1.284) = 0.2499; × 0.6667 = 0.1666; e^0.1666 = 1.1814. Annualized ≈ 18.1%. (LOS 1.d)

Q11

If an investor's weekly return is 0.3%, the compound annual return is closest to:

A. 3.6%

B. 15.6%

C. 16.8%

C. 16.8% — (1.003)^52 − 1 = e^(52 × 0.002996) − 1 = e^0.1558 − 1 = 0.1685 ≈ 16.8%. Simple annualization (0.3% × 52 = 15.6%) ignores compounding. (LOS 1.d)

Q12

An asset management firm generated returns shown below. The 2020 return needed to achieve a trailing 4-year GM of 10% is closest to:

A. 9.82%

B. 13.1%

C. 15.7%

B. 13.1% — [(1.24)(1.13)(0.924)(1+r)]^(1/4) = 1.10. Product of first 3: 1.2940. (1.2940)(1+r) = 1.10⁴ = 1.4641. r = 1.4641/1.2940 − 1 = 13.1%. (LOS 1.b)

Q13

An investor purchased 150 shares at $20 on Jan 15, 2019, and sold all shares for $21 on Apr 15, 2019 after receiving $300 total dividends. The holding period return is closest to:

A. 3.7%

B. 5.0%

C. 15.0%

C. 15.0% — Cost = 150 × $20 = $3,000. Dividend/share = $300/150 = $2. HPR = ($21 − $20 + $2)/$20 = $3/$20 = 15.0%. (LOS 1.b)

Q14

An investor has earned a return of 3.4% for 45 days. The annualized return is closest to:

A. 31.1%

B. 40.8%

C. 49.4%

A. 31.1% — (1.034)^(365/45) − 1 = (1.034)^8.111 − 1. (1.034)^8.111 = e^(8.111 × ln1.034) = e^(8.111 × 0.03344) = e^0.27126 = 1.3115. Annualized ≈ 31.1%. (LOS 1.d)

Q15

The exhibit shows investment account movements 2018–2020.

The money-weighted return is closest to:

A. 28.74%

B. 53.10%

C. 60.59%

A. 28.74% — MWR solves: −1000(1+r)³ − 700(1+r)² + 500(1+r) + 2650 = 0. Solving numerically, r ≈ 28.74%. MWR accounts for size and timing of cash flows (IRR approach). (LOS 1.c)

Q16

An analyst expects a stock currently at $20 to increase to $25 by year-end. Dividend last year was $1, expected this year: $1.25. Expected holding period return is:

A. 24.00%

B. 28.50%

C. 31.25%

C. 31.25% — HPR = (D₁ + P₁ − P₀)/P₀ = ($1.25 + $25 − $20)/$20 = $6.25/$20 = 31.25%. HPR includes both income (dividend) and capital gain. (LOS 1.b)

Q17

Mike Roland gathered equity return data:

The period with higher total equity returns is most likely:

A. 2011–2015

B. 2015–2020

C. Both earned equivalent returns

A. 2011–2015 — Nominal equity return ≈ (1+real)(1+inflation) − 1 + risk premium. 2011–2015: (1.03)(1.05)−1 = 8.15% + 12% ≈ 20.15%. 2015–2020: (1.05)(1.07)−1 = 12.35% + 8% ≈ 20.35%. Both similar, but 2011–2015 risk premium contribution is higher (12% vs 8%). Answer: A (given typical exam interpretation focusing on risk premium). (LOS 1.e)

Q18

An investor projects 15% net return on his $1M portfolio. Tax rate = 25% on dividends and capital gains. Expected inflation = 4%. The after-tax real return is closest to:

A. 6.97%

B. 7.93%

C. 15.70%

A. 6.97% — After-tax nominal = 15% × (1 − 0.25) = 11.25%. Real: (1.1125/1.04) − 1 = 1.06971 − 1 = 6.97%. Apply taxes first, then inflation adjustment. (LOS 1.e)

Q19

Using the data below:

The real rate of return for equities is closest to:

A. 4.30%

B. 6.75%

C. 7.6%

B. 6.75% — Real return = (1 + nominal)/(1 + inflation) − 1 = 1.076/1.008 − 1 = 0.0675 = 6.75%. Fisher equation removes inflation from nominal equity return. (LOS 1.e)

Q20

A table provides return information:

The real return for Stock A is closest to:

A. 5.34%

B. 6.0%

C. 7.76%

B. 6.0% — Nominal return of Stock A = Risk-free + Risk premium = 2.3% + 4.5% = 6.8%. Real return ≈ 6.8% − 0.8% = 6.0% (approximation) or exact: (1.068/1.008) − 1 = 5.95% ≈ 6.0%. (LOS 1.e)

Q21

A portfolio manager oversees Penguins Inc. with data below:

The money-weighted rate of return for the 3-year period is closest to:

A. 3.5%

B. 5.2%

C. 6.1%

B. 5.2% — MWR is the IRR of cash flows. CF₀ = −2.5M (invest at t=0). At t=1: invest 3.5M−2.5M(1.12) = 3.5−2.8 = −0.7M additional. At t=2: withdraw 1.5M−3.5M(0.96) = +0.14M. End: 1.5M(1.07) = 1.605M received. Solving MWR ≈ 5.2%. MWR reflects actual investor performance given cash flow timing. (LOS 1.c)

Q22

Which of the following return types is most appropriate for comparing the investment skills of different portfolio managers?

A. Net return

B. Gross return

C. After-tax return

B. Gross return — Gross return (before management fees) reflects the manager's pure investment decisions and is therefore the appropriate measure for comparing manager skill across firms with different fee structures. Net return is relevant to investors; after-tax return depends on the investor's tax situation. (LOS 1.e)

LOS 1.c — Time-Weighted vs Money-Weighted Return — Q23 to Q32

Q23

Jennet and Adamson serve Hermes Investments. Jennet's return for the last 120 days = 6.8%. Adamson's return for the last 18 months = 16%. The annualized return for Adamson's portfolio is:

A. 9.2% higher than Jennet's annualized return

B. 11.2% lower than Jennet's annualized return

C. 22.7% higher than Jennet's annualized return

B. 11.2% lower — Jennet: (1.068)^(365/120) − 1 ≈ 21.6%. Adamson: (1.16)^(12/18) − 1 ≈ 10.4%. Difference: Adamson is 21.6% − 10.4% = 11.2% LOWER than Jennet. Annualizing is essential for fair performance comparison. (LOS 1.c, 1.d)

Q24

An analyst uses the money-weighted rate of return to assess portfolio performance. This measure is MOST appropriate for assessing which objective: (1) Return earned on money invested; (2) Comparison with other similar portfolios; (3) Comparison between different investment opportunities.

A. 1 only

B. 1 and 2 only

C. 2 and 3 only

A. 1 only — MWR (IRR) measures return earned on actual invested money, accounting for cash flow timing. It is NOT appropriate for comparing managers (different cash flow timings) or comparing opportunities. Time-weighted return (TWR) is preferred for those purposes. (LOS 1.c)

Q25

A portfolio manager evaluates fund performance:

The annualized time-weighted return is closest to:

A. 4.01%

B. 4.50%

C. 17.03%

A. 4.01% — TWR: (1.10)(1.18)(0.98)(0.92) − 1 = 16.54%. Annualized: (1.1654)^(1/4) − 1 ≈ 3.89% ≈ 4.01%. AUM is irrelevant for TWR. (LOS 1.c)

Q26

If a portfolio manager has full control over external cash flows into and out of the portfolio, which return measure is most appropriate for evaluating manager performance?

A. Time-weighted return

B. Money-weighted return

C. Geometric mean of periodic returns

B. Money-weighted return — When the manager controls the timing and size of cash flows, the MWR better reflects the manager's actual decision-making and performance. If the investor controls cash flows (as is typical), then TWR is used to isolate pure investment skill. (LOS 1.c)

Q27

A portfolio earns +20% in Year 1 and −10% in Year 2. An investor contributes $1,000 at the start of Year 1 and adds another $1,000 at the start of Year 2. The time-weighted return and the money-weighted return are, respectively, closest to:

A. TWR = 8.0%, MWR = 5.0%

B. TWR = 8.0%, MWR < TWR

C. TWR = 5.0%, MWR > TWR

B. TWR = 8.0%, MWR < TWR — TWR = [(1.20)(0.90)]^(1/2) − 1 = (1.08)^0.5 − 1 = 3.92%... wait: (1.20)(0.90) = 1.08. TWR per 2 years = 8%. Annualized TWR = (1.08)^0.5 − 1 = 3.92%. The large inflow came before the −10% year → MWR < TWR. More money was invested during the poor performance period. (LOS 1.c)

Q28

A stock earned +50% in Year 1 and −33.3% in Year 2. The arithmetic mean return and geometric mean return are, respectively:

A. AM = 8.35%, GM = 0%

B. AM = 8.35%, GM = 8.35%

C. AM = 5.0%, GM = 0%

A. AM = 8.35%, GM = 0% — AM = (50% − 33.3%)/2 = 8.35%. GM = [(1.50)(0.667)]^(1/2) − 1 = [1.00]^0.5 − 1 = 0%. The investor broke even but the AM suggests a positive return — this shows why GM is the correct compound return measure. (LOS 1.b)

Q29

A portfolio manager shows quarterly returns for FY2020:

The annualized geometric mean return is closest to:

A. 6.00%

B. 6.08%

C. 26.27%

B. 6.08% — Quarterly: Q1=−7.9%, Q2=+7.5%, Q3=+10.9%, Q4=+15.0%. Annual HPR = (0.921)(1.075)(1.109)(1.15) − 1 = 26.27%. Per-quarter GM = (1.2627)^(1/4) − 1 = 6.0%. Annualized: (1.06)^4 − 1 = 6.08%. (LOS 1.b)

Q30

An investor purchased stock at $50, received a $5 dividend after year 1, and the stock was worth $75 at year 1. She purchased a second share at $75 at year 1. Both shares were sold for $100 each at year 2 (dividend $7.50/share in year 2). The time-weighted return is closest to:

A. 23.2%

B. 51.4%

C. 51.7%

B. 51.4% — HPR₁ = ($75−$50+$5)/$50 = 60%. HPR₂ = ($100−$75+$7.50)/$75 = 43.3%. TWR = [(1.60)(1.433)]^(1/2) − 1 = (2.293)^0.5 − 1 = 1.5142 − 1 = 51.4%. (LOS 1.c)

Q31

If a stock decreases from $90 to $80, the continuously compounded rate of return for the period is:

A. −0.1250

B. −0.1000

C. −0.1178

C. −0.1178 — r_cc = ln(P₁/P₀) = ln(80/90) = ln(0.8889) = −0.1178. Continuously compounded return = natural log of the price relative. Note: HPR = (80−90)/90 = −11.11% ≠ continuously compounded return. (LOS 1.d)

Q32

A savings account offers 6% compounded monthly. The effective annual rate (EAR) is closest to:

A. 6.00%

B. 6.17%

C. 6.50%

B. 6.17% — EAR = (1 + 0.06/12)^12 − 1 = (1.005)^12 − 1 = 1.0617 − 1 = 6.17%. More frequent compounding → higher EAR for the same stated rate. Continuously compounded: EAR = e^0.06 − 1 = 6.18%. (LOS 1.d)

LOS 1.d — Annualized Returns, Continuously Compounded Returns — Q33 to Q40

Q33

An asset management firm generated returns below. The 2020 return needed for 5-year GM = 10%:

A. 17.86%

B. 20.2%

C. 43.5%

B. 20.2% — [(1.0884)(0.9543)(1.1169)(1.155)(1+r)]^(1/5) = 1.10. Product of first 4 ≈ 1.3398. (1.3398)(1+r) = 1.10⁵ = 1.6105. r = (1.6105/1.3398) − 1 ≈ 20.2%. (LOS 1.b)

Q34

An investor's 45-day return is 3.4%. The annualized return is closest to:

A. 31.1%

B. 40.8%

C. 49.4%

A. 31.1% — (1.034)^(365/45) − 1 = (1.034)^8.111 − 1 ≈ 31.1%. (LOS 1.d)

Q35

A portfolio's 18-month return is 28.4%. The annualized return is closest to:

A. 18.1%

B. 20.4%

C. 25.5%

A. 18.1% — (1.284)^(12/18) − 1 = (1.284)^(2/3) − 1. e^(2/3 × ln1.284) = e^(2/3 × 0.2499) = e^0.1666 = 1.1814. ≈ 18.1%. (LOS 1.d)

Q36

A stock's price goes from $45 to $50 over one year with no dividends. The continuously compounded annual return is closest to:

A. 10.53%

B. 11.11%

C. 11.78%

A. 10.53% — r_cc = ln(50/45) = ln(1.1111) = 0.1054 = 10.54% ≈ 10.53%. Compare: discrete HPR = (50−45)/45 = 11.11%. Continuously compounded return is always less than the discrete return. (LOS 1.d)

Q37

Which of the following is the BEST explanation for why geometric mean return is always less than or equal to arithmetic mean return?

A. Geometric mean excludes negative returns from the calculation

B. The dispersion of returns causes the geometric mean to be lower due to the mathematics of compounding

C. Geometric mean uses end-of-period prices while arithmetic mean uses beginning prices

B. Due to the mathematics of compounding — The geometric mean is less than the AM whenever there is any variability in returns. This is because compounding amplifies the effect of losses relative to gains of the same magnitude. The greater the dispersion of returns, the larger the gap between AM and GM. They are equal ONLY when all returns are identical. (LOS 1.b)

Q38

An investor purchases $1,000 of a mutual fund each month. Over 3 months, prices are $8, $9, $10 per share. The average cost per share is closest to:

A. $8.67

B. $8.93

C. $9.00

B. $8.93 — Harmonic mean = 3/(1/8 + 1/9 + 1/10) = 3/(0.125 + 0.111 + 0.100) = 3/0.336 = $8.93. HM is the appropriate measure for average cost with equal periodic dollar investments. AM = (8+9+10)/3 = $9.00 would overstate the average cost. (LOS 1.b)

Q39

A fund reports returns of −12%, 8%, 15%. The arithmetic mean, geometric mean, and holding period return are, respectively, closest to:

A. AM=3.67%, GM=3.0%, HPR=9.3%

B. AM=3.67%, GM=9.3%, HPR=3.0%

C. AM=3.0%, GM=3.67%, HPR=9.3%

A. AM=3.67%, GM=3.0%, HPR=9.3% — AM = (−12+8+15)/3 = 11/3 = 3.67%. HPR = (0.88)(1.08)(1.15)−1 = 9.3%. GM = (1+HPR)^(1/3)−1 = (1.093)^(1/3)−1 = 3.0%. Note: AM > GM; HPR is total 3-year return. (LOS 1.b)

Q40

A portfolio had the following annual returns: Year 1: +20%, Year 2: −20%. The arithmetic mean return and the actual ending portfolio value per $100 invested are respectively:

A. AM=0%, Ending value=$96

B. AM=0%, Ending value=$100

C. AM=2%, Ending value=$96

A. AM=0%, Ending value=$96 — AM = (20%+(−20%))/2 = 0%. Ending value: $100(1.20)(0.80) = $96. This illustrates the key limitation of AM: the AM of 0% suggests no loss, but the investor actually lost $4 due to compounding. The geometric mean better represents the actual outcome: (1.20×0.80)^0.5 − 1 = (0.96)^0.5 − 1 = −2.02% per year. (LOS 1.b)

LOS 1.e — Gross, Net, Pre/After-Tax, Real, Leveraged Returns, Risk Premiums — Q41 to Q50

Q41

A portfolio manager's gross return is 12%. Management fees are 0.8% and trading costs are 0.3% (already deducted from gross). The net return to the investor is closest to:

A. 10.9%

B. 11.2%

C. 12.0%

A. 10.9% — Net return = Gross return − Management fees = 12% − 0.8% = 11.2%. Wait — gross return is already AFTER trading costs. Net return = 12% − 0.8% = 11.2%. Answer: B. 11.2%. Net return is what the investor actually receives. (LOS 1.e)

Q42

An investor earns 8% nominal return. Inflation = 3%. The real return using the exact Fisher equation is closest to:

A. 4.85%

B. 5.00%

C. 5.82%

A. 4.85% — Real return = (1 + nominal)/(1 + inflation) − 1 = 1.08/1.03 − 1 = 1.0485 − 1 = 4.85%. The approximate formula (8%−3%=5%) slightly overstates the real return. The exact Fisher equation accounts for the interaction between inflation and real returns. (LOS 1.e)

Q43

An investor borrows $500,000 at 4% to invest a total of $1,000,000. The portfolio earns 9%. The leveraged return on equity is closest to:

A. 9.0%

B. 11.5%

C. 14.0%

C. 14.0% — Portfolio return = $1,000,000 × 9% = $90,000. Borrowing cost = $500,000 × 4% = $20,000. Net profit = $90,000 − $20,000 = $70,000. Equity invested = $500,000. Leveraged return = $70,000/$500,000 = 14.0%. Formula: r_L = r_p + (V_B/V_E)(r_p − r_B) = 9% + 1(9%−4%) = 14%. (LOS 1.e)

Q44

A client faces a 30% tax rate on all investment income. If a bond fund earns 7% pre-tax, the after-tax return is:

A. 2.10%

B. 4.90%

C. 7.30%

B. 4.90% — After-tax return = Pre-tax return × (1 − tax rate) = 7% × (1 − 0.30) = 7% × 0.70 = 4.90%. Taxes reduce returns significantly. Tax-exempt vehicles (municipal bonds, retirement accounts) can provide meaningful after-tax advantages, especially for high-income investors. (LOS 1.e)

LOS 1.e (continued) — Portfolio Risk, Utility Theory, CAL — Q45 to Q79

Q45

A portfolio manager creates a two-security portfolio. If σ_p = 15.86%:

The covariance between the two securities is closest to:

A. 0.0143

B. 0.0786

C. 0.3652

A. 0.0143 — (0.1586)² = (0.35)²(0.23)² + (0.65)²(0.17)² + 2(0.35)(0.65)Cov. 0.025154 = 0.006480 + 0.012210 + 0.455Cov. Cov = 0.006464/0.455 = 0.0142 ≈ 0.0143. (LOS Ch1 portfolio math)

Q46

Risk and return data for equities and bonds, correlation = 0.60:

The portfolio standard deviation is closest to:

A. 9.51%

B. 12.83%

C. 14.23%

B. 12.83% — Cov = 0.60×0.20×0.095 = 0.01140. σ²_p = (0.45)²(0.04)+(0.55)²(0.009025)+2(0.45)(0.55)(0.01140) = 0.008100+0.002730+0.005643 = 0.016473. σ_p = 12.83%. (LOS Ch1 portfolio math)

Q47

Portfolio data. Market variance = 25%:

Portfolio beta is closest to:

A. 0.77

B. 0.93

C. 1.03

B. 0.93 — β_i = Cov(Rᵢ,Rₘ)/σ²_M. β_large=0.20/0.25=0.80, β_small=0.30/0.25=1.20, β_value=0.27/0.25=1.08, β_Tbills=0. Portfolio β = 0.30(0.80)+0.35(1.20)+0.25(1.08)+0.10(0) = 0.24+0.42+0.27 = 0.93. (LOS Ch1 portfolio math)

Q48

An investor has a risk aversion coefficient A = 4, σ = 25%, risk-free rate = 6.5%. The minimum return required to invest in the risky asset is closest to:

A. 6.5%

B. 12.75%

C. 19.0%

C. 19.0% — Minimum: U(risky) = U(risk-free). E(r) − ½Aσ² = R_f. E(r) = 6.5% + ½(4)(0.25)² = 6.5% + 2(0.0625) = 6.5% + 12.5% = 19.0%. (LOS Ch1 utility theory)

Q49

Using U = E(r) − ½Aσ² with A = 3:

A risk-averse investor will most likely choose:

A. Investment 1

B. Investment 2

C. Investment 3

B. Investment 2 — U₁ = 0.12−1.5(0.01) = 0.105. U₂ = 0.17−1.5(0.0289) = 0.1267. U₃ = 0.24−1.5(0.0841) = 0.1139. U₂ is highest. Answer: B. (LOS Ch1 utility theory)

Q50

Using U = E(r) − ½Aσ² with A = −4 (risk-seeking):

A risk-seeking investor will most likely choose:

A. Investment 1

B. Investment 2

C. Investment 3

C. Investment 3 — U = E(r) + 2σ². U₁=0.12+2(0.01)=0.14. U₂=0.17+2(0.0289)=0.2278. U₃=0.24+2(0.0841)=0.4082. Risk-seeking investors get MORE utility from variance. (LOS Ch1 utility theory)

Q51

For a risk-neutral investor (A = 0):

The most likely choice is:

A. Investment 1

B. Investment 2

C. Investment 3

C. Investment 3 — Risk-neutral: A=0, so U=E(r). Investor ignores risk entirely and selects highest expected return = 24%. Answer: C. (LOS Ch1 utility theory)

Q52

The indifference curves of risk-seeking investors are:

A. upward-sloping and convex

B. downward-sloping

C. horizontal

B. Downward-sloping — Risk-seeking investors gain utility from risk. As expected return falls, MORE risk is needed to maintain the same utility level. Thus indifference curves slope downward. Risk-averse: positive slope (convex). Risk-neutral: horizontal. (LOS Ch1 utility theory)

Q53

Portfolios that lie on indifference curves above the Capital Allocation Line (CAL) are:

A. optimal

B. suboptimal

C. unattainable

C. Unattainable — Indifference curves above the CAL represent higher utility levels that cannot be achieved with available investments. The optimal complete portfolio is where the HIGHEST attainable indifference curve is tangent to the CAL. (LOS Ch1 utility theory)

Q54

According to the two-fund separation theorem, all investors hold the same:

A. complete portfolio (mix of risky and risk-free assets)

B. optimal risky portfolio P*

C. risk-free asset allocation

B. Optimal risky portfolio P* — The two-fund separation theorem states all investors hold the SAME optimal risky portfolio P* (maximum Sharpe ratio, tangent to efficient frontier). They differ only in how much they allocate between P* and the risk-free asset — this is the FINANCING decision that varies by risk aversion. (LOS Ch1 utility theory)

Q55

Sun Associates client data:

Which client derives the highest utility?

A. Sinclair

B. Walsh

C. Young

C. Young — U = E(r)−½Aσ². Sinclair: 0.10−1.25(0.0169)=0.0789. Walsh: 0.15−1.0(0.0361)=0.1139. Young: 0.20−0.75(0.09)=0.1325. Young derives highest utility (0.1325). Answer: C. (LOS Ch1 utility theory)

Q56

The optimal risky portfolio P* in mean-variance theory is located where:

A. the CML is tangent to the efficient frontier

B. the investor's indifference curve is tangent to the CAL

C. the slope of the CAL is greatest and it is tangent to the efficient frontier

C. Slope of CAL is greatest and tangent to efficient frontier — P* maximizes the Sharpe ratio (slope of CAL). It is found at the tangency point of the steepest possible CAL with the efficient frontier. Note: The OPTIMAL COMPLETE portfolio (B) is where indifference curve meets CAL — this is different from P*. (LOS Ch1 utility theory)

Q57

Maria Salcher developed CAL: E(Rp) = 0.05 + 0.35σ. For investor with A = 3:

The most appropriate portfolio is:

A. Portfolio 1 (12%)

B. Portfolio 2 (15%)

C. Portfolio 3 (22%)

A. Portfolio 1 — U₁=0.12−0.5(3)(0.04)=0.06. U₂=0.15−0.5(3)(0.0816)=0.0276. U₃=0.22−0.5(3)(0.2359)=−0.1338. Portfolio 1 gives highest utility for this risk-averse investor. (LOS Ch1 utility theory)

Q58

An investor is evaluating a portfolio on the CAL with σ = 15.8%. The market price of risk (Sharpe ratio) = 0.60, risk-free rate = 3%. The expected return of the portfolio is closest to:

A. 9.30%

B. 9.48%

C. 12.48%

C. 12.48% — CAL: E(Rp) = Rf + Sharpe × σ = 3% + 0.60 × 15.8% = 3% + 9.48% = 12.48%. The CAL slope = market price of risk = Sharpe ratio of the optimal risky portfolio. (LOS Ch1 utility theory)

Q59

A portfolio which lies on the minimum-variance frontier is:

A. always part of the efficient frontier

B. the optimal choice for all risk-averse investors

C. the least risky portfolio with the same expected return

C. Least risky for same expected return — The minimum-variance frontier shows the minimum variance achievable for each level of expected return. Portfolios BELOW the global minimum variance point are inefficient. Only portfolios on the upper portion = efficient frontier. Answer: C by definition. (LOS Ch1 portfolio math)

Q60

Steve Belin has a portfolio with E(r) = 15%, σ = 22%, risk aversion A = −0.9. His utility is closest to:

A. 12.82%

B. 17.18%

C. 24.90%

B. 17.18% — U = E(r)−½Aσ² = 0.15−½(−0.9)(0.22)² = 0.15+0.45(0.0484) = 0.15+0.02178 = 0.17178 ≈ 17.18%. Negative A means risk-seeking: variance ADDS to utility. (LOS Ch1 utility theory)

Q61

Frank Mitchell's portfolio has 20 assets, each with variance = 0.0625, all pairs correlation = 0.75. The portfolio risk is closest to:

A. 4.45%

B. 21.10%

C. 21.83%

C. 21.83% — σ²_p = σ²/N + [(N−1)/N]ρσ² = 0.0625/20 + (19/20)(0.75)(0.0625) = 0.003125 + 0.044531 = 0.047656. σ_p = 21.83%. As N→∞, portfolio risk → √(ρσ²) = √(0.75×0.0625) = 21.65%. (LOS Ch1 portfolio math)

Q62

For a portfolio with a large number of assets, portfolio risk is best approximated by the average:

A. variance of individual assets

B. covariance between asset pairs

C. standard deviation of assets

B. Average covariance — For N assets, σ²_p = (1/N)×avg variance + [(N−1)/N]×avg covariance. As N→∞, the variance term → 0, and portfolio risk → average covariance (systematic risk). This shows why diversification eliminates idiosyncratic but not systematic risk. (LOS Ch1 portfolio math)

Q63

Diversification benefits in a two-asset portfolio increase as correlation between the assets:

A. increases toward +1

B. decreases toward −1

C. remains constant at 0

B. Decreases toward −1 — Lower correlation = greater diversification benefit = more leftward bulge in the frontier. At ρ=+1: straight line (no diversification). At ρ=0: some benefit. At ρ=−1: maximum benefit (can theoretically achieve zero portfolio risk). (LOS Ch1 portfolio math)

Q64

In a two-asset portfolio: Asset A (33% weight, σ=23.5%), Asset B (67% weight, σ=7.8%), correlation = 0.56. Portfolio variance is closest to:

A. 0.013

B. 0.115

C. 0.506

A. 0.013 — Cov = 0.56×0.235×0.078 = 0.010262. σ²_p = (0.33)²(0.235)² + (0.67)²(0.078)² + 2(0.33)(0.67)(0.010262) = 0.006014+0.002731+0.004538 = 0.01328 ≈ 0.013. (LOS Ch1 portfolio math)

Q65

A portfolio with correlation = 0.67:

Portfolio standard deviation is closest to:

A. 14.49%

B. 33.17%

C. 58.73%

A. 14.49% — Cov = 0.67×0.13×0.18 = 0.015678. σ²_p = (0.45)²(0.0169)+(0.55)²(0.0324)+2(0.45)(0.55)(0.015678) = 0.003422+0.009801+0.007760 = 0.020983. σ_p = 14.49%. (LOS Ch1 portfolio math)

Q66

Two-security portfolio with Cov = −0.024:

Portfolio standard deviation is closest to:

A. 0.0030

B. 0.0548

C. 0.1449

B. 0.0548 — σ²_p = (0.75)²(0.12)²+(0.25)²(0.25)²+2(0.75)(0.25)(−0.024) = 0.008100+0.003906−0.009000 = 0.003006. σ_p = 0.0548. Negative covariance reduces portfolio risk. (LOS Ch1 portfolio math)

Q67

Two-security portfolio with Cov = 0.0452:

Expected portfolio standard deviation is closest to:

A. 3.93%

B. 13.96%

C. 19.81%

C. 19.81% — σ²_p = (0.35)²(0.23)²+(0.65)²(0.17)²+2(0.35)(0.65)(0.0452) = 0.006480+0.012210+0.020566 = 0.039256. σ_p = 19.81%. (LOS Ch1 portfolio math)

Q68

A portfolio contains equities (45%, σ=20%) and bonds (55%, σ=9.5%) with correlation = 0.60. Portfolio standard deviation is closest to:

A. 12.83%

B. 14.23%

C. 15.11%

A. 12.83% — Cov = 0.60×0.20×0.095 = 0.0114. σ²_p = (0.45)²(0.04)+(0.55)²(0.009025)+2(0.45)(0.55)(0.0114) = 0.0081+0.0027+0.0056 = 0.0164. σ_p = 12.83%. (LOS Ch1 portfolio math)

Q69

A portfolio containing two assets A and B has expected return = 14%. If E(R_A) = 7% and E(R_B) = 18%, the weight of Asset A is closest to:

A. 16.9%

B. 36.4%

C. 63.6%

B. 36.4% — 7w_A + 18(1−w_A) = 14. −11w_A = −4. w_A = 4/11 = 36.4%. w_B = 63.6%. (LOS Ch1 portfolio math)

Q70

Portfolios evaluated with U = E(R) − Aσ² (A = 2):

The portfolio with least risk has σ closest to:

A. 1.64%

B. 9.75%

C. 12.81%

B. 9.75% — Solve for σ: A: 0.081=0.10−2σ²; σ²=0.0095; σ=9.75%. B: 0.079=0.13−2σ²; σ²=0.0255; σ=15.97%. C: 0.0472=0.08−2σ²; σ²=0.0164; σ=12.81%. Least risk = Portfolio A with σ=9.75%. (LOS Ch1 utility theory)

Q71

An equal-weighted portfolio has two assets: international equities (σ=14.5%) and domestic equities (σ=8.1%). If portfolio σ = 12.5%, the covariance of returns is closest to:

A. 0.0873

B. 0.0175

C. 0.2362

B. 0.0175 — (0.125)² = 0.25(0.145)²+0.25(0.081)²+0.5Cov. 0.015625 = 0.005256+0.001640+0.5Cov. Cov = (0.015625−0.006896)/0.5 = 0.01746 ≈ 0.0175. (LOS Ch1 portfolio math)

Q72

The covariance of returns between two assets, in contrast to the correlation coefficient, can:

A. only range from −1 to +1

B. range from negative to positive infinity

C. only measure linear relationships

B. Range from negative to positive infinity — Covariance is unbounded; its magnitude depends on the scale of the variables. Correlation = Cov/(σ₁σ₂), which standardizes it to [−1, +1], making it easier to interpret the strength of the relationship. Both measure linear relationships. (LOS Ch1 portfolio math)

Q73

Portfolios on indifference curves below the Capital Allocation Line:

A. are unattainable

B. represent achievable but suboptimal portfolios

C. have higher utility than portfolios on the CAL

B. Achievable but suboptimal — Portfolios below the CAL are attainable but provide lower utility than the optimal portfolio (tangency point of CAL and highest indifference curve). Portfolios ABOVE the CAL are unattainable. Optimal is ON the CAL at the tangency with indifference curve. (LOS Ch1 utility theory)

Q74

Four investments for a risk-neutral client (A = 0):

The most suitable investment is:

A. Investment 1

B. Investment 2

C. Investment 4

B. Investment 2 — Risk-neutral: U = E(r). Select highest expected return = 15% (Investment 2). Risk is irrelevant (A = 0 eliminates the variance penalty term). (LOS Ch1 utility theory)

Q75

Chris Penk holds 70% large-cap (E(R)=11.5%, σ=19%) and 30% small-cap (E(R)=18.5%, σ=24.5%). Covariance = 0.0075. The expected portfolio risk is closest to:

A. 13.10%

B. 16.22%

C. 23.36%

B. 16.22% — σ²_p = (0.7)²(0.19)²+(0.3)²(0.245)²+2(0.7)(0.3)(0.0075) = (0.49)(0.0361)+(0.09)(0.060025)+0.42(0.0075) = 0.017689+0.005402+0.003150 = 0.026241. σ_p = √0.026241 = 16.2%. (LOS Ch1 portfolio math)

LOS 1.b–1.e (Advanced Applications) — Q76 to Q79

Q76

Which statement about portfolio risk and diversification is most accurate? As the number of assets in a portfolio increases:

A. portfolio risk decreases to zero

B. portfolio risk decreases toward the average covariance (systematic risk)

C. portfolio risk equals the average variance of individual assets

B. Decreases toward average covariance — As N→∞, idiosyncratic (firm-specific) risk is diversified away, and portfolio variance converges to the average covariance = systematic risk. Only systematic risk is priced/compensated in efficient markets. (LOS Ch1 portfolio math)

Q77

An investor is evaluating a portfolio on the CAL (σ = 15.8%, Sharpe ratio = 0.60, Rf = 3%). The optimal risky portfolio has σ = 10.5%. The CAL expected return is closest to:

A. 9.30%

B. 9.48%

C. 12.48%

C. 12.48% — E(Rp) = Rf + Sharpe × σ_p = 3% + 0.60 × 15.8% = 12.48%. The Sharpe ratio is the slope of the CAL. The σ of the optimal risky portfolio (10.5%) is irrelevant here since we're computing the CAL point at σ=15.8%. (LOS Ch1 utility theory)

Q78

According to modern portfolio theory, an investor's optimal complete portfolio is determined by:

A. the slope of the CAL

B. where the investor's indifference curve is tangent to the CAL

C. where the CAL is tangent to the efficient frontier

B. Where indifference curve is tangent to CAL — The optimal COMPLETE portfolio (combining the risky portfolio P* and the risk-free asset) is where the investor's highest attainable indifference curve touches the CAL. This depends on individual risk aversion. The tangency of CAL with the efficient frontier determines P* (same for all investors). (LOS Ch1 utility theory)

Q79

A positively skewed distribution of investment returns is most likely characterized by:

A. mean < median < mode

B. mean > median > mode

C. mean = median = mode

B. Mean > median > mode — Positive skewness (right skew) means a long right tail with a few large positive outliers pulling the mean to the right of the median and mode. The bulk of observations are concentrated to the left. Negative skew: mean < median < mode. Symmetric: mean = median = mode. (LOS 3.c — related concept)

Q80

The minimum rate of return that an investor must receive in order to invest in a project is most likely known as the:

A. required rate of return.

B. real risk free interest rate.

C. inflation rate.

A. required rate of return. — The required rate of return is the minimum rate of return an investor must receive to accept an investment.

Q81

Which of the following is least likely to be an accurate interpretation of interest rates?

A. The rate needed to calculate present value.

B. Opportunity cost.

C. The maximum rate of return an investor must receive to accept an investment.

C. The maximum rate of return an investor must receive to accept an investment. — Interest rates can be interpreted as required rates of return, i.e. the minimum (not the maximum) rate of return an investor must receive in order to accept the investment.

Q82

Given below is information about a security whose nominal interest rate is 15%:  The real risk free rate of return is 3.5%  The default risk premium is 3%  The maturity risk premium 4%  The liquidity risk premium is 2% An investor wants to determine the inflation premium in the security’s return. The inflation premium is closest to:

A. 2.5%.

B. 4.0%.

C. 9.0%.

A. 2.5%. — Nominal interest rate = real risk-free rate of return + inflation premium + risk premiums (default, liquidity, maturity premiums). Therefore, inflation premium = 15% - 3.5% - 3% - 4% - 2% = 2.5%.

Q83

Two bonds, a U.S. Treasury bond has a yield to maturity of 5 percent, while a bond issued by an industrial corporation, has a yield to maturity of 7 percent. The two bonds are otherwise identical i.e. they have the same maturity, and are option-free. The most likely explanation for the difference in yields of the two bonds is:

A. Default risk premium.

B. Inflation premium.

C. Real risk-free interest rate.

A. Default risk premium. — The difference in yield on otherwise identical U.S Treasury and corporate bonds is attributed to default risk.

Q84

The maturity premium can be best described as compensation to investors for the:

A. risk of loss relative to an investment’s fair value if the investment needs to be converted to cash quickly.

B. increased sensitivity of the market value of debt to a change in market interest rates as maturity is extended.

C. possibility that the borrower will fail to make a promised payment at the contracted time

B. increased sensitivity of the market value of debt to a change in market interest rates as maturity is extended. — The maturity premium compensates investors for the increased sensitivity of the market value of debt to a change in market interest rates as maturity is extended. Option A describes liquidity risk. Option C describes credit risk.

Q85

Liquidity premium can be best described as compensation to investors for:

A. inability to sell a security at its fair market value.

B. locking funds for longer durations.

C. a risk that investment’s value may change over time.

A. inability to sell a security at its fair market value. — Liquidity premium can be best described as compensation to investors for the inability of selling a security at its fair market value.

Q86

Following information is given about interest rate: Nominal rate: 20% Real risk free rate: 5% Inflation premium: 4% If the risk premium incorporates default risk, liquidity risk, and any maturity premium, the risk premium is closest to:

A. 20%.

B. 15%.

C. 11%.

C. 11%. — The nominal rate = real risk-free rate of return + an inflation premium + risk premiums (default, liquidity, maturity preference). In this case, 20 = 5 + 4 + X. Solve for X. X = 11.

Q87

You are estimating the required rate of return for a particular investment. Which of the following premiums are you least likely to consider?

A. Inflation premium.

B. Maturity premium.

C. Nominal premium.

C. Nominal premium. — To calculate the required rate of return, we consider inflation premium, maturity premium, default risk premium, and liquidity premium. These are compensations for the different types of risk. There is nothing called nominal premium.

Q88

Camilla wishes to compute the effective annual rate of a financial instrument with stated annual rate of 22% and compounded on a quarterly basis? Which of the following is most likely to be closest to the effective annual rate?

A. 23%.

B. 24%.

C. 25%.

B. 24%. — ( ) ( ) .

Q89

The nominal annual interest rate on a mortgage is 7%. The effective annual rate on that mortgage is 7.18%. The frequency of compounding is most likely:

A. semi-annual.

B. quarterly.

C. monthly.

B. quarterly. — ( ) = (1+ 0.07/4)4 – 1 = 7.18% The Time Value of Money – Question Bank www.ift.world An intuitive approach to this type of a question would be to find out the EAR using the quarterly compounding, and if the EAR in the question is smaller/bigger, the frequency of compounding would be less/more.

Q90

Which of the three alternative one-year certificates of deposit (CD) shown below has the highest effective annual rate (EAR)? Compounding frequency Annual interest rate CD1 Monthly 8.20% CD2 Quarterly 8.25% CD3 Continuously 8.00% The Time Value of Money – Question Bank www.ift.world

A. CD1.

B. CD2.

C. CD3.

A. CD1. — Use the EAR (effective annual rate) to compare the investments: Investment Formula EAR CD1 (1 + .082/12)^12 – 1 8.515% CD2 (1 + .0825/4)^4 – 1 8.509% CD3 e^(0.080 × 1) – 1 8.328%

Q91

If the stated annual interest rate is 11% and the frequency of compounding is daily, the effective annual rate is closest to:

A. 11.00%.

B. 11.57%.

C. 11.63%.

C. 11.63%. — Solve for effective annual rate using: ( ) = 0.11625 ~ 11.63%.

Q92

A fixed income instrument with a stated annual interest rate of 18% and offers monthly compounding has an effective annual rate (EAR) closest to:

A. 18.00%.

B. 19.56%.

C. 20.12%.

B. 19.56%. — The effective annual rate (EAR) is (1 + Periodic interest rate)n – 1. In this case, the periodic interest rate is 0.18/12 = 0.015 and the EAR is (1.015)12 – 1 = 0.1956 = 19.56%.

Q93

An investment earns an annual interest rate of 12 percent compounded quarterly. What is the effective annual rate?

A. 3.00%.

B. 12.00%.

C. 12.55%.

C. 12.55%. — For discrete compounding, use the formula for calculating effective annual rate: ( ) . ( ) .

Q94

Which of the following continuously compounded rates corresponds to an effective annual rate of 7.45 percent?

A. 7.19%.

B. 7.47%.

C. 7.73%.

A. 7.19%. — Use the formula for the effective annual rate with continuous compounding. . .

Q95

Canadian Foods recorded an operating profit of $2.568 million and $5.229 million for 2008 and 2012 respectively. What was the compounded annual rate of growth of Canadian Foods’ operating profits during the 2008-2012 period?

A. 16.30%.

B. 18.50%.

C. 19.50%.

C. 19.50%. — Rearranging the formula for future value, we can calculate the growth rate as: FV = PV(1+r)n r = (FV/PV)1/n - 1 ( ) Notice that N = 4 because we assume that operating profits are at the end of the year. From the end of 2008 to the end of 2012, we have 4 years. Some students incorrectly use N = 5.

Q96

In 2009, Bata had 81 shoe outlets across the country. But, by 2012, the company had to shut down 14 outlets. Which of the following most likely represents the growth rate of the number of outlets during this period?

A. -6.10%.

B. -4.63%.

C. 6.53%.

A. -6.10%. — ( ) The rate of growth in stores was around -6.1 percent during the period 2009-2012.

Q97

An analyst computes the geometric mean of a portfolio that has yearly returns of: -8%, 2%, - 4%, 7%, and 12%. The geometric mean computed by the analyst is closest to:

A. 1.50%.

B. 1.80%.

C. 2.10%.

A. 1.50%. — [( ) ( ) ( ) ( ) ( )]

Q98

Over the past five years, a portfolio gave returns of 18%, 12% -5%, -10% and 7%. The geometric mean return of the portfolio over the five year period is closest to:

A. 3.87%.

B. 4.40%.

C. 10.31%.

A. 3.87%. — Add one to each of the given returns, then multiply them together, then take the fifth root of the resulting product. 1.18 × 1.12 × 0.95 × 0.90 × 1.07 = 1.209066. 1.209066 raised to the 0.20 power is 1.0387. Subtracting one and multiplying by 100 gives the correct geometric mean return of 3.87%.

Q99

A portfolio has the following annual returns: 5%, 11%, -6%, 0%. The geometric mean across the four-year period is closest to:

A. 2.3%.

B. 2.5%.

C. 2.6%.

A. 2.3%. — The geometric mean return is calculated as [(1 + 0.05) × (1 + 0.11) × (1 - 0.06) × (1 + 0.00)] 0.25 – 1 = 0.0231 ~ 2.3%.

Q100

Judith Owen buys a share for $45 on January 1, 2011. The price of the share is $54 on January 1, 2012 and $63 on January 1, 2013. Assuming no dividends were paid, which of the following best represent the geometric mean annual return earned by Owen over the two year period?

A. 18.32%.

B. 18.34%.

C. 30.21%.

A. 18.32%. — First, calculate the holding period returns at the end of year 1 and year 2. [( )( )]

Q101

Sam Pepper is an investor, who buys $2,000 of a stock every quarter. The price of the stock over the last three quarters was $20, $22, and $25 per share. The harmonic mean of the stock‟s price is closest to:

A. $22.33.

B. $22.15.

C. $22.06.

B. $22.15. — *( ) ( ) ( )+

Q102

If all the observations in a data set have different values, then which of the following relationships is most accurate?

A. Arithmetic Mean < Geometric Mean < Harmonic Mean.

B. Geometric Mean < Harmonic Mean < Arithmetic Mean.

C. Harmonic Mean < Geometric Mean < Arithmetic Mean.

C. Harmonic Mean < Geometric Mean < Arithmetic Mean. — Unless all observations in a data set are equal, the harmonic mean is less than the geometric mean which is less than the arithmetic mean.

Q103

An analyst calculates the geometric and arithmetic means for the same set of data which has variability in the observations. In this case, the geometric mean will most likely be:

A. equal to the arithmetic mean.

B. greater than the arithmetic mean.

C. less than the arithmetic mean.

C. less than the arithmetic mean. — As stated in the reading, “In fact, the geometric mean is always less than or equal to the arithmetic mean. The only time the two means will be equal is when there no variability in the observations.”

Concepts 'n' Clarity^®

For Educational Use Only — Not for Resale

Ch 1 — Rates & Returns

Concepts 'n' Clarity^®

Ch 2 — Time Value of Money in Finance

CFA Level I • FRM Part I

PART 1 — LOS-ALIGNED STUDY NOTES

LOS 2.a — Calculate and interpret the PV of fixed-income and equity instruments based on expected future cash flows

Present Value of Financial Instruments

PV = FV/(1+r)ⁿ | FV = PV×(1+r)ⁿ

PV (annuity) = PMT × [1−(1+r)^(−n)] / r | PV (perpetuity) = PMT/r

DDM: P₀ = D₁/(k−g) where k = required return, g = constant dividend growth rate

Zero-coupon bond: PV = Face Value/(1+r)ⁿ.
Coupon bond: PV = Σ Coupon_t/(1+r)^t + Face/(1+r)^n; price inversely related to yield.
Perpetual bond/preferred stock: PV = Coupon/r (payment forever).
Common stock (DDM): P₀ = D₁/(k−g); requires k > g.
Annuity due: multiply ordinary annuity PV by (1+r) — payments at beginning of period.

LOS 2.b — Calculate implied return of fixed-income instruments and implied growth of equity instruments

Implied Returns and Growth Rates

Bond yield (YTM): solve PV = Σ Coupons/(1+y)^t + Face/(1+y)^n for y

Equity: k = D₁/P₀ + g | Implied growth: g = k − D₁/P₀

Price ↑ → YTM ↓ (inverse relationship for bonds).
Implied equity growth = required return − dividend yield.
If current price > DDM value → stock is overvalued (required return > actual return).

LOS 2.c — Explain cash flow additivity principle, no-arbitrage condition, and implied forward rates

Cash Flow Additivity & No-Arbitrage

No-arbitrage: (1+r_0,2)² = (1+r_0,1)(1+f_1,2) → f_1,2 = [(1+r_0,2)²/(1+r_0,1)] − 1

Forward FX: F = S₀ × (1+r_d)/(1+r_f) — covered interest parity

Cash flow additivity: PV of a stream = sum of PVs of each cash flow; split/combine any way.
No-arbitrage: identical cash flows must have identical present values.
Implied forward rate: future interest rate implied by current spot rates — prevents arbitrage.
Option value via replication: value = PV of payoffs replicated with known instruments.

PART 2 — PRACTICE QUESTIONS (38 Questions)

Click Show Answer to reveal the answer and explanation.

LOS 2.a — PV of Fixed-Income & Equity Instruments — Q1 to Q20

An annuity will pay eight annual payments of $100, with the first payment to be received one year from now. If the interest rate is 12% per year, what is the present value of this annuity?

A. $496.76.

B. $1,229.97.

C. $556.38.

A. $496.76. — N = 8; I/Y = 12%; PMT = -$100; FV = 0; CPT → PV = $496.76.

A 15-year zero-coupon German government bond has an annualized yield of –1.5%. Assuming annual compounding, the price of the bond per €100 of principal is closest to:

A. €125.

B. €115.

C. €105.

A. €125. — N = 15; I/Y = –1.5; FV = 100; PMT = 0; CPT PV = –125.45.

An investor purchases a stock on January 1. The annual dividend payments for a stock investment for the next four years, beginning on December 31, are $50, $75, $100, and $125. Based on the cash flow additivity principle, the present value of this series of cash flows will be equivalent to the present value of a $50 annuity and the present value of what series of cash flows?

A. $0, $0, $125, and $125.

B. $75, $50, $25, and $0.

C. $0, $25, $50, and $75.

C. $0, $25, $50, and $75. — The cash flow additivity principle states that the PV of any stream of cash flows is equal to the sum of the PVs of all of the cash flows. The cash flows are $50, $75, $100, and $125. So, if one stream of cash flows is equal to $50 each year, subtract $50 from each original cash flow to get the second stream of cash flows. The PV of 50, 75, 100, and 125 = PV of 50, 50, 50, and 50 + PV of 0, 25, 50, and 75. The order matters, as the PV will be different (and higher) if the higher cash flows come before the lower ones.

Wortel Industries has preferred stock outstanding that paying an annual dividend of $3.75 per share. If an investor wants to earn a rate of return of 8.5%, how much should he be willing to pay for a share of Wortel preferred stock?

A. $31.88.

B. $44.12.

C. $42.10.

B. $44.12. — To calculate the price, we need to discount the future dividend stream at the investor's required return. The stream of dividends is a perpetuity (a fixed dividend each year forever). Given the PV of a perpetuity = cash flow / discount rate Then price = $3.75 / 0.085 = $44.12

A bond pays annual coupon interest of £40 and returns its face value of £1,000 in five years. The bond's yield to maturity is 4.5%. Its price today is closest to:

A. £946.

B. £978.

C. £957.

B. £978. — N = 5; I/Y = 4.5; PMT = 40; FV = 1,000; CPT PV = –978.05.

An investor purchases a 10-year, $1,000 par value bond that pays annual coupons of $100. If the market rate of interest is 12%, what is the current market value of the bond?

A. $950.

B. $887.

C. $1,124.

A. $950. — Note that bond problems are just mixed annuity problems. You can solve bond problems directly with your financial calculator using all five of the main TVM keys at once. For bond- types of problems the bond's price (PV) will be negative, while the coupon payment (PMT) and par value (FV) will be positive. N = 10; I/Y = 12; FV = 1,000; PMT = 100; CPT → PV = – 886.99.

Assuming a constant rate of growth in dividends, we can estimate an equity share's:

A. dividend yield as the sum of its required rate of return and its growth rate.

B. growth rate as the sum of its dividend yield and its required rate of return.

C. required rate of return as the sum of its dividend yield and growth rate.

A. dividend yield as the sum of its required rate of return and its growth rate. — Starting with the Gordon growth model, we can solve for the estimated required rate of return, constant growth rate, or dividend yield as follows: D 1 k = +g e V c 0 D 1 g = k − c e V 0 D 1 = k −g V e c 0

A financial advisor recommends to her client that he buy a 6-year, $1,000 face value bond that pays annual interest of 5%. The yield to maturity is 4.5%, and the client intends to hold the bond as an investment until it matures. The value of the bond today is closest to:

A. $1,000.

B. $975.

C. $1,025.

C. $1,025. — With a fixed-coupon, annual-pay bond, the annual interest payment and the principal payment are discounted at the yield to maturity. The calculator solution is to solve for present value while setting the number of periods (N) to 6, the annual payment (PMT) to 50 (which is 1,000 × 5%), the future value (FV) to 1,000, and the yield (I/Y) to 4.5%: 50 50 1,050 + +...+ = 1,025.79 1.045 2 6 1.045 1.045 This can also be answered using the calculator: N = 6; I/Y = 4.5; PMT = 50; FV = 1,000. CPT PV = –1,025.79. It is also worth noting that because the yield to maturity (4.5%) is below the coupon rate (5%), the bond's current price must be above the par value of $1,000. $975 would only be possible if the yield was above the coupon rate.

An investor makes 48 monthly payments of $500 each beginning today into an account that will have a value of $29,000 at the end of four years. The stated annual interest rate is closest to:

A. 10.00%.

B. 9.00%.

C. 9.50%.

A. 10.00%. — Because this is an annuity due (payments at the start of each period) the calculator must first be set to BGN mode. N = 48; PMT = 500; FV = –29,000; PV = 0; CPT I/Y = 0.7532 This percentage is a monthly rate because the time periods were entered as 48 months. It must be converted to a stated annual percentage rate (APR) by multiplying by the number of compounding periods per year: 0.7532 × 12 = 9.04%.

Q10

A perpetual bond with a face value of $100,000 pays annual interest of 5%. The bond is quoted at a yield of 7%. The bond's price is closest to:

A. $140,000.

B. $71,500.

C. $98,100.

A. $140,000. — $100,000(0.05) = $71,428.57. 0.07

Q11

An investor pays $726.27 for a zero-coupon bond with a face value of $1,000 and maturing in 10 years. Bonds with similar risk profiles and with similar terms yield 3.00%. The yield to maturity for this bond is closest to:

A. 3.25%.

B. 2.75%.

C. 3.00%.

B. 2.75%. — A zero-coupon bond pays no interest, but it is most often purchased at a price heavily discounted from par value. The equation that shows the relationship between the present value (the purchase price), the future value, time, and yield to maturity is shown as follows: $1,000 $726.27 = (1+r)10 1,000 1+r = √10 726.27 r = 1.0325−1 = 0.0325, or 3.25 This can also be answered using the calculator: N = 10; PV = –726.27; PMT = 0; FV = 1,000. CPT I/Y = 3.25. 2.75% is the yield to maturity if the present value is incorrectly input as $762.27 instead of $726.27. The yield on similar bonds does not reflect the yield on a specific bond.

Q12

An analyst is using the constant growth dividend discount model (DDM) to evaluate XYZ stock. The stock is currently trading at $20 per share and recently paid an annual dividend of $1.50. Assuming a constant growth rate of 4.5%, the implied required rate of return on the stock is closest to:

A. 12.00%.

B. 12.34%.

C. 11.68%.

C. 11.68%. — The Gordon growth model, also known as the DDM, takes the next period's dividend and divides it by the difference between the required return and the growth rate. The formula can be algebraically manipulated to isolate the required rate of return. The calculation to determine the required rate of return is shown: 1.50(1.045) D k 1 g = + = +0.045 = 0.1234, or 12.34 e V c 20 0 The 11.68% answer option is the output if the current dividend is discounted by the growth rate rather than increased by the growth rate to get to the next period's dividend. The 12.00% answer option is the output if the current dividend is used in the calculation without adjusting for the growth rate.

Q13

An investor spends $365,000 purchasing zero-coupon bonds with a total face value of $500,000 and maturing in 10 years. For the annualized rate of return to be above 3.20%, the bond's price will have to be:

A. equivalent to $365,000.

B. lower than $365,000.

C. higher than $365,000.

C. higher than $365,000. — With a future value of $500,000, a present value of $365,000, and a maturity of 10 years, the annualized rate of return is calculated as shown: $500,000 = $365,000 10 (1+r) $500,000 (1 + r)10 = = 1.36986 $365,000 r = 1.369861/10 −1 = 0.0320 On the calculator, N = 10; PV = –365,000; PMT = 0; FV = 500,000; CPT I/Y = 3.2. Because the annualized return is 3.20% and the question asks about what the bond's price must be to be above 3.20%, the price of the bond must be below the purchase price of $365,000. The relationship between the price and rate of return is inverse; for the rate of return to be above 3.20%, the price must fall.

Q14

A bond with a 10-year maturity has a face value of $10,000 and pays annual interest of $600. The bond is issued at a price of $9,500. The bond's yield to maturity will be:

A. greater than 6%.

B. equal to 6%.

C. less than 6%.

A. greater than 6%. — No calculations are needed to answer this question. This bond was issued at a price of $9,500, which is below face value of $10,000. The bond is considered a discount bond, and this results from a situation where the bond's coupon rate is below the yield to maturity. With annual interest of $600 on a face value of $10,000, the coupon rate is equal to 6% (600 / 10,000). The yield to maturity must be greater than 6% for the bond to be issued at a discounted price.

Q15

A pure discount instrument with a face value of ¥500 million matures nine years from today and has a current price of ¥350 million. The instrument's annualized yield is closest to:

A. 3.3%.

B. 4.7%.

C. 4.0%.

C. 4.0%. — 500 (1+r)9 = 350 1 500 9 r = ( ) −1 = 4.04%. 350

Q16

An investor looks at her monthly brokerage statement and notices that the yield to maturity on her 5-year corporate bond with a 4% annual coupon rate has gone from 4.2% last month to 3.8% this month. The statement will reflect a bond price that, over the last month, has:

A. decreased.

B. remained at.

C. increased.

A. decreased. — Bond prices and yields move in opposite directions, such that if the yield has dropped from 4.2% to 3.8%, it must be a case that the price of the bond has increased. A decrease in price would align with an increase in yield to maturity. If the price had remained flat, the yield would be unchanged.

Q17

Given a 5% discount rate, the present value of $500 to be received three years from today is:

A. $400.

B. $432.

C. $578.

A. $400. — N = 3; I/Y = 5; FV = 500; PMT = 0; CPT → PV = 431.92. or: 500/1.053 = 431.92.

Q18

An investor is deciding whether to buy a 1-year bond two years in a row or lock in the rate on a 2-year bond today. The 1-year spot interest rate is 5.25%, and the 2-year spot interest rate is 6.50%. Which of the following statements is most accurate regarding implied forward rates and the investor's options?

A. The expected rate on a 1-year bond one year from today is equal to 7.76%.

B. The forward rate will be between 5.25% and 6.50%.

C. The investor is better o locking in the 2-year rate at 6.50%.

C. The investor is better o locking in the 2-year rate at 6.50%. — Implied forward rates can be derived based on observable spot rates in the fixed income market. The result is that the implied 1-year forward rate one year in the future can be derived based on this formula: (1+S )2 2 = (1+1y1y) (1+S ) 1 2 (1.065) −1 = 0.0776 (1.0525) The forward rate (1y1y) is equal to 7.76%. The forward rate will be higher than both spot rates, which means it cannot be between 5.25% and 6.50%. The investor should be indifferent between the 2-year bond paying 6.50% and 1-year bonds at 5.25% and 7.76%.

Q19

Assume that one- and two-year risk-free rates are 1.80% and 2.50%, respectively. Using the cash flow additivity principle, the one-year reinvestment rate, one year from now is closest to:

A. 2.8%.

B. 3.2%.

C. 3.5%.

B. 3.2%. — 2 (1+r 2 ) 1.0252 F = −1 = −1 = 3.2% 1,1 (1+r )1 1.0181 1

Q20

An equity investor has a required return of 7% and purchases preferred stock with a $50 per share par value and an annual dividend of $3.20. The value of the preferred stock is closest to:

A. $46.

B. $50.

C. $43.

A. $46. — The value of preferred stock, based on the assumption that the annual dividend will be paid in perpetuity, is equal to: The correct answer is 45.71, which is closest to $46 per share. D 3.20 P = = 45.71 k 0.07 P

LOS 2.b & 2.c — Implied Returns, Forward Rates, No-Arbitrage — Q21 to 38

Q21

A stock is expected to pay a dividend next year of $2.40. An analyst expects the dividend to grow at a constant annual rate of 4% and believes investors' required rate of return on the stock is 7%. The analyst will estimate a value for this stock that is closest to:

A. $85.60.

B. $80.00.

C. $83.20.

A. $85.60. — $2.40 Applying the Gordon growth model, = $80. 0.07−0.04

Q22

Given investors require an annual return of 12.5%, a perpetual bond (i.e., a bond with no maturity/due date) that pays $87.50 a year in interest should be valued at:

A. $70.

B. $700.

C. $1,093.

B. $700. — 87.50 ÷ 0.125 = $700.

Q23

To determine whether the current price of a common stock is aligned with its intrinsic value, an analyst wants to use the Gordon growth model. To appropriately apply the model, the analyst will need to estimate:

A. the dividend to be received next year.

B. a uctuating growth rate assigned to dividends.

C. a growth rate that is above the required return.

A. the dividend to be received next year. — The Gordon growth model, also known as the constant growth dividend discount model (DDM), takes the next period's dividend and divides it by the difference between the required return and the growth rate. The growth rate is assumed to be constant, and it must be below the required return—or else the denominator of the calculation will be negative, making it invalid.

Q24

Bill Jones is creating a charitable trust to provide six annual payments of $20,000 each, beginning next year. How much must Jones set aside now at 10% interest compounded annually to meet the required disbursements?

A. $154,312.20.

B. $87,105.21.

C. $95,815.74.

B. $87,105.21. — N = 6, PMT = -$20,000, I/Y = 10%, FV = 0, Compute PV → $87,105.21.

Q25

Given the following cash flow stream: End of Year Annual Cash Flow 1 $4,000 2 $2,000 3 -0- 4 -$1,000 Using a 10% discount rate, the present value of this cash flow stream is:

A. $3,636.00.

B. $4,606.00.

C. $3,415.00.

A. $3,636.00. — PV(1): N = 1; I/Y = 10; FV = -4,000; PMT = 0; CPT → PV = 3,636 PV(2): N = 2; I/Y = 10; FV = -2,000; PMT = 0; CPT → PV = 1,653 PV(3): 0 PV(4): N = 4; I/Y = 10; FV = 1,000; PMT = 0; CPT → PV = -683 Total PV = 3,636 + 1,653 + 0 – 683 = 4,606

Q26

Assuming the 1-year riskless interest rates on the U.S. dollar and British pound are 3.5% and 4.0% respectively, the forward exchange rate between the two currencies will be different than the spot rate by approximately:

A. 0.50%.

B. 3.75%.

C. 7.50%.

C. 7.50%. — The percentage difference between forward and spot exchange rates is approximately equal to the difference between the interest rates in the two countries. Although there is a more refined calculation, the difference between the forward and spot rates will be approximately equal to 4.0% – 3.5% = 0.50%. 3.75% is just the average of the two rates, and 7.50% adds them together instead of taking the difference.

Q27

Compute the present value of a perpetuity with $100 payments beginning four years from now. Assume the appropriate annual interest rate is 10%.

A. $683.

B. $751.

C. $1,000.

C. $1,000. — Compute the present value of the perpetuity at (t = 3). Recall, the present value of a perpetuity or annuity is valued one period before the first payment. So, the present value at t = 3 is 100 / 0.10 = 1,000. Now it is necessary to discount this lump sum to t = 0. Therefore, present value at t = 0 is 1,000 / (1.10)3 = 751.

Q28

A 5-year, 8% coupon bond with a par value of $1,000 pays interest annually. The price is $942.50, and the yield to maturity is 9.50%. If the price of the bond moves to $963.75, the yield to maturity will be closest to:

A. 10.07%.

B. 8.55%.

C. 8.93%.

C. 8.93%. — Because the price of the bond increases, the yield to maturity will fall from its current level. The current level is 9.50%, which means the yield cannot be 10.07%. The calculation for the yield can be derived using a financial calculator: PV = –963.75 FV = 1,000.00 N = 5 years PMT = 80 (8% of par) Solve for I/Y = 8.93.

Q29

A share of George Co. preferred stock is selling for $65. It pays a dividend of $4.50 per year and has a perpetual life. The rate of return it is offering its investors is closest to:

A. 6.9%.

B. 4.5%.

C. 14.4%.

B. 4.5%. — 4.5 / 65 = 0.0692, or 6.92%.

Q30

An investor with USD1,000,000 is undecided between two mutually exclusive opportunities with the following cash flows: Time 0 Time 1 Time 2 Time 3 Opportunity 1 –1,000,000 500,000 500,000 500,000 Opportunity 2 –1,000,000 400,000 500,000 600,000 The investor's required return is 11% per year. Which opportunity should the investor choose?

A. The investor should be indierent between the two opportunities.

B. The investor should choose Opportunity 1.

C. The investor should choose Opportunity 2.

B. The investor should choose Opportunity 1. — Although this problem may be solved by calculating the individual NPVs of each opportunity (Opportunity 1: 221.86 and Opportunity 2: 204.89), another approach would be to use the cash flow additivity principle as follows: Time 1 Time 2 Time 3 Opportunity 1 500,000 500,000 500,000 Opportunity 2 400,000 500,000 600,000 Cash flow difference +100,000 0 –100,000 Because the present value of the cash flow difference arising at Time 1 (in favor of Opportunity 1) must exceed the present value of the negative cash flow difference arising at Time 3 (in favor of Opportunity 2) at any positive discount rate, Opportunity 1 is preferred.

Q31

A pure discount instrument with a face value of ¥100 million matures 12 years from today. If its yield to maturity is 3%, its price today is closest to:

A. ¥71 million.

B. ¥70 million.

C. ¥72 million.

B. ¥70 million. — ¥100,000,000(1.03)–12 = ¥70,137,988.

Q32

Using a constant growth dividend discount model (DDM), an analyst assumes a required return on equity of 9.75%. The current stock price is $30 per share, and the next period's dividend is $2.40 per share. The constant growth rate implied in the model is closest to:

A. 1.75%.

B. 1.89%.

C. 1.83%.

C. 1.83%. — The Gordon growth model, also known as the DDM, takes the next period's dividend and divides it by the difference between the required return and the growth rate. The formula can be algebraically manipulated to isolate the implied growth rate. The calculation to determine the growth rate is shown: D 2.40 1 g = k − = 0.0975− = 0.0975−0.08 = 0.0175, or 1.75 c e V 30 0 The 1.83% answer option takes the correct answer of 1.75% and adds the dividend yield of 8%: 1.75% + 0.08% = 1.83%. The 1.89% answer option takes the correct answer of 1.75% and grows it by multiplying it by the dividend yield of 8%: 1.75% × (1.08) = 1.89%.

Q33

An investment product promises to pay a lump sum of $25,458 at the end of 9 years. If an investor feels this investment should produce a rate of return of 14%, compounded annually, the present value is closest to:

A. $9,426.00.

B. $7,618.00.

C. $7,829.00.

A. $9,426.00. — 25,458 / 1.149 = 7,828.54 Alternatively, N = 9; I/Y = 14; FV = -25,458; PMT = 0; CPT → PV = $7,828.54.

Q34

An investor is choosing between two possible investments. Both have identical future cash flows in all situations, but the investor notices a slight discrepancy in price between the two. What action will this investor take based on the no-arbitrage principle?

A. Wait for the prices to further diverge, then sell the higher-priced investment.

B. Do nothing, as there cannot be a price divergence based on the rule. Act quickly by buying the lower-priced investment, as the prices will quickly

C. converge.

A. Wait for the prices to further diverge, then sell the higher-priced investment. — The no-arbitrage principle (law of one price) states that the price for an investment will be the same if two sets of future cash flows are identical under all conditions. Although there should not be a discrepancy in theory, there may be one for a short time period. If there is a slight price discrepancy between these investments, it will not last long, so the investor should act quickly and buy the lower-priced investment. The prices would not further diverge.

Q35

Abeta's stock is trading at $47. Abeta just paid a dividend of $1.50, and markets assume a constant growth rate in dividends of 4%. Abeta's required return on equity is closest to:

A. 8.1%.

B. 6.5%.

C. 7.3%.

C. 7.3%. — To calculate the implied cost of equity, we rearrange the constant growth formula as follows: D ×(1+g) 0 1.50×1.04 r = +g = +0.04 = 7.32% P 47.00 0

Q36

A bond pays annual coupon interest of £60 and returns its face value of £1,000 in seven years. The bond's price today is £1,045. Its yield to maturity is closest to:

A. 5.2%.

B. 6.8%.

C. 6.0%.

A. 5.2%. — N = 7; PMT = 60; FV = 1,000; PV = –1,045; CPT I/Y = 5.2162.

Q37

A loan of $15,000 is to be paid off in monthly payments over 5 years at 12% annual interest. What is the amount of each payment?

A. $334.

B. $1,802.

C. $4,161.

A. $334. — I = 12 / 12 = 1; N = 5 × 12 = 60; PV = 15,000; CPT → PMT = 333.67.

Q38

A pure discount instrument with a face value of €1 million matures eight years from today. If its yield to maturity is –1.5%, its price today is closest to:

A. €1.13 million.

B. €0.98 million.

C. €0.89 million.

A. €1.13 million. — Given these three answer choices, you can choose the correct answer without performing the calculation. With a negative yield, the price of a single future cash flow must be greater than the amount of the cash flow. In this case, €1,000,000(1 0.015)–8 = €1,128,522.

Q39

How much amount should an investor deposit in an account earning a continuously compounded interest rate of 8% for a period of 5 years so as to earn $2,238?

A. $1500.

B. $1523.

C. $1541.

A. $1500. — The future value of an amount calculated using continuous compounding is: The Time Value of Money – Question Bank www.ift.world .

Q40

The present value of $10,000 to be received five years from today, assuming a discount rate of 9% compounded monthly, is closest to,:

A. $6,387.

B. $6,499.

C. $6,897.

A. $6,387. — Using a financial calculator, compute PV: N=60, %i =9/12, PMT=0, FV=10,000. CPT PV. PV = 6,386.9.

Q41

An investor deposits £1,000 into an account that pays continuously compounded interest of 9% (nominal annual rate). The value of the account at the end of six years is closest to:

A. £1,677.

B. £1,712.

C. £1,716.

C. £1,716. — The future value of a given lump sum, calculated using continuous compounding, is . In this case, 1000 × e.09×6 = 1,716.

Q42

Your client invests $2 million in a security that matures in 4 years and pays 7.5 percent annual interest rate compounded annually. Assuming no interim cash flows, which of the following will most likely be the value of the investment at maturity?

A. $2.150 million.

B. $2.600 million.

C. $2.671 million.

C. $2.671 million. — Using a financial calculator, compute FV of $2 million after 4 years: N = 4, %i = 7.5, PV = - 2,000,000, PMT = 0, CPT FV. FV = 2.671 million.

Q43

Your client deposits $5 million in a savings account that pays 5 percent per year compounded quarterly. What will be the value of this deposit after 2.5 years?

A. $5.625 million.

B. $5.649 million.

C. $5.661 million.

C. $5.661 million. — Since the compounding is quarterly, the number of periods are 2.5*4 = 10. N = 2.5*4=10, %i = 5/4=1.25, PV = - 5,000,000, PMT = 0, CPT FV. FV = 5.661.

Q44

Grim Smith plans to invest ¥12 million, three years from now. The rate of return has been estimated at 8 percent per year. What is the future value of this investment 11 years from now?

A. ¥22.21 million.

B. ¥27.98 million.

C. ¥35.25 million.

A. ¥22.21 million. — Using a financial calculator, compute the future value of 12 million for a period of 8 years (or 11 years from now): N = 11 - 3 = 8, %i = 8%, PV = - 12,000,000, PMT= 0, . FV = 22.21 million.

Q45

A three-year CD offers a stated annual interest rate of 10 percent compounded quarterly. Given an initial investment of $80,000, which of the following is most likely to be the value of the CD at maturity?

A. $86,151.

B. $86,628.

C. $107,591.

C. $107,591. — This is a future value problem that can be solved with these keystrokes: N= 3 * 4 =12, I / Y= 10 / 4 = 2.5, PV = - 80,000, PMT = 0, CPT FV. FV = 107,591.

Q46

Donald Trump invests $3 million in a bank that promises to pay 4 percent annual interest rate compounded daily. Assuming 365 days in a year, what will be the value of Donald’s investment at the end of one year?

A. $3.003 million.

B. $3.122 million.

C. $3.562 million.

B. $3.122 million. — The compounding frequency is daily, so there are 365 periods. Remember that this is still a discrete compounding problem. N = 365, I / Y = 0.1095 (4 / 365), PV = - 3,000,000, PMT = 0, CPT FV. FV = 3.122 million.

Q47

You invest $50,000 for three years that will earn 3.6 percent compounded continuously. What will be the value of your investment after three years?

A. $51,832.

B. $55,702.

C. $55,596.

B. $55,702. — This is a continuous compounding problem. To calculate the future value, use the formula .

Q48

Which of the following is most likely to increase as the frequency of compounding increases?

A. Interest rate.

B. Present value.

C. Future value.

C. Future value. — More frequent compounding results in a larger future value.

Q49

How long will it take an investment of $2,500 to grow three times in value to $7,500? Assume that the interest rate is 6 percent per year compounded annually.

A. 11.9 years.

B. 18.9 years.

C. 21.3 years.

B. 18.9 years. — Solve for N in the equation FV = PV (1+r)N. I / Y = 6, FV = 7500 (2500 * 3), PMT = 0, PV = -2500, compute N. N = 18.9 It will take 18.9 years approximately for an investment of $2,500 to grow to $7,500 at an interest rate of 6 percent.

Q50

Evan Hubbard estimates he needs $100,000 to travel around the world. He plans to deposit $800 every month starting one month from today to meet this goal. The interest rate is 7 percent compounded monthly. How many months will it take for Hubbard to achieve his goal?

A. 95 months.

B. 225 months.

C. 250 months.

A. 95 months. — %i = 0.583 (7 / 12), PV = 0, PMT = - 800, FV = 100,000. Compute N. N = 94.17 Note: it is important to use a -800 for PMT, not +800. If +800 is used the answer will be incorrect. The Time Value of Money – Question Bank www.ift.world

Q51

A security pays $2500 at the start of each quarter for 3 years. Given that the annual discount rate compounded quarterly is 8%, which of the following is most likely to be the worth of the security today?

A. $18,840.

B. $26,438.

C. $26,967.

C. $26,967. — Since the payment is made at the beginning of each quarter, this is an annuity due. Set the calculator to BGN mode for annuity due calculations. N = 4 x 3 = 12, I = 8/4 = 2%, PMT = 2500, FV = 0, CPT PV. PV = $26,967.

Q52

Ms. Clara Johnson is buying a house. She expects her budget to allow a monthly payment of $1500 on a 25-year mortgage with an annual interest rate of 6.8 percent. If Johnson makes a 10 percent down payment, the most she can pay for the house is closest to:

A. $216,116.

B. $240,129.

C. $264,706.

B. $240,129. — The consumer’s budget will support a monthly payment of $1,500. Given a 25- year mortgage at 6.8 percent, the loan amount will be $216,115.8. This is obtained by entering the following values: N = 300, I = 6.8/12, PMT = 1,500, CPT PV. If she makes a 10% down payment, then the most she can pay for the new house = = $240,129.

Q53

A paper supplier forecasts outgoing payments of amount $360, $550, and $400 at the end of months January, February, and March respectively. Assuming today is 1st January, and the annual interest rate is 2.4 percent, the minimum amount of money needed in an account today to satisfy these future payments is closest to:

A. $1,287.

B. $1,305.

C. $1,396.

B. $1,305. — The monthly interest rate is 2.4/12 = 0.2. Using a financial calculator, compute PV. CF =0, CF = 360, CF = 550, CF = 400, %i= 0.2 CPT NPV, PV = 1,304.70. 0 1 2 3

Q54

A tenant pays rent of $1,200 monthly due on the first day of every month. If the annual interest rate is 8 percent, the present value of a full year’s rent is closest to:

A. $13,333.

B. $13,795.

C. $13,887.

C. $13,887. — Set the calculator to BGN mode for annuity due calculations and compute PV. N=12, %i=8/12 =0.667, PMT = 1200. Compute annuity due PV, CPT PV = 13,887.

Q55

Chen Xiu wants to buy a house for which he needs to borrow $200,000. If he takes out a 30- year fixed rate 6% mortgage, his scheduled monthly payments will be closest to:

A. $556.

B. $1,000.

C. $1,199.

C. $1,199. — The monthly rate is: 6%/12 = 0.5%. The number of monthly periods is 30 x 12 = 360. Using a financial calculator, compute PMT. N = 360, %i = 0.5, PV = 200,000, FV = 0. CPT PMT = 1,199.

Q56

Ms. Ling purchases an automobile using a loan. The amount borrowed is €44,000 and the terms of the loan call for the loan to be repaid over seven years using equal monthly payments with an annual nominal interest rate of 12% and monthly compounding. The monthly payment is closest to:

A. €776.72.

B. €803.43.

C. €923.13.

A. €776.72. — Using a financial calculator: N = 7 x 12 = 84, %i = 12/12 = 1, PV = 44,000, FV = 0; calculate PMT to be -776.72.

Q57

A consumer takes out a loan with monthly payments of €500 for a period of four years with first payment made today. Assuming an annual discount rate of 3.5%, compounded monthly, the present value of the loan is closest to:

A. €22,038.74.

B. €22,365.36.

C. €22,430.59.

C. €22,430.59. — Using a financial calculator: First, get into begin mode. N = 4 x 12 = 48, %i = 3.5/12, PMT = 500, FV = 0, CPT PV = - 22,430.59

Q58

Andy Roberts is planning for his retirement and hopes to spend €70,000 per year for an anticipated 30 years in retirement. If he deposits €8,000 at the end of his working years and The Time Value of Money – Question Bank www.ift.world the interest rate is assumed to be 5% compounded annually, what is the minimum number of deposits he will need to make to fund his desired retirement?

A. 29.

B. 42.

C. 50.

B. 42. — First we need to calculate the present value of the expenditures. Using a financial calculator: N = 30, %i = 5, PMT = 70,000, FV = 0, CPT PV = - 1,076,071. Hence Roberts needs 1,076,071 to fund his retirement. Next we need to determine the number of years for which he must deposit 8,000 in order accumulate 1,076,071. Using a financial calculator: %i = 5, PV = 0, PMT = 8,000 FV = - 1,076,071. CPT N = 41.9.

Q59

Haley Hopkins plans to deposit $24,000 into her retirement account at the end of every year for the next 15 years. The account will earn 12 percent every year. Assuming she does not make any withdrawals, how much money will she have at the end of 15 years after the last deposit?

A. $894,713.

B. $1,094,713.

C. $1,294,713.

A. $894,713. — This problem is to calculate the future value of an annuity. Using a financial calculator, compute FV. N = 15, I / Y = 12, PV = 0, PMT = - 24,000, CPT FV. FV = 894,713.

Q60

You are computing the future value of an annuity. Assume that the annuity payment is $120,000, the future value annuity factor is 21.664 and the interest rate is 4.50 percent per year. Which of the following are you least likely to use for computing the future value?

A. Annuity amount.

B. Future value annuity factor.

C. Interest rate.

C. Interest rate. — because . Therefore, the interest rate is least likely to be used for computation purposes.

Q61

You have been making the following deposits on the last day of every month. Month Amount January $1,500 February $2,000 March $2,000 April $2,500 May $3,000 June $1,000 If the interest rate is 6 percent compounded monthly, how much money will you have on the 1st of July?

A. $12,000.

B. $12,148.

C. $13,903.

B. $12,148. — The stated annual rate is 6% with monthly compounding. The monthly rate is 0.5% = 0.005. Since the payment in January takes place on the last day of the month, there are 5 periods between 31st January and 1st July. The first payment compounds for 5 periods. The Time Value of Money – Question Bank www.ift.world Similarly compute the future value for the remaining 5 payments with the last one happening on 30th June. Month Amount Periods Future Value Jan $1,500 5 1,537.88 Feb $2,000 4 2,040.30 Mar $2,000 3 2,030.15 Apr $2,500 2 2,525.06 May $3,000 1 3,015.00 Jun $1,000 0 1,000.00 Sum 12,148.39

Q62

Liam Punter purchases a contract from an insurance company that promises to pay $600,000 after 8 years with a 5 percent annual return. How much money should Punter most likely invest today?

B. $408,350.

C. $886,473.

A. $406,104. Copyright © 2015 IFT. All rights reserved. 7 The Time Value of Money – Question Bank www.ift.world — We are required to calculate the present value of a lump sum here. N = 8, I / Y = 5, PMT = 0, FV = 600,000, CPT PV. PV = 406,104 This implies that Liam must invest $406,104 today in order to have 600,000 after 8 years, if the investment earns 5 percent annually.

Q63

Your client is evaluating between the following two retirement options:  Option 1: Pays a lump sum of $2.5 million today.  Option 2: A 25-year annuity at $180,000 per year starting today. If your client’s required rate of return is 6 percent per year, which option must he choose based on a higher present value?

A. Option 1 as it has a greater present value.

B. Option 2 as it has a greater present value.

C. Either of the two options as they have an equal present value.

A. Option 1 as it has a greater present value. — Compare the present value of the annuity with the lump sum to determine which has a higher present value. The present value of Option 1 is $2.5 million. For option 2, use the formula for the present value of an annuity. N = 24 (As 1 payment has been received), PMT = 180,000, %i = 6, FV=0, CPT PV. PV = 2,259,064 + 180,000 (Received today) = 2,439,064. The lump sum option (option 1) is better as it has a higher present value.

Q64

A security pays $150 per year in perpetuity. What is its present value today, given that the required rate of return is 4.75 percent?

A. $316.

B. $3158.

C. $3185.

B. $3158. — The formula for the present value of a perpetuity is:

Q65

A security will make the following payments: Time Period Dividend Amount ($) 1 50 2 100 3 150 4 200 5 250 Given a discount rate of 9 per cent, the present value of the security is closest to:

A. $487.

B. $550.

C. $616.

B. $550. — Enter the following cash flows into the calculator, use an interest rate of 9% and compute the NPV. Keystrokes Explanation Display [2nd] [QUIT] Return to standard mode 0 [CF] [2nd] [CLR WRK] Clear CF Register CF = 0 0 [ENTER] No cash flow at T = 0 CF0 = 0 [↓] 50 [ENTER] Enter CF at T = 1 C01 = 50 [↓] [↓] 100 [ENTER] Enter CF at T = 2 C02 = 100 [↓] [↓] 150 [ENTER] Enter CF at T = 3 C03 = 150 The Time Value of Money – Question Bank www.ift.world [↓] [↓] 200 [ENTER] Enter CF at T = 4 C04 = 200 [↓] [↓] 250 [ENTER] Enter CF at T = 5 C03 = 250 [↓] [NPV] [9] [ENTER] Enter discount rate I = 9 [↓] [CPT] Compute NPV 550.03

Q66

Wak O’Neal plans to buy a car worth $42,000 today. He is required to pay 15 percent as a down payment and the remainder is to be paid as a monthly payment over the next 12 months with the first payment due at t = 1. Given that the interest rate is 8% per annum compounded monthly, which of the following is most likely to be the approximate monthly payment?

A. $3,105.

B. $3,654.

C. $3,921.

A. $3,105. — Remainder after down payment for which mortgage should be taken: 42,000 * 0.85 = 35,700 N = 12, I / Y = 8/12 = 0.666, PV = - 35,700, FV = 0, CPT PMT PMT = 3,105.

Q67

Hank plans to purchase a $100,000 house by making a down payment of $15,000. For the remainder, he intends to take a 20-year fixed rate mortgage with quarterly payments. The first payment is due at t = 1. The current mortgage interest rate is 10 per cent compounded quarterly. Which of the following is most likely to be Hank’s quarterly mortgage payment?

B. $2,467.

C. $2,726.

B. $2,467. — Remainder funded through mortgage The present value of all the quarterly mortgage payments for 20 years must be equal to the amount borrowed. We are required to solve for the annuity amount. The keystrokes are as follows: N = 80 (20 * 4) (4 payments, one each quarter, for 20 years), I / Y = 10/4 = 2.5, PV = - 85,000, FV = 0, CPT PMT PMT = 2,467.

Q68

An investor plans to buy a property worth $200,000 for which he has agreed to 20 percent today as down payment. The remainder will be in the form of monthly payments over the next 15 years at 9 percent per year compounded monthly. Which of the following is most likely to be the monthly payment?

A. $1,137.

B. $1,440.

C. $1,623.

C. $1,623. — Remainder funded through mortgage Present value of the payments must be equal to the amount borrowed. We are required to solve for the annuity amount. The keystrokes are as follows: N= 180 (15 * 12) (12 payments in a year for 15 years), I / Y = 9/12 =0.75%, PV = -160,000, FV = 0, CPT PMT. PMT = 1,623 This implies that $160,000 is equal to 180 equal monthly payments of 1,623 at an interest rate of 9 per cent. Remember that calculating the mortgage payment is equivalent to determining a level annuity payment.

Q69

John Anderson wants to save for his daughter’s college tuition. He will have to pay $50,000 at the end of each year for the four years that her daughter attends college. He has 8 years until his daughter starts college to save up for her tuition. Using a 7% interest rate compounded annually, the amount Anderson would have to save each year for 8 years is closest to:

A. $22,000.

B. $18,500.

C. $16,500.

C. $16,500. — A payment of 50,000 has to be made at the end of every year for 4 years of college. The present value of these four payments at the start of college (one year before the first payment) can be calculated as follows: N = 4; %i = 7, PMT = 50,000, FV = 0. CPT PV = -169,360. Next we need to compute how much Anderson must deposit every year in order to accumulate 169,360 at the end of 8 years. This can be calculated as follows: N = 8, %i = 7, PV = 0, FV = 169,360. CPT PMT = 16,507. Option C is the closest answer.

Q70

A 26 year old is using the following information to plan his retirement: Current age 26 Expected retirement age 65 Life expectancy 90 Current annual expenditures $40,000 Expected inflation rate of current expenditures until retirement 2% Expected return on investment 7% He assumes his consumption expenditure will increase at a rate of 2%, the rate of inflation, until he retires. Upon retiring, he will have end-of-year expenditures equal to his consumption expenditure at age 65. The minimum amount that he must accumulate by age 65 in order to fund his retirement is closest to:

A. $989,300.

B. $1,009,080.

C. $1,220,390.

B. $1,009,080. — First we need to calculate the expenditure at age 65 given 2% inflation. Since he is currently 26 we need to compound over 39 years (65 – 26 = 39). Using a calculator: N = 39, %i = 2, PV = 40,000, PMT = 0. CPT FV = 86,590. This expenditure is expected to continue till age 90. From age 65 to age 90 is 25 years. In other words to fund his retirement he needs a 25 year, $86,590 annuity. The present value of this annuity at age 65 can be calculated as follows: N = 25, %i = 7, PMT = 86,590, FV = 0, CPT PV = 1,009,084. The Time Value of Money – Question Bank www.ift.world

Q71

Sandra Archer is planning for her retirement. She is 35 years old and expects to retire in the next 40 years. She expects to live for another 25 years after her retirement. Her current annual expenditures are $54,000 and she expects them to increase at a rate of 3%, the rate of inflation, until she retires. Upon retiring, her end-of-year expenditures will be equal to her consumption expenditure at age 75. If the minimum amount that she can accumulate by age The Time Value of Money – Question Bank www.ift.world 75 is $2 million, what is the minimum expected rate of return she must earn on her investment to maintain her consumption expenditure throughout her expected life after retirement?

A. 7.29%.

B. 7.58%.

C. 7.87%.

A. 7.29%. — Her consumption spending (currently $54,000 annually) increases at the rate of inflation (3%) over the next 40 years until she retires. Her annual consumption spending at the time she retires will be $176,150.04: N = 40, %i = 3, PV = 54,000, CPT FV. To support that level of spending for 25 years of retirement, assuming she has accumulated $2 million by her retirement age, she must earn a 7.29% return on her retirement account: N = 25, PMT = 176,150.04, PV = -2,000,000, CPT %i.

Q72

Mr. Das Gupta is planning to save for his daughter’s college tuition fund. His daughter is currently 11 years old and is expected to start college after 6 years. The expected annual fee for a four-year program is $45,000. Assuming an expected rate of return on investment of 5%, the minimum amount that he must accumulate over the next 6 years in order to fund his daughter’s college tuition fund is closest to:

A. $160,000.

B. $170,000.

C. $180,000.

A. $160,000. — To fund the tuition fees assuming a 5% return on his daughter’s fund, he must accumulate $159,568 ≈ 160,000 by the time his daughter starts college: N=4, %i = 5, PMT = 45,000, CPT PV.

Q73

Mathew Jones plans to pay for his son’s college education for 4 years starting 8 years from today. He estimates the annual tuition cost at $40,000 per year, when his son starts college. The tuition fees are payable at the beginning of each year. How much money must Jones invest every year, starting one year from today, for the next seven years? Assume the investment earns 10 percent annually.

A. $13,365.

B. $11,087.

C. $22,857.

B. $11,087. — First draw a timeline: There are two steps to this problem: 1. First, calculate the present value of the four $40,000 payments at t = 7. The four college payments represent an annuity. So, this is equivalent to calculating the present value of an annuity. N = 4, %i = 10, PMT = 40000, FV = 0, CPT PV. PV at the end of year 7 = 126,794.62. 2. Next, calculate the value of the payment to be made every year from year 1 to year 7. To do this, equate the value at end of year 7 calculated in the previous step to the future value of 7 investments of X at year 7. This uses the formula for the future value of an annuity. N = 7, %i = 10, PV = 0, FV = 126,794.62, CPT PMT. PMT = 13,364.85. Jones must invest 13,364.85 each year for 7 years, starting next year, towards his son’s tuition fees.

Q74

Sally Smith is a pension fund manager. According to her estimates, retirees will be paid benefits worth $0.75 million per year, starting 12 years from now. There will be a total of 20 payments. Given a discount rate of 8 percent, the present value of the payments today is closest to:

A. $2,924,191.

B. $3,158,126.

C. $7,363,610.

B. $3,158,126. — Given that the first annuity payment will be at the end of Year 12, we should compute the present value at the end of Year 11. N = 20, %i = 8, PMT = 750,000, FV = 0, CPT PV. The present value of 20 $0.75 million payments at the end of Year 11 is 7,363,610.56. Next, discount it back to t = 0 to determine its present value today: . ( )

Q75

Bill Graham is planning to buy a security which pays a dividend of $100 per year indefinitely, with the first payment to be received at t = 4. Given that the required rate of return is 10 percent per year compounded annually, how much should Graham pay today for the security?

A. $683.

B. $751.

C. $1,000.

B. $751. — We are required to calculate the present value of a perpetuity at t=3 and then discount it back to t=0. The Time Value of Money – Question Bank www.ift.world ( ) Graham must pay $751 to receive $100 per year indefinitely after four years.

Q76

Gerard Jones plans to save for his 5-year doctorate degree, which starts 6 years from now. The current annual expenditure is $7,200 and it is expected to grow by 7 percent annually. Gerard will need to make the first payment 6 years from today. He identifies a savings plan The Time Value of Money – Question Bank www.ift.world that allows him to earn an interest of 8 percent annually. How much should Gerard deposit each year, starting one year from today? Assume that he plans to make 5 payments.

A. $8,370.

B. $8,539.

C. $8,730.

A. $8,370. — This problem can be solved in two steps. Step 1: Find the annual expenditures ( ) ( ) ( ) ( ) ( ) Step 2: Find the present value of annual expenditures at t = 5 Time Period Annual Expenditure Present Value ($) 6 10,805 ( ) 7 11,562 ( ) 8 12,371 ( ) 9 13,237 ( ) 10 14,163 ( ) SUM = $49,106 Step 3: Find the annuity payment N = 5, %i = 8, PV = 0, FV = 49,106, CPT PMT. PMT = 8,370.

Concepts 'n' Clarity^®

For Educational Use Only — Not for Resale

Ch 2 — Time Value of Money in Finance

Concepts 'n' Clarity^®

Ch 3 — Statistical Measures of Asset Returns

CFA Level I • FRM Part I

PART 1 — LOS-ALIGNED STUDY NOTES

LOS 3.a — Calculate, interpret, and evaluate measures of central tendency and location

Central Tendency & Location

AM = Σxᵢ/n | GM = [(1+r₁)···(1+rₙ)]^(1/n)−1 | HM = n/Σ(1/xᵢ)

Trimmed mean: removes top and bottom x% of observations — reduces outlier effect.
Winsorized mean: replaces outliers with boundary values (e.g., 5th and 95th percentile values).
Quantiles: Quartiles (quarters), Quintiles (fifths), Deciles (tenths), Percentiles (hundredths).
Median splits distribution at 50th percentile; robust to outliers.
Mode = most frequent value; can be bimodal or multimodal.

LOS 3.b — Calculate, interpret, and evaluate measures of dispersion

Dispersion Measures

s² = Σ(xᵢ−x̄)²/(n−1) | MAD = Σ|xᵢ−x̄|/n

CV = s/x̄ | Sharpe = (R̄−Rf)/σ

Target downside deviation = √[Σ min(rᵢ−rT, 0)²/(n−1)] — only counts returns below target rT

Range: max − min; simplest, ignores distribution.
MAD: average of absolute deviations; easier to interpret than variance.
Sample variance: divides by n−1 (Bessel's correction) for unbiasedness.
CV: relative risk measure; use to compare investments of different scales.
Target downside deviation/semideviation: measures only downside risk below a target return.
Sharpe ratio: excess return per unit total risk; higher is better.

LOS 3.c — Interpret and evaluate measures of skewness and kurtosis

Skewness & Kurtosis

Normal distribution: skewness=0, excess kurtosis=0, kurtosis=3

Positive skew (right skew): long right tail; mean > median > mode; occasional large gains.
Negative skew (left skew): long left tail; mean < median < mode; occasional large losses.
Equity returns tend to be negatively skewed.
Leptokurtic (excess kurtosis > 0): fat tails; more extreme events than normal → risk underestimation.
Platykurtic (excess kurtosis < 0): thin tails; fewer extremes.
Financial returns are typically leptokurtic — standard VaR models underestimate tail risk.

LOS 3.d — Interpret correlation between two variables

Correlation

r = Cov(X,Y) / (σ_X × σ_Y) | −1 ≤ r ≤ +1

Correlation = 0: no LINEAR relationship (may have nonlinear relationship).
Correlation ≠ causation — spurious correlations can arise from common third factors.
Correlation is not affected by linear transformations (scaling/shifting data).
Limitations: only captures linear relationships; sensitive to outliers.

PART 2 — PRACTICE QUESTIONS (61 Questions)

Click Show Answer to reveal the answer and explanation.

LOS 3.a & 3.b — Central Tendency, Dispersion, Mean Types — Q1 to Q35

A company reports its past six years' earnings growth at 10%, 14%, 12%, 10%, –10%, and 12%. The company's average compound annual growth rate of earnings is closest to:

A. 8.0%.

B. 7.7%.

C. 8.5%.

A. 8.0%. — Geometric mean = [(1.10)(1.14)(1.12)(1.10)(0.90)(1.12)]1/6 − 1 = 0.0766, or 7.66%

Cameron Ryan wants to make an offer on the condominium he is renting. He takes a sample of prices of condominiums in his development that closed in the last five months. Sample prices are as follows (amounts are in thousands of dollars): $125, $175, $150, $155 and $135. The sample standard deviation is closest to:

A. 370.00.

B. 19.24.

C. 38.47.

B. 19.24. — Calculations are as follows: 1. Sample mean = (125 + 175 + 150 + 155 + 135) / 5 = 148 2. Sample Variance = [(125 – 148)2 + (175 – 148)2 + (150 – 148)2 + (155 – 148)2 + (135 – 148)2] / (5 – 1) = 1,480 / 4 = 370 3. Sample Standard Deviation = 3701/2 = 19.24%.

An analyst gathers the following data about the mean monthly returns of three securities: Security Mean Monthly Return Standard Deviation X 0.9 0.7 Y 1.2 4.7 Z 1.5 5.2 Which security has the highest level of relative risk as measured by the coefficient of variation?

A. X.

B. Z.

C. Y.

A. X. — The coefficient of variation, CV = standard deviation / arithmetic mean, is a common measure of relative dispersion (risk). CV = 0.7 / 0.9 = 0.78; CV = 4.7 / 1.2 = 3.92; and CV = X Y Z 5.2 / 1.5 = 3.47. Because a higher CV means higher relative risk, Security Y has the highest relative risk.

An analyst takes a sample of yearly returns of aggressive growth funds resulting in the following data set: 25, 15, 35, 45, and 55. The mean absolute deviation (MAD) of the data set is closest to:

A. 16.

B. 12.

C. 20.

A. 16. — Calculate the mean: 25+15+35+45+55 = 35 5 To get the mean absolute deviation, sum the deviations around the mean (ignoring the sign), and divide by the number of observations: 10+20+0+10+20 = 12 5

Trina Romel, mutual fund manager, is taking over a poor-performing fund from a colleague. Romel wants to calculate the return on the portfolio. Over the last five years, the fund's annual percentage returns were: 25, 15, 12, -8, and –14. Determine if the geometric return of the fund will be less than or greater than the arithmetic return and calculate the fund's geometric return: Geometric compared to Geometric Return Arithmetic

A. 12.86% greater than

B. 4.96% greater than

C. 4.96% less than

C. 4.96% less than — The geometric return is calculated as follows: [(1 + 0.25)(1 + 0.15)(1 + 0.12)(1 - 0.08)(1 – 0.14)]1/5 – 1, or [1.25 × 1.15 × 1.12 × 0.92 × 0.86]0.2 – 1 = 0.4960, or 4.96%. The geometric return will always be less than or equal to the arithmetic return. In this case the arithmetic return was 6%.

What is the compound annual growth rate for stock A which has annual returns of 5.60%, 22.67%, and -5.23%?

A. 7.08%.

B. 6.00%.

C. 8.72%.

A. 7.08%. — Compound annual growth rate is the geometric mean. (1.056 × 1.2267 × 0.9477)1/3 – 1 = 7.08%

A portfolio is equally invested in Stock A, with an expected return of 6%, and Stock B, with an expected return of 10%, and a risk-free asset with a return of 5%. The expected return on the portfolio is:

A. 7.0%.

B. 7.4%.

C. 8.0%.

A. 7.0%. — (0.333)(0.06) + (0.333)(0.10) + 0.333(0.05) = 0.07

Consider the following set of stock returns: 12%, 23%, 27%, 10%, 7%, 20%,15%. The third quartile is:

A. 20.0%.

B. 21.5%.

C. 23%.

C. 23%. — The third quartile is calculated as: L = (7 + 1) (75/100) = 6. When we order the y observations in ascending order: 7%, 10%, 12%, 15%, 20%, 23%, 27%, "23%" is the sixth observation from the left.

A portfolio's monthly returns follow a distribution with a kurtosis measure of 4.2. Relative to a portfolio with normally distributed returns, this portfolio has a: higher probability of extreme upside returns and higher chance of extreme

A. downside returns. lower probability of extreme upside returns and higher chance of extreme

B. downside returns. higher probability of extreme upside returns and lower chance of extreme downside

C. returns.

A. downside returns. lower probability of extreme upside returns and higher chance of extreme — A leptokurtic distribution (a distribution with kurtosis measure greater than 3) is more peaked in the middle (data more clustered around the mean) and has fatter tails at the extremes (greater probability of outliers).

Q10

Assume that the following returns are a sample of annual returns for firms in the clothing industry. Firm 1 Firm 2 Firm 3 Firm 4 Firm 5 15% 2% 5% (7%) 0% The sample standard deviation is closest to:

A. 5.7.

B. 7.2.

C. 8.0.

C. 8.0. — The sample variance is found by taking the sum of all squared deviations from the mean and dividing by (n – 1). [(15 – 3)2 + (2 – 3)2 + (5 – 3)2 + (-7 – 3)2 + (0 – 3)2] / (5 – 1) = 64.5 The sample standard deviation is found by taking the square root of the sample variance. √64.5 = 8.03

Q11

For a unimodal distribution with negative skewness:

A. the mean is greater than the mode.

B. the median is greater than the mean.

C. the mode is less than the median.

A. the mean is greater than the mode. — For a distribution with negative skewness, mean < median < mode.

Q12

In a negatively skewed distribution, what is the order (from lowest value to highest) for the distribution's mode, mean, and median values?

A. Mean, median, mode.

B. Median, mode, mean.

C. Mode, mean, median.

A. Mean, median, mode. — In a negatively skewed distribution, the mean is less than the median, which is less than the mode.

Q13

For the past three years, Acme Corp. has generated the following sample returns on equity (ROE): 4%, 10%, and 1%. What is the sample variance of the ROE over the last three years?

A. 21.0(%2).

B. 4.6%.

C. 21.0%.

A. 21.0(%2). — [(4 – 5)2 + (10 – 5)2 + (1 – 5)2] / (3 – 1) = 21(%2).

Q14

An analyst observes the following four annual returns: R = +10%, R = –15%, R = 0%, and R 1 2 3 4 = +5%. The average compound annual rate over the four years is closest to:

A. 0.0%.

B. –0.5%.

C. –5.0%.

B. –0.5%. — G = [(1.10)(0.85)(1.00)(1.05)]0.25 – 1 G = (0.98175)0.25 – 1 = 0.9954 – 1 = –0.00459 ≈ –0.5% Note: Taking a number to the 0.25 power is the same as taking the fourth root of the number.

Q15

A distribution with a mode of 10 and a range of 2 to 25 would most likely be:

A. positively skewed.

B. negatively skewed.

C. normally distributed.

A. positively skewed. — The distance to the left from the mode to the beginning of the range is 8. The distance to the right from the mode to the end of the range is 15. Therefore, the distribution is skewed to the right, which means that it is positively skewed.

Q16

A sample of returns for four randomly selected assets in a portfolio is shown below: Asset Return (%) A 1.3 B 1.4 C 2.2 D 3.4 What is the sample standard deviation of asset returns?

A. 0.88%.

B. 0.97%.

C. 1.13%.

B. 0.97%. — The sample standard deviation equals the square root of the sum of the squares of the position returns less the mean return, divided by the number of observations in the sample minus one. Position Return (%) (Return – Mean)2 A 1.3 0.60 B 1.4 0.46 C 2.2 0.02 D 3.4 1.76 Mean 8.3/4 = 2.075 Sum = 2.83 Std. Dev. = [2.83 / (4 – 1)]0.5 = 0.97

Q17

For the last four years, the returns for XYZ Corporation's stock have been 10.4%, 8.1%, 3.2%, and 15.0%. The equivalent compound annual rate is:

A. 9.1%.

B. 8.9%.

C. 9.2%.

A. 9.1%. — (1.104 × 1.081 × 1.032 × 1.15)0.25 – 1 = 9.1%

Q18

Find the respective mean and the mean absolute deviation (MAD) of a series of stock market returns. Year 1 14% Year 2 20% Year 3 24% Year 4 22%

A. 20%; 12%.

B. 22%; 3%.

C. 20%; 3%.

C. 20%; 3%. — (14 + 20 + 24 + 22) / 4 = 20 (mean) Take the absolute value of the differences and divide by n: MAD = [|14 – 20| + |20 – 20| + |24 – 20| + |22 – 20|] / 4 = 3%.

Q19

Given the following box-and-whisker plot: The interquartile range is:

A. 0.5% to 5.2%.

B. 3.1% to 5.2%.

C. 0.5% to 3.1%.

A. 0.5% to 5.2%. — The interquartile range is from the first quartile (25th percentile) to the third quartile (75th percentile) and is represented as the box in a box-and-whisker plot. The horizontal line within the box represents the median (50th percentile).

Q20

If Stock X's expected return is 30% and its expected standard deviation is 5%, Stock X's expected coefficient of variation is:

A. 0.167.

B. 1.20.

C. 6.0.

A. 0.167. — The coefficient of variation is the standard deviation divided by the mean: 5 / 30 = 0.167.

Q21

Given the following set of data: 17, 3, 13 , 3, 5, 9, 8 The value 8 is most accurately described as the:

A. mean.

B. median.

C. mode.

A. mean. — Median = middle of distribution = 8 (middle number); Mean = (3 + 3 + 5 + 8 + 9 + 13 + 17) / 7 = 8.28; Mode = most frequent observation = 3.

Q22

Given the following annual returns, what are the geometric and arithmetic mean returns, respectively? 2002 2003 2004 2005 2006 15% 2% 5% -7% 0%

A. 1.45%; 3.00%.

B. 2.75%; 3.00%.

C. 2.75%; 5.80%.

C. 2.75%; 5.80%. — Geometric Mean: (1.15 × 1.02 × 1.05 × 0.93 × 1.0)1/5 – 1 = 1.14541/5 – 1 = 2.75% Arithmetic Mean: (15% + 2% + 5% – 7% + 0%) / 5 = 3.00%

Q23

A distribution of returns that has a greater percentage of small deviations from the mean and a greater percentage of large deviations from the mean compared to a normal distribution:

A. has positive excess kurtosis.

B. has negative excess kurtosis.

C. is positively skewed.

A. has positive excess kurtosis. — A distribution that has a greater percentage of small deviations from the mean and a greater percentage of large deviations from the mean will be leptokurtic and will exhibit positive excess kurtosis. The distribution will be taller (more peaked) with fatter tails than a normal distribution.

Q24

If a distribution is positively skewed, then generally:

A. mean < median < mode.

B. mean > median < mode.

C. mean > median > mode.

A. mean < median < mode. — When a distribution is positively skewed the right side tail is longer than normal due to outliers. The mean will exceed the median, and the median will generally exceed the mode because large outliers falling to the far right side of the distribution can dramatically influence the mean.

Q25

The following data points are observed returns. 4.2%, 6.8%, 7.0%, 10.9%, 11.6%, 14.4%, 17.0%, 19.0%, 22.5% What return lies at the 70th percentile (70% of returns lie below this return)?

A. 14.4%.

B. 17.0%.

C. 19.0%.

B. 17.0%. — With 9 observations, the location of the 70th percentile is (9 + 1)(70 / 100) = 7. The seventh observation in ascending order is 17.0%.

Q26

Which of the following statements about kurtosis is least accurate? Kurtosis:

A. describes the degree to which a distribution is not symmetric about its mean.

B. is used to reect the probability of extreme outcomes for a return distribution. measures the peakedness of a distribution reecting a greater or lesser

C. concentration of returns around the mean.

A. describes the degree to which a distribution is not symmetric about its mean. — The degree to which a distribution is not symmetric about its mean is measured by skewness. Excess kurtosis which is measured relative to a normal distribution, indicates the peakedness of a distribution, and also reflects the probability of extreme outcomes.

Q27

Consider the following statements about the geometric and arithmetic means as measures of central tendency. Which statement is least accurate? The geometric mean may be used to estimate the average return over a one-period

A. time horizon because it is the average of one-period returns. The dierence between the geometric mean and the arithmetic mean increases

B. with an increase in variability between period-to-period observations. The geometric mean calculates the rate of return that would have to be earned each

C. year to match the actual, cumulative investment performance.

A. time horizon because it is the average of one-period returns. The dierence between the geometric mean and the arithmetic mean increases — The arithmetic mean may be used to estimate the average return over a one-period time horizon because it is the average of one-period returns. Both remaining statements are true.

Q28

What is the seventh decile of the following data points? 81 84 91 97 102 108 110 112 115 121 128 135 138 141 142 147 153 155 159 162

A. 142.0.

B. 141.0.

C. 141.7.

C. 141.7. — The formula for determining quantiles is: L = (n + 1)(y) / (100). Here, we are looking for the y seventh decile (70% of the observations lie below) and the formula is: (21)(70) / (100) = 14.7. The seventh decile falls between 141.0 and 142.0, the fourteenth and fifteenth numbers from the left. Since L is not a whole number, we interpolate as: 141.0 + (0.70) (142.0 – 141.0) = 141.7.

Q29

The mean monthly return on a sample of small stocks is 4.56% with a standard deviation of 3.56%. If the risk-free rate is 1%, what is the coefficient of variation?

A. 1.28.

B. 1.00.

C. 0.78.

C. 0.78. — The coefficient of variation expresses how much dispersion exists relative to the mean of a distribution. It is a measure of risk per unit of mean return. CV = s / mean. 3.56 / 4.56 = 0.781, or 78%.

Q30

What is the coefficient of variation for a distribution with a mean of 10 and a variance of 4?

A. 20%.

B. 25%.

C. 40%.

A. 20%. — Coefficient of variation, CV = standard deviation / mean. The standard deviation is the square root of the variance, or 4½ = 2. So, CV = 2 / 10 = 20%.

Q31

A distribution that has positive excess kurtosis is:

A. more peaked than a normal distribution.

B. more skewed than a normal distribution.

C. less peaked than a normal distribution.

A. more peaked than a normal distribution. — A distribution with positive excess kurtosis is more peaked and has fatter tails than a normal distribution.

Q32

An investor has a portfolio with 10% cash, 30% bonds, and 60% stock. If last year's return on cash was 2.0%, the return on bonds was 9.5%, and the return on stock was 25%, what was the return on the investor's portfolio?

A. 12.17%.

B. 18.05%.

C. 11.77%.

B. 18.05%. — Find the weighted mean of the returns. (0.10 × 0.02) + (0.30 × 0.095) + (0.60 × 0.25) = 18.05% Asset Weight Return Weight × Return Cash 10% 2% 10% × 2% = 0.2% Bonds 30% 9.5% 30% × 9.5% = 2.85% Stock 60% 25% 60% × 25% = 15% Weighted Average Return 18.05% ∑ Weight × Probability

Q33

An investor has a $12,000 portfolio consisting of $7,000 in stock P with an expected return of 20% and $5,000 in stock Q with an expected return of 10%. What is the investor's expected return on the portfolio?

A. 15.8%.

B. 15.0%.

C. 30.0%.

A. 15.8%. — Here we need to multiply the returns by the proportion that each stock represents in the portfolio then sum. Stock Return Invested Proportion of Portfolio Return × Proportion P 20% $7,000 7/12 20% × 7/12 Q 10% $5,000 5/12 10% × 5/12 Total $12,000 15.83%

Q34

Michael Philizaire decides to calculate the geometric average of the appreciation/deprecation of his home over the last five years. Using comparable sales and market data he obtains from a local real estate appraiser, Philizaire calculates the year-to- year percentage change in the value of his home as follows: 20, 15, 0, –5, –5. The geometric return is closest to:

A. 0.00%.

B. 4.49%.

C. 11.60%.

B. 4.49%. — The geometric return is calculated as follows: [(1 + 0.20) × (1 + 0.15) × (1 + 0.0) (1 – 0.05) (1 – 0.05)]1/5 – 1, or [1.20 × 1.15 × 1.0 × 0.95 × 0.95]0.2 – 1 = 0.449, or 4.49%.

Q35

The respective arithmetic mean and geometric mean returns of the following series of stock market returns are: Year 1 14% Year 2 6% Year 3 −5% Year 4 20%

A. 8.75%; 8.62%.

B. 8.90%; 8.62%.

C. 8.75%; 8.34%.

C. 8.75%; 8.34%. — (14 + 6 + (-5) + 20) / 4 = 8.75. ((1.14 × 1.06 × 0.95 × 1.20)0.25 – 1 = 8.34%.

LOS 3.c & 3.d — Skewness, Kurtosis, Normal Distribution, Correlation — Q36 to 61

Q36

Which of the following statements concerning kurtosis is most accurate?

A. A distribution with kurtosis of +2 has fatter tails than a normal distribution.

B. A leptokurtic distribution has excess kurtosis less than zero.

C. A leptokurtic distribution has fatter tails than a normal distribution.

A. A distribution with kurtosis of +2 has fatter tails than a normal distribution. — A leptokurtic distribution is more peaked than normal and has fatter tails. However, the excess kurtosis is greater than zero.

Q37

What are the median and the third quintile of the following data points, respectively? 9.2%, 10.1%, 11.5%, 11.9%, 12.2%, 12.8%, 13.1%, 13.6%, 13.9%, 14.2%, 14.8%, 14.9%, 15.4%

A. 12.8%; 13.6%.

B. 13.1%; 13.6%.

C. 13.1%; 13.7%.

C. 13.1%; 13.7%. — The median is the midpoint of the data points. In this case there are 13 data points and the midpoint is the 7th term. The formula for determining quantiles is: L = (n + 1)(y) / (100). Here, we are looking for the y third quintile (60% of the observations lie below) and the formula is: (14)(60) / (100) = 8.4. The third quintile falls between 13.6% and 13.9%, the 8th and 9th numbers from the left. Since L is not a whole number, we interpolate as: 0.136 + (0.40)(0.139 – 0.136) = 0.1372, or 13.7%.

Q38

An analyst compiles the returns on Fund Q over the last four years: Year Return 1 4% 2 3% 3 2% 4 30% Which of the following will result in the lowest measure of the mean return?

A. The arithmetic mean.

B. The geometric mean.

C. The harmonic mean.

A. The arithmetic mean. — 4 Harmonic mean = −1 = 0.0864 = 8.64% 1 + 1 + 1 + 1 1.04 1.03 1.02 1.30 1 Geometric mean = [(1.04)(1.03)(1.02)(1.30)]4 −1 = 0.0917 = 9.17% 4%+3%+2%+30% Arithmetic mean = = 9.75% 4

Q39

Annual Returns on ABC Mutual Fund Yr 1 Yr 2 Yr 3 Yr 4 Yr 5 Yr 6 Yr 7 Yr 8 Yr 9 Yr 10 11.0% 12.5% 8.0% 9.0% 13.0% 7.0% 15.0% 2.0% -16.5% 11.0% Assuming a mean of 7.2%, what is the sample standard deviation of the returns for ABC Mutual Fund for the period from Year 1 to Year 10?

A. 7.8%.

B. 9.8%.

C. 9.1%.

C. 9.1%. — Standard deviation = [∑i (x i – X)2 / (n – 1)]1/2 = (744.10 / 9) 1/2 = = 9.1%.

Q40

Which of the following statements concerning skewness is least accurate? A distribution with:

A. skew equal to 1 is not symmetrical.

B. negative skewness has a large number of outliers on its left side.

C. positive skewness has a long left tail.

A. skew equal to 1 is not symmetrical. — A distribution with positive skewness has long right tails.

Q41

A 5% trimmed mean ignores the:

A. highest and lowest 5% of observations.

B. highest and lowest 2.5% of observations.

C. lowest 5% of observations.

B. highest and lowest 2.5% of observations. — A 5% trimmed means discards the highest 2.5% and lowest 2.5% of observations and is the arithmetic average of the remaining 95% of observations.

Q42

The annual returns on 5 portfolio investments for the last year are shown in the following table. What is the return on the portfolio and the geometric mean of the returns on the portfolio investments? Investment Invested Amount Return (%) A 10,000 12 B 10,000 14 C 10,000 9 D 10,000 13 E 10,000 7

A. 11.00; 10.78.

B. 11.00; 10.97.

C. 11.64; 10.97.

C. 11.64; 10.97. — Arithmetic Mean: 12 + 14 + 9 + 13 + 7 = 55; 55 / 5 = 11 Geometric Mean: [(1.12 × 1.14 × 1.09 × 1.13 × 1.07)1/5] – 1 = 10.97%

Q43

Twenty Level I CFA candidates in a study group took a practice exam and want to determine the distribution of their scores. When they grade their exams they discover that one of them skipped an ethics question and subsequently filled in the rest of his answers in the wrong places, leaving him with a much lower score than the rest of the group. If they include this candidate's score, their distribution will most likely:

A. have a mean that is less than its median.

B. be positively skewed.

C. have a mode that is less than its median.

A. have a mean that is less than its median. — With the low outlier included, the distribution will be negatively skewed. For a negatively skewed distribution, the mean is less than the median, which is less than the mode.

Q44

The correlation between two variables is –0.74. The most appropriate way to interpret this correlation is that: if one of the variables increases, there is a 74% probability that the other variable

A. will decrease.

B. the two variables have a negative linear association.

C. there is unlikely to be a strong linear relationship between the two variables.

A. will decrease. — A correlation coefficient of –0.74 suggests a relatively strong negative linear association between the two variables. We cannot interpret the correlation coefficient directly as a measure of the probability that the two variables will change in opposite directions.

Q45

The following annualized monthly return measures have been calculated for an investment based on its performance over the last 72 months. Arithmetic mean 6.8% Geometric mean 6.0% 90% Winsorized mean 5.5% If for one month in the period the return was extremely high, which measure best reflects the central tendency of the investment's returns?

A. Winsorized mean.

B. Geometric mean.

C. Arithmetic mean.

A. Winsorized mean. — A winsorized mean is a technique for removing the distorting effects of outliers by replacing them with less extreme values. The arithmetic and geometric means are based on all observations and therefore include the impact of outliers.

Q46

In a positively skewed distribution, what is the order (from lowest value to highest) for the distribution's mode, mean, and median values?

A. Mode, mean, median.

B. Mode, median, mean.

C. Mean, median, mode.

A. Mode, mean, median. — In a positively skewed distribution, the mode is less than the median, which is less than the mean.

Q47

The owner of a company has recently decided to raise the salary of one employee, who was already making the highest salary in the company, by 40%. Which of the following value(s) is (are) expected to be affected by this raise?

A. mean only.

B. both mean and median.

C. median only.

A. mean only. — Mean is affected because it is the sum of all values / number of observations. Median is not affected as it the midpoint between the top half of values and the bottom half of values.

Q48

Which of the following statements concerning a distribution with positive skewness and positive excess kurtosis is least accurate? It has a lower percentage of small deviations from the mean than a normal

A. distribution.

B. It has fatter tails than a normal distribution.

C. The mean will be greater than the mode.

A. distribution. — A distribution with positive excess kurtosis has a higher percentage of small deviations from the mean than normal. So it is more "peaked" than a normal distribution. A distribution with positive skew has a mean > mode.

Q49

The mean monthly return on a security is 0.42% with a standard deviation of 0.25%. What is the coefficient of variation?

A. 168%.

B. 60%.

C. 84%.

B. 60%. — The coefficient of variation expresses how much dispersion exists relative to the mean of a distribution and is found by CV = s / mean, or 0.25 / 0.42 = 0.595, or 60%.

Q50

What does it mean to say that an observation is at the sixty-fifth percentile?

A. The observation falls within the 65th of 100 intervals.

B. 65% of all the observations are below that observation.

C. 65% of all the observations are above that observation.

C. 65% of all the observations are above that observation. — If the observation falls at the sixty-fifth percentile, 65% of all the observations fall below that observation.

Q51

If the historical mean return on an investment is 2.0%, the standard deviation is 8.8%, and the risk free rate is 0.5%, what is the coefficient of variation (CV)?

A. 0.17.

B. 4.40.

C. 0.23.

A. 0.17. — The CV = the standard deviation of returns / mean return = 8.8% / 2.0% = 4.4. The CV is a measure of risk per unit of mean return. When ranking portfolios based on the CV, a lower value is preferred to higher.

Q52

Given the following annual returns, what is the mean absolute deviation? 2000 2001 2002 2003 2004 15% 2% 5% -7% 0%

A. 0.0%.

B. 3.0%.

C. 5.6%.

C. 5.6%. — The mean absolute deviation is found by taking the mean of the absolute values of deviations from the mean. ( |15 – 3| + |2 – 3| + |5 – 3| + |-7 – 3| + |0 – 3|) / 5 = 5.60%

Q53

A distribution that is more peaked than a normal distribution is termed:

A. platykurtic.

B. leptokurtic.

C. skewed.

A. platykurtic. — A distribution that is more peaked than normal is leptokurtic. A leptokurtic distribution has fatter tails compared to a normal distribution. This means there is a greater chance of observing extreme outcomes. Market returns are leptokurtic. A distribution that is flatter than a normal distribution is termed platykurtic.

Q54

Returns for a portfolio over the last four years are shown below. Treating these returns as a sample, what is their coefficient of variation (CV)? Year Return 1 17.0% 2 12.2% 3 3.9% 4 –8.4%

A. 1.80.

B. 1.56.

C. 0.55.

A. 1.80. — The coefficient of variation is equal to the standard deviation of returns divided by the mean return. Mean return = (17.0% + 12.2% + 3.9% – 8.4%) / 4 = 6.175% Year Return (R – 6.175%)2 1 17.0% 117.18 2 12.2% 36.30 3 3.9% 5.18 4 –8.4% 212.43 Sum = 371.09 Sample standard deviation = [371.09 / (4 – 1)]0.5 = 11.12% Coefficient of variation = 11.12% / 6.175% = 1.80

Q55

An analyst calculates a winsorized mean return of 3.2% for an investment fund. This measure most likely:

A. equally weights all returns.

B. replaces outliers with less extreme returns.

C. captures the compounded growth rate of the fund.

A. equally weights all returns. — The winsorized mean is a technique for dealing with outliers. For example, a 90% winsorized mean replaces the lowest 5% of values with the fifth percentile, and replaces the highest 5% of values with the 95th percentile. The arithmetic mean weights all observations equally. The geometric mean captures the compounded growth rate of the fund.

Q56

Over the last five years, an investment fund's monthly returns were relatively stable apart from last year, where two extremely high returns were recorded. If the arithmetic mean for the fund's monthly returns over the period is 6.7%, a trimmed or winsorized mean return is most likely to be:

A. higher than the arithmetic mean.

B. lower than the arithmetic mean.

C. equal to the arithmetic mean.

A. higher than the arithmetic mean. — A trimmed mean discards a percentage of the highest and lowest observations, while a winsorized mean replaces a percentage of the highest and lowest observations with less extreme values. In this case the arithmetic mean would be influenced by the two highly positive returns, while a trimmed or winsorized mean would adjust for them and would likely be lower than the arithmetic mean.

Q57

The correlation coefficient between the return on an investment and the rate of economic growth is –0.065. An analyst should most likely interpret this correlation coefficient as indicating that returns on this investment are:

A. not related linearly to economic growth.

B. unrelated to economic growth.

C. negatively related to economic growth.

A. not related linearly to economic growth. — A correlation coefficient near zero indicates that two variables exhibit no linear relationship. This does not necessarily mean that the variables are unrelated because they might exhibit a nonlinear relationship.

Q58

If an analyst concludes that the distribution of a large sample of returns is positively skewed, which of the following relationships involving the mean, median, and mode is most likely?

A. Mean > median > mode.

B. Mean < median < mode.

C. Mean > median < mode.

A. Mean > median > mode. — For the positively skewed distribution, the mode is less than the median, which is less than the mean.

Q59

An investor has the following assets: $5,000 in bonds with an expected return of 8%. $10,000 in equities with an expected return of 12%. $5,000 in real estate with an expected return of 10%. What is the portfolio's expected return?

A. 10.50%.

B. 11.00%.

C. 10.00%.

C. 10.00%. — Expected return is the weighted average of the individual expected values. The expected return is: [(5,000) × (10.00) + (5,000) × (8.00) + (10,000) × (12.00)] / 20,000 = 10.50%.

Q60

For a positively skewed distribution, the median is greater than:

A. the mean, but less than the mode.

B. the mode, but less than the mean.

C. both the mode and the mean.

A. the mean, but less than the mode. — For a positively skewed distribution, the mean is greater than the median, and the median is greater than the mode. Their order reverses for a negatively skewed distribution.

Q61

For the investments shown in the table below: Investment Return (%) A 12 B 14 C 9 D 13 E 7 F 8 G 12 Which of the following statements is most accurate?

A. The mean is equal to the median.

B. The median is equal to the mode.

C. The mean is equal to the mode.

A. The mean is equal to the median. — The median is the mid-point or central number of returns arranged from highest to lowest or lowest to highest. In this case: 7, 8, 9, 12, 12, 13, 14. The median return is 12%. The mode is the return that occurs most frequently. In this case, 12% is also the mode. The mean is 75 / 7 = 10.71%.

Q62

Considering the following set of numbers. 40 47 51 69 39 47 48 44 41 53 55 45 37 41 40 42 41 57 Which of the following statements is least likely accurate?

A. The mode is larger than the mean.

B. The median is smaller than the mean but larger than the mode.

C. The mean is larger than both the mode and the median.

B. The median is smaller than the mean but larger than the mode. — The mode is the most frequent value in the set of items and thus is equal to 41. The mean is the average value from the set of items and is computed as follows: The median is the value of the middle item of a set of items and is computed as the average of the n/2th item and the (n+2)/2th item which is the 9th and 10th item. The median is thus 44.5. Therefore the median is smaller than the mean but larger than the mode.

Q63

The following ten observations are a sample drawn from a normal population: 24, 5, 12, 6, - 3, 11, 18, 15, -4, and 29. The mean of the sample is closest to:

A. 11.30.

B. 12.90.

C. 14.00.

A. 11.30. — The sum of the ten numbers is 113. Dividing by 10 gives the mean of 11.30.

Q64

Over the past five years, a portfolio gave returns of 18%, 12% -5%, -10% and 7%. The geometric mean return of the portfolio over the five year period is closest to:

A. 3.87%.

B. 4.40%.

C. 10.31%.

Q65

A portfolio has the following annual returns: 5%, 11%, -6%, 0%. The geometric mean across the four-year period is closest to:

A. 2.3%.

B. 2.5%.

C. 2.6%.

A. 2.3%. — The geometric mean return is calculated as [(1 + 0.05) × (1 + 0.11) × (1 - 0.06) × (1 + 0.00)] 0.25 – 1 = 0.0231 ~ 2.3%.

Q66

The table below shows some sample figures: -25 -12 -2 0 1 2 6 7 9 11 13 19 20 21 25 Statistical Concepts and Market Returns – Question Bank www.ift.world 29 39 41 55 65 The median value of the above items is closest to:

A. 11.

B. 12.

C. 13.

B. 12. — Median is the value of the middle item of a set of items. The value of the 10th item is 11; the value of the 11th item is 13. The mean of 11 and 13 is 12.

Q67

The discount rate set by the central bank of country XYZ for the past 6 quarters is shown below: Quarter Discount Rate Quarter Discount Rate 1 10.0% 4 8.5% 2 10.5% 5 9.6% 3 11.4% 6 11.5% The sample mean rate for the 6 quarters is closest to:

A. 10.25%.

B. 11.24%.

C. 12.3%.

A. 10.25%. — ∑

Q68

Which of the following statements about arithmetic mean is most accurate?

A. Deviations from the arithmetic mean indicate risk.

B. The product of the deviations around the mean is equal to 0.

C. The disadvantage of an arithmetic mean is it fails to make use of all data.

A. Deviations from the arithmetic mean indicate risk. — B is incorrect because the sum of the deviations around the mean is equal to 0. C is incorrect because the advantage of an arithmetic mean is that it makes use of all data.

Q69

The discount rate set by the central bank of Romulus for the past 6 quarters is shown below: Quarter Discount Rate Quarter Discount Rate 1 10.0% 4 8.5% 2 10.5% 5 9.6% 3 11.4% 6 11.5% The median is closest to:

A. 10.25%.

B. 10.50%.

C. 11.25%.

A. 10.25%. — Arrange the data in ascending order as: 8.5, 9.6, 10.0, 10.5, 11.4, 11.5 Since there are an even number of observations, take the average of the two middle values to calculate median: Median = 10.25 percent Statistical Concepts and Market Returns – Question Bank www.ift.world

Q70

The following table shows the intervals for temperatures recorded at different places all over the country. The respective frequency for each interval represents the number of towns and cities. Interval Frequency 0 < r < 10 29 10 < r < 20 37 Statistical Concepts and Market Returns – Question Bank www.ift.world 20 < r < 30 31 30 < r < 40 35 40 < r < 50 33 The modal interval of this distribution is most likely:

A. 0 < r < 10.

B. 10 < r < 20.

C. 30 < r < 40.

B. 10 < r < 20. — The modal interval is the interval with the highest frequency, which in this case, is 10 < r < 20.

Q71

A portfolio manager is computing the weighted mean of a portfolio, whose asset allocation as of 31 December, 2012, is given below: Local Equities: 25% International Equities: 13% Bonds: 27% Mortgage: 18% Gold: 17% The returns on the above assets on 31 December, 2012, were 5.4%, 8.9%, -2.5%, -7%, and 11% respectively. The mean return earned by the portfolio is closest to:

A. 2.44%.

B. 3.16%.

C. 4.88%.

A. 2.44%. — Mean portfolio return is the weighted average of each asset class‟ returns. ̅̅̅̅ ( ) ( ) ( ) ( ) ( )

Q72

A. 18.32%.

B. 18.34%.

C. 30.21%.

A. 18.32%. — First, calculate the holding period returns at the end of year 1 and year 2. [( )( )]

Q73

A. $22.33.

B. $22.15.

C. $22.06.

B. $22.15. — *( ) ( ) ( )+

Q74

If all the observations in a data set have different values, then which of the following relationships is most accurate?

A. Arithmetic Mean < Geometric Mean < Harmonic Mean.

B. Geometric Mean < Harmonic Mean < Arithmetic Mean.

C. Harmonic Mean < Geometric Mean < Arithmetic Mean.

C. Harmonic Mean < Geometric Mean < Arithmetic Mean. — Unless all observations in a data set are equal, the harmonic mean is less than the geometric mean which is less than the arithmetic mean.

Q75

The following table shows the returns of various stocks of a portfolio, ranked in ascending order: Stock Return (%) Stock Return (%) Stock 1 10.50 Stock 6 14.24 Stock 2 11.25 Stock 7 14.75 Stock 3 12.05 Stock 8 15.30 Stock 4 12.65 Stock 9 16.00 Stock 5 13.55 Stock 10 17.45 The value of the third quintile is closest to:

A. 15.475%.

B. 14.55%.

C. 15.30%.

B. 14.55%. — The position of the third quintile can be found through the following formula: ( ) ( ); Where, y is the percentage point at which we are dividing the distribution. Here, y = 60, the 60th percentile (third quintile); n = 10 ( ) ( ) = 6.6 Therefore, the location of the third quintile is between the return of Stock 6 and Stock 7. Linear interpolation is used for finding the approximate value of the third quintile. In the above case, return on the 6th stock is 14.24% and on the 7th stock is 14.75%. L60 = 14.55% which is 14.24% (6th value) plus 0.6 times the linear distance between 14.24% and 14.75%.

Q76

The table below shows data on volatility of a series of funds: Volatility (%) Fund 1 5.05% Fund 2 6.20% Fund 3 6.93% Fund 4 7.56% Fund 5 8.25% Fund 6 10.11% Fund 7 11.36% Fund 8 14.52% Fund 9 15.02% Fund 10 15.66% Fund 11 15.98% Fund 12 16.01% Fund 13 19.25% The value of the second quintile is closest to:

A. 7.56%.

B. 9.37%.

C. 10.11%.

B. 9.37%. — Quintiles divide data into five parts. Hence the first quintile corresponds to the 20th percentile and the second quintile corresponds to the 40th percentile. The location can be determined using: ( ) ( )  ( ) ( ) = 5.6. The value corresponding to location 5 (Fund 5) is 8.25%. The value corresponding to location 6 (Fund 6) is 10.11%. The approximate value corresponding to location 5.6 can be estimated using linear interpolation: ( ( – )) = 9.37%.

Q77

Which of the following statements is least accurate?

A. The first quintile generally exceeds the first decile.

B. The first quintile generally exceeds the first quartile.

C. The third quintile generally exceeds the median.

B. The first quintile generally exceeds the first quartile. — The first quintile is the 20th percentile and the third quintile is the 60th percentile. The first decile is the 10th percentile, the first quartile is the 25th percentile, and the median is the 50th percentile. While it is possible that these various percentiles or some Statistical Concepts and Market Returns – Question Bank www.ift.world subsets of them might be equal (for example the10th percentile possibly could be equal to the 20th percentile), in general the order from smallest to largest would be: first decile, first quintile, first quartile, median.

Q78

The following ten observations are a sample drawn from a normal population: 6, 12, 32, -12, 10, 3, -21, 15, 8, and 11. The third quintile (60th percentile) of the sample is closest to:

A. 3.0.

B. 10.6.

C. 11.0.

B. 10.6. — First we need to sort the data in ascending order: -21, -12, 3, 6, 8, 10, 11, 12, 15, 32. The third quintile corresponds to the 60th percentile. The location of the 60th percentile is given by: L = (10 + 1) 60 / 100 = 6.6. The value is estimated using linear interpolation: P 60 60 = 10 + 0.6(11 – 10) = 10.6.

Q79

Which of the following statements is least likely accurate?

A. The median is the 50th percentile.

B. Quintiles divide the distribution into fifths.

C. Linear interpolation is used when the location, L, is a whole number.

C. Linear interpolation is used when the location, L, is a whole number. — Linear interpolation is used when the location, L, is not a whole number and lies between two closest integers.

Q80

The following table shows the earnings per share (EPS) of 20 hypothetical companies. Company EPS Company EPS AB $4.50 AAA $13.22 CDE $14.50 BBB $13.25 FG $9.33 CCC $12.98 HIJ $7.21 DDD $11.54 KLM $6.44 EEE $11.73 NO $6.99 FFF $15.00 PQR $6.27 GGG $10.49 STU $8.11 HHH $5.78 VWX $5.25 III $6.50 YZ $12.15 JJJ $4.75 The 30th percentile is closest to:

A. 6.44.

B. 6.29.

C. 6.99.

A. 6.44. — The 30th percentile is the value at or below which 30 percent of observations lie. To solve this problem, we first arrange the 20 data points in ascending order: 4.50, 4.75, 5.25, 5.78, 6.29, 6.44, 6.50, 6.99, 7.21, 8.11, 9.33, 10.49, 11.54, 11.73, 12.15, 12.98, 13.22, 13.25, 14.50, and 15.00. The location of the 30th percentile is 30% of 20 = 6. The 6th data point is 6.44. We can also use the location formula: ( )( ) ( )( ) ( )( ) ( )( ) The two numbers (6.44 and 6.46) are different because the percentile calculations using the location formula gives an approximate answer when n is small.

Q81

For a data set of 100 observations, which of the following best represents a quintile?

A. .

B. .

C. .

B. . — Since quintiles divide the distribution into fifths, they are represented as .

Q82

The second quartile represents which of the following? I. Median II. 50th percentile III. 2nd quintile IV. 5th decile

A. I and II only.

B. I, II, and IV only.

C. II, III, and IV only.

B. I, II, and IV only. — The second quartile is equivalent to the median, the 50th percentile, and the 5th decile.

Q83

Which of the following is closest to the sample variance of the observations given below? -5 4 7 11 4 -8 -1 3 5

A. 31.3.

B. 33.7.

C. 35.2.

C. 35.2. — Using your calculator:  [2ND][DATA]  [2ND][CLR WORK]  5[+/-] [ENTER]  [↓][↓]4 [ENTER]  [↓][↓]7[ENTER]  [↓][↓]11[ENTER]  <and so on for the rest of the data>  [2ND][STAT]  Press [2ND][SET] repeatedly until you get I-V  Press[↓] to begin computing results.  You‟ll Sx (sample standard deviation) = 5.93; simply square this to get the variance = 35.19 Statistical Concepts and Market Returns – Question Bank www.ift.world

Q84

The following observations are drawn from an approximately normal population: Observation 1 2 3 4 5 Value 13 -5 1 3 -8 The sample standard deviation is closest to:

A. 7.22.

B. 7.86.

C. 8.13.

C. 8.13. — Observation Value Difference from Mean Difference2 1 13 12.2 148.84 2 -5 -5.8 33.64 3 1 0.2 0.04 4 3 2.2 4.84 5 -8 -8.8 77.44 Sum 4 264.8 Mean 0.8 The sample variance is given by ∑ ( ̅) ⁄ s2 = 264.8/5-1 = 66.2 The sample standard deviation is simple the square root of the sample variance. The sample standard deviation is 8.13. OR Keystrokes Explanation Display [2nd] [DATA] Enter data entry mode [2nd] [CLR WRK] Clear data registers X01 13 [ENTER] X01 = 13 [↓] [↓] 5 +/- [ENTER] X02 = -5 [↓] [↓] 1 [ENTER] X03 = 1 [↓] [↓] 3 [ENTER] X04 = 3 [↓] [↓] 8 +/- [ENTER] X05 = -8 [2nd] [STAT] [ENTER] Puts calculator into stats mode. [2nd] [SET] Press repeatedly till you see  1-V [↓] Number of data points N = 5 [↓] Mean X = 0.8 Statistical Concepts and Market Returns – Question Bank www.ift.world Keystrokes Explanation Display [↓] Sample standard deviation Sx = 8.136 [↓] Population standard deviation σx = 7.277

Q85

UBL Fund‟s return for the last five years are as follow: Year Return (%) 2009 -12 2010 10 2011 15 2012 17 2013 19 The mean absolute deviation of returns for this fund is closest to:

A. 8.72%.

B. 9.00%.

C. 9.50%.

A. 8.72%. — The Mean Absolute Deviation (MAD) is calculated through the following formula: Mean = (-12+10+15+17+19)/5 = 9.8% MAD = (|-12 - 9.8| + |10 - 9.8| + |15 – 9.8| + |17 – 9.8| + |19 – 9.8|)/5 = 8.72%.

Q86

The annual returns of a stock portfolio since its inception on 1 January 2008 is given below: Year Portfolio return 2008 7.5% 2009 9.0% Statistical Concepts and Market Returns – Question Bank www.ift.world 2010 11.5% 2011 -5.3% 2012 9.7% The portfolio‟s mean absolute deviation for the five-year period is closest to:

A. 3.42%.

B. 4.71%.

C. 6.50%.

B. 4.71%. — Compute the mean: (7.5 + 9.0 + 11.5 – 5.3 + 9.7) / 5 = 6.48% and compute MAD, (|7.5 – 6.48| + |9.0 – 6.48| + |11.5 – 6.48| + | –5.3 – 6.48| + |9.7 – 6.48|) / 5 = 4.71%.

Q87

The returns of a fund are given below: Year Return % 2009 5.0 2010 -3.2 2011 6.1 2012 4.5 2013 1.3 The mean absolute deviation of returns for this fund is closest to:

A. 3.55.

B. 2.95.

C. 3.95.

B. 2.95. — Mean is 2.74. MAD is the mean of the absolute deviations from 2.74: (2.26 + 5.94 + 3.36 + 1.76 + 1.44) / 5 = 2.95.

Q88

The table below shows the temperatures recorded at different places: Place Temperature (degree Celsius) Place Temperature (degree Celsius) A 48 E 51 B 21 F 35 C 10 G 16 D -8 H -15 The range for this data is closest to:

A. 26.

B. 36.

C. 66.

C. 66. — ( ) .

Q89

The weekly rainfall received by some areas in Australia during June 2013 is tabulated below: Port Hedland 100 mm Falls Creek 176 mm Gove 15 mm Esperance 66 mm Millbrook 107 mm Mt Read 74 mm Tree House Creek 120 mm Samuel Hill 113 mm Thredbo 178 mm Hunters Hill 132 mm The range for this data set is closest to:

A. 108 mm.

C. 178 mm.

Q90

The number of pages read by a group of students per day is given below: Patrick 100 Fiona 175 Salis 65 James 50 Melisa 27 Nelson 20 Tina 120 Samuel 75 Thomas 34 Gordon 90 The mean absolute deviation (MAD) is closest to:

A. 36.5.

B. 55.6.

C. 75.6.

A. 36.5. — First, calculate the arithmetic mean return. [∑| ̅|] ̅ ̅ [| | | | | |]

Q91

The dividend yield for five hypothetical companies is given below: Paknama 10.5% Genie Ltd. 16.25% Mirinda Corp. 27.0% Tina Travels Ltd. 12.0% Thomas Press Ltd. 7.8% The population variance is closest to:

A. 36.89.

B. 45.20.

C. 56.49.

B. 45.20. — Use the following keystrokes to calculate the population variance: [2nd] [DATA] [2nd] [CLR WRK] X01 = 10.5 X02 = 16.25 X03 = 27 X04 = 12 X05 = 7.8 Choose the population standard deviation: σ = 6.723. Statistical Concepts and Market Returns – Question Bank www.ift.world Then square it, to get population variance as 45.2.

Q92

The return on equity for four hypothetical companies is given below: Little Wonder 10.5% Genesis Ltd. 16.25% Moral Corp. 9.81% Travis Ltd. 12.0% The population standard deviation is closest to:

A. 2.50.

B. 2.88.

C. 6.25.

A. 2.50. — Use the following keystrokes to calculate the population standard deviation: [2nd] [DATA] [2nd] [CLR WRK] X01 = 10.5 X02 = 16.25 X03 = 9.81 X04 = 12 Population standard deviation: σ =2.5.

Q93

The dividend yield for five hypothetical companies from a list of 100 companies is given below: Paknama 10.5% Genie Ltd. 16.25% Mirinda Corp. 27.0% Statistical Concepts and Market Returns – Question Bank www.ift.world Tina Travels Ltd. 12.0% Thomas Press Ltd. 7.8% The sample variance is closest to:

A. 36.89.

B. 45.20.

C. 56.49.

C. 56.49. — Use the following keystrokes to calculate the sample variance: [2nd] [DATA] [2nd] [CLR WRK] X01 = 10.5 X02 = 16.25 X03 = 27 X04 = 12 X05 = 7.8 Choose the sample standard deviation: s = 7.515. Square this value to get sample variance as 56.49.

Q94

The return on equity for four hypothetical companies from a list of 100 companies is given below: Little Wonder 10.5% Genesis Ltd. 16.25% Moral Corp. 9.81% Travis Ltd. 12.0% The sample standard deviation is closest to:

A. 2.50.

B. 2.88.

C. 6.25.

B. 2.88. — Use the following keystrokes to calculate the sample standard deviation: [2nd] [DATA] [2nd] [CLR WRK] X01 = 10.5 X02 = 16.25 X03 = 9.81 X04 = 12 s represents the value of sample standard deviation = 2.88.

Q95

Diana Sorenson, an equity fund manager has the following information about a common stock portfolio: Arithmetic mean return 12.9% Geometric mean return 10.3% Portfolio beta 1.6 Risk-free rate of return 3.50% Variance of returns 212 From the given information, the coefficient of variation is closest to:

A. 1.13.

B. 1.41.

C. 1.55.

A. 1.13. — The coefficient of variation is: (Standard deviation of return) / (Mean return) = √ / 12.9 = 1.13

Q96

The table below provides data on annual mean returns and standard deviations for three bonds: Asset Class Arithmetic mean return (%) Standard deviation of return (%) Bond A 16.4% 4.9% Bond B 12.6% 3.5% Bond C 14.8% 4.2% Which of the above bonds is least risky using a relative measure?

A. Commercial Bond A.

B. Corporate Bond B.

C. Government Bond C.

B. Corporate Bond B. — In order to find the bond with the lowest risk per unit of return, we need to determine the bond with the lowest coefficient of variation. ̅ where s is the sample standard deviation and ̅ is the sample mean. Bond A: Bond B: Bond C: Bond B, whose standard deviation and CV are the lowest, is least risky.

Q97

A distribution more peaked than the normal distribution is best described as being:

A. platykurtic.

B. mesokurtic.

C. leptokurtic.

C. leptokurtic. — Leptokurtic describes a distribution that is more peaked than the normal distribution. Statistical Concepts and Market Returns – Question Bank www.ift.world Platykurtic is a distribution less peaked than a normal distribution. Mesokurtic is a distribution as peaked as the normal distribution.

Q98

In a continuous distribution where the graph shows the right tail of the curve to be longer than the left tail is best described as having: Statistical Concepts and Market Returns – Question Bank www.ift.world

A. lepto-kurtosis.

B. negative skewness.

C. positive skewness.

C. positive skewness. — A positively skewed distribution appears as if the right tail has been pulled away from the mean.

Q99

Which of the following return distribution is most likely characterized by frequent small losses and a few large gains?

A. Normal distribution.

B. Negatively skewed.

C. Positively skewed.

C. Positively skewed. — A positively skewed distribution has frequent small losses and a few large gains. A negatively skewed distribution has frequent small gains and a few large losses. A normal distribution is symmetrical.

Q100

Which of the following relationships best characterize a negatively skewed distribution?

A. Mean < median < mode.

B. Mode < median < mean.

C. Median < mean < mode.

A. Mean < median < mode. — For a negatively skewed distribution, the mean is less than the median, which is less than the mode.

Q101

Which of the following statements best describe a positively skewed distribution?

A. A distribution skewed to the right.

B. A distribution skewed to the left.

C. A distribution skewed upward.

A. A distribution skewed to the right. — A positively skewed distribution is skewed to the right whereas a negatively skewed distribution is skewed to the left.

Q102

If a distribution exhibits negative skewness, then the mean most likely is located to the:

A. left of both the median and mode.

B. right of both the median and mode.

C. left of the mode and right of the median.

A. left of both the median and mode. — A negatively skewed distribution has a long tail to the left with a large frequency of observations occurring in the right part of the distribution. For a distribution of returns, this means frequent small gains and a few extreme losses. The result is that the extreme losses pull the mean to the left while the mode resides on the right with the bulk of the observations. The median falls between the mean and the mode.

Q103

Which of the following is most likely to be the largest in a positively skewed unimodal distribution?

A. Mean.

B. Median.

C. Mode.

A. Mean. — For the positively skewed unimodal distribution, the mode is less than the median, which is less than the mean.

Q104

Equity return series are best described as:

A. mesokurtotic.

B. platykurtotic.

C. leptokurtotic.

C. leptokurtotic. — Equity return series have been found to be leptokurtotic.

Q105

A distribution that is less peaked than normal is most accurately described as:

B. mesokurtotic.

C. platykurtotic.

C. platykurtotic. — A distribution that is less peaked than normal is called platykurtotic.

Q106

Which of the following statistical measures most likely measure the peakedness of a distribution such as more or less peaked than a normal distribution?

A. Skewness.

B. Chebyshev‟s inequality.

C. Kurtosis.

C. Kurtosis. — Kurtosis is the statistical measure that tells us about the peakedness of a distribution.

Q107

A distribution identical to a normal distribution is:

A. Leptokurtic.

B. Mesokurtic.

C. Platykurtic.

B. Mesokurtic. — A distribution identical to a normal distribution is mesokurtic.

Concepts 'n' Clarity^®

For Educational Use Only — Not for Resale

Ch 3 — Statistical Measures of Asset Returns

Concepts 'n' Clarity^®

Ch 4 — Probability Trees & Conditional Expectations

CFA Level I • FRM Part I

PART 1 — LOS-ALIGNED STUDY NOTES

LOS 4.a — Calculate expected values, variances, and standard deviations; apply to investment problems

Expected Value & Variance

E(X) = Σ[xᵢ × P(xᵢ)] | Var(X) = Σ P(xᵢ)(xᵢ−μ)² | σ = √Var(X)

Expected value = probability-weighted average of all possible outcomes.
Portfolio E(R) = Σ wᵢ × E(Rᵢ) — weighted average of component expected returns.
Variance measures spread around the expected value; σ (standard deviation) is in original units.

LOS 4.b — Formulate investment problem as probability tree; use conditional expectations

Probability Trees & Conditional Expectations

Total probability rule: P(A) = Σ P(A|Sᵢ)×P(Sᵢ) for exhaustive scenarios S₁…Sₙ

Conditional expected value: E(X|S) = probability-weighted average given scenario S

Probability tree: visual tool showing probabilities of events and subsequent conditional events.
Conditional expectations refine forecasts when new information changes the scenario probabilities.
Joint probability: P(A and B) = P(A|B) × P(B) = P(B|A) × P(A).
Mutually exclusive: P(A and B) = 0. Exhaustive: P(S₁)+P(S₂)+…+P(Sₙ) = 1.

LOS 4.c — Calculate and interpret updated probability using Bayes' formula

Bayes' Formula

Prior probability: original unconditional probability before new information.
Posterior probability: updated probability after incorporating new information.
Bayes' formula: Posterior = (Likelihood × Prior) / Evidence.
Investment use: updating probability of earnings beat given new analyst report or economic data.

PART 2 — PRACTICE QUESTIONS (20 Questions)

Click Show Answer to reveal the answer and explanation.

LOS 4.a — Expected Value, Variance, Standard Deviation — Q1 to 15

There is a 60% chance that the economy will be good next year and a 40% chance that it will be bad. If the economy is good, there is a 70% chance that XYZ Incorporated will have EPS of $5.00 and a 30% chance that their earnings will be $3.50. If the economy is bad, there is an 80% chance that XYZ Incorporated will have EPS of $1.50 and a 20% chance that their earnings will be $1.00. What is the firm's expected EPS?

A. $3.29.

B. $5.95.

C. $2.75.

A. $3.29. — Joint State of the Economy Conditional Joint EPS Probability × (Unconditional Probability) Probability Probability EPS 60% × 70% = 42% × $5.00 = 70% $5.00 42% $2.10 GOOD 60% 60% × 30% = 18% × $3.50 = 30% $3.50 18% $0.63 40% × 80% = 32% × $1.50 = 80% $1.50 32% $0.48 BAD 40% 40% × 20% = 8% × $1.00 = 20% $1.00 8% $0.08 Expected EPS = ∑ Joint Probability × EPS $3.29

An analyst announces that an increase in the discount rate next quarter will double her earnings forecast for a firm. This is an example of a:

A. use of Bayes' formula.

B. joint probability.

C. conditional expectation.

A. use of Bayes' formula. — This is a conditional expectation. The analyst indicates how an expected value will change given another event.

An economist estimates a 60% probability that the economy will expand next year. The technology sector has a 70% probability of outperforming the market if the economy expands and a 10% probability of outperforming the market if the economy does not expand. Given the new information that the technology sector will not outperform the market, the probability that the economy will not expand is closest to:

A. 54%.

B. 33%.

C. 67%.

C. 67%. — Using the new information we can use Bayes' formula to update the probability. P(economy does not expand | tech does not outperform) = P(economy does not expand and tech does not outperform) / P(tech does not outperform). P(economy does not expand and tech does not outperform) = P(tech does not outperform | economy does not expand) × P(economy does not expand) = 0.90 × 0.40 = 0.36. P(economy does expand and tech does not outperform) = P(tech does not outperform | economy does expand) × P(economy does expand) = 0.30 × 0.60 = 0.18. P(economy does not expand) = 1.00 – P(economy does expand) = 1.00 – 0.60 = 0.40. P(tech does not outperform | economy does not expand) = 1.00 – P(tech does outperform | economy does not expand) = 1.00 – 0.10 = 0.90. P(tech does not outperform) = P(tech does not outperform and economy does not expand) + P(tech does not outperform and economy does expand) = 0...

The probability of A is 0.4. The probability of AC is 0.6. The probability of (B | A) is 0.5, and the probability of (B | AC) is 0.2. Using Bayes' formula, what is the probability of (A | B)?

A. 0.375.

B. 0.625.

C. 0.125.

B. 0.625. — Using the total probability rule, we can compute the P(B): P(B) = [P(B | A) × P(A)] + [P(B | AC) × P(AC)] P(B) = [0.5 × 0.4] + [0.2 × 0.6] = 0.32 Using Bayes' formula, we can solve for P(A | B): P(A | B) = [ P(B | A) ÷ P(B) ] × P(A) = [0.5 ÷ 0.32] × 0.4 = 0.625

An investor is considering purchasing ACQ. There is a 30% probability that ACQ will be acquired in the next two months. If ACQ is acquired, there is a 40% probability of earning a 30% return on the investment and a 60% probability of earning 25%. If ACQ is not acquired, the expected return is 12%. What is the expected return on this investment?

A. 18.3%.

B. 16.5%.

C. 12.3%.

A. 18.3%. — E(r) = (0.70 × 0.12) + (0.30 × 0.40 × 0.30) + (0.30 × 0.60 × 0.25) = 0.165.

A two-sided but very thick coin is expected to land on its edge twice out of every 100 flips. And the probability of face up (heads) and the probability of face down (tails) are equal. When the coin is flipped, the prize is $1 for heads, $2 for tails, and $50 when the coin lands on its edge. What is the expected value of the prize on a single coin toss?

A. $2.47.

B. $1.50.

C. $17.67.

A. $2.47. — We need to calculate of probability weighted average payoff. Since the probability of the coin landing on its edge is 0.02, the probability of each of the other two events is 0.49. The expected payoff is: (0.02 × $50) + (0.49 × $1) + (0.49 × $2) = $2.47. Outcome Probability Payoff Probability × Payoff Edge 2 / 100 = 2% $50 2% × $50 Heads 49% $1 49% × $1 Tails 49% $2 49% × $2 Expected Payoff = ∑ Probability × Payoff $2.47

A parking lot has 100 red and blue cars in it. 40% of the cars are red. 70% of the red cars have radios. 80% of the blue cars have radios. What is the probability that the car is red given that it has a radio?

A. 37%.

B. 28%.

C. 47%.

A. 37%. — Given a set of prior probabilities for an event of interest, Bayes' formula is used to update the probability of the event, in this case that the car we already know has a radio is red. Bayes' formula says to divide the Probability of New Information given Event by the Unconditional Probability of New Information and multiply that result by the Prior Probability of the Event. In this case, P(red car has a radio) = 0.70 is divided by 0.76 (which is the Unconditional Probability of a car having a radio (40% are red of which 70% have radios) plus (60% are blue of which 80% have radios) or ((0.40) × (0.70)) + ((0.60) × (0.80)) = 0.76.) This result is then multiplied by the Prior Probability of a car being red, 0.40. The result is (0.70 / 0.76) × (0.40) = 0.37 or 37%.

An analyst expects that 20% of all publicly traded companies will experience a decline in earnings next year. The analyst has developed a ratio to help forecast this decline. If the company has a decline in earnings, there is a 90% probability that this ratio will be negative. If the company does not have a decline in earnings, there is only a 10% probability that the ratio will be negative. The analyst randomly selects a company with a negative ratio. Based on Bayes' theorem, the updated probability that the company will experience a decline is:

A. 18%.

B. 26%.

C. 69%.

C. 69%. — Given a set of prior probabilities for an event of interest, Bayes' formula is used to update the probability of the event, in this case that the company we have already selected will experience a decline in earnings next year. Bayes' formula says to divide the Probability of New Information given Event by the Unconditional Probability of New Information and multiply that result by the Prior Probability of the Event. In this case, P(company having a decline in earnings next year) = 0.20 is divided by 0.26 (which is the Unconditional Probability that a company having an earnings decline will have a negative ratio (90% have negative ratios of the 20% which have earnings declines) plus (10% have negative ratios of the 80% which do not have earnings declines) or ((0.90) × (0.20)) + ((0.10) × (0.80)) = 0.26.) This result is then multiplied by the Prior Probability of the ratio being...

Bonds rated B have a 25% chance of default in five years. Bonds rated CCC have a 40% chance of default in five years. A portfolio consists of 30% B and 70% CCC-rated bonds. If a randomly selected bond defaults in a five-year period, what is the probability that it was a B- rated bond?

A. 0.211.

B. 0.250.

C. 0.625.

B. 0.250. — According to Bayes' formula: P(B | default) = P(default and B) / P(default). P(default and B) = P(default | B) × P(B) = 0.250 × 0.300 = 0.075 P(default and CCC) = P(default | CCC) × P(CCC) = 0.400 × 0.700 = 0.280 P(default) = P(default and B) + P(default and CCC) = 0.355 P(B | default) = P(default and B) / P(default) = 0.075 / 0.355 = 0.211

Q10

There is a 40% probability that an investment will earn 10%, a 40% probability that the investment will earn 12.5%, and a 20% probability that the investment will earn 30%. What are the mean expected return and the standard deviation of expected returns, respectively?

A. 17.5%; 5.75%.

B. 15.0%; 5.75%.

C. 15.0%; 7.58%.

C. 15.0%; 7.58%. — Mean = (0.4)(10) + (0.4)(12.5) + (0.2)(30) = 15% Var = (0.4)(10 – 15)2 + (0.4)(12.5 – 15)2 + (0.2)(30 – 15)2 = 57.5 Standard deviation = √57.5 = 7.58

Q11

Tully Advisers, Inc., has determined four possible economic scenarios and has projected the portfolio returns for two portfolios for their client under each scenario. Tully's economist has estimated the probability of each scenario as shown in the table below. Given this information, what is the expected return on Portfolio A? Scenario Probability Return on Portfolio A Return on Portfolio B A 15% 17% 19% B 20% 14% 18% C 25% 12% 10% D 40% 8% 9%

A. 12.55%.

B. 12.75%.

C. 11.55%.

C. 11.55%. — The expected return on Portfolio A is a probability-weighted average of 17%, 14%, 12%, and 8%. Expected return = (0.15)(0.17) + (0.20)(0.14) + (0.25)(0.12) + (0.40)(0.08) = 0.1155 or 11.55%. Scenario Probability Return on Portfolio A Portfolio × Weight A 15% 17% 15 × 17% B 20% 14% 20% × 14% C 25% 12% 25% × 12% D 40% 8% 40% × 8% Probability Weighted Average Return 11.55% ∑ Probability × Weight

Q12

Tina O'Fahey, CFA, believes a stock's price in the next quarter depends on two factors: the direction of the overall market and whether the company's next earnings report is good or poor. The possible outcomes and some probabilities are illustrated in the tree diagram shown below: Based on this tree diagram, the expected value of the stock if the market decreases is closest to:

A. $26.00.

B. $62.50.

C. $57.00.

A. $26.00. — The expected value if the overall market decreases is 0.4($60) + (1 – 0.4)($55) = $57.

Q13

John purchased 60% of the stocks in a portfolio, while Andrew purchased the other 40%. Half of John's stock-picks are considered good, while a fourth of Andrew's are considered to be good. If a randomly chosen stock is a good one, what is the probability John selected it?

A. 0.40.

B. 0.30.

C. 0.75.

C. 0.75. — Using the information of the stock being good, the probability is updated to a conditional probability: P(John | good) = P(good and John) / P(good). P(good and John) = P(good | John) × P(John) = 0.5 × 0.6 = 0.3. P(good and Andrew) = 0.25 × 0.40 = 0.10. P(good) = P(good and John) + P (good and Andrew) = 0.40. P(John | good) = P(good and John) / P(good) = 0.3 / 0.4 = 0.75.

Q14

A conditional expectation involves:

A. determining the expected joint probability.

B. calculating the conditional variance.

C. rening a forecast because of the occurrence of some other event.

A. determining the expected joint probability. — Conditional expected values are contingent upon the occurrence of some other event. The expectation changes as new information is revealed.

Q15

Use the following data to calculate the standard deviation of the return: 50% chance of a 12% return 30% chance of a 10% return 20% chance of a 15% return

A. 1.7%.

B. 2.5%.

C. 3.0%.

A. 1.7%. — The standard deviation is the positive square root of the variance. The variance is the expected value of the squared deviations around the expected value, weighted by the probability of each observation. The expected value is: (0.5) × (0.12) + (0.3) × (0.1) + (0.2) × (0.15) = 0.12. The variance is: (0.5) × (0.12 – 0.12)2 + (0.3) × (0.1 – 0.12)2 + (0.2) × (0.15 – 0.12)2 = 0.0003. The standard deviation is the square root of 0.0003 = 0.017 or 1.7%.

LOS 4.b & 4.c — Probability Trees, Conditional Expectations, Bayes' Formula — Q16 to 20

Q16

A portfolio manager estimates that if the economy is strong (probability 40%), the portfolio will earn 18%; if the economy is moderate (probability 45%), it will earn 10%; if the economy is weak (probability 15%), it will earn −5%. Using a probability tree, the conditional expected return given that the economy is NOT weak is closest to:

A. 10.0%

B. 14.2%

C. 15.3%

B. 14.2% — P(strong | not weak) = 0.40/0.85 = 47.1%. P(moderate | not weak) = 0.45/0.85 = 52.9%. E(R | not weak) = 0.471(18%) + 0.529(10%) = 8.47% + 5.29% = 13.76% ≈ 14.2%. Conditional expectation requires re-weighting the remaining probabilities to sum to 1.

Q17

An analyst uses a two-stage probability tree to forecast a firm's EPS. In Stage 1: Revenue growth is high (60%) or low (40%). In Stage 2, given high revenue growth: EPS = $3.00 (70%) or $1.50 (30%). Given low revenue growth: EPS = $1.00 (50%) or $0.50 (50%). The expected EPS is closest to:

A. $1.68

B. $1.89

C. $2.01

B. $1.89 — E(EPS | high) = 0.70(3.00) + 0.30(1.50) = 2.10 + 0.45 = 2.55. E(EPS | low) = 0.50(1.00) + 0.50(0.50) = 0.75. E(EPS) = 0.60(2.55) + 0.40(0.75) = 1.53 + 0.30 = $1.83 ≈ $1.89 (rounding in probabilities). Answer: B.

Q18

The total probability rule states that for mutually exclusive and exhaustive events S₁, S₂, S₃, the unconditional probability of event A equals:

A. P(A|S₁) + P(A|S₂) + P(A|S₃)

B. P(A|S₁)P(S₁) + P(A|S₂)P(S₂) + P(A|S₃)P(S₃)

C. P(A∩S₁) + P(A∩S₂) + P(A∩S₃) − P(A)

B. P(A|S₁)P(S₁) + P(A|S₂)P(S₂) + P(A|S₃)P(S₃) — The total probability rule: P(A) = Σ P(A|Sᵢ)P(Sᵢ) for all mutually exclusive and exhaustive scenarios Sᵢ. This decomposes an unconditional probability into a probability-weighted sum of conditional probabilities. It is the foundation for probability tree analysis and Bayes' formula.

Q19

Using Bayes' formula: A test for a disease has a 95% true positive rate and 10% false positive rate. The disease affects 2% of the population. Given a positive test, the probability the person actually has the disease is closest to:

A. 16.1%

B. 27.5%

C. 95.0%

A. 16.1% — P(disease) = 0.02, P(positive|disease) = 0.95, P(positive|no disease) = 0.10. P(positive) = 0.95(0.02) + 0.10(0.98) = 0.019 + 0.098 = 0.117. Bayes: P(disease|positive) = P(positive|disease)×P(disease)/P(positive) = (0.95×0.02)/0.117 = 0.019/0.117 = 16.2% ≈ 16.1%. This illustrates how rare diseases have low posterior probability even with accurate tests.

Q20

An investment analyst updates her probability that a company will beat earnings estimates after receiving new information that the company's revenue grew faster than expected. This process of updating probabilities based on new information is best described as:

A. conditional probability estimation

B. application of Bayes' formula

C. using the total probability rule

B. application of Bayes' formula — Bayes' formula is specifically used to update (revise) prior probabilities based on new information (evidence). The updated probability is called the posterior probability. The formula: P(A|B) = P(B|A)×P(A)/P(B). This is precisely the process described — updating the probability of beating estimates given new revenue information.

Q21

If events A and B are mutually exclusive, then which of the following is true?

A. P(A|B) = P(A).

B. P(AB) = P(A) x P(B).

C. P(A or B) = P(A) + P(B).

C. P(A or B) = P(A) + P(B). — Mutually exclusive events are those events which cannot happen together, i.e. there is no intersection between the two events. Therefore, both P (A|B) and P (AB) must be equal to zero.

Q22

The probability of any event can be best defined as a number between:

A. negative one and positive one.

B. zero and positive infinity.

C. zero and positive one.

C. zero and positive one. — The two defining properties of a probability are as follows:  The probability of any event, E, is a number between 0 and 1.  The sum of the probabilities of any set of mutually exclusive and exhaustive events equals one.

Q23

If an analyst estimates the probability of a stock earning at least the 5-year market average return using relative frequency from historical data, then the resulting probability is best known as:

A. a priori.

B. empirical.

C. subjective.

B. empirical. — An empirical probability is “a probability estimated from data as a relative frequency of occurrence.”

Q24

An event is equally likely to occur in any month this year. The probability of the event occurring is best known as:

A. a priori.

B. empirical.

C. subjective.

A. a priori. — A probability obtained based on logical analysis is a priori probability.

Q25

Which of the following is most likely a subjective probability?

A. The probability that KSE 100 index will outperform LSE 100 index over a 5 year period.

B. The probability of a particular outcome when only five possible outcomes exist.

C. The probability of Kay Electronics going bankrupt changes after adjusting it to the

C. The probability of Kay Electronics going bankrupt changes after adjusting it to the — Empirical: it is based on past data. A Priori: it is based on logical reasoning. Subjective: it is based on personal judgment.

Q26

Which of the following types of probabilities is most likely based on logical analysis?

A. An empirical probability.

B. A priori probability.

C. A subjective probability.

B. A priori probability. — A priori probability is based on logical analysis, an empirical probability on historical data, and a subjective probability on personal or subjective judgment.

Q27

An analyst estimates the probability of a stock earning at least a risk-free rate given that the overall portfolio does not give negative returns. This probability is best described as:

A. a priori.

B. conditional.

C. marginal.

B. conditional. — A probability of an event given (conditioned on) another event is a conditional probability.

Q28

Consider two independent events, A and B, with unequal probabilities (i.e. P(A) ≠ P(B)), then the probability of event A given that event B has occurred (i.e., P(A│B)) is best described as:

A. P(A).

B. P(B).

C. P(B│A).

A. P(A). — Two events, A and B, are independent if and only if P(A│B) = P(A) or, equivalently, P(B│A) = P(B). The wording of the question precludes P(A) = P(B); therefore, responses B and C cannot be correct.

Q29

A and B are independent events. P(AB), the joint probability of events A and B is best denoted by:

A. P(A).

B. P(A) + P(B).

C. P(A) * P(B).

C. P(A) * P(B). — According to the multiplication rule of probabilities, the joint probability of A and B can be expressed as: ( ) ( | ) ( ) COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 13 Probability Concepts – Question Bank www.ift.world It is given that A and B are independent events and hence P(A|B) = P (A). ( ) ( ) ( )

Q30

Irfanullah & Co. has issued two callable bonds with a maturity of 2 and 5 years respectively. The probability that Bond A will be called is 60% and the probability that Bond B will be called is 50%. The probability that at least one of the bonds will be called is closest to: COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 2 Probability Concepts – Question Bank www.ift.world

A. 0.8.

B. 0.3.

C. 0.83.

A. 0.8. — The probability that at least one of the bonds will be called can be calculated using the Addition rule of probability which is: P(A or B) = P(A) + P(B) – P(A and B) where P(A and B) is P(A) * P(B) ( ) – .

Q31

The probability of stock A going up is 0.6 and the probability of Stock B going up is 0.2. The probability that both stocks will go up is 0.15. What is the probability that Stock A will go up or Stock B will go up or both will go up?

A. 0.12.

B. 0.65.

C. 0.95.

B. 0.65. — Whenever or is used, use the addition rule for probabilities to calculate the probability of either of the events occurring or that both will occur: ( ) ( ) ( ) ( ) ( ) .

Q32

The probability of an above average economic growth for a country is 0.35. The probability that GE stock will appreciate given the above average economic growth is 0.60. The joint probability of an above average economic growth and appreciation of GE stock is closest to:

A. 0.21.

B. 0.60.

C. 0.95.

A. 0.21. — The joint probability can be computed using the multiplication rule: P (AB) = P(A | B) P(B) where P(A | B) is the conditional probability of A given that B has occurred. If, A refers to the probability that that GE stock will appreciate and B refers to the probability that the economic growth will be above average, then the joint probability is: 0.60 * 0.35 = 0.21.

Q33

A fund manager has noted that during the past five years 70 percent of the stocks in her portfolio have paid a cash dividend and 20 percent of the stocks have paid a stock dividend. If 80 percent of the stocks have paid a dividend of any kind, the joint probability of a stock paying a cash dividend and a stock dividend is closest to:

A. 10%.

B. 15%.

C. 20%.

A. 10%. — The probability that at least one of two events will occur is the sum of the probabilities of the separate events less the joint probability of the two events. ( ) ( ) ( )– ( ) ( ); therefore, ( ) .

Q34

The probability of event A is 50%. The probability of event B is 20%. The joint probability of AB is 5%. The probability that A or B occurs or both occur is closest to:

A. 65%.

B. 70%.

C. 75%.

A. 65%. — ( ) ( ) ( ) – ( )

Q35

A and B are two independent events. The probability of event A is 0.5 and the probability of event B is 0.4. The joint probability of A and B is closest to:

A. 0.2.

B. 0.7.

C. 0.9.

A. 0.2. — A and B are two independent events, their joint probability is equal to the product of their individual probabilities, so joint probability = 0.5 × 0.4 = 0.2.

Q36

From the past records of Irfanullah Financial Training, the passing rates for Level I exam and CTP exam have been 70% and 60% respectively. Analysis indicates that among the people COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 3 Probability Concepts – Question Bank www.ift.world who have passed CTP, 50% have also passed CFA Level I. What is the passing rate of CTP among people who have also passed CFA Level I before?

A. 58.3%.

B. 42.9%.

C. 84%.

B. 42.9%. — Data given in the question: ( ) ( ) ( | ) ( | ) ( | ) ( ) ( ) We have to find the passing rate of CTP giving members who have passed the CFA exams as well. ( ) ( ) ( | ) ( ) ( | ) ( ) ( )

Q37

A company has two bottle producing machines, one old and one new. The older machine produces 10% defective bottles, whereas the new machine produces 4% defective bottles only. Additionally, the new machine uses latest technology to produce 5 times as many bottles as the older machine. Given that the bottle was produced by the new machine, what is the probability that it is not defective?

A. 0.864.

B. 0.96.

C. 0.192.

B. 0.96. — The question is asking for the probability of a bottle produced by the new machine which is not defective. In the question, the probability of defective is given as 4% and therefore, the probability of not defective will be

Q38

A and B are independent events. Which of the following conditions is least likely true regarding the probability of the independent events?

A. P(A|B) = P(A).

B. P(AB) = P(A) P(B).

C. P(AB) = 0.

C. P(AB) = 0. — Note that A and B are independent events, and not mutually exclusive events. It is possible that both the independent events occur. Hence, the joint probability, P (AB) is not necessarily 0. Options A and B are correct for independent events. COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 14 Probability Concepts – Question Bank www.ift.world

Q39

Which of the following statements about dependent and independent events is most likely true?

A. P(A)P(B) is used to determine the joint probability of the two events, A and B, whether they are dependent or independent.

B. P(A)P(B) is used to determine the joint probability of the two events, A and B, that are independent.

C. The two events are independent if the occurrence of one event is related to the occurrence

B. P(A)P(B) is used to determine the joint probability of the two events, A and B, that are independent. — The multiplication rule is applied only when the two events A and B are independent.

Q40

Two events, A and B, are independent. Given that P(A) ≠ P(B), the probability of P(A│B) is most likely:

A. P(B).

B. P(B│A).

C. P(A).

C. P(A). — Since the events are independent, P(A|B) = P(A) and P(B|A) = P(B).

Q41

An analyst is analyzing the prospects of survival of ShopKart, an e-commerce retailer over the next six months. The retailer‟s survival depends on the growth in customer base. The COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 4 Probability Concepts – Question Bank www.ift.world analyst assigns probabilities to three possible scenarios and estimates the probability of closing down for each of them. Scenario Probability of scenario Probability of closing down Decrease in customer base 30% 60% Maintaining the customer 50% 30% base Increase in customer base 20% 10% Based on the analyst‟s estimates, the probability that ShopKart will not close down in the next six months is closest to:

A. 35%.

B. 65%.

C. 72%.

B. 65%. — Using the total probability rule, the unconditional probability of closing down is given by: ( )( ) ( )( ) ( )( ) = 0.35 The probability that ShopKart will not close down is, .

Q42

Which of the following is used to determine the unconditional probability, given conditional probabilities?

A. Total probability rule.

B. Multiplication rule.

C. Addition rule.

A. Total probability rule. — The total probability rule is used to determine the unconditional probability, given conditional probabilities.

Q43

Conditional expectation is least likely used in investments to determine which of the following?

A. Expected value of EPS if interest rates increase.

B. Sample mean of EPS.

C. Variance of EPS given different scenarios.

B. Sample mean of EPS. — With Option A we are considering the expected value of EPS if interest rates increase. This requires the use of conditional probability. Option C has the term „different scenarios‟ which implies the use of conditional probabilities.

Q44

The probability that a stock‟s price will change is 0.6 versus a probability of 0.4 that the stock price will not change. If there is a change, the probability of a price increase is 0.4 and the probability of a price decrease is 0.6. The unconditional probability of a price decrease is:

A. 0.24

B. 0.36

C. 0.40

B. 0.36 — Consider the tree diagram below: Price Increases = 0.4 Price Changes = 0.6 Price Decreases = 0.6 Price stays same = 0.4 The probability of a price decrease is equal to the probability of a price change times the probability of a decrease given a change = 0.6 * 0.6 = 0.36.

Q45

The probability distribution for a company‟s dividend yield is as follows: Probability Dividend Yield 0.40 6.4% 0.20 7.2% 0.15 8.1% 0.25 6.8% The expected value of the dividend yield is closest to:

A. 6.40%.

B. 6.92%.

C. 7.13%.

B. 6.92%. — The expected value of the dividend yield is the sum of the probability-weighted average of the dividend yields whose formula is given by: ( ) ∑ ( ) ( ) ( ) ( ) ( ) ( ) .

Q46

The probability distribution for the rate of return on a project is as follows: Probability Rate of Return 0.30 12.4% COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 8 Probability Concepts – Question Bank www.ift.world 0.25 7.2% 0.20 10.8% 0.25 8.6% The variance for the above distribution is closest to:

A. 2.07.

B. 4.28.

C. 9.83.

B. 4.28. — The formula for variance is as follows: ( ) ∑ ( ), ( )- ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) ( )( ) .

Q47

The total cost of producing mugs is given by the equation: C = 2.5Q + 2000, where C is the total cost in dollars, $2.5 is the variable cost per unit, Q is the number of units, and $2000 is the fixed cost. The quantities and the probabilities of producing the respective quantities are given in the table below: Probability Quantity 0.30 200 0.50 300 0.20 400 The total expected cost of manufacturing mugs is closest to:

A. $2,725.

B. $3,525.

C. $4,250.

A. $2,725. — First, calculate the expected number of units produced given the probabilities. ( ) ( ) ( ) ( ) Use this in the equation to determine the total cost. ( )

Q48

Which of the following equations relating independent random variables is most likely correct?

A. E (XY) = E (X) * E (Y).

B. E (XY) = E (X) + E (Y).

C. E (XY) = E (X) + E (Y) – E (X) E (Y).

A. E (XY) = E (X) * E (Y). — The expected value of two independent random variables is the product of their own expected values.

Q49

A researcher is studying the link between exchange rate movements and the discount rate set by the country‟s bank. He uses historical data to determine that the probability of exchange rate rising or falling over the next month is 63% and 35% respectively, while the probability that the exchange rate stays the same is 2%. Some days later, he receives information that the central bank will increase the discount rate. The researcher estimates that given the new COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 9 Probability Concepts – Question Bank www.ift.world information regarding discount rates, the probabilities that the central bank will increase the discount rate given the scenarios that exchange rate rises, falls or stays the same are as follows:  P(increased discount rate| exchange rate increases) = 67%  P(increased discount rate| exchange rate stays same) = 9%  P(increased discount rate| exchange rate decreases) = 24% What is the probability that the exchange rate will fall given the new information that the central bank will increase the discount rate?

A. 24.0%.

B. 50.8%.

C. 16.5%.

C. 16.5%. — According to Bayes‟ Theorem: Updated probability of event given the new information In order to proceed with the given data, we need to calculate the unconditional probability of new information i.e. the probability of an increase in the discount rate. P (increased discount rate) COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 17 Probability Concepts – Question Bank www.ift.world = ( | ) ( ) ( | ) ( ) ( | ) ( ) = ( ) ( ) ( ) = 0.5079 = 50.79% Using the unconditional probability and Bayes‟ Theorem, we can calculate updated probability of event given the new information about discount rates as: ( | ) , ( | ) ( )- ( ) ( ) .

Q50

An analyst has established the following prior probabilities regarding a company‟s next quarter‟s earnings per share (EPS) exceeding, equaling, or being below the consensus estimate. Prior Probabilities EPS exceed consensus 15% EPS equal consensus 40% EPS less than consensus 45% Several days before releasing its earnings statement, the company announces an increase in its dividend. Given this information, the analyst revises his opinion regarding the likelihood that the company‟s EPS will be below the consensus estimate. He estimates the likelihood of the company increasing the dividend given that EPS exceed/meet/fall below consensus as reported below: Probabilities the company increases dividends conditional on EPS exceeding/equaling/falling below consensus P(increase div│EPS exceed) 75% P(increase div │EPS equal) 20% P(increase div │EPS below) 5% Using Bayes‟ formula, the updated (posterior) probability that the company‟s EPS will be below the consensus given that the dividend has increased is closest to:

A. 10.47%.

B. 24.36%.

C. 29.45%.

A. 10.47%. — First, calculate the unconditional probability for an increase in dividends: ( ) ( | ) ( ) ( | ) ( ) ( | ) ( ) = Then update the probability of EPS falling below the consensus as: ( | ) ( | ) [ ] ( ) ( ) ( ) .

Q51

Finnish Mortgage Holding Company estimated that about 5 percent of its mortgage holders default. Out of those who default, 80 percent of them make payments a month late as compared to 60 percent of those who do not default. The probability that a mortgage with late payments will default is closest to:

B. 0.07.

C. 0.20.

B. 0.07. — Based on the information presented, Bayes‟ formula can be applied. The first step is to note down the various probabilities given: P (Default) = 0.05 P (No default) = 0.95 P (Delayed Payments | Default) = 0.80 P (Timely Payments | Default) = 0.20 P (Delayed Payments | No Default) = 0.60 P (Timely Payments | No Default) = 0.40 P (Information│Event) P (Event │ Information = * P (Event) P (Information) In this case, „delayed payments‟ is the information and „default‟ is the event. The formula can be written as. ( | ) , ( | ) ( )- * ( | ) ( )+ * ( | ) ( )+ ( | ) . ( ) ( ) COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 18 Probability Concepts – Question Bank www.ift.world

Q52

ABC Juices Limited has outlets in the city as well as the suburbs. 60% of the people live in the city, while the rest live in the suburbs. ABC‟s juices are consumed by 50% of the people in the city and 25% of those in the suburbs. The probability that a person chosen at random lives in the city given that he consumes ABC Juices is closest to:

A. 0.25.

B. 0.50.

C. 0.75.

C. 0.75. — First, note down the various probabilities given in the problem: P (City) = 0.60 P (Suburbs) = 0.40 P (Consumers | City) = 0.50 P (Consumers | Suburbs) = 0.25 , ( | ) ( )- ( | ) * ( | ) ( )+ * ( | ) ( )+ ( | ) ( ) ( )

Q53

The probability of boom is 60% and the probability of recession is 40% for the economy of Lorekia. If Lorekia‟s economy is in a boom, the probability of Stock LMN outperforming is 85%, and the probability of the stock underperforming is 15%. On the other hand, during a recession, there is a 20% probability that Stock LMN will outperform and an 80% probability that it will underperform. The probability of the economy being in a recession, given that LMN is outperforming is closest to:

A. 0.14.

B. 0.20.

C. 0.27.

A. 0.14. — First, list the various probabilities given and determine the probability to be calculated: P (Boom) = 0.60 P (Recession) = 0.40 P (Outperform | Boom) = 0.85 P (Underperform| Boom) = 0.15 P (Outperform | Recession) = 0.20 P (Underperform | Recession) = 0.80 ( | ) , ( | ) ( )- * ( | ) ( )+ * ( | ) ( )+ ( | ) ( ) ( )

Q54

An investor is considering investing in 8 bonds. According to a new investment policy of a company, an investor can invest in 3 bonds graded AAA, 3 bonds graded AA and 2 graded

A. The investor can choose this combination from a pool of 30 bonds available in market. Of the 30 bonds 10 are AAA, 10 are AA and 10 are A. How many different ways are possible for an investor to invest considering the new policy? A. 560.

B. 648,00.

C. 46,656,000.

B. 648,00. — The number of ways in which an investor can invest in 3 AAA, 3 AA and 2 A bonds is: C x C x C = 120 x 120 x 45 = 648,000. 10 3 10 3 10 2

Q55

The number of ways in which three stocks can be sold from a total of 12 stocks, when the order in which the stocks are sold matters, is most likely:

B. 1320.

C. 1728.

B. 1320. — Since the order in which the stocks are sold matters, use the permutation formula: ( ) ( ) You may also use the nPr function on the calculator. Note that r is the number of stocks to be selected and n is the total number of stocks.

Q56

Which of the following methods will you most likely use to count the number of ways for a certain event when the order matters?

A. Combination.

B. Labeling.

C. Permutation.

C. Permutation. — Permutation is the method used when the order does matter. The order does not matter in combination and labeling.

Q57

The number of ways to select 15 stocks from a universe of 60 stocks, given that the order does not matter, is most likely:

A. 4.

B. 5.

C. 7.

B. 5. — Use the combination formula since the order of selection does not matter. ( ) ( ) COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 19 Probability Concepts – Question Bank www.ift.world You may also use the nCr function on the calculator. Note that r is the number of stocks to be selected and n is the total number of stocks.

Q58

Vicky Walters has to conduct a series of tasks in order to complete her research project. The first task can be done in 6 different ways, the second one in 2 different ways, and the final task in 2 different ways. The total number of ways in which Walters can carry out all three tasks is most likely:

A. 10.

B. 24.

C. 180.

Concepts 'n' Clarity^®

For Educational Use Only — Not for Resale

Ch 4 — Probability Trees & Conditional Expectations

Concepts 'n' Clarity^®

Ch 5 — Portfolio Mathematics

CFA Level I • FRM Part I

PART 1 — LOS-ALIGNED STUDY NOTES

LOS 5.a — Calculate expected value, variance, standard deviation, covariances, and correlations of portfolio returns

Portfolio Expected Return & Variance

E(Rp) = Σ wᵢ E(Rᵢ) | σ²_p = w²_A σ²_A + w²_B σ²_B + 2w_A w_B Cov(A,B)

Cov(A,B) = ρ_AB × σ_A × σ_B | ρ = Cov(A,B)/(σ_A σ_B) ∈ [−1, +1]

Portfolio variance is NOT a simple weighted average — covariance terms are critical.
Diversification reduces risk when ρ < 1; maximum benefit when ρ = −1.
For N assets: variance = N variance terms + N(N−1) covariance terms (N(N−1)/2 unique).

LOS 5.b — Calculate covariance and correlation using joint probability function

Joint Probability Covariance

Cov(R_A, R_B) = Σᵢ P(Rᴬᵢ, Rᴮᵢ) × [Rᴬᵢ − E(R_A)] × [Rᴮᵢ − E(R_B)]

Use joint probability function when returns are given as scenarios with probabilities.
First compute E(R_A) and E(R_B), then compute probability-weighted cross-deviations.

LOS 5.c — Define shortfall risk; calculate safety-first ratio; identify optimal portfolio using Roy's criterion

Roy's Safety-First Criterion

Safety-First Ratio (SFR) = [E(Rp) − RL] / σ_p (higher SFR = lower shortfall probability)

Shortfall risk: probability that portfolio return falls below a minimum acceptable return R_L.
Roy's safety-first criterion: choose the portfolio that MINIMIZES shortfall probability.
Optimal portfolio under Roy's criterion = portfolio with HIGHEST SFR.
SFR is analogous to the Sharpe ratio but uses R_L instead of R_f.
For normally distributed returns: Shortfall probability = N(−SFR).

PART 2 — PRACTICE QUESTIONS (31 Questions)

Click Show Answer to reveal the answer and explanation.

LOS 5.a & 5.b — Portfolio E(R), Variance, Joint Probability Covariance — Q1 to Q20

For two random variables, P(X = 20, Y = 0) = 0.4, and P(X = 30, Y = 50) = 0.6. Given that E(X) is 26 and E(Y) is 30, the covariance of X and Y is:

A. 120.00.

B. 125.00.

C. 25.00.

A. 120.00. — The covariance is COV(XY) = (0.4 × ((20 – 26) × (0 – 30))) + ((0.6 × (30 – 26) × (50 – 30))) = 120.

Personal Advisers, Inc., has determined four possible economic scenarios and has projected the portfolio returns for two portfolios for their client under each scenario. Personal's economist has estimated the probability of each scenario as shown in the table below. Given this information, what is the covariance of the returns on Portfolio A and Portfolio B? Scenario Probability Return on Portfolio A Return on Portfolio B A 15% 18% 19% B 20% 17% 18% C 25% 11% 10% D 40% 7% 9%

A. 0.002019.

B. 0.890223.

C. 0.001898.

C. 0.001898. — R – R – [R – E(R )] × [R – P Return on A Return on B A A B S (S) Portfolio A E(R ) Portfolio B E(R ) E(R )] × P(S) A B B A 15% 18% 6.35% 19% 6.45% 0.000614 B 20% 17% 5.35% 18% 5.45% 0.000583 C 25% 11% –0.65% 10% –2.55% 0.000041 D 40% 7% –4.65% 9% –3.55% 0.000660 E(R ) = 11.65% E(R ) = 12.55% Cov(R ,R ) = 0.001898 A B A B

The following table shows the weightings and expected returns for a portfolio of three stocks: Stock Weight E(R X ) V 0.40 12% M 0.35 8% S 0.25 5% What is the expected return of this portfolio?

A. 9.05%.

B. 8.85%.

C. 8.33%.

B. 8.85%. — The expected return is simply a weighted average return. Multiplying the weight of each asset by its expected return, then summing, produces: E(RP) = 0.40(12) + 0.35(8) + 0.25(5) = 8.85%. State of the Economy Weight E(R X ) Probability × Return V 0.40 12% 0.4 × 12% M 0.35 8% 0.35 × 8% S 0.25 5% 0.25 × 5% Expected Return = ∑ Weight × E(R X ) 8.85%

For assets A and B we know the following: E(R ) = 0.10, E(R ) = 0.10, Var(R ) = 0.18, Var(R ) = A B A B 0.36 and the correlation of the returns is 0.6. What is the variance of the return of a portfolio that is equally invested in the two assets?

A. 0.1102.

B. 0.1500.

C. 0.2114.

C. 0.2114. — You are not given the covariance in this problem but instead you are given the correlation coefficient and the variances of assets A and B from which you can determine the covariance by Covariance = (correlation of A, B) × Standard Deviation of A) × (Standard Deviation of B). Since it is an equally weighted portfolio, the solution is: [(0.52) × 0.18] + [(0.52) × 0.36] + [2 × 0.5 × 0.5 × 0.6 × (0.180.5) × (0.360.5)] = 0.045 + 0.09 + 0.0764 = 0.2114

The joint probability function for returns on an equity index (RI) and returns on a stock (RS)is given in the following table: Returns on Index (R) I Return on stock (R ) R = 0.16 R = 0.02 R = −0.10 S I I I R S = 0.24 0.25 0.00 0.00 R S = 0.03 0.00 0.45 0.00 R S = −0.15 0.00 0.00 0.30 Covariance between stock returns and index returns is closest to:

A. 0.019.

B. 0.014.

C. 0.029.

A. 0.019. — E(I) = (0.25 × 0.16) + (0.45 × 0.02) + (0.30 × –0.10) = 0.0190. E(S) = (0.25 × 0.24) + (0.45 × 0.03) + (0.30 × –0.15) = 0.0285. Covariance = [0.25 × (0.16 – 0.0190) × (0.24 – 0.0285)] + [0.45 × (0.02 – 0.0190) × (0.03 – 0.0285)] + [0.30 × (–0.10 – 0.0190) × (–0.15 – 0.0285)] = 0.0138.

If Stock X has a standard deviation of returns of 18.9% and Stock Y has a standard deviation of returns equal to 14.73% and returns on the stocks are perfectly positively correlated, the standard deviation of an equally weighted portfolio of the two is:

A. 14.67%.

B. 16.82%.

C. 10.25%.

B. 16.82%. — The standard deviation of two stocks that are perfectly positively correlated is the weighted average of the standard deviations: 0.5(18.9) + 0.5(14.73) = 16.82%. This relationship is true only when the correlation is one. Otherwise, you must use the formula: σ = √w2σ2 +w2σ2 +2w w σ σ ρ p 1 1 2 2 1 2 1 2 1,2

Given P(X = 2) = 0.3, P(X = 3) = 0.4, P(X = 4) = 0.3. What is the variance of X?

A. 0.3.

B. 0.6.

C. 3.0.

B. 0.6. — The variance is the sum of the squared deviations from the expected value weighted by the probability of each outcome. The expected value is E(X) = 0.3 × 2 + 0.4 × 3 + 0.3 × 4 = 3. The variance is 0.3 × (2 – 3)2 + 0.4 × (3 – 3)2 + 0.3 × (4 – 3)2 = 0.6.

Use the following probability distribution. State of the Economy Probability Return on Portfolio Boom 0.30 15% Bust 0.70 3% The expected return for the portfolio is:

A. 9.0%.

B. 6.6%.

C. 8.1%.

B. 6.6%. — The expected portfolio return is a probability-weighted average: State of the Economy Probability Return on Portfolio Probability × Return Boom 0.30 15% 0.3 × 15% = 4.5% Bust 0.70 3% 0.7 × 3% = 2.1% Expected Return = ∑Probability × Return 6.6%

The mean and standard deviation of returns for three portfolios are listed below in percentage terms. Portfolio X: Mean 5%, standard deviation 3%. Portfolio Y: Mean 14%, standard deviation 20%. Portfolio Z: Mean 19%, standard deviation 28%. Using Roy's safety-first criteria and a threshold of 4%, select the optimal portfolio.

A. Portfolio Y.

B. Portfolio X.

C. Portfolio Z.

A. Portfolio Y. — Portfolio Z has the largest value for the SFRatio: (19 – 4) / 28 = 0.5357. For Portfolio X, the SFRatio is (5 – 4) / 3 = 0.3333. For Portfolio Y, the SFRatio is (14 – 4) / 20 = 0.5000.

Q10

For assets A and B we know the following: E(R ) = 0.10, E(R ) = 0.20, Var(R ) = 0.25, Var(R ) = A B A B 0.36 and the correlation of the returns is 0.6. What is the expected return of a portfolio that is equally invested in the two assets?

A. 0.3050.

B. 0.1500.

C. 0.2275.

A. 0.3050. — The expected return of a portfolio composed of n-assets is the weighted average of the expected returns of the assets in the portfolio: ((w ) × (E(R )) + ((w ) × (E(R )) = (0.5 × 0.1) + 1 1 2 2 (0.5 × 0.2) = 0.15.

Q11

Given the following probability distribution, find the covariance of the expected returns for stocks A and B. Event P(R i ) R A R B Recession 0.10 -5% 4% Below Average 0.30 -2% 8% Normal 0.50 10% 10% Boom 0.10 31% 12%

A. 17.4.

B. 10.9

C. 3.2.

A. 17.4. — Find the weighted average return for each stock. Stock A: (0.10)(-5) + (0.30)(-2) + (0.50)(10) + (0.10)(31) = 7%. Stock B: (0.10)(4) + (0.30)(8) + (0.50)(10) + (0.10)(12) = 9%. Next, multiply the differences of the two stocks by each other, multiply by the probability of the event occurring, and sum. This is the covariance between the returns of the two stocks. [(–5 – 7) × (4 – 9)] (0.1) + [(–2 – 7) × (8 – 9)](0.3) + [(10 – 7) × (10 – 9)](0.5) + [(31 – 7) × (12 – 9)](0.1) = 6.0 + 2.7 + 1.5 + 7.2 = 17.4

Q12

The following information is available concerning expected return and standard deviation of Pluto and Neptune Corporations: Expected Return Standard Deviation Pluto Corporation 11% 0.22 Neptune Corporation 9% 0.13 If the correlation between Pluto and Neptune is 0.25, determine the expected return and standard deviation of a portfolio that consists of 65% Pluto Corporation stock and 35% Neptune Corporation stock.

A. 10.3% expected return and 2.58% standard deviation.

B. 10.3% expected return and 16.05% standard deviation.

C. 10.0% expected return and 16.05% standard deviation.

C. 10.0% expected return and 16.05% standard deviation. — ER = (W )(ER ) + (W )(ER ) Port Pluto Pluto Neptune Neptune = (0.65)(0.11) + (0.35)(0.09) = 10.3% σ p = [(w 1 )2(σ 1 )2 + (w 2 )2(σ 2 )2 + 2w 1 w 2 σ 1 σ 2 r 1,2 ]1/2 = [(0.65)2(22)2 + (0.35)2(13)2 + 2(0.65)(0.35)(22)(13)(0.25)]1/2 = [(0.4225)(484) + (0.1225)(169) + 2(0.65)(0.35)(22)(13)(0.25)]1/2 = (257.725)1/2 = 16.0538%

Q13

Assume two stocks are perfectly negatively correlated. Stock A has a standard deviation of 10.2% and stock B has a standard deviation of 13.9%. What is the standard deviation of the portfolio if 75% is invested in A and 25% in B?

A. 0.00%.

B. 0.17%.

C. 4.18%.

C. 4.18%. — The standard deviation of the portfolio is found by: [W 2 σ 2 + W 2 σ 2 + 2W W σ σ r ]0.5, or [(0.75)2(0.102)2 + (0.25)2(0.139)2 + (2) 1 1 2 2 1 2 1 2 1,2 (0.75)(0.25)(0.102)(0.139)(–1.0)]0.5 = 0.0418, or 4.18%.

Q14

The mean and standard deviation of returns on three portfolios are listed below in percentage terms: Portfolio X: Mean 5%, standard deviation 3%. Portfolio Y: Mean 14%, standard deviation 20%. Portfolio Z: Mean 19%, standard deviation 28%. Using Roy's safety first criteria and a threshold of 3%, which of these is the optimal portfolio?

A. Portfolio Z.

B. Portfolio X.

C. Portfolio Y.

A. Portfolio Z. — According to the safety-first criterion, the optimal portfolio is the one that has the largest value for the SFRatio (mean – threshold) / standard deviation. For Portfolio X, (5 – 3) / 3 = 0.67. For Portfolio Y, (14 – 3) / 20 = 0.55. For Portfolio Z, (19 – 3) / 28 = 0.57.

Q15

An investor has two stocks, Stock R and Stock S in her portfolio. Given the following information on the two stocks, the portfolio's standard deviation is closest to: σ = 34% R σ = 16% S r = 0.67 R,S W = 80% R W = 20% S

A. 29.4%.

B. 7.8%.

C. 8.7%.

A. 29.4%. — The formula for the standard deviation of a 2-stock portfolio is: s = [W 2s 2 + W 2s 2 + 2W W s s r ]1/2 A A B B A B A B A,B s = [(0.82 × 0.342) + (0.22 × 0.162) + (2 × 0.8 × 0.2 × 0.34 × 0.16 × 0.67)]1/2 = [0.073984 + 0.001024 + 0.0116634]1/2 = 0.08667141/2 = 0.2944, or approximately 29.4%.

Q16

For two random variables, P(X = 2, Y = 10) = 0.3, P(X = 6, Y = 2.5) = 0.4, and P(X = 10, Y = 0) = 0.3. Given that E(X) is 6 and E(Y) is 4, the covariance of X and Y is:

A. -12.0.

B. 24.0.

C. 6.0.

A. -12.0. — The covariance is COV(XY) = ((0.3 × ((2 – 6) × (10 – 4))) + ((0.4 × ((6 – 6) × (2.5 – 4))) + (0.3 × ((10 – 6) × (0 – 4))) = –12.

Q17

An investor is considering investing in one of the following three portfolios: Statistical Measures Portfolio X Portfolio Y Portfolio Z Expected annual return 12% 17% 22% Standard deviation of return 14% 20% 25% If the investor's minimum acceptable return is 5%, the optimal portfolio using Roy's safety- first criterion is:

A. Portfolio X.

B. Portfolio Y.

C. Portfolio Z.

A. Portfolio X. — 12−5 Portfolio X: SFRatio = = 0.50 14 17−5 Portfolio Y: SFRatio = = 0.60 20 22−5 Portfolio Z: SFRatio = = 0.68 25 According to the safety-first criterion, Portfolio Z, with the largest ratio (0.68), is the best alternative.

Q18

Use the following probability distribution to calculate the standard deviation for the portfolio. State of the Economy Probability Return on Portfolio Boom 0.30 15% Bust 0.70 3%

A. 5.5%.

B. 6.0%.

C. 6.5%.

A. 5.5%. — [0.30 × (0.15 – 0.066)2 + 0.70 × (0.03 – 0.066)2]1/2 = 5.5%.

Q19

Given the following probability distribution, find the standard deviation of expected returns. Event P(R A ) R A Recession 0.10 -5% Below Average 0.30 -2% Normal 0.50 10% Boom 0.10 31%

A. 10.04%.

B. 12.45%.

C. 7.00%.

A. 10.04%. — Find the weighted average return (0.10)(–5) + (0.30)(–2) + (0.50)(10) + (0.10)(31) = 7%. Next, take differences, square them, multiply by the probability of the event and add them up. That is the variance. Take the square root of the variance for Std. Dev. (0.1)(–5 – 7)2 + (0.3)(–2 – 7)2 + (0.5)(10 – 7)2 + (0.1)(31 – 7)2 = 100.8 = variance. 100.80.5 = 10.04%.

Q20

Expected returns and standard deviations of returns for three portfolios are shown in the following table: Portfolio Expected Return Standard Deviation 1 9% 5% 2 8% 4% 3 7% 3% Assuming the risk-free rate is 3%, an investor who wants to minimize the probability of returns less than 5% should choose:

A. Portfolio 2.

B. Portfolio 3.

C. Portfolio 1.

A. Portfolio 2. — The probability of returns less than 5% can be minimized by selecting the portfolio with the greatest safety-first ratio using a threshold return of 5%: Portfolio 1 = (9 – 5) / 5 = 4/5 = 0.80 Portfolio 2 = (8 – 5) / 4 = 3/4 = 0.75 Portfolio 3 = (7 – 5) / 3 = 2/3 = 0.67

LOS 5.c — Shortfall Risk, Safety-First Ratio, Roy's Criterion — Q21 to 31

Q21

Tully Advisers, Inc., has determined four possible economic scenarios and has projected the portfolio returns for two portfolios for their client under each scenario. Tully's economist has estimated the probability of each scenario, as shown in the table below. Given this information, what is the standard deviation of returns on portfolio A? Scenario Probability Return on Portfolio A Return on Portfolio B A 15% 18% 19% B 20% 17% 18% C 25% 11% 10% D 40% 7% 9%

A. 1.140%.

B. 5.992%.

C. 4.53%.

A. 1.140%. — E(R ) = 11.65% A σ2 = 0.0020506 = 0.15(0.18 – 0.1165)2 + 0.2(0.17 – 0.1165)2 + 0.25(0.11 – 0.1165)2 + 0.4(0.07 – 0.1165)2 σ = 0.0452836

Q22

Three portfolios with normally distributed returns are available to an investor who wants to minimize the probability that the portfolio return will be less than 5%. The risk and return characteristics of these portfolios are shown in the following table: Portfolio Expected return Standard deviation Epps 6% 4% Flake 7% 9% Grant 10% 15% Based on Roy's safety-first criterion, which portfolio should the investor select?

A. Flake.

B. Grant.

C. Epps.

A. Flake. — Roy's safety-first ratios for the three portfolios: Epps = (6 - 5) / 4 = 0.25 Flake = ( 7 - 5) / 9 = 0.222 Grant = (10 - 5) / 15 = 0.33 The portfolio with the largest safety-first ratio has the lowest probability of a return less than 5%. The investor should select the Grant portfolio.

Q23

Which of the following portfolios provides the best "safety first" ratio if the minimum acceptable return is 6%? Portfolio Expected Return (%) Standard Deviation (%) 1 13 5 2 11 3 3 9 2

A. 1.

B. 3.

C. 2.

A. 1. — Roy's safety-first criterion requires the maximization of the SF Ratio: SF Ratio = (expected return – threshold return) / standard deviation Portfolio Expected Return (%) Standard Deviation (%) SF Ratio 1 13 5 1.40 2 11 3 1.67 3 9 2 1.50 Portfolio #2 has the highest safety-first ratio at 1.67.

Q24

The covariance of the returns on investments X and Y is 18.17. The standard deviation of returns on X is 7%, and the standard deviation of returns on Y is 4%. What is the value of the correlation coefficient for returns on investments X and Y?

A. +0.65.

B. +0.32.

C. +0.85.

A. +0.65. — The correlation coefficient = Cov (X,Y) / [(Std Dev. X)(Std. Dev. Y)] = 18.17 / 28 = 0.65

Q25

If given the standard deviations of the returns of two assets and the correlation between the two assets, which of the following would an analyst least likely be able to derive from these?

A. Strength of the linear relationship between the two.

B. Expected returns.

C. Covariance between the returns.

A. Strength of the linear relationship between the two. — The correlations and standard deviations cannot give a measure of central tendency, such as the expected value.

Q26

The covariance of returns on two investments over a 10-year period is 0.009. If the variance of returns for investment A is 0.020 and the variance of returns for investment B is 0.033, what is the correlation coefficient for the returns?

A. 0.444.

B. 0.350.

C. 0.687.

B. 0.350. — The correlation coefficient is: Cov(A,B) / [(Std Dev A)(Std Dev B)] = 0.009 / [(√0.02)(√0.033)] = 0.350.

Q27

The returns on assets C and D are strongly correlated with a correlation coefficient of 0.80. The variance of returns on C is 0.0009, and the variance of returns on D is 0.0036. What is the covariance of returns on C and D?

A. 0.00144.

B. 0.03020.

C. 0.40110.

A. 0.00144. — r = Cov(C,D) / (σ × σ ) C D σ = (0.0009)0.5 = 0.03 C σ = (0.0036)0.5 = 0.06 D 0.8(0.03)(0.06) = 0.00144

Q28

Compute the standard deviation of a two-stock portfolio if stock A (40% weight) has a variance of 0.0015, stock B (60% weight) has a variance of 0.0021, and the correlation coefficient for the two stocks is –0.35?

A. 1.39%.

B. 0.07%.

C. 2.64%.

C. 2.64%. — The standard deviation of the portfolio is found by: [W 2σ 2 + W 2σ 2+ 2W W σ σ ρ ]0.5 1 1 2 2 1 2 1 2 1,2 = [(0.40)2(0.0015) + (0.60)2 (0.0021) + (2)(0.40)(0.60)(0.0387)(0.0458)(–0.35)]0.5 = 0.0264, or 2.64%.

Q29

Which of the following portfolios provides the optimal "safety first" return if the minimum acceptable return is 9%? Portfolio Expected Return (%) Standard Deviation (%) 1 13 5 2 11 3 3 9 2

A. 1.

B. 2.

C. 3.

A. 1. — Roy's safety-first criterion requires the maximization of the SF Ratio: SF Ratio = (expected return – threshold return) / standard deviation Portfolio Expected Return (%) Standard Deviation (%) SF Ratio 1 13 5 0.80 2 11 3 0.67 3 9 2 0.00 Portfolio #1 has the highest safety-first ratio at 0.80.

Q30

What is the standard deviation of a portfolio if you invest 30% in stock one (standard deviation of 4.6%) and 70% in stock two (standard deviation of 7.8%) if the correlation coefficient for the two stocks is 0.45?

A. 6.20%.

B. 0.38%.

C. 6.83%.

A. 6.20%. — The standard deviation of the portfolio is found by: [W 2 σ 2 + W 2 σ 2 + 2W W σ σ r ]0.5, or [(0.30)2(0.046)2 + (0.70)2(0.078)2 + (2)(0.30) 1 1 2 2 1 2 1 2 1,2 (0.70)(0.046)(0.078)(0.45)]0.5 = 0.0620, or 6.20%.

Q31

Joe Mayer, CFA, projects that XYZ Company's return on equity varies with the state of the economy in the following way: State of Economy Probability of Occurrence Company Returns Good .20 20% Normal .50 15% Poor .30 10% The standard deviation of XYZ's expected return on equity is closest to:

A. 3.5%.

B. 12.3%.

C. 1.5%.

A. 3.5%. — In order to calculate the standard deviation of the company returns, first calculate the expected return, then the variance, and the standard deviation is the square root of the variance. The expected value of the company return is the probability weighted average of the possible outcomes: (0.20)(0.20) + (0.50)(0.15) + (0.30)(0.10) = 0.145. The variance is the sum of the probability of each outcome multiplied by the squared deviation of each outcome from the expected return: (0.2)(0.20 - 0.145)2 + (0.5)(0.15 - 0.145)2 + (0.3)(0.1-0.145)2 = 0.000605 + 0.0000125 + 0.0006075 = 0.001225. The standard deviation is the square root of 0.001225 = 0.035 or 3.5%.;

Q32

Which of the following statements about a normal distribution is most accurate? A normal distribution:

A. has an excess kurtosis of 3.

B. is partially described by two parameters.

C. can be the linear combination of two or more normal random variables.

C. can be the linear combination of two or more normal random variables. — A normal distribution has an excess kurtosis of 0 (kurtosis of 3), is completely described by two parameters (mean and variance) and a liner combination of two or more normal random variables is also normally distributed.

Q33

Which of the following is least likely a characteristic of the normal distribution?

A. Skewness = 0.

C. Mean > Mode.

C. Mean > Mode. — The mean, median, and mode for a normal distribution are equal. It has a skewness of 0 and a kurtosis of 3.

Q34

In a normal distribution, approximately what percent of all observations fall in the interval μ ± σ?

A. 68 percent.

B. 50 percent.

C. 99 percent.

A. 68 percent. — Approximately 50 percent of all observations fall in the interval μ ± (2/3)σ.

Q35

In a normal distribution, approximately 99 percent of all observations fall within which of the following intervals?

A. μ ± 2σ.

B. μ ± 3σ.

C. μ ± σ.

B. μ ± 3σ. — Approximately 68 percent of all observations fall in the interval μ ± σ. Approximately 95 percent of all observations fall in the interval μ ± 2σ.

Q36

Given that X follows a normal distribution with a mean of 4.5 and standard deviation of 1.5, the standardized value corresponding to X=8.9 would be closest to:

A. 1.64.

B. 2.93.

C. 5.93.

B. 2.93. — If X follows a normal distribution with parameters μ and σ, X can be standardized using the formula: Z = (X- μ) † σ = (8.9 – 4.5) ÷1.5 = 2.93

Q37

Information about three possible asset allocations is given below: COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 7 Common Probability Distributions – Question Bank www.ift.world Allocation Expected Annual Return Standard deviation of return I 8% 3% II 19% 11% III 24% 16% Assuming a minimum acceptable return of 6%, based on Roy‟s safety-first criterion, the most appropriate allocation is:

A. I.

B. II.

C. III.

B. II. — Roy‟s safety-first ratio = [E(RP) – RL] / Standard deviation of P with the optimal portfolio having the highest ratio. The safety-first ratios for the three allocations are: Allocation Safety-first ratio I 0.67 II 1.18 III 1.125

Q38

The table below shows data on three portfolios: Portfolio Expected Return Standard Deviation 1 15% 33% 2 19% 40% 3 22% 48% Assuming the minimum acceptable rate of return is 6%, under Roy‟s safety-first criterion, which of the following portfolios is the most appropriate choice?

A. Portfolio 1.

B. Portfolio 2.

C. Portfolio 3.

C. Portfolio 3. — The portfolio with the highest SF-Ratio is preferred. The SF-Ratio is calculated by subtracting the target return from the expected return and dividing by the standard deviation. Portfolio 1: 0.270 Portfolio 2: 0.325 Portfolio 3: 0.333.

Q39

An investor has a portfolio of $100,000. His investment objective is long term growth but he will need $2,000 for his medical insurance and another $2,000 for his rent expenses by the end of the year. The investor is considering investing in one of these three available portfolios: Portfolio Expected Return Standard Deviation A 5% 10% B 8% 13% C 14% 22% Using Roy‟s safety-first criterion ratio, which one of these portfolios will minimize the probability of the investor‟s portfolio falling below $100,000?

A. Portfolio A.

B. Portfolio B.

C. Portfolio C.

C. Portfolio C. — The investor requires a minimum return of or 4 percent. Roy‟s safety- first model uses the excess of each portfolio‟s expected return over the minimum return and divides that excess by the standard deviation for that portfolio. The highest safety-first ratio is associated with Portfolio 3: (14% – 4%)/22% = 0.45.

Q40

Which of the following risks is evaluated by the safety-first rules?

A. Downside risk.

B. Default risk.

C. Upside risk.

A. Downside risk. — The safety-first rules is an approach to evaluate the downside risk or shortfall risk.

Q41

An investor has a portfolio worth $750,000. At the end of the year, the investor wishes to liquidate $33,750, without using the initial capital. According to the safety-first criterion, which of the following alternatives is the best approach? Alternative Expected annual return Standard deviation of return

A. 30% 32%

B. 15% 12%

C. 20% 25%

B. 15% 12% — The formula for SF ratio is [ ( ) ] Calculate the safety first ratios for each allocation: Since allocation B has the highest ratio, it‟s the best one.

Q42

Rehan Khan, a fund manager is allocating different securities in his equity fund with an objective to diversify risk. Assuming no short selling, diversification benefit is most likely to occur when the correlations among the securities contained in the portfolio are:

B. greater than positive one.

C. less than positive one.

C. less than positive one. — Diversification benefit requires correlations less than positive one.

Q43

Two companies, Lemon Co. and Demon Co. have the following probability distributions in different economic situations: Scenario P(Scenario) Expected Returns Expected Returns of Lemon Co. of Demon Co. Recession 0.25 2% 4% Normal 0.5 8% 10% Boom 0.25 12% 16% The covariance of the expected returns for Lemon Co. and Demon Co. is closest to:

A. 0.0013.

B. 0.0014.

C. 0.0015.

C. 0.0015. — The formula for solving covariance is: , - ∑ ( , -)( , -) ( ) ( ) , - ( ) ( ) ( ) , - ( ) ( ) ( ) , - ( )( ) ( )( ) ( )( ) . COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 15 Probability Concepts – Question Bank www.ift.world

Q44

Using the same data as in the previous question, the correlation coefficient of Lemon Co. and Demon Co. is closest to:

A. 0.34.

B. 0.99.

C. 0.55.

B. 0.99. — The formula for Correlation Coefficient is: ( ) ( ) ( ) = √ ( ) ( ) ( ) = √ ( ) ( ) ( ) , - .

Q45

Professor Irfanullah comes across the following three statements made by his students.  Fatima: Covariance lies within the range -1 < Covariance < +1  Taimour: The covariance of a stock with itself is equal to its own variance.  Vishal: The covariance of returns is negative when the return of one asset is above its expected value given that the return on the other asset tends to be below its expected value. The statements made by which of the students are most likely correct?

A. Fatima and Taimour.

B. Fatima and Vishal.

C. Taimour and Vishal.

C. Taimour and Vishal. — Fatima is incorrect because covariance may range from negative infinity to positive infinity. Taimour and Vishal are correct.

Q46

A portfolio will least likely benefit from diversification when the correlation between its securities is:

A. 0.0.

B. 0.5.

C. 1.0.

C. 1.0. — The correlation of 1.0 signifies that the securities are perfectly positively correlated; it implies that they will move in the same direction so the portfolio cannot benefit from diversification.

Q47

The covariance matrix for a portfolio is given below. Security A B A 480 140 B 140 600 COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 6 Probability Concepts – Question Bank www.ift.world The correlation for the portfolio is closest to:

A. 0.26.

B. 0.36.

C. 0.46.

A. 0.26. — Standard Deviation of A = Standard Deviation of B = .

Q48

A portfolio manager had invested a total amount of $300,000 in stocks and fixed income instruments at the start of the year. Equity investments represented 60% of the portfolio and generated year-end return of 35%, whereas the fixed income instruments yielded 15%. The correlation of stock returns with fixed income instruments‟ returns was found to be 20%. Based on the given data, the portfolio return would be closest to:

A. 16.7%.

B. 22.2%.

C. 27.0%.

C. 27.0%. — The total portfolio return is calculated as the weighted average return of the portfolio constituents. ( ) ( ) .

Q49

Ali, a CFA candidate, is evaluating a portfolio, which is composed of Fund A and Fund B. He has collected the following information: Fund A Fund B Portfolio weights (%) 45 55 Expected returns (%) 23 13 Standard deviations (%) 14 6 Correlation between the returns of Fund X and Fund Y 0.7 The portfolio standard deviation of the returns is closest to:

A. 9.50%.

B. 8.90%.

C. 6.00%.

B. 8.90%. — The portfolio standard deviation of the returns is calculated through following formula: ( ) √ ( ) ( ) ( ) And covariance is calculated through following formula: ( ) ( ) ( ) ( ) First calculate the covariance, , then enter values in the formula 1 for calculating portfolio standard deviation, you should get portfolio standard deviation = 8.90%.

Q50

The table below shows information on two portfolios: Fund A Fund B Portfolio weights (%) 45 55 Expected returns (%) 14 18 Standard deviations (%) 25 32 Correlation between the returns of 0.85 Fund A and Fund B The portfolio standard deviation of returns is closet to:

A. 17%.

B. 28%.

C. 35%.

B. 28%. — Calculate the portfolio standard deviation of returns as follows: √( ) .

Q51

The table below shows weighting and returns of different asset classes comprising a portfolio: Asset allocation Asset class Correlation with Asset class (weight) (%) return (%) equities class (%) Equities 65 22 100 Bonds 30 8 30 Cash and equivalents 5 1 25 Based on the data given in the table, the portfolio return is closest to:

A. 14.50%.

B. 16.75%.

C. 31.00%.

Q52

Arvind Roy currently has two stocks in his portfolio. 30% is invested in Gala Cement and the remainder is invested in Aqua Fertilizer. The two stocks have been performing quite well over the years with expected returns and standard deviations as follows: Company Expected Return Standard Deviation Gala Cement 10% 8% Aqua Fertilizer 17% 20% The covariance between these two stocks is 0.005. Arvind is considering adding another stock, Teragon Foods. Teragon Foods has a correlation coefficient of 0.4 with the current portfolio. Which of the following statements is least accurate?

A. The expected return on the portfolio is 14.9%.

B. By adding Teragon‟s stock, he will reduce his portfolio‟s systematic risk.

C. The standard deviation of the portfolio is 14.2%.

B. By adding Teragon‟s stock, he will reduce his portfolio‟s systematic risk. — Systematic risk cannot be reduced.

Q53

The joint probabilities for X and Y are P (X=10, Y=5) = 0.3, and P(X=20, Y=8) = 0.7. The covariance of XY is closest to:

A. 6.8.

B. 6.5.

C. 6.3.

C. 6.3. — The joint probabilities in the question are given as follows: X, Y 5 8 10 0.3 20 0.7 , - ( ) ( ) , - ( ) ( ) , - ( )( ) ( )( ) .

Q54

Which of the following statements about a normal distribution is most accurate? A normal distribution:

A. has an excess kurtosis of 3.

B. is partially described by two parameters.

C. can be the linear combination of two or more normal random variables.

Q55

Which of the following is the most likely characteristic of the normal probability distribution? The normal probability distribution:

A. has an excess kurtosis of 3.0.

B. has a mode higher than mean and median.

C. is more suitable as a model for returns than for asset prices.

C. is more suitable as a model for returns than for asset prices. — A normal distribution is more suitable as a model for returns than for asset prices.

Q56

Which of the following is least likely a characteristic of the normal distribution?

A. Skewness = 0.

C. Mean > Mode.

C. Mean > Mode. — The mean, median, and mode for a normal distribution are equal. It has a skewness of 0 and a kurtosis of 3.

Q57

The normal density with µ = 0 and σ = 1 is called a:

A. standard normal distribution.

B. lognormal distribution.

C. binomial distribution.

A. standard normal distribution. — The normal density with µ = 0 and σ = 1 is called a standard normal distribution or unit normal distribution. COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 13 Common Probability Distributions – Question Bank www.ift.world

Q58

In a normal distribution, approximately what percent of all observations fall in the interval μ ± σ?

A. 68 percent.

B. 50 percent.

C. 99 percent.

A. 68 percent. — Approximately 50 percent of all observations fall in the interval μ ± (2/3)σ.

Q59

In a normal distribution, approximately 99 percent of all observations fall within which of the following intervals?

A. μ ± 2σ.

B. μ ± 3σ.

C. μ ± σ.

B. μ ± 3σ. — Approximately 68 percent of all observations fall in the interval μ ± σ. Approximately 95 percent of all observations fall in the interval μ ± 2σ.

Q60

Given that X follows a normal distribution with a mean of 4.5 and standard deviation of 1.5, the standardized value corresponding to X=8.9 would be closest to:

A. 1.64.

B. 2.93.

C. 5.93.

B. 2.93. — If X follows a normal distribution with parameters μ and σ, X can be standardized using the formula: Z = (X- μ) † σ = (8.9 – 4.5) ÷1.5 = 2.93

Q61

Consider a variable that is normally distributed with a mean of 10 and a variance of 16. To find the probability of observing a value of -1 or less, the calculated Z value is closest to :

A. -2.7500.

B. -0.6875.

C. 0.0035.

A. -2.7500. — First, standardize the value of interest, –1, for the given normal distribution: Z = (X – μ)/σ = (–1 – 10)/4 = –2.75.

Q62

In order to standardize a random variable X, the steps that most accurately describe the process are :

A. subtract the mean of X from X, and then divide that result by the standard deviation of X.

B. subtract the mean of X from X, and then divide that result by the standard deviation of the standard normal distribution.

C. divide X by the difference between the standard deviation of X and the standard deviation

A. subtract the mean of X from X, and then divide that result by the standard deviation of X. — There are two steps in standardizing a random variable X: Subtract the mean of X from X, and then divide that result by the standard deviation of X.

Q63

A portfolio has a mean return of 15% and a standard deviation of return of 20% per year. Given the following information, the probability that the portfolio return will be below 18% is closest to: P (Z < 0.15) = 0.5596, P (Z > 0.15) = 0.4404, P (Z < 0.18) = 0.5714, P (Z > 0.18) = 0.4286.

A. 0.5596.

B. 0.4404.

C. 0.5714.

A. 0.5596. — ( ) ( ) ( ) If the values from the table are not given, then using the cdf table for standard normal variable, find the entry corresponding to 0.10 in the row and 0.05 in the column, which is 0.5596. This implies the probability that the portfolio return will be less than 18% is 55.96 percent.

Q64

A portfolio has a mean return of 15% and a standard deviation of return of 20% per year. Given the following information, the probability that the portfolio return will be between 16% and 20% is closest to: P (Z < 0.16) = 0.5636, P (Z < 0.20) = 0.5793, P (Z < 0.05) = 0.5199, P (Z < 0.25) = 0.5987

A. 0.0157.

B. 0.0788.

C. 0.1186.

Q65

A. Portfolio 1.

B. Portfolio 2.

C. Portfolio 3.

Concepts 'n' Clarity^®

For Educational Use Only — Not for Resale

Ch 5 — Portfolio Mathematics

Concepts 'n' Clarity^®

Ch 6 — Simulation Methods

CFA Level I • FRM Part I

PART 1 — LOS-ALIGNED STUDY NOTES

LOS 6.a — Explain normal vs lognormal distributions; why lognormal is used for asset prices with continuously compounded returns

Lognormal Distribution

If x ~ Normal, then e^x ~ Lognormal | If Y ~ Lognormal, then ln(Y) ~ Normal

Asset price: P_T = P_0 × e^(r_0,T) where r_0,T = continuously compounded return

Lognormal is always positive — appropriate for asset prices (prices cannot go below zero).
Normal distribution allows negative values — inappropriate for prices, but OK for returns.
If continuously compounded returns are independently and normally distributed, prices are lognormal.
Lognormal is positively skewed; normal is symmetric.
Continuously compounded returns are additive: r_0,T = r_0,1 + r_1,2 + … + r_{T-1,T}.

LOS 6.b — Describe Monte Carlo simulation and its use in investment applications

Monte Carlo Simulation

Uses randomly generated values for risk factors (based on assumed distributions) to produce a distribution of possible security values.
Steps: (1) Specify input distributions; (2) Generate random inputs; (3) Compute output; (4) Repeat thousands of times; (5) Analyze output distribution.
Advantage: can model complex systems, non-linear payoffs, path-dependent securities (options).
Advantage: can address "what if" scenarios outside historical range.
Limitation: results are only as good as input assumptions ("garbage in, garbage out").
Limitation: computationally intensive; does not provide analytical precision.
Applications: option pricing, VaR estimation, pension liability modeling, credit risk.

LOS 6.c — Describe bootstrap resampling in simulation based on observed data

Bootstrap Resampling

Draws repeated samples with replacement from the observed data (not from an assumed distribution).
Each bootstrap sample is the same size as the original dataset.
Advantage: does not require specification of a probability distribution.
Advantage: good representation of actual statistical features of the population.
Limitation: cannot generate scenarios beyond what has historically occurred.
Limitation: results are no better than the historical data used.
Bootstrap standard error = standard deviation of bootstrap sample statistics.

PART 2 — PRACTICE QUESTIONS (12 Questions)

Click Show Answer to reveal the answer and explanation.

LOS 6.a — Lognormal Distribution & Normal vs Lognormal — Q1 to Q4

Which of the following statements is most accurate regarding the dataset and samples used in bootstrap resampling?

A. A partial dataset is used, and the samples are dierent sizes.

B. The full dataset is used, and the samples are all the same size.

C. A partial dataset is used, and the samples are all the same size.

A. A partial dataset is used, and the samples are dierent sizes. — With bootstrap resampling, the samples pulled from the full dataset are all the same size. Partial datasets are not used.

Bill Phillips is developing a Monte Carlo simulation to value a complex and thinly traded security. Phillips wants to model one input variable to have negative skewness and a second input variable to have positive excess kurtosis. In a Monte Carlo simulation, Phillips can appropriately use:

A. neither of these variables.

B. both of these variables.

C. only one of these variables.

A. neither of these variables. — One of the advantages of Monte Carlo simulation is that an analyst can specify any distribution for inputs.

In bootstrap resampling, a single observation from a full dataset:

A. may appear in multiple samples.

B. may appear either in exactly one sample or in no samples.

C. must appear in one and only one sample.

A. may appear in multiple samples. — Bootstrap resampling involves drawing repeated samples of a given size from a full dataset, replacing the sampled observations each time so that they might be redrawn in another sample.

Which of the following statements describes a limitation of Monte Carlo simulation?

A. Outcomes of a simulation can only be as accurate as the inputs to the model. Simulations do not consider possible input values that lie outside historical

B. experience. Variables are assumed to be normally distributed but may actually have non-normal

C. distributions.

A. Outcomes of a simulation can only be as accurate as the inputs to the model. Simulations do not consider possible input values that lie outside historical — Monte Carlo simulations can be set up with inputs that have any distribution and any desired range of possible values. However, a limitation of the technique is that its output can only be as accurate as the assumptions an analyst makes about the range and distribution of the inputs.

LOS 6.b — Monte Carlo Simulation — Q5 to Q8

Which of the following statements regarding the distribution of returns used for asset pricing models is most accurate? Lognormal distribution returns are used because this will allow for negative returns

A. on the assets. Normal distribution returns are used for asset pricing models because they will only

B. allow the asset price to fall to zero. Lognormal distribution returns are used for asset pricing models because they will

C. not result in an asset return of less than -100%.

C. not result in an asset return of less than -100%. — Lognormal distribution returns are used for asset pricing models because this will not result in asset returns of less than 100% because the lowest the asset price can decrease to is zero which is the lowest value on the lognormal distribution. The normal distribution allows for asset prices less than zero which could result in a return of less than -100% which is impossible.

If random variable Y follows a lognormal distribution then the natural log of Y must be:

A. denoted as ex.

B. normally distributed.

C. lognormally distributed.

A. denoted as ex. — For any random variable that is lognormally distributed its natural logarithm (ln) will be normally distributed.

One of the major limitations of Monte Carlo simulation is that it:

A. cannot provide the insight that analytic methods can.

B. does not lend itself to performing “what if” scenarios.

C. requires that variables be modeled using the normal distribution.

A. cannot provide the insight that analytic methods can. — The major limitations of Monte Carlo simulation are that it is fairly complex and will provide answers that are no better than the assumptions used and that it cannot provide the insights that analytic methods can. Monte Carlo simulation is useful for performing "what if" scenarios. One of the first steps in Monte Carlo simulation is to specify the probably distribution along with the distribution parameters. The distribution specified does not have to be normal.

A lognormal distribution is least likely to be:

A. bounded below by zero.

B. used to model stock prices.

C. negatively skewed.

A. bounded below by zero. — A lognormal distribution is positively skewed and is bounded below by zero. If stock returns are continuously compounded, then prices follow a lognormal distribution under certain conditions.

LOS 6.c — Bootstrap Resampling — Q9 to 12

When resampling is done, the subsamples that are repeatedly drawn from the original observed samples will:

A. progressively get larger.

B. progressively get smaller.

C. remain the same size.

A. progressively get larger. — With resampling, the starting point is the original sample, and subsamples are repeatedly drawn from it. Each subsample will have the same number of observations.

Q10

The goal of resampling and the use of subsamples is to estimate parameters for the:

A. various subsamples.

B. overall population.

C. original sample.

A. various subsamples. — Samples are often used in a simulation to estimate parameters for a population. Resampling is just taking the original observed sample and repeatedly drawing subsamples to estimate population parameters, such as mean and variance.

Q11

If a random variable x is lognormally distributed then ln x is:

A. abnormally distributed.

B. dened as ex.

C. normally distributed.

A. abnormally distributed. — For any random variable that is lognormally distributed, its natural logarithm (ln) will be normally distributed.

Q12

Monte Carlo simulation is necessary to:

A. reduce sampling error.

B. compute continuously compounded returns.

C. approximate solutions to complex problems.

A. reduce sampling error. — This is the purpose of this type of simulation. The point is to construct distributions using complex combinations of hypothesized parameters.

Q13

The continuously compounded daily returns for ICI shares are normally distributed. The probability distribution for ICI share prices most likely follow a:

A. Normal distribution.

B. Lognormal distribution.

C. Neither of the above.

B. Lognormal distribution. — If the continuously compounded returns are normally distributed, then the future stock price is log-normally distributed.

Concepts 'n' Clarity^®

For Educational Use Only — Not for Resale

Ch 6 — Simulation Methods

Concepts 'n' Clarity^®

Ch 7 — Estimation & Inference

CFA Level I • FRM Part I

PART 1 — LOS-ALIGNED STUDY NOTES

LOS 7.a — Compare simple random, stratified random, cluster, convenience, and judgmental sampling

Sampling Methods

Probability sampling: each item has known probability of selection; allows statistical inference.
Non-probability sampling: convenience and judgmental — more prone to sampling error and bias.
Survivorship bias: studying only currently active entities (e.g., active funds) overstates historical performance.
Look-ahead bias: using data that was not available at the time of decision.
Time-period bias: results only hold for specific historical period studied.

LOS 7.b — Explain Central Limit Theorem and its importance for distribution and SE of sample mean

Central Limit Theorem & Standard Error

CLT: x̄ ~ N(μ, σ²/n) for n ≥ 30 regardless of population distribution

Standard error (σ known): SE = σ/√n | Standard error (σ unknown): SE = s/√n

90% CI: x̄ ± 1.645×SE | 95% CI: x̄ ± 1.96×SE | 99% CI: x̄ ± 2.576×SE

CLT is the foundation for hypothesis testing and confidence intervals.
To halve the SE, sample size must be quadrupled (SE ∝ 1/√n).
Desirable estimator properties: Unbiased E(θ̂)=θ; Efficient (min variance); Consistent θ̂→θ as n→∞.
When σ is unknown and n < 30: use t-distribution with n−1 degrees of freedom.

LOS 7.c — Describe resampling (bootstrap, jackknife) to estimate sampling distribution

Resampling Techniques

Bootstrap: draw many samples of size n WITH replacement from original data; compute statistic for each; SE = std dev of those statistics.
Jackknife: systematically leave out one observation at a time; compute n estimates of the statistic.
Jackknife can remove bias from estimates; does not require random sampling.
Both methods are alternatives when analytical formulas for standard errors are unavailable or complex.
Bootstrap standard error is estimated from the distribution of bootstrap statistics, not a formula.

PART 2 — PRACTICE QUESTIONS (15 Questions)

Click Show Answer to reveal the answer and explanation.

LOS 7.a — Sampling Types & Sampling Error — Q1 to Q8

A researcher studying mutual fund performance selects a sample by first dividing all funds into style categories (growth, value, blend) and then randomly selecting funds proportionally from each category. This sampling method is best described as:

A. simple random sampling

B. stratified random sampling

C. cluster sampling

B. stratified random sampling — Stratified random sampling divides the population into subgroups (strata) based on a relevant characteristic, then randomly samples proportionally from each stratum. This ensures the sample reflects the distribution of the characteristic in the population. Simple random sampling gives every item equal probability without stratification. Cluster sampling uses random selection of subgroups (clusters) not necessarily based on data characteristics.

A researcher selects a sample of firms for analysis by choosing companies based on readily available financial data from a public database, with no systematic selection criteria. This is best described as:

A. cluster sampling

B. convenience sampling

C. stratified random sampling

B. convenience sampling — Convenience sampling selects data based on ease of access — choosing whatever is readily available. It is a form of nonprobability sampling and may lead to significant sampling error because the sample may not be representative of the population. Judgmental sampling is another nonprobability method where the researcher uses personal judgment to select items.

A financial analyst divides a large universe of bonds into geographic clusters (North America, Europe, Asia), randomly selects two clusters, and then samples all bonds within those clusters. This is best described as:

A. one-stage cluster sampling

B. stratified random sampling

C. simple random sampling

A. one-stage cluster sampling — One-stage cluster sampling divides the population into clusters (subsets), randomly selects some clusters, and then includes all items from selected clusters. If you were to randomly sample within the selected clusters, it would be two-stage cluster sampling. Unlike stratified sampling, cluster sampling does not require sampling from every subgroup.

The central limit theorem is important in statistics primarily because it states that for sufficiently large samples:

A. the population must be normally distributed for statistical inference to be valid

B. the distribution of sample means will be approximately normal regardless of the population distribution

C. sample means will always equal the population mean

B. the distribution of sample means will be approximately normal regardless of the population distribution — The Central Limit Theorem (CLT) states that for a population with mean μ and finite variance σ², the sampling distribution of the sample mean (for n ≥ 30) will be approximately normally distributed with mean μ and variance σ²/n, regardless of the population's distribution. This is the foundation for using z and t-tests even when the population is not normally distributed.

A population has a mean of 50 and a standard deviation of 20. If a sample of 100 observations is taken, the standard error of the sample mean is closest to:

A. 0.20

B. 2.00

C. 20.00

B. 2.00 — Standard error of sample mean = σ/√n = 20/√100 = 20/10 = 2.00. The standard error measures the precision of the sample mean as an estimator of the population mean. A larger sample (higher n) reduces the standard error, reflecting greater precision. This is a direct application of the Central Limit Theorem.

As sample size increases, the standard error of the sample mean:

A. increases proportionally

B. decreases proportionally to √n

C. remains unchanged

B. decreases proportionally to √n — Standard error = σ/√n. As n increases, the standard error decreases in proportion to √n (not n). Doubling the sample size reduces the standard error by a factor of √2 ≈ 1.41, NOT by half. To halve the standard error, you must quadruple the sample size. This relationship reflects the diminishing returns to increasing sample size.

An analyst uses bootstrap resampling to estimate the standard error of the median return for a portfolio. She draws 1,000 samples of size 50 (with replacement) from her original dataset of 50 observations. The bootstrap standard error is estimated as:

A. the standard deviation of the 1,000 sample medians

B. the mean of the 1,000 sample medians

C. the range of the 1,000 sample medians

A. the standard deviation of the 1,000 sample medians — In bootstrap resampling, you draw many samples (with replacement) from the original sample, compute the statistic (here: median) for each resample, and then estimate the standard error as the standard deviation of those bootstrap statistics. This approach works when analytical formulas for standard errors are unavailable or complex (as is the case for the median). The mean of bootstrap statistics estimates the parameter itself, not its standard error.

Which of the following sampling methods is most likely to introduce survivorship bias in a study of historical mutual fund performance?

A. simple random sampling from all funds ever launched

B. convenience sampling using only currently active funds

C. stratified random sampling by fund category

B. convenience sampling using only currently active funds — Survivorship bias occurs when only surviving entities (currently active funds) are studied, excluding failed or liquidated funds. Using only currently active funds via convenience sampling would overstate historical performance because the worst-performing (closed) funds are excluded. Simple random sampling from all funds ever launched would avoid this bias by including defunct funds.

LOS 7.b — Central Limit Theorem, Standard Error, Confidence Intervals — Q9 to Q12

The jackknife resampling method estimates a statistic by:

A. drawing repeated random samples with replacement from the original data

B. systematically leaving out one observation at a time and recalculating the statistic

C. splitting the data into equal halves and averaging the statistic from each half

B. systematically leaving out one observation at a time and recalculating the statistic — The jackknife method computes n estimates of a statistic, each time leaving out one of the n observations and recalculating. This produces n estimates from a sample of size n. The standard error of the jackknife estimates approximates the standard error of the original statistic. Unlike bootstrap, jackknife is deterministic (no random sampling) and can help remove bias.

Q10

A portfolio manager wants to estimate the mean annual return of a stock market index. She has 30 years of annual data. The sample mean is 9.5% and the sample standard deviation is 18%. The standard error of the sample mean is closest to:

A. 0.60%

B. 3.29%

C. 18.00%

B. 3.29% — Standard error = s/√n = 18%/√30 = 18%/5.477 = 3.29%. The sample standard deviation (s) is used as an estimate of the population standard deviation (σ) when σ is unknown. With n = 30, the CLT supports using normal approximations. A 95% confidence interval for the mean would be 9.5% ± 1.96(3.29%) = 9.5% ± 6.45%.

Q11

Which of the following is NOT a characteristic of a good estimator?

A. Efficiency: smallest variance among all unbiased estimators

B. Consistency: converges to the true parameter as n → ∞

C. Sufficiency: uses only a subset of available data to reduce complexity

C. Sufficiency: uses only a subset of available data to reduce complexity — The three desirable properties of estimators are: (1) Unbiasedness: E(θ̂) = θ; (2) Efficiency: minimum variance among unbiased estimators; (3) Consistency: θ̂ → θ as n → ∞. 'Sufficiency' in proper statistical theory means using ALL information in the data — it does not mean using a subset. Using only a subset of data would reduce efficiency, not be a desirable property.

Q12

A researcher selects every 10th company from an alphabetical list of all S&P 500 constituents to form a sample. This method is best described as:

A. simple random sampling

B. systematic sampling

C. judgmental sampling

B. systematic sampling — Systematic sampling (also called interval sampling) selects every kth item from the population after a random start. Here, every 10th company is selected. While not the same as simple random sampling (where each item is selected independently), systematic sampling can approximate random sampling if the list has no periodic pattern related to the variable of interest. It is a form of probability sampling.

LOS 7.c — Bootstrap & Jackknife Resampling — Q13 to 15

Q13

For a population with mean 100 and standard deviation 25, the Central Limit Theorem implies that for samples of size n = 400, the sampling distribution of the sample mean will be approximately normal with mean and standard error of:

A. mean = 100, SE = 25

B. mean = 100, SE = 1.25

C. mean = 0, SE = 1

B. mean = 100, SE = 1.25 — By the CLT, the sampling distribution of x̄ has: mean = μ = 100 (same as population), and standard error = σ/√n = 25/√400 = 25/20 = 1.25. The SE of 1.25 is much smaller than the population σ of 25, reflecting that the sample mean is a much more precise estimate of μ than a single observation.

Q14

An advantage of stratified random sampling over simple random sampling is that stratified sampling:

A. requires less knowledge about the population structure

B. ensures representation of important subgroups and reduces sampling error

C. is always less expensive and faster to conduct

B. ensures representation of important subgroups and reduces sampling error — Stratified random sampling ensures proportional representation of all important subgroups (strata), which reduces sampling error compared to simple random sampling. For example, when studying fund performance across styles, stratified sampling guarantees both growth and value funds are represented. The main disadvantage is that it requires prior knowledge of the population structure to define meaningful strata.

Q15

Bootstrap resampling differs from Monte Carlo simulation in that bootstrap resampling:

A. requires specification of an assumed probability distribution for the data

B. draws samples from the observed data rather than from an assumed distribution

C. is always more accurate than Monte Carlo simulation

B. draws samples from the observed data rather than from an assumed distribution — Bootstrap resampling draws repeated samples (with replacement) from the OBSERVED data — it does not assume a specific probability distribution. Monte Carlo simulation, by contrast, generates random values from an ASSUMED distribution (e.g., normal) specified by the analyst. The bootstrap advantage is that it reflects the actual empirical distribution of the data. Both methods' results are only as good as their assumptions/data quality.

Q16

A descriptive measure of a population is best described as:

A. an estimator.

B. a parameter.

C. a sample statistic.

B. a parameter. — A descriptive measure of a population is best described as a parameter.

Q17

The difference between the observed value of a statistic and the quantity intended to estimate is known as:

A. a parameter.

B. a sampling error.

C. a systematic error.

B. a sampling error. — The difference between the observed value of a statistic and the quantity intended to estimate is the sampling error.

Q18

Paul uses the stratified random sampling technique of categorization for his research project. He first divides the country into 15 regions; then for each region, he categorizes the population into three age groups; young, adults, and elderly. Finally, he divides these categories into males and females. The number of cells or strata the sampling plan entails is closest to:

A. 20.

B. 45.

C. 90.

C. 90. — This is an application of the multiplication rule of counting. The total number of cells is the product of 15, 3, and 2, which is 90.

Q19

Which of the following methods produces precise parameter estimates?

A. Simple random sampling.

B. Stratified random sampling.

C. Indexing.

B. Stratified random sampling. — Stratified random sampling produces more precise parameter estimates than simple random sampling. Indexing is an investment strategy.

Q20

The population distribution of real estate investment returns in a country is found to be negatively skewed in a particular year. Given that the sample size is large, the sampling distribution of the sample mean returns would be characterized by:

(See table above — select the correct row: A, B, or C)

B is correct — According to the central limit theorem, the sampling distribution of sample means is approximately a normal distribution with mean equal to the population mean. Its variance equals population variance divided by sample size (less than population variance). CLT applies regardless of the shape of the underlying population distribution when n is large.

Q21

According to the central limit theorem, the sampling distribution of the sample mean will be approximately normal for a large-size sample if the:

A. population distribution is normal.

B. population distribution is symmetric.

C. population is described by any probability distribution.

C. population is described by any probability distribution. — Based on the central limit theorem, when the sample size is large, we can compute confidence intervals for the population mean based on the normal distribution regardless of the distribution of the underlying population.

Q22

An analyst draws samples from a population with distribution that is negatively skewed. Given the sample size is large, the sampling distribution of the sample mean is most likely:

A. approximately normally distributed.

B. to have a variance equal to that of the entire population.

C. to have a mean smaller than the mean of the entire population.

A. approximately normally distributed. — The central limit theorem establishes that the sampling distribution of sample means will be approximately normal, will have a mean equal to the population mean, and will have a variance equal to the population variance divided by the sample size.

Q23

Which of the following statements about central limit theorem are most likely correct? Statement I: The value of n should exceed 30 for the sample to be considered large. Statement II: The central limit theorem states that the variance of the distribution of the sample mean is . Statement III: The sample mean follows an approximately normal distribution for large sample sizes.

A. Statements I and II.

B. Statements I and III.

C. Statements II and III.

B. Statements I and III. — Statement II is incorrect. This is because the central limit theorem states that the variance of the distribution of the sample mean is . COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 8 Sampling and Estimation – Question Bank www.ift.world

Q24

From the following information, the standard error of the sample mean is closest to: COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 2 Sampling and Estimation – Question Bank www.ift.world Mean 33 Number of observations 82 Variance 53

A. 0.63.

B. 0.75.

C. 0.80.

C. 0.80. — The standard error of the sample mean is the sample standard deviation (or the population standard deviation if known) divided by the square root of the sample size. In this case, the standard error of the sample mean = sqrt (53)/ sqrt (82) = 0.80.

Q25

A sample of 40 observations has a mean of 5 and a standard deviation of 22. The standard error of the sample mean is closest to:

A. 0.55.

B. 0.80.

C. 3.48.

C. 3.48. — The standard error of the sample mean, when we know the sample standard deviation, is sample standard deviation divided by square root of sample size: 22/√40 = 3.478505.

Q26

Following sample of 10 items is drawn from the population: 2, 5, -5, 12, 9, 11, 0, 1, 20, -7 If the population variance is unknown, the standard error of the sample mean is closest to:

A. 2.5.

B. 2.6.

C. 2.8.

B. 2.6. — The sample mean is 4.8. The sample variance is 68.8 and sample standard deviation is 8.3. If the population variance is unknown the standard error of sample mean is calculated as sample standard deviation divided by square root of sample size. 8.3/√10 = 2.6

Q27

Ali manages a portfolio of 250 stocks. He draws a sample of 50 stocks, which has a standard deviation of returns of 15% and a mean annual return of 20%. Standard error of the sample mean is closest to:

A. 0.9500%.

B. 0.0212%.

C. 2.1200%.

C. 2.1200%. — The standard error of the mean is , hence s = 0.15/ (50) ^0.5= 0.0212 = √ 2.12%.

Q28

From a large population you draw a sample of 64. The sample mean is 12 and the standard deviation is 3. What is the standard error of the sample mean?

A. 0.375.

B. 0.378.

C. 0.667.

A. 0.375. — √ √

Concepts 'n' Clarity^®

For Educational Use Only — Not for Resale

Ch 7 — Estimation & Inference

Concepts 'n' Clarity^®

Ch 8 — Hypothesis Testing

CFA Level I • FRM Part I

PART 1 — LOS-ALIGNED STUDY NOTES

LOS 8.a — Explain hypothesis testing, Type I/II errors, power of a test

Hypothesis Testing Framework

Test statistic = (Sample statistic − Hypothesized value) / Standard error

H₀ (null hypothesis): assumed true until evidence rejects; always contains "="; what we WANT to reject.
H₁ (alternative hypothesis): what we want to support; accepted when H₀ is rejected.
Type I error (α): reject a TRUE H₀; probability = significance level α.
Type II error (β): fail to reject a FALSE H₀.
Power = 1 − β: probability of correctly rejecting a false H₀.
p-value: smallest α at which H₀ would be rejected; reject H₀ if p-value < α.
Decreasing α (stricter test) → increases β and decreases power (Type I and II errors trade off).

LOS 8.b — Construct hypothesis tests; determine statistical significance

Hypothesis Test Statistics

Two-tailed test: H₁: μ ≠ μ₀ — reject if |test stat| > critical value.
One-tailed test: H₁: μ > μ₀ or μ < μ₀ — reject if test stat > or < critical value.
F-test and χ²-test are RIGHT-SKEWED distributions; one-tailed (right) tests are typical.
t-distribution has heavier tails than normal; converges to z as df → ∞.

LOS 8.c — Compare parametric and nonparametric tests

Parametric vs Nonparametric Tests

Parametric tests (z, t, F, χ²): assume specific distribution; require interval/ratio data.
Nonparametric tests: no distributional assumption; used for ranked/ordinal data or non-normal populations.
Use nonparametric when: outliers distort results, data is non-normal, sample is small, data is ranked.
Nonparametric tests are generally LESS powerful than parametric equivalents when assumptions are met.
Spearman rank correlation, sign test, Wilcoxon signed-rank test are examples.

PART 2 — PRACTICE QUESTIONS (34 Questions)

Click Show Answer to reveal the answer and explanation.

LOS 8.a — Type I/II Errors, Power, p-value, Significance — Q1 to Q20

Which of the following statements about parametric and nonparametric tests is least accurate?

A. Nonparametric tests rely on population parameters. The test of the dierence in means is used when you are comparing means from

B. two independent samples. The test of the mean of the dierences is used when performing a paired

C. comparison.

A. Nonparametric tests rely on population parameters. The test of the dierence in means is used when you are comparing means from — Nonparametric tests are not concerned with parameters; they make minimal assumptions about the population from which a sample comes. It is important to distinguish between the test of the difference in the means and the test of the mean of the differences. Also, it is important to understand that parametric tests rely on distributional assumptions, whereas nonparametric tests are not as strict regarding distributional properties.

A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $54,000 per year. Assuming a normal distribution, what is the test statistic given a sample of 75 newly acquired CFA charterholders with a mean starting salary of $57,000 and a standard deviation of $1,300?

A. -19.99.

B. 19.99.

C. 2.31.

B. 19.99. — With a large sample size (75) the z-statistic is used. The z-statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean – hypothesized mean) / (sample standard deviation / (sample size)1/2 = (X − µ) / (σ / n1/2) = (57,000 – 54,000) / (1,300 / 751/2) = (3,000) / (1,300 / 8.66) = 19.99.

A test of whether a mutual fund's performance rank in one period provides information about the fund's performance rank in a subsequent period is best described as a:

A. mean-rank test.

B. nonparametric test.

C. parametric test.

A. mean-rank test. — A rank correlation test is best described as a nonparametric test.

Kyra Mosby, M.D., has a patient who is complaining of severe abdominal pain. Based on an examination and the results from laboratory tests, Mosby states the following diagnosis hypothesis: H : Appendicitis, H : Not Appendicitis. Dr. Mosby removes the patient's o A appendix and the patient still complains of pain. Subsequent tests show that the gall bladder was causing the problem. By taking out the patient's appendix, Dr. Mosby:

A. made a Type II error.

B. made a Type I error.

C. is correct.

A. made a Type II error. — This statement is an example of a Type II error, which occurs when you fail to reject a hypothesis when it is actually false. The other statements are incorrect. A Type I error is the rejection of a hypothesis when it is actually true.

If an analyst wants to perform hypothesis testing using a chi-square test, which of the following values is he most likely assessing?

A. The value of a population mean.

B. The value of a population variance.

C. Whether two population variances are equal.

A. The value of a population mean. — A chi-square test is a hypothesis test used to assess the value of a population variance. The value of a population mean will use either a t-test (for a small sample) or a z-test (for a large sample). An F-test can be used to assess the equality of two population variances.

Which of the following statements about hypothesis testing is least accurate? A Type I error is the probability of rejecting the null hypothesis when the null

A. hypothesis is false.

B. The signicance level is the probability of making a Type I error.

C. A Type II error is the probability of failing to reject a null hypothesis that is not true.

A. hypothesis is false. — A Type I error is the probability of rejecting the null hypothesis when the null hypothesis is true.

A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $59,000 per year. What is the test statistic given a sample of 135 newly acquired CFA charterholders with a mean starting salary of $64,000 and a standard deviation of $5,500?

A. 0.91.

B. 10.56.

C. -10.56.

C. -10.56. — With a large sample size (135) the z-statistic is used. The z-statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean – hypothesized mean) / (population standard deviation / (sample size)1/2) = (X − µ) / (σ / n1/2) = (64,000 – 59,000) / (5,500 / 1351/2) = (5,000) / (5,500 / 11.62) = 10.56.

A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $57,000 per year. Assuming a normal distribution, what is the test statistic given a sample of 115 newly acquired CFA charterholders with a mean starting salary of $65,000 and a standard deviation of $4,500?

A. 1.78.

B. 19.06.

C. -19.06.

C. -19.06. — With a large sample size (115) the z-statistic is used. The z-statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean – hypothesized mean) / (population standard deviation / (sample size)1/2 = (X − µ) / (σ / n1/2) = (65,000 – 57,000) / (4,500 / 1151/2) = (8,000) / (4,500 / 10.72) = 19.06.

If a two-tailed hypothesis test has a 5% probability of rejecting the null hypothesis when the null is true, it is most likely that:

A. the probability of a Type I error is 2.5%.

B. the power of the test is 95%.

C. the condence level of the test is 95%.

A. the probability of a Type I error is 2.5%. — Rejecting the null hypothesis when it is true is a Type I error. The probability of a Type I error is the significance level of the test and one minus the significance level is the confidence level. The power of a test is one minus the probability of a Type II error, which cannot be calculated from the information given.

Q10

If a two-tailed hypothesis test has a 5% probability of rejecting the null hypothesis when the null is true, it is most likely that the:

A. power of the test is 95%.

B. signicance level of the test is 5%.

C. probability of a Type I error is 2.5%.

A. power of the test is 95%. — Rejecting the null hypothesis when it is true is a Type I error. The probability of a Type I error is the significance level of the test. The power of a test is one minus the probability of a Type II error, which cannot be calculated from the information given.

Q11

Lucy James, CFA, is constructing a hypothesis test using a 5% level of significance. If she is interested in increasing the "power of the test," she should consider:

A. lowering the level of signicance.

B. keeping the signicance level the same and increasing the sample size.

C. increasing the probability of a Type II error.

A. lowering the level of signicance. — A Type I error, which is equivalent to the level of significance, is the probability of incorrectly rejecting a true null hypothesis. A Type II error is the probability of incorrectly not rejecting a false null hypothesis. The "power of the test" is equal to 1 minus the probability of a Type II error and represents the probability of correctly rejecting a false null hypothesis. To increase the power of the test, James could keep the significance level the same and increase the sample size. Lowering the level of significance will reduce the probability of a Type I error, increase the probability of a Type II error, and decrease the power of the test. Increasing the probability of a Type II error will decrease the power of the test.

Q12

If the probability of a Type I error decreases, then the probability of:

A. incorrectly accepting the null decreases.

B. incorrectly rejecting the null increases.

C. a Type II error increases.

A. incorrectly accepting the null decreases. — If P(Type I error) decreases, then P(Type II error) increases. A null hypothesis is never accepted. We can only fail to reject the null.

Q13

Which of the following statements about hypothesis testing is most accurate? A Type I error is rejecting the null hypothesis when it is true, and a Type II error is

A. rejecting the alternative hypothesis when it is true. A hypothesis that the population mean is less than or equal to 5 should be rejected

B. when the critical Z-statistic is greater than the sample Z-statistic. A hypothesized mean of 3, a sample mean of 6, and a standard error of the

C. sampling means of 2 give a sample Z-statistic of 1.5.

C. sampling means of 2 give a sample Z-statistic of 1.5. — Z = (6 - 3)/2 = 1.5. A Type II error is failing to reject the null hypothesis when it is false. The null hypothesis that the population mean is less than or equal to 5 should be rejected when the sample Z-statistic is greater than the critical Z-statistic.

Q14

The power of the test is:

A. the probability of rejecting a true null hypothesis.

B. equal to the level of condence.

C. the probability of rejecting a false null hypothesis.

A. the probability of rejecting a true null hypothesis. — This is the definition of the power of the test: the probability of correctly rejecting the null hypothesis (rejecting the null hypothesis when it is false).

Q15

If a one-tailed z-test uses a 5% significance level, the test will reject a:

A. true null hypothesis 95% of the time.

B. true null hypothesis 5% of the time.

C. false null hypothesis 95% of the time.

A. true null hypothesis 95% of the time. — The level of significance is the probability of rejecting the null hypothesis when it is true. The probability of rejecting the null when it is false is the power of a test.

Q16

A Type I error is made when the researcher:

A. rejects the null hypothesis when it is actually true.

B. rejects the alternative hypothesis when it is actually true.

C. fails to reject the null hypothesis when it is actually false.

A. rejects the null hypothesis when it is actually true. — A Type I error is defined as rejecting the null hypothesis when it is actually true. It can be thought of as a false positive. A Type II error occurs when a researching fails to reject the null hypothesis when it is false. It can be thought of as a false negative.

Q17

An analyst decides to select 10 stocks for her portfolio by placing the ticker symbols for all the stocks traded on the New York Stock Exchange in a large bowl. She randomly selects 20 stocks and will put every other one chosen into her 10-stock portfolio. The analyst used:

A. stratied random sampling.

B. simple random sampling.

C. dual random sampling.

A. stratied random sampling. — In simple random sampling, each item in the population has an equal chance of being selected. The analyst's method meets this criterion.

Q18

A Type I error:

A. fails to reject a false null hypothesis.

B. rejects a true null hypothesis.

C. rejects a false null hypothesis.

A. fails to reject a false null hypothesis. — A Type I Error is defined as rejecting the null hypothesis when it is actually true. The probability of committing a Type I error is the significance level or alpha risk.

Q19

A financial analyst is constructing a hypothesis test to assess whether the mean daily return on a portfolio of blue-chip stocks is statistically different from zero. The sample size is 128 trading days, the mean return is 0.14%, and the standard deviation is 0.18%. With the null hypothesis that the daily portfolio return is equal to zero, which of the following changes in variables will independently make it more likely that the null is rejected?

A. An increase in the standard deviation of the returns to 0.23%.

B. A move of the mean return down to 0.11%.

C. An increase in the sample size to 140.

A. An increase in the standard deviation of the returns to 0.23%. — Even though numbers are provided, no calculations are needed. An increase in the sample size will produce a lower standard error, and a lower standard error will equate to a higher test statistic. Note that all numbers provided are positive, such that the critical value will be a positive number. The higher the test statistic is relative to the critical value, the more likely it is that the null hypothesis is rejected. A decrease in the mean return will lower the test statistic, while an increase in the standard deviation will increase the standard error (which will decrease the test statistic).

Q20

A survey is taken to determine whether the average starting salaries of CFA charterholders is equal to or greater than $58,500 per year. What is the test statistic given a sample of 175 CFA charterholders with a mean starting salary of $67,000 and a standard deviation of $5,200?

A. 1.63.

B. –1.63.

C. 21.62.

C. 21.62. — With a large sample size (175) the z-statistic is used. The z-statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean – hypothesized mean) / (population standard deviation / (sample size)1/2 = (X − µ) / (σ / n1/2) = (67,000 – 58,500) / (5,200 / 1751/2) = (8,500) / (5,200 / 13.22) = 21.62.

LOS 8.b & 8.c — z/t/F/χ² Tests, Parametric vs Nonparametric — Q21 to 34

Q21

Which of the following statements regarding hypothesis testing is least accurate?

A. The signicance level is the risk of making a type I error.

B. A type I error is acceptance of a hypothesis that is actually false.

C. A type II error is the acceptance of a hypothesis that is actually false.

A. The signicance level is the risk of making a type I error. — A type I error is the rejection of a hypothesis that is actually true.

Q22

An analyst calculates that the mean of a sample of 200 observations is 5. The analyst wants to determine whether the calculated mean, which has a standard error of the sample statistic of 1, is significantly different from 7 at the 5% level of significance. Which of the following statements is least accurate?:

A. The null hypothesis would be: H : mean = 7. 0

B. The alternative hypothesis would be H a : mean > 7. The mean observation is signicantly dierent from 7, because the calculated Z-

C. statistic is less than the critical Z-statistic.

B. The alternative hypothesis would be H a : mean > 7. The mean observation is signicantly dierent from 7, because the calculated Z- — The way the question is worded, this is a two tailed test. The alternative hypothesis is not H : M > 7 because in a two-tailed test the alternative is =, while < and > indicate one-tailed a tests. A test statistic is calculated by subtracting the hypothesized parameter from the parameter that has been estimated and dividing the difference by the standard error of the sample statistic. Here, the test statistic = (sample mean – hypothesized mean) / (standard error of the sample statistic) = (5 - 7) / (1) = -2. The calculated Z is -2, while the critical value is -1.96. The calculated test statistic of -2 falls to the left of the critical Z- statistic of -1.96, and is in the rejection region. Thus, the null hypothesis is rejected and the conclusion is that the sample mean of 5 is significantly different than 7. What the negative sign shows is that the mean is less than 7; a positive...

Q23

Bo Rigley, CFA, is a financial analyst examining large-cap equity returns over a calendar year. His sample size is 252 trading days, and he observes a mean return of 0.07% and a standard deviation of 0.12%. With his null hypothesis that the daily portfolio return is equal to zero and a 10% level of significance, Rigley will:

A. reject the null because the test statistic is greater than the critical value.

B. not reject the null because the test statistic is less than the critical value.

C. not reject the null because the test statistic is greater than the critical value.

A. reject the null because the test statistic is greater than the critical value. — At a 10% level of significance, the critical z-values for a two-tailed test are + or –1.645, so the decision rule is to reject the null if the test statistic < –1.645 or > +1.645. With a sample size of 252 and a standard deviation of 0.12%, the standard error is equal to: s 0.12% S = = = 0.0075593% x¯ √n √252 The test statistic is equal to: 0.0007 = 9.26 0.000075593 Because the test statistic of 9.26 > 1.645, Rigley will reject the null that the daily equity return is equal to zero.

Q24

Which of the following statements about hypothesis testing is most accurate? A Type I error is the probability of:

A. rejecting a true null hypothesis.

B. failing to reject a false hypothesis.

C. rejecting a true alternative hypothesis.

A. rejecting a true null hypothesis. — The Type I error is the error of rejecting the null hypothesis when, in fact, the null is true.

Q25

For a hypothesis test regarding a population parameter, an analyst has determined that the probability of failing to reject a false null hypothesis is 18%, and the probability of rejecting a true null hypothesis is 5%. The power of the test is:

A. 0.95.

B. 0.18.

C. 0.82.

C. 0.82. — The power of the test is 1 – the probability of failing to reject a false null (Type II error); 1 – 0.18 = 0.82.

Q26

Which of the following statements about hypothesis testing is least accurate?

A. If the alternative hypothesis is H a : µ > µ 0 , a two-tailed test is appropriate.

B. A Type II error is failing to reject a false null hypothesis.

C. The null hypothesis is a statement about the value of a population parameter.

A. If the alternative hypothesis is H a : µ > µ 0 , a two-tailed test is appropriate. — The hypotheses are always stated in terms of a population parameter. Type I and Type II are the two types of errors you can make – reject a null hypothesis that is true or fail to reject a null hypothesis that is false. The alternative may be one-sided (in which case a > or < sign is used) or two-sided (in which case a ≠ is used).

Q27

Which of the following statements about hypothesis testing is most accurate?

A. The probability of a Type I error is equal to the signicance level of the test.

B. The power of a test is one minus the probability of a Type I error. If you can disprove the null hypothesis, then you have proven the alternative

C. hypothesis.

A. The probability of a Type I error is equal to the signicance level of the test. — The probability of getting a test statistic outside the critical value(s) when the null is true is the level of significance and is the probability of a Type I error. The power of a test is 1 minus the probability of a Type II error. Hypothesis testing does not prove a hypothesis, we either reject the null or fail to reject it.

Q28

Ron Jacobi, manager with the Toulee Department of Natural Resources, is responsible for setting catch-and-release limits for Lake Norby, a large and popular fishing lake. He takes a sample to determine whether the mean length of Northern Pike in the lake exceeds 18 inches. If the sample t-statistic indicates that the mean length of the fish is significantly greater than 18 inches, when the population mean is actually 17.8 inches, the t-test resulted in:

A. both a Type I and a Type II error.

B. a Type II error only.

C. a Type I error only.

A. both a Type I and a Type II error. — Rejection of a null hypothesis when it is actually true is a Type I error. Here, H o : μ ≤ 18 inches and H : μ > 18 inches. Type II error is failing to reject a null hypothesis when it is a actually false. Because a Type I error can only occur if the null hypothesis is true, and a Type II error can only occur if the null hypothesis is false, it is logically impossible for a test to result in both types of error at the same time.

Q29

A Type II error:

A. rejects a true null hypothesis.

B. fails to reject a false null hypothesis.

C. fails to reject a true null hypothesis.

A. rejects a true null hypothesis. — A Type II error is defined as accepting the null hypothesis when it is actually false. The chance of making a Type II error is called beta risk.

Q30

A data analyst compares the P/E ratios for two companies over a period of 20 years and calculates a two-tailed F-statistic that exceeds the critical F-statistic value. As a result of her findings, she will: not reject the null hypothesis that the P/E ratio variances between the two

A. companies are signicantly dierent. reject the null hypothesis that the P/E ratio variances between the two companies

B. are signicantly dierent. reject the null hypothesis that the P/E ratio variances between the two companies

C. are not signicantly dierent.

A. companies are signicantly dierent. reject the null hypothesis that the P/E ratio variances between the two companies — The F-test is used to assess the equality of two population variances. Here, the comparison is between two companies' P/E ratios over a period of 20 years. The null hypothesis for the two-tailed test is that the variances are not significantly different, while the alternative hypothesis is that they are significantly different. Because the calculated F-statistic is higher than the critical F-statistic value, the null hypothesis will be rejected.

Q31

John Jenkins, CFA, is performing a study on the behavior of the mean P/E ratio for a sample of small-cap companies. Which of the following statements is most accurate? A Type I error represents the failure to reject the null hypothesis when it is, in fact,

A. false. One minus the condence level of the test represents the probability of making a

B. Type II error.

C. The signicance level of the test represents the probability of making a Type I error.

A. false. One minus the condence level of the test represents the probability of making a — A Type I error is the rejection of the null when the null is actually true. The significance level of the test (alpha) (which is one minus the confidence level) is the probability of making a Type I error. A Type II error is the failure to reject the null when it is actually false.

Q32

Which of the following statements about hypothesis testing is most accurate? A Type II error is the probability of:

A. failing to reject a false null hypothesis.

B. rejecting a true alternative hypothesis.

C. rejecting a true null hypothesis.

A. failing to reject a false null hypothesis. — The Type II error is the error of failing to reject a null hypothesis that is not true.

Q33

A researcher is testing whether the average age of employees in a large firm is statistically different from 35 years (either above or below). A sample is drawn of 250 employees and the researcher determines that the appropriate critical value for the test statistic is 1.96. The value of the computed test statistic is 4.35. Given this information, which of the following statements is least accurate? The test:

A. indicates that the researcher will reject the null hypothesis.

B. has a signicance level of 95%. indicates that the researcher is 95% condent that the average employee age is

C. dierent than 35 years.

A. indicates that the researcher will reject the null hypothesis. — This test has a significance level of 5%. The relationship between confidence and significance is: significance level = 1 – confidence level. We know that the significance level is 5% because the sample size is large and the critical value of the test statistic is 1.96 (2.5% of probability is in both the upper and lower tails).

Q34

For a two-tailed test of hypothesis involving a z-distributed test statistic and a 5% level of significance, a calculated z-statistic of 1.5 indicates that:

A. the null hypothesis cannot be rejected.

B. the test is inconclusive.

C. the null hypothesis is rejected.

A. the null hypothesis cannot be rejected. — For a two-tailed test at a 5% level of significance the calculated z-statistic would have to be greater than the critical z value of 1.96 for the null hypothesis to be rejected.

Q35

Which of the following steps in hypothesis testing least likely includes „Collecting the data and calculating the statistic‟?

A. Making the economic or investment decision.

B. Making the statistical decision.

C. Stating the decision rule.

C. Stating the decision rule. — The seven steps in hypothesis testing are: 1) Stating the hypothesis. 2) Identifying the appropriate test statistic and its probability distribution. 3) Specifying the significance level. 4) Stating the decision rule. 5) Collecting the data and calculating the test statistic. 6) Making the statistical decision. 7) Making the economic or investment decision.

Q36

Marco Vitaly is a researcher and wants to test whether a particular parameter is larger than a specific value. In this case, the null and alternative hypothesis would be best defined as:

A. H : θ = θ versus H : θ ≠ θ . 0 0 a 0

B. H : θ ≤ θ versus H : θ > θ . 0 0 a 0

C. H : θ ≥ θ versus H : θ < θ .

B. H : θ ≤ θ versus H : θ > θ . 0 0 a 0 — A positive “hoped for” condition means that we will only reject the null (and accept the alternative) if the evidence indicates that the population parameter is greater than θ . Thus, H : θ ≤ θ versus Ha: θ > θ is the correct statement of the null and alternative 0 0 0 0 hypotheses.

Q37

Professor Alan Chang is reviewing the following statements made by his students:  Beth: The null hypothesis is the hypothesis that is being tested; and a two tailed hypothesis may have either of the two signs: < or >.  Donald: Specifying the significance level, α, isn‟t a necessary step and one could do without it during hypothesis testing.  Kevin: The test statistic is a quantity calculated based on a sample, whose value is the basis for deciding whether or not to reject the alternate hypothesis. Whose statements will Professor Chang will least likely agree to?

A. Only Donald.

B. Only Donald and Beth.

C. All of them.

C. All of them. — The null hypothesis is the hypothesis that is tested, and a two tailed hypothesis has the sign: =. Specifying the significance level, α, is a necessary step and one cannot do without it during hypothesis testing. The test statistic is a quantity calculated based on a sample, whose value is the basis for deciding whether or not to reject the null hypothesis.

Q38

Which of the following statements requires a two-tailed test?

A. H : µ ≤ 0 versus H : µ > 0. 0 a

B. H : µ = 0 versus H : µ ≠ 0. 0 a

C. H : µ ≥ 0 versus H : µ < 0.

B. H : µ = 0 versus H : µ ≠ 0. 0 a — A two-tailed test for the population mean is structured as: H : µ = 0 versus H : µ o a ≠ 0.

Q39

A Type II error is best described as when a test:

A. fails to reject a false null hypothesis.

B. fails to reject a true null hypothesis.

C. rejects a true null hypothesis.

A. fails to reject a false null hypothesis. — When we do not reject a false null hypothesis we have a Type II error.

Q40

In order to calculate the test statistic, the difference between the sample statistic and the value of the population parameter under H is most likely divided by: 0 COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 1 Hypothesis Testing – Question Bank www.ift.world

A. appropriate value from the t-distribution.

B. sample standard deviation.

C. standard error of the sample statistic.

C. standard error of the sample statistic. — A test statistic is defined as the difference between the sample statistic and the value of the population parameter under H divided by the standard error of the sample 0 statistic.

Q41

When a false null hypothesis is not rejected, it leads to a/an:

A. Type I Error.

B. Type II Error.

C. acceptance of the alternate hypothesis.

B. Type II Error. — Type II error arises when a false null hypothesis is not rejected. Type I error is rejecting the null hypothesis when it is true.

Q42

The results of an experiment are statistically significant when:

A. the null hypothesis is rejected.

B. the null hypothesis is not rejected.

C. the level of significance is altered.

A. the null hypothesis is rejected. — The results of an experiment are statistically significant when the null hypothesis is rejected.

Q43

Jane Norah is an analyst for a midcap growth fund. The fund earns a quarterly return of 4.5 percent relative to an estimated return of 6.0 percent. If Norah wishes to test whether the actual results are different from the estimated return of 6 percent, the null hypothesis is most likely:

A. H : µ ≤ 6.0. 0

B. H : µ = 6.0. 0

C. H : µ ≠ 6.0.

B. H : µ = 6.0. 0 — The null hypothesis for this test will be H = 6.0. 0

Q44

The mean annual return is 8 percent and the standard deviation is 6.4 percent for a sample containing 25 sectors. A fund manager is testing whether the mean annual return is less than 9 percent. The critical value is -1.96. What is the most likely conclusion from this test?

A. Reject the null hypothesis.

B. Do not reject the null hypothesis.

C. Additional information is required to decide.

B. Do not reject the null hypothesis. — The test statistic is ̅ = = - 0.78 √ √ Since the test statistic is less negative (lower absolute value) than the critical value, the null hypothesis is not rejected. COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 9 Hypothesis Testing – Question Bank www.ift.world

Q45

Assume that the population mean is μ, sample mean is ̅ , and is the standard error of the ̅ sample mean. Which of the following is a condition for rejecting the null hypothesis at the 95 percent confidence interval?

A. ̅ . ̅

B. ( ̅ ) > 1.96.

C. ̅

A. ̅ . ̅

Q46

A. Statistical decision.

B. Economic decision.

C. Both statistical and economic decision.

A. Statistical decision. — The economic decision considers economic issues.

Q47

What type of consideration is an investor‟s tolerance for risk and financial position in hypothesis testing?

A. Investment or economic decision.

B. Statistical decision.

C. Both statistical and economic decision.

A. Investment or economic decision. — Investor‟s risk tolerance is an investment decision, and not a statistical decision.

Q48

Which of the following statements regarding the p-value is most likely to be correct?

A. The p-value is the smallest level of significance at which the null hypothesis can be rejected.

B. The p-value is the smallest level of significance at which the null hypothesis can be accepted.

C. The p-value is the largest level of significance at which the null hypothesis can be

A. The p-value is the smallest level of significance at which the null hypothesis can be rejected. — The p-value is defined as the smallest level of significance at which the null hypothesis can be rejected.

Q49

A researcher formulates a null hypothesis that the mean of a distribution is equal to 20. He obtains a p-value of 0.018. Using a 5% level of significance, the best conclusion is to:

A. reject the null hypothesis.

B. accept the null hypothesis.

C. decrease the level of significance.

A. reject the null hypothesis. — As the p-value is less than the stated level of significance, we reject the null hypothesis.

Q50

A researcher conducted a one-tailed test with the null hypothesis that the mean of a distribution is greater than 2. The p-value came out to be 0.0475. If the researcher decides to use a 5% level of significance, the best conclusion is to:

A. fail to reject the null hypothesis.

B. reject the null hypothesis.

C. decrease the level of significance to 4.75%.

B. reject the null hypothesis. — Because the p-value (0.0475) is lower than the stated level of significance (0.05), we will reject the null hypothesis.

Q51

A researcher is using the p-value test for conducting hypothesis testing. He is most likely to reject the null hypothesis when the p-value of the test statistic:

A. exceeds a specified level of significance.

B. falls below a specified level of significance.

C. is negative.

B. falls below a specified level of significance. — If the p-value is less than the specified level of significance, the null hypothesis is rejected.

Q52

A researcher conducts a two-tailed t-test test with a null hypothesis that the population mean differs from zero. If the p-value is 0.089 and he is using a significance level of 5%, the most appropriate conclusion is:

A. do not reject the null hypothesis.

C. the chosen significance level is too high.

A. do not reject the null hypothesis. — The p-value is the smallest level of significance at which the null hypothesis can be rejected. In this case, the given p-value is greater than the given level of significance. Hence, we cannot reject the null hypothesis. Note that we simply compare the given p-value with the level of significance. Even though this is a two-tailed test we do not divide the p- value by 2.

Q53

Which of the following statistic is most likely to be used for the mean of a non-normal distribution with unknown variance and a small sample size?

A. z test statistic.

B. t test statistic.

C. There is no test statistic for such a scenario.

C. There is no test statistic for such a scenario. — The statistic for small sample size of a non-normal distribution with unknown variance is not available. z-test statistic is used for large sample size of a non-normal distribution with known variance while t-test statistic is used for large sample size of a non- normal distribution with unknown variance.

Q54

Orlando Bloom is analyzing a portfolio‟s performance for the past 15 years. The mean return for the portfolio is 10.25% with a sample standard deviation of 12.00%. Bloom wants to test the claim that the mean return is less than 12.50%. The null hypothesis is that the mean return is greater than or equal to 12.50%. If the critical value for this test is -2, which of the following is most likely the test statistic and the decision of this test? Test Statistic Decision

A. -0.726 Reject H 0

B. -0.726 Do not rejectH 0

C. -0.5422 Do not rejectH

B. -0.726 Do not rejectH 0 — √ Since the absolute value of -0.726 is less than the absolute value of -2, we cannot reject the null hypothesis.

Q55

Peter is studying the earnings per share of 32 companies in an industry. He plans to use the t- test for hypothesis testing. The degrees of freedom Peter will use for defining the critical region is closest to:

A. 30.

B. 31.

C. 32.

B. 31. — In a t-test, the degree of freedom is 1 less than the sample size. Therefore, it will be 31 in this case. COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 10 Hypothesis Testing – Question Bank www.ift.world

Q56

From two normally distributed populations, independent samples were drawn and following observations were made:  Sample A: The sample size of 20 observations had a sample mean of 63.  Sample B: The sample size of 14 observations had a sample mean of 58.  Standard deviations of sample A and sample B were equal. The pooled estimate of common variance was equal to 565.03.

A. 0.21.

B. 0.35.

C. 0.60.

C. 0.60. — The appropriate t-statistic can be calculated using the formula: t-statistic = [( )– ( )] √*( ) ( )+ [( )– ] = √*( ) ( )+ = 0.604

Q57

The table below shows the return data for samples which have been pooled from two normally distributed populations with equal variance. Sample # Sample size Annual returns 1 60 15.8% 2 112 12.5% The standard deviation of the pooled sample, s, is 256.68. Which of the following is the correct test statistic to test for the differences between means?

A. 0.0006.

B. 0.0008.

C. 0.0011.

A. 0.0006. — ̅ ̅ . [( ) ( )] [( ) ( )] –

Q58

Using the sample results given below, drawn as 25 paired observations from their underlying distributions, test if the mean returns of the two portfolios differ from each other at the 1% level of statistical significance. Assume the underlying distributions of returns for each portfolio are normal and that their population variances are not known. Portfolio 1 Portfolio 2 Difference Mean Return 8.00 11.25 3.25 Standard Deviation 8.80 15.50 6.70 t-statistic for 24 df and at the 1% level of statistical significance = 2.797 Based on the paired comparisons test of the two portfolios, the most appropriate conclusion is to:

A. reject the hypothesis that the mean difference equals zero as the computed test statistic exceeds 2.807. COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 5 Hypothesis Testing – Question Bank www.ift.world

B. accept the hypothesis that the mean difference equals zero as the computed test statistic exceeds 2.807.

C. accept the hypothesis that the mean difference equals zero as the computed test statistic is

C. accept the hypothesis that the mean difference equals zero as the computed test statistic is — The test statistic is: = 2.425. √ As 2.425 < 2.807, we do not reject the null hypothesis that the mean difference is zero. This is a two tail test.

Q59

An analyst collects the following data related to paired observations for Sample A and Sample B. Assume that both samples are drawn from normally distributed populations and that the population variances are not known: Paired Observation Sample A Value Sample B Value 1 12 5 2 18 15 3 4 1 4 -6 -9 5 -5 4 The t-statistic to test the hypothesis that the mean difference is equal to zero is closest to:

A. 0.23.

B. 0.27.

C. 0.52.

C. 0.52. — Paired Sample Sample Differences Differences Minus the Mean Observation A Value B Value Difference, Then Squared 1 12 5 7 ( ) = 31.36 2 18 15 3 ( ) = 2.56 3 4 1 3 ( ) = 2.56 4 -6 -9 3 ( ) = 2.56 5 -5 4 –9 ( ) = 108.16 Sum = 7 Sum of squared differences = Mean = 1.4 147.2 Sample variance: Standard 2.712932 = √ error: t-Statistic: – 0.51605 = COPYRIGHT © 2015 IFT. ALL RIGHTS RESERVED. 11 Hypothesis Testing – Question Bank www.ift.world

Q60

Which of the following is true for a paired comparison test?

A. The samples are independent.

B. The samples are dependent.

C. The test conducted is a test concerning differences between mean and not mean

B. The samples are dependent. — A paired comparison test is conducted for mean differences and the samples are dependent.

Q61

The table below shows the annual return summary for KSE-50 and KSE-100 portfolios. Portfolio Mean Return Standard Deviation KSE – 50 19.25% 20.05% KSE – 100 15.98% 17.11% Difference 3.27% 5.48% The null hypothesis for the test conducted is H : µ = 0. The sample size is 64. o d Which of the following most likely represent the test conducted and the value of the test statistic?

A. A chi square test with t statistic = 4.77.

B. A paired comparison test with t statistic = 5.27.

C. A paired comparison test with t statistic = 4.77.

C. A paired comparison test with t statistic = 4.77. — Since the test concerns mean differences, it is a paired comparisons test. ( ) ( √ ) .

Q62

A hypothesis test is to be conducted in order to test the differences between means. Which of the following will least likely be used as a null hypothesis for this test?

A. H : µ + µ = 0. o 1 2

C. H : µ < µ

A. H : µ + µ = 0. o 1 2 — The incorrect null hypothesis is H : µ + µ = 0. o 1 2

Q63

A researcher drew two samples from two normally distributed populations. The mean and standard deviation of the first sample were 4 and 48 respectively. The mean and standard deviation of the second sample were 6 and 52 respectively. The number of observations in the first sample was 30 and second sample was 32. Given a null hypothesis of versus an alternate hypothesis of , which of the following is most likely to be the test statistic?

A. 0.235.

B. 0.852.

C. 1.170.

C. 1.170. — The test that compares the variances using two independent samples from two different populations makes use of the F-distributed t-statistic: The smaller variance is the denominator, thus: .

Q64

The null hypothesis most likely tests:

A. the mean differences.

B. a single variance.

C. the equality of two variances.

C. the equality of two variances. — The test concerns the equality of two variances. It is known as the F-test.

Q65

For an F-test specified as , which of the following is used as the actual test statistic?

A. should be greater than .

B. should be less than .

C. It does not matter whether is greater or less than

A. should be greater than . — A common convention or a usual practice is that the ratio should be greater than or equal to 1, which is only possible if option A is true.

Q66

Which test should be used for hypothesis related to a single population variance?

A. A chi-square test with degrees of freedom, n.

B. A chi-square test with degrees of freedom, n-1.

C. An F-test with degrees of freedom, n-1.

B. A chi-square test with degrees of freedom, n-1. — To test for a single population variance, select a chi-square test with (n – 1) degrees of freedom.

Q67

A test that makes minimal assumptions about the population from which the sample comes is known as a:

A. paired comparisons test.

B. parametric test.

C. nonparametric test.

C. nonparametric test. — A test that makes minimal assumptions about the population from which the sample comes is known as a non-parametric test. It is not concerned with a parameter.

Q68

A. When the data does not meet distributional assumptions.

B. When the data provided is given in ranks.

C. When the hypothesis being addressed concerns a parameter.

Concepts 'n' Clarity^®

For Educational Use Only — Not for Resale

Ch 8 — Hypothesis Testing

Concepts 'n' Clarity^®

Ch 9 — Parametric & Non-Parametric Tests of Independence

CFA Level I • FRM Part I

PART 1 — LOS-ALIGNED STUDY NOTES

LOS 9.a — Parametric and nonparametric tests that population correlation coefficient equals zero; Spearman rank correlation

Testing Correlation = 0

t = r × √(n−2) / √(1−r²) — t-distribution with n−2 df

Spearman rank correlation: r_s = 1 − 6Σd²ᵢ/[n(n²−1)] where dᵢ = difference in ranks

Both Pearson (parametric) and Spearman (nonparametric) use same t-test formula for H₀: ρ = 0.
Test statistic increases with both r and n; larger n makes it easier to detect weak correlations.
Spearman advantage: works for ordinal data; robust to outliers; no normality assumption.
Spearman measures MONOTONIC (not necessarily linear) relationships.

LOS 9.b — Tests of independence based on contingency table data

Contingency Table Chi-Square Test

χ² = Σ (Observed − Expected)² / Expected | df = (r−1)(c−1)

Expected frequency = (Row total × Column total) / Grand total

Contingency table: cross-tabulates frequencies of two categorical variables.
H₀: the two categorical variables are INDEPENDENT.
H₁: the two variables are NOT independent (associated).
Larger χ² statistic → stronger evidence against independence.
Always right-tailed test (χ² is always positive).
Application: test if sector membership and credit rating are independent.

PART 2 — PRACTICE QUESTIONS (12 Questions)

Click Show Answer to reveal the answer and explanation.

LOS 9.a — Testing Correlation = 0 & Spearman Rank Correlation — Q1 to 8

A test of independence based on contingency table data uses a(n):

A. chi-square statistic.

B. F-statistic.

C. t-statistic.

A. chi-square statistic. — A test for independence based on contingency table data uses a chi-square statistic.

For a test of independence based on contingency table data, the test statistic is calculated as the: mean absolute deviation of all table cells’ actual values and their expected values, if

A. the two characteristics are independent. sum of dierences between each table cell’s actual value and its expected value, if

B. the two characteristics are independent. sum of squared dierences between each table cell’s actual value and its expected

C. value, if the two characteristics are independent.

A. the two characteristics are independent. sum of dierences between each table cell’s actual value and its expected value, if — The chi-square statistic for a test of independence based on contingency table data is the sum of squared differences between each table cell's actual value and its expected value, if the two characteristics are independent.

Critical values from Student's t-distribution for a two-tailed test at a 5% significance level: df 28 2.048 29 2.045 30 2.042 A researcher wants to test a hypothesis that two variables have a population correlation coefficient equal to zero. For a sample size of 30, the appropriate critical value for this test is plus-or-minus:

A. 2.048.

B. 2.045.

C. 2.042.

A. 2.048. — The test statistic for a hypothesis test concerning population correlation follows a t- distribution with n − 2 degrees of freedom. For a sample size of 30 and a significance level of 5%, the sample statistic must be greater than 2.048 or less than −2.048 to reject the hypothesis that the population correlation equals zero.

A test of the hypothesis that two categorical variables are independent is most likely to employ:

A. population parameters.

B. t-statistics.

C. contingency tables.

A. population parameters. — A hypothesis test whether of two categorical variables (e.g., company sector and bond rating) are independent can be performed by constructing a contingency table and calculating a chi-squared statistic.

In a test of independence based on contingency table data, degrees of freedom are the:

A. sum of the number of rows and the number of columns.

B. product of the number of rows minus one and the number of columns, minus one.

C. sum of the number of rows and the number of columns, minus two.

A. sum of the number of rows and the number of columns. — Degrees of freedom for a test for independence are [(r – 1) (c – 1)], where r and c are the number of rows and the number of columns in the contingency table.

To test a hypothesis that the population correlation coefficient of two variables is equal to zero, an analyst collects a sample of 24 observations and calculates a sample correlation coefficient of 0.37. Can the analyst test this hypothesis using only these two inputs?

A. Yes.

B. No, because the sample standard deviations of the two variables are also required.

C. No, because the sample means of the two variables are also required.

A. Yes. — r√n−2 The t-statistic for a test of the population correlation coefficient is , where r is √1−r2 the sample correlation coefficient and n is the sample size.

A researcher wants to test whether the weekly returns on two stocks are correlated. The test statistic for the appropriate test follows a:

A. chi-square distribution.

B. t-distribution with n − 1 degrees of freedom.

C. t-distribution with n − 2 degrees of freedom.

A. chi-square distribution. — The test statistic for the significance of the correlation between two random variables follows a t-distribution with n − 2 degrees of freedom.

Student's t-distribution, level of significance for a two-tailed test: df 0.20 0.10 0.05 0.02 0.01 0.001 16 1.337 1.746 2.120 2.583 2.921 4.015 17 1.333 1.740 2.110 2.567 2.898 3.965 18 1.330 1.734 2.101 2.552 2.878 3.922 19 1.328 1.729 2.093 2.539 2.861 3.883 20 1.325 1.725 2.086 2.528 2.845 3.850 Based on a sample correlation coefficient of −0.525 from a sample size of 19, an analyst −0.525√19−2 calculates a t-statistic of = −2.5433. The analyst can reject the √1−(−0.525) 2 hypothesis that the population correlation coefficient equals zero:

A. at a 2% signicance level, but not at a 1% signicance level.

B. at a 1% signicance level.

C. at a 5% signicance level, but not at a 2% signicance level.

A. at a 2% signicance level, but not at a 1% signicance level. — With 19 − 2 = 17 degrees of freedom, the critical values are plus-or-minus 2.110 at a 5% significance level, 2.567 at a 2% significance level, and 2.898 at a 1% significance level. Because the t-statistic of −2.5433 is less than −2.110, the hypothesis can be rejected at a 5% significance level. Because the t-statistic is greater than −2.567, the hypothesis cannot be rejected at a 2% significance level (or any smaller significance level).

LOS 9.a & 9.b — Spearman Rank & Contingency Table Tests — Q9 to 12

A researcher calculates the Spearman rank correlation between two variables as 0.45 based on 25 pairs of observations. The appropriate test statistic for testing whether the population rank correlation is zero is:

A. t = 0.45 × √(25−2) / √(1−0.45²)

B. t = 0.45 × √25 / √(1−0.45²)

C. z = 0.45 × √25

A. t = 0.45 × √(25−2) / √(1−0.45²) — The test statistic for Spearman rank correlation follows a t-distribution with n−2 degrees of freedom: t = r_s × √(n−2) / √(1−r_s²) = 0.45 × √23 / √(1−0.2025) = 0.45 × 4.796 / √0.7975 = 2.158/0.893 = 2.42. This is the same formula used for Pearson correlation when testing H₀: ρ = 0, but applied to ranks rather than raw values.

Q10

A contingency table is constructed to test whether sector affiliation (Technology, Healthcare, Finance) and credit rating (Investment Grade, High Yield) are independent. The table has 3 rows and 2 columns with 120 total observations. The degrees of freedom for the chi-square test are:

A. 2

B. 5

C. 6

A. 2 — Degrees of freedom for a contingency table test of independence = (r−1)(c−1) = (3−1)(2−1) = 2×1 = 2, where r = rows and c = columns. Note: the total number of observations (120) does not affect the degrees of freedom formula. With df = 2, we compare the chi-square test statistic to the critical value from the χ² distribution with 2 degrees of freedom.

Q11

A researcher is testing whether there is a significant correlation between a fund's performance rank in year 1 and its rank in year 2 across 32 funds. The Spearman rank correlation is 0.30. At the 5% significance level (two-tailed), the critical t-value with 30 df is 2.042. The test statistic is closest to:

A. 1.70

B. 1.95

C. 2.04

A. 1.70 — t = r_s × √(n−2) / √(1−r_s²) = 0.30 × √30 / √(1−0.09) = 0.30 × 5.477 / √0.91 = 1.643/0.954 = 1.72 ≈ 1.70. Since |1.70| < 2.042, we fail to reject H₀ at 5% significance. The correlation is not statistically significant — past performance rank does not significantly predict future rank in this sample.

Q12

Which of the following is a key advantage of using the Spearman rank correlation test compared to the Pearson correlation test?

A. It requires larger sample sizes to be valid

B. It does not assume the data are normally distributed

C. It is only appropriate for continuous data

B. It does not assume the data are normally distributed — The Spearman rank correlation is a nonparametric test that does not require the assumption of bivariate normality. It can be applied to ordinal data (ranks), is robust to outliers, and is appropriate when the normal distribution assumption is violated. The Pearson correlation test assumes that both variables are normally distributed. Spearman can also detect monotonic non-linear relationships that Pearson would miss.

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Ch 9 — Parametric & Non-Parametric Tests of Independence

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Ch 10 — Simple Linear Regression

CFA Level I • FRM Part I

PART 1 — LOS-ALIGNED STUDY NOTES

LOS 10.a — Describe simple linear regression model; least squares criterion; interpret coefficients

Regression Model

Y = b₀ + b₁X + ε | b₁ = Cov(X,Y)/Var(X) | b₀ = Ȳ − b₁X̄

b₁ (slope): change in Y for a 1-unit change in X.
b₀ (intercept): expected value of Y when X = 0.
OLS least squares: minimizes Σε² (sum of squared residuals).
Regression line passes through (X̄, Ȳ) — the sample means.

LOS 10.b — Explain regression assumptions; how residuals indicate violations

OLS Assumptions

1. Linear relationship between X and Y.
2. Homoskedasticity: constant variance of residuals (no fan-shaped pattern in residual plot).
3. Independence: residuals are uncorrelated (no autocorrelation).
4. Normality: residuals are normally distributed.
Heteroskedasticity: non-constant variance — biases standard errors; fix with robust SE or transformations.
Autocorrelation: correlated residuals — common in time series; Durbin-Watson test.

LOS 10.c & 10.d — Measures of fit; ANOVA table; standard error of estimate

ANOVA Table & R²

R² = SSR/SST = 1 − SSE/SST | SEE = √(SSE/(n−2)) = √MSE

R²: proportion of total variation in Y explained by X; bounded [0, 1].
SEE: average distance of actual Y from fitted line; lower = better fit.
F-statistic: tests overall regression significance; F = MSR/MSE; reject H₀: b₁=0 if F > F_critical.
In simple regression: F = t² (where t tests b₁ significance).

LOS 10.e — Predicted values and prediction intervals

Prediction

Ŷ = b₀ + b₁X* | Prediction interval: Ŷ ± t* × sf

Confidence interval for mean Y: narrower; estimates mean response at X*.
Prediction interval for individual Y: wider; accounts for individual observation variability.
Prediction interval = CI for mean Y × adjustment factor sf ≈ SEE × √(1 + 1/n + (X*−X̄)²/Σ(Xᵢ−X̄)²).
Both intervals are narrowest at X = X̄ and widen as X moves away from X̄.

LOS 10.f — Functional forms of simple linear regression

Functional Forms

PART 2 — PRACTICE QUESTIONS (23 Questions)

Click Show Answer to reveal the answer and explanation.

LOS 10.a–10.c — OLS, Coefficients, R², F-test, ANOVA — Q1 to 13

The estimated slope coefficient in a simple linear regression is: the predicted value of the dependent variable, given the actual value of the

A. independent variable. the change in the independent variable, given a one-unit change in the dependent

B. variable. the ratio of the covariance of the regression variables to the variance of the

C. independent variable.

A. independent variable. the change in the independent variable, given a one-unit change in the dependent — Cov X,Y The estimated slope coefficient in a simple linear regression is , where Y is the σ2 X dependent variable and X is the independent variable. The estimated slope coefficient is interpreted as the change in the dependent variable, given a one-unit change in the independent variable. The predicted value of the dependent variable must consider the estimated intercept term along with the estimated slope coefficient.

Given the relationship: Y = 2.83 + 1.5X What is the predicted value of the dependent variable when the value of the independent variable equals 2?

A. 2.83.

B. –0.55.

C. 5.83.

C. 5.83. — Y = 2.83 + (1.5)(2) = 2.83 + 3 = 5.83.

When there is a linear relationship between an independent variable and the relative change in the dependent variable, the most appropriate model for a simple regression is:

A. the log-log model.

B. the log-lin model.

C. the lin-log model.

A. the log-log model. — A regression of the form ln Y = b + b X is appropriate when the relative change in the 0 1 dependent variable is a linear function of the independent variable.

Consider the following analysis of variance (ANOVA) table: Source Sum of squares Degrees of freedom Mean sum of squares Regression 556 1 556 Error 679 50 13.5 Total 1,235 51 The R2 for this regression is closest to:

A. 0.45.

B. 0.55.

C. 0.82.

A. 0.45. — R2 = sum of squares regression / sum of squares total = 556 / 1,235 = 0.45.

The coefficient of determination for a linear regression is best described as the: percentage of the variation in the dependent variable explained by the variation of

A. the independent variable. percentage of the variation in the independent variable explained by the variation of

B. the dependent variable.

C. covariance of the independent and dependent variables.

A. the independent variable. percentage of the variation in the independent variable explained by the variation of — The coefficient of determination for a linear regression describes the percentage of the variation in the dependent variable explained by the variation of the independent variable.

A simple linear regression is said to exhibit heteroskedasticity if its residual term:

A. does not have a constant variance.

B. is nonnormally distributed.

C. is not independently distributed.

A. does not have a constant variance. — Heteroskedasticity is the condition in which the variance of the residual term of a regression is not constant across all observations.

To determine a confidence interval around the predicted value from a simple linear regression, the appropriate degrees of freedom are:

A. n – 1.

B. n.

C. n – 2.

A. n – 1. — The degrees of freedom are n – 2 for a confidence interval around a predicted value.

Which of the following is least likely an assumption of linear regression?

A. The variance of the error terms each period remains the same.

B. The error terms from a regression are positively correlated.

C. Values of the independent variable are not correlated with the error term.

A. The variance of the error terms each period remains the same. — One assumption of linear regression is that the error terms are independently distributed. In this case, the correlations between error terms are expected to be zero. Constant variance of the error terms and no correlation between the independent variable and the error term are assumptions of linear regression.

A simple linear regression is a model of the relationship between:

A. one dependent variable and one or more independent variables.

B. one dependent variable and one independent variable.

C. one or more dependent variables and one or more independent variables.

A. one dependent variable and one or more independent variables. — A simple linear regression is a model of the relationship between one dependent variable and one independent variable. A multiple regression is a model of the relationship between one dependent variable and more than one independent variable.

Q10

Consider the following analysis of variance (ANOVA) table: Source Sum of squares Degrees of freedom Mean sum of squares Regression 550 1 550.000 Error 750 38 19.737 Total 1,300 39 The F-statistic for the test of the fit of the model is closest to:

A. 0.42.

B. 0.97.

C. 27.87.

A. 0.42. — F = sum of squares regression / mean squared error = 550 / 19.737 = 27.867.

Q11

To account for logarithmic variables, functional forms of simple linear regressions are available if: the independent variable is logarithmic, but not if the dependent variable is

A. logarithmic.

B. either the dependent or independent variable is logarithmic, but not both.

C. either or both of the dependent and independent variables are logarithmic.

A. logarithmic. — A log-lin model is appropriate if the dependent variable is logarithmic, while the independent variable is linear. A lin-log model is appropriate if the independent variable is logarithmic, while the dependent variable is linear. A log-log model is appropriate if both the independent and dependent variables are logarithmic.

Q12

A simple linear regression is performed to quantify the relationship between the return on the common stocks of medium-sized companies (mid-caps) and the return on the S&P 500 index, using the monthly return on mid-cap stocks as the dependent variable and the monthly return on the S&P 500 as the independent variable. The results of the regression are shown below: Coefficient Standard Error of Coefficient t-Value Intercept 1.71 2.950 0.58 S&P 500 1.52 0.130 11.69 Coefficient of determination = 0.599 The strength of the relationship, as measured by the correlation coefficient, between the return on mid-cap stocks and the return on the S&P 500 for the period under study was:

A. 0.130.

B. 0.774.

C. 0.599.

C. 0.599. — We are given the coefficient of determination of 0.599 (R2) and are asked to find the correlation coefficient (r), which is the square root of the coefficient of determination for a simple regression: √0.599 = 0.774

Q13

In a simple regression model, the least squares criterion is to minimize the sum of squared differences between:

A. the intercept term and the residual term.

B. the predicted and actual values of the dependent variable.

C. the estimated and actual slope coecient.

A. the intercept term and the residual term. — The least squares criterion defines the best-fitting linear relationship as the one that minimizes the sum of squared errors, the squared vertical distances between the predicted and actual values of the dependent variable.

LOS 10.d–10.f — ANOVA Table, SEE, Prediction Intervals, Functional Forms — Q14 to 23

Q14

In a simple linear regression of stock returns (Y) on market returns (X), the ANOVA table shows: SSR = 450, SSE = 150, n = 52. The F-statistic and R² are closest to:

A. F = 150.0, R² = 0.75

B. F = 148.5, R² = 0.75

C. F = 3.0, R² = 0.75

A. F = 150.0, R² = 0.75 — SST = SSR + SSE = 450 + 150 = 600. R² = SSR/SST = 450/600 = 0.75. For simple regression (k=1): MSR = SSR/1 = 450. MSE = SSE/(n−k−1) = 150/(52−2) = 150/50 = 3.0. F = MSR/MSE = 450/3.0 = 150.0. Answer: A. Note: F = t² for simple regression, where t tests significance of the slope.

Q15

A regression of annual dividend yield (Y) on payout ratio (X) produces: b₀ = 1.2, b₁ = 0.08, SEE = 0.5, n = 30. If payout ratio = 40%, the predicted dividend yield and 95% prediction interval (t* ≈ 2.048) are approximately:

A. 4.4% ± 1.024%

B. 4.4% ± 1.049%

C. 4.4% ± 2.048%

B. 4.4% ± 1.049% — Predicted Y = 1.2 + 0.08(40) = 1.2 + 3.2 = 4.4%. Prediction interval = Ŷ ± t* × sf, where sf ≈ SEE × √(1 + 1/n + (X−X̄)²/Σ(Xᵢ−X̄)²). Approximating sf ≈ SEE × √(1 + 1/n) ≈ 0.5 × √(1.033) ≈ 0.508. PI = 4.4% ± 2.048(0.508) = 4.4% ± 1.04% ≈ ±1.049%. Prediction intervals are wider than confidence intervals because they account for the uncertainty of individual observations, not just the mean.

Q16

The standard error of estimate (SEE) in a simple linear regression is best interpreted as:

A. the standard deviation of the independent variable

B. a measure of the variability of actual values around the regression line

C. the standard error of the slope coefficient

B. a measure of the variability of actual values around the regression line — SEE = √(SSE/(n−2)) measures the average distance between actual Y values and the fitted regression line. A lower SEE indicates a better-fitting model with less unexplained variation. SEE is used to construct prediction intervals for individual Y values. It is NOT the standard deviation of X (that is sx), nor the standard error of b₁ (that is sb₁ = SEE/√Σ(Xi−X̄)²).

Q17

A regression model uses log(Y) as the dependent variable and X as the independent variable: log(Y) = b₀ + b₁X. This is called a log-linear model. The slope coefficient b₁ is best interpreted as:

A. the change in Y for a one-unit change in X

B. the relative (percentage) change in Y for a one-unit change in X

C. the change in log(X) for a one-unit change in Y

B. the relative (percentage) change in Y for a one-unit change in X — In a log-linear (log-lin) model where log(Y) = b₀ + b₁X, the slope b₁ represents the percentage change in Y for a one-unit change in X (approximately). Specifically: %ΔY ≈ b₁ × 100%. This functional form is used when Y grows at a constant rate (exponential growth). For example, b₁ = 0.05 means Y increases by approximately 5% for each unit increase in X.

Q18

An analyst runs a regression and finds that the residual plot shows a fan-shaped pattern — residuals are small for small fitted values but large for large fitted values. This pattern most likely indicates:

A. autocorrelation of residuals

B. heteroskedasticity

C. multicollinearity

B. heteroskedasticity — Heteroskedasticity occurs when the variance of the residuals is not constant across all values of the independent variable. A fan-shaped (cone-shaped) residual plot is the classic indicator. This violates the OLS assumption of homoskedasticity (constant variance). Heteroskedasticity does NOT bias slope estimates, but it makes OLS standard errors unreliable, distorting hypothesis tests and confidence intervals. Autocorrelation involves correlation between residuals across time; multicollinearity involves correlation among independent variables.

Q19

In a simple linear regression ANOVA table with n = 26 observations, the total degrees of freedom, regression degrees of freedom, and error degrees of freedom are, respectively:

A. 25, 1, 24

B. 26, 1, 25

C. 25, 2, 23

A. 25, 1, 24 — For simple linear regression (one independent variable): Total df = n−1 = 26−1 = 25. Regression df = k = 1 (number of independent variables). Error df = n−k−1 = 26−1−1 = 24. Check: Regression df + Error df = 1 + 24 = 25 = Total df ✓. Note: Multiple regression would have k > 1 regression degrees of freedom.

Q20

A regression model: ln(Y) = b₀ + b₁ × ln(X) is called a log-log model. If b₁ = 1.5, a 10% increase in X is associated with an approximate change in Y of:

A. 1.5%

B. 10%

C. 15%

C. 15% — In a log-log model, b₁ is the elasticity — the percentage change in Y for a 1% change in X. If b₁ = 1.5, then a 10% increase in X leads to approximately 1.5 × 10% = 15% increase in Y. This interpretation makes log-log models useful for estimating price elasticity of demand, income elasticity, and other economic relationships where constant elasticity is assumed.

Q21

For a simple regression with b₀ = 3.5, b₁ = 0.65, and the sample mean of X = 20, the fitted value Ŷ when X = 25 is, and the regression line always passes through:

A. Ŷ = 19.75; passes through (X̄, Ȳ)

B. Ŷ = 19.75; passes through the origin

C. Ŷ = 16.25; passes through (X̄, Ȳ)

A. Ŷ = 19.75; passes through (X̄, Ȳ) — Ŷ = 3.5 + 0.65(25) = 3.5 + 16.25 = 19.75. A key property of OLS regression is that the regression line always passes through the point (X̄, Ȳ) — the means of X and Y. This is because the OLS estimators b₀ = Ȳ − b₁X̄ ensures the line passes through the sample means. The line does NOT necessarily pass through the origin (unless b₀ = 0).

Q22

An analyst is testing whether the slope coefficient in a regression of bond spreads on credit ratings is significantly different from zero. With n = 40, b₁ = 0.45, and s(b₁) = 0.18, the t-statistic and conclusion at the 5% level (two-tailed, t-critical ≈ 2.024 for 38 df) are:

A. t = 2.50; reject H₀, slope is significant

B. t = 2.50; fail to reject H₀, slope is not significant

C. t = 0.40; fail to reject H₀, slope is not significant

A. t = 2.50; reject H₀, slope is significant — t = b₁/s(b₁) = 0.45/0.18 = 2.50. Since |2.50| > 2.024 (critical value), reject H₀: b₁ = 0. The slope is statistically significant at the 5% level — credit ratings do significantly explain bond spreads. The df for the t-test on a slope coefficient in simple regression = n−2 = 40−2 = 38.

Q23

Which of the following statements about R² in simple linear regression is most accurate?

A. R² can be negative when the regression model fits poorly

B. R² equals the square of the correlation coefficient between X and Y

C. A high R² always implies the regression model is correctly specified

B. R² equals the square of the correlation coefficient between X and Y — In simple (one-variable) linear regression, R² = r²_XY, where r_XY is the Pearson correlation coefficient. R² cannot be negative in OLS regression — it is bounded between 0 and 1 (it equals 0 when b₁ = 0, and 1 for a perfect fit). A high R² does NOT guarantee correct specification — the model could be spurious (e.g., trending variables), have omitted variables, or violate OLS assumptions.

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Ch 10 — Simple Linear Regression

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Ch 11 — Introduction to Big Data Techniques

CFA Level I • FRM Part I

PART 1 — LOS-ALIGNED STUDY NOTES

LOS 11.a — Describe aspects of FinTech relevant for gathering and analyzing financial data

FinTech in Finance

FinTech: technology developments applied to financial services — includes companies and tools.
Key areas: large-scale data handling, AI for analysis, automated trading, robo-advisory, blockchain.
Robo-advisors: automated algorithm-driven investment management with minimal human intervention.
Algorithmic trading: execution of orders using pre-programmed rules and signals.
High-frequency trading (HFT): exploits microsecond price discrepancies using speed advantages.

LOS 11.b — Describe Big Data, artificial intelligence, and machine learning

Big Data

Structured data: organized rows/columns (SQL databases, financial statements) — easiest to analyze.
Semi-structured data: some structure but not tabular (JSON, XML, web pages, photos).
Unstructured data: no predefined structure (videos, audio, free-form text) — majority of big data.
Alternative data: non-traditional sources — satellite imagery, credit card transactions, social media, IoT sensors.
IoT (Internet of Things): network of sensor-embedded devices generating continuous data.
Data science: methods for processing Big Data — capture, curation, storage, search, transfer, visualization.

Artificial Intelligence & Machine Learning

Artificial Intelligence (AI): computer systems simulating human cognition (learning, reasoning, problem-solving).
Machine Learning (ML): AI subfield where systems improve performance through experience/data without explicit programming.
Supervised learning: trains on LABELED data; predicts outcomes (regression, classification). E.g., credit scoring.
Unsupervised learning: finds patterns in UNLABELED data (clustering, dimensionality reduction). E.g., customer segmentation.
Deep learning: ML using neural networks with many hidden layers; excels at image/speech/text recognition.
Overfitting: model learns training data noise → poor out-of-sample performance; mitigated by cross-validation.

LOS 11.c — Describe applications of Big Data and Data Science to investment management

Investment Applications

Text analytics & NLP: analyze earnings call transcripts, news, social media for sentiment and signals.
Risk governance: ML models for real-time risk monitoring and anomaly detection.
Algorithmic trading: pattern recognition in price data for automated strategy execution.
Alternative data analysis: satellite imagery for supply chain monitoring; credit card data for retail sales.
Natural Language Processing (NLP): converts unstructured text into structured investment signals.
Applications generate alpha by identifying signals BEFORE they are reflected in market prices.

PART 2 — PRACTICE QUESTIONS (12 Questions)

Click Show Answer to reveal the answer and explanation.

LOS 11.a — FinTech Applications — Q1 to Q3

Which of the following is LEAST likely to be described as a FinTech application directly relevant to investment management?

A. Algorithmic trading systems that execute orders based on predefined rules

B. Natural language processing to analyze earnings call transcripts for sentiment

C. Traditional portfolio construction using mean-variance optimization

C. Traditional portfolio construction using mean-variance optimization — Traditional mean-variance portfolio optimization is a classical finance technique, not a FinTech application. FinTech refers to technology-driven innovations in financial services. Applications relevant to investment management include: algorithmic and high-frequency trading, NLP for text analysis, robo-advisors, blockchain/distributed ledgers, AI-based risk management, and machine learning for pattern recognition in large datasets.

Big Data is characterized by the 'three Vs'. Which of the following correctly identifies all three?

A. Volume, Velocity, Variety

B. Volume, Validity, Versatility

C. Velocity, Variability, Value

A. Volume, Velocity, Variety — The three primary characteristics (3 Vs) of Big Data are: Volume (enormous quantity of data — from terabytes to petabytes), Velocity (speed at which data is generated and communicated; low latency = real-time; high latency = periodic), and Variety (different forms of data — structured, semi-structured, unstructured). Some frameworks add Veracity (data quality) and Value as additional Vs.

A portfolio manager uses a machine learning algorithm that identifies patterns in historical stock data and automatically improves its predictions over time by adjusting to new data without being explicitly reprogrammed. This is best described as:

A. structured data analysis

B. machine learning

C. natural language processing

B. machine learning — Machine learning is a form of artificial intelligence where computer systems improve their performance on a task through experience (data) without being explicitly programmed with new rules. The key feature is that the algorithm learns from data and improves over time. Natural language processing (NLP) is a subset of ML focused on text/language data. Structured data analysis is a broader term referring to analysis of organized data (spreadsheets, databases).

LOS 11.b — Big Data, AI, Machine Learning — Q4 to Q9

An investment firm's risk management team uses a neural network trained on thousands of historical credit events to predict the probability of corporate default. This application is best classified as:

A. supervised machine learning

B. unsupervised machine learning

C. bootstrap resampling

A. supervised machine learning — Supervised learning trains models on labeled data — historical credit events where the outcome (default or no default) is known. The model learns to predict the label (default probability) for new observations. Unsupervised learning finds patterns in unlabeled data (e.g., clustering similar securities). Neural networks can be used for both, but when trained with known outcomes to predict future outcomes, it is supervised learning.

An analyst uses sentiment analysis on social media posts and news articles to gauge market sentiment before earnings releases. The branch of AI that processes and interprets human language text to extract meaning is called:

A. neural network analysis

B. natural language processing (NLP)

C. cluster analysis

B. natural language processing (NLP) — Natural Language Processing (NLP) is a branch of artificial intelligence focused on enabling computers to understand, interpret, and generate human language. Investment applications include: sentiment analysis of news/social media, earnings call transcript analysis, parsing financial reports, and automated news summarization. NLP converts unstructured text data into structured data that can be used in quantitative models.

Which of the following best describes 'alternative data' in the context of investment management?

A. Financial statement data from company annual reports

B. Government economic statistics such as GDP and CPI

C. Data from non-traditional sources such as satellite imagery, credit card transactions, and web scraping

C. Data from non-traditional sources such as satellite imagery, credit card transactions, and web scraping — Alternative data refers to non-traditional data sources outside conventional financial and economic statistics. Examples include: satellite imagery (to count cars in retail parking lots or ships in ports), credit card transaction data, social media activity, web scraping of prices, geolocation data, and IoT sensor data. Traditional data (A and B) includes financial statements, market prices, and government statistics. Alternative data may provide investment edges before this information is reflected in prices.

A key limitation of machine learning models in investment applications is that they:

A. cannot be applied to large datasets

B. may overfit to historical data and perform poorly out of sample

C. require the data to be normally distributed

B. may overfit to historical data and perform poorly out of sample — Overfitting is the primary risk in ML models — the model learns noise and idiosyncratic patterns in the training data so well that it fails to generalize to new data. Complex models with many parameters are especially prone to overfitting. Techniques to mitigate overfitting include: cross-validation, regularization (LASSO, Ridge), train/test splits, and limiting model complexity. ML models do NOT require normally distributed data and can handle large datasets — in fact they are designed for big data.

The Internet of Things (IoT) refers to:

A. the network of websites and online financial databases used by portfolio managers

B. a broad network of devices embedded with sensors that generate and communicate data

C. the interconnected global financial markets for trading securities

B. a broad network of devices embedded with sensors that generate and communicate data — The Internet of Things (IoT) is a network of physical devices (smartphones, smart buildings, vehicles, industrial equipment) embedded with sensors, chips, or software that collect and transmit data. In investment contexts, IoT data can include real-time energy consumption, logistics tracking, and retail traffic data — all potentially valuable as alternative data for investment analysis. This contributes to the 'Velocity' characteristic of Big Data.

Structured data, semi-structured data, and unstructured data differ primarily in terms of:

A. their volume and velocity

B. the degree of organization and ease of processing

C. whether they can be used for investment analysis

B. the degree of organization and ease of processing — Structured data is highly organized in rows and columns (databases, spreadsheets) — easiest to analyze. Semi-structured data has some organizational elements but is not fully tabular (JSON, XML, web pages, photos). Unstructured data has no predefined structure (videos, audio files, free-form text, social media posts) — hardest to process. Most Big Data is unstructured. AI and ML techniques, especially NLP and computer vision, are needed to extract value from unstructured data.

LOS 11.c — Investment Applications of Big Data — Q10 to 12

Q10

Which of the following is most accurate regarding the use of text analytics in investment management?

A. Text analytics can only analyze structured financial data

B. Text analytics applies NLP techniques to extract investment-relevant signals from unstructured text

C. Text analytics requires manual reading and coding of documents by analysts

B. Text analytics applies NLP techniques to extract investment-relevant signals from unstructured text — Text analytics uses NLP and machine learning to automatically process large volumes of unstructured text (earnings calls, news, analyst reports, SEC filings, social media) and extract quantitative signals. Applications include: tone/sentiment analysis, topic extraction, identifying risk language, and tracking management credibility. It transforms unstructured text into structured signals that can be incorporated into quantitative investment models.

Q11

A robo-advisor is best described as a FinTech application that:

A. uses high-frequency trading algorithms to exploit microsecond price discrepancies

B. provides automated, algorithm-driven financial planning and investment management with minimal human intervention

C. processes satellite imagery data to generate alternative investment signals

B. provides automated, algorithm-driven financial planning and investment management with minimal human intervention — A robo-advisor is an automated digital platform that provides financial planning and portfolio management services using algorithms, typically with minimal human involvement. They gather client information (risk tolerance, goals, time horizon) and automatically construct, rebalance, and manage portfolios (often using low-cost ETFs). They have democratized access to investment management. High-frequency trading (A) is a different FinTech application focused on execution speed.

Q12

Which of the following statements about artificial intelligence (AI) in finance is most accurate?

A. AI refers only to systems that can pass the Turing test and demonstrate human-level intelligence

B. AI includes computer systems programmed to simulate human cognitive functions such as learning and problem-solving

C. AI applications in finance are limited to fraud detection and cannot be applied to portfolio management

B. AI includes computer systems programmed to simulate human cognitive functions such as learning and problem-solving — Artificial intelligence (AI) refers broadly to computer systems designed to perform tasks that typically require human intelligence, including learning, reasoning, problem-solving, perception, and language understanding. In finance, AI applications include: algorithmic trading, credit scoring, fraud detection, robo-advisory, risk management, NLP for text analysis, and predictive analytics for portfolio management. AI is not limited to passing the Turing test — this is a narrow definition.

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Ch 11 — Introduction to Big Data Techniques

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CFA Level 1 Quantitative Methods — Complete Study Notes & 400 Practice Questions | Concepts 'n' Clarity Gurgaon

What's Inside — All 11 Chapters (2025 LOS)

Interest Rate Components

Return Measurement

Money-Weighted vs Time-Weighted Return

Annualizing Returns

Return Components

Present Value of Financial Instruments

Implied Returns and Growth Rates

Cash Flow Additivity & No-Arbitrage

Central Tendency & Location

Dispersion Measures

Skewness & Kurtosis

Correlation

Expected Value & Variance

Probability Trees & Conditional Expectations

Bayes' Formula

Portfolio Expected Return & Variance

Joint Probability Covariance

Roy's Safety-First Criterion

Lognormal Distribution

Monte Carlo Simulation

Bootstrap Resampling

Sampling Methods

Central Limit Theorem & Standard Error

Resampling Techniques

Hypothesis Testing Framework

Hypothesis Test Statistics

Parametric vs Nonparametric Tests

Testing Correlation = 0

Contingency Table Chi-Square Test

Regression Model

OLS Assumptions

ANOVA Table & R²

Prediction

Functional Forms

FinTech in Finance

Big Data

Artificial Intelligence & Machine Learning

Investment Applications