An annuity pays €8 000 at the end of each year for 6 years. The interest rate is 8% per year. Find the present value PV of the annuity.
A = €8 000i = 8%n = 6
PV = €
2.Time value of moneyHard
A loan of €150 000 is repaid with equal monthly payments over 25 years. The interest rate is 3.6% per year, compounded monthly. Find the monthly payment A.
P = €150 0003.6% / 12N = 300
A = €
3.NPV & IRRMedium
A project costs €100 000 today. It generates €7 000 at the end of the first year, and the cash flow then grows by 2% per year forever. The cost of capital is 8%. Find the net present value NPV.
C0 = €100 000CF1 = €7 000g = 2%r = 8%
NPV = €
4.NPV & IRRHard
A project costs €50 000 and returns €34 500, €26 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −50 000CF1 = 34 500CF2 = 26 000
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 3 years and pays annual coupons. It sells for €1 028 when the yield to maturity is 3.5%. Find the coupon rate c in percent.
F = €1 000n = 3P = €1 028YTM = 3.5%
c = %
6.Bond pricing & yieldHard
A bond trades at €1 005 with a yield to maturity of 3% (annual compounding) and a Macaulay duration of 11.1 years. Use the modified duration to estimate the new price P if the yield falls by 25 basis points.
P0 = €1 005y = 3%D = 11.1Δy = −25 bp
P = €
7.CAPM & WACCMedium
A company has a debt-to-equity ratio (at market values) of 0.25. Its cost of equity is 13%, its pre-tax cost of debt is 6% and the tax rate is 28%. Find the WACC.
D/E = 0.25re = 13%rd = 6%T = 28%
WACC = %
8.CAPM & WACCHard
A company has 2 000 000 shares at €55.00 each and 20 000 bonds trading at €1 020 each. The pre-tax cost of debt is 3% and the tax rate is 25%. The share’s beta is 1.2, the risk-free rate 2% and the market return 6.5%. Find the WACC.
β = 1.2rf = 2%rm = 6.5%rd = 3%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 45, 20, 54, 47, 47.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 12. What proportion P of values is above 85.6?
μ = 100σ = 12x = 85.6
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.15, 0.25, 0.35, 0.25
Var(X) =
12.ProbabilityHard
A game costs €1 to play. You win €5 with probability 0.15 and €50 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€5: p = 0.15€50: p = 0.01
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.2661. Find z to two decimals.
A = 0.2661
z =
14.Normal distributionHard
Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
A sample of 21 items has mean 118.5 and standard deviation s = 17. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.
n = 21x̄ = 118.5s = 1799%t(20) = 2.845
upper limit =
16.Confidence intervalsHard
The population standard deviation is 14. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 2.5?
σ = 14E = 2.595%
n =
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 9 students, mean score 105.5, standard deviation 10. Group 2: 8 students, mean score 100.4, standard deviation 11. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).
n₁ = 9x̄₁ = 105.5s₁ = 10n₂ = 8x̄₂ = 100.4s₂ = 11
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 25 measurements has mean 48.3. The population standard deviation is 4. Find the lower limit L of the 95% confidence interval for the mean.
x̄ = 48.3σ = 4n = 2595%
L =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 25?
n = 5ŷ = 25
x =
3Solutions
1.Time value of money
Answer: PV = 36983 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Annuity factor
(1 − (1 + i)−n)/i = (1 − 0.6302)/0.08 = 4.623
Present value
PV = 8 000·4.623 = €36 983.04
Answer
PV = €36 983.04
2.Time value of money
Answer: A = 759 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Monthly rate and number of payments
i = 0.036000000000000004/12 = 0.003
N = 12·25 = 300
Payment
The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.
(1 + i)−N = 0.4071
A = 150 000·0.003/(1 − 0.4071) = €759.00
Answer
A = €759.00
3.NPV & IRR
Answer: NPV = 16667 €
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
Value of the cash flows
PV = 7 000/(0.08 − 0.02) = 116 666.67
NPV
NPV = 116 666.67 − 100 000 = €16 666.67
Answer
NPV = €16 666.67
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 14.44 %
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
Bracket the IRR
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
NPV(10%) = 2 851.24
NPV(15%) = −340.26
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
Bond pricing: a bond’s price is the present value of its cash flows, discounted at the yield to maturity (YTM): the coupons form an annuity and the face value is paid back at maturity. P = C·(1 − (1 + y)⁻ⁿ)/y + F/(1 + y)ⁿ. When the yield is above the coupon rate the bond trades below face value, and the other way round.
Present value of the face value
1000·(1 + 0.035)−3 = €901.94
Coupon
C = (1028 − 901.94)/2.802 = €44.99
Coupon rate
c = 44.99/1000 = 4.499%
Answer
c = 4.499%
6.Bond pricing & yield
Answer: P = 1032 €
Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.
Modified duration
Dmod = 11.1/1.03 = 10.78
Estimated price change
%ΔP ≈ −10.78·(−0.0025) = 2.694%
P ≈ 1005·(1 + 0.02694) = €1 032.08
Answer
P = €1 032.08
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 11.26 %
Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).
Weights from D/E
E/V = 1/(1 + 0.25) = 0.8
D/V = 0.25/(1 + 0.25) = 0.2
WACC
WACC = 0.8·13% + 0.2·6%·(1 − 0.28) = 11.26%
Answer
WACC = 11.26%
8.CAPM & WACC
Answer: WACC = 6.594 %
Cost of capital: CAPM gives the return shareholders require: re = rf + β·(rm − rf). The weighted average cost of capital mixes equity and debt by their market values: WACC = E/V·re + D/V·rd·(1 − T), with V = E + D. Interest is tax-deductible, hence the (1 − T).
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = −1.2
Standard normal table
Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849
Answer
P = 0.8849
11.Probability
Answer: Var(X) = 4.04
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Net outcomes
G: 4 (p = 0.15), 49 (p = 0.01), −1 (p = 0.84)
Expected value
E(G) = 4·0.15 + 49·0.01 − 1·0.84 = 0.25 €
Answer
E(G) = 0.25 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = 0.34
From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.
Area to the left of z
Φ(z) = (1 + 0.2661)/2 = 0.6331
Read the z-table backwards
Find 0.6331 inside the table and read off z.
z = 0.34
Answer
z = 0.34
14.Normal distribution
Answer: P = 0.7881
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the difference
μ = 52 − 48 = 4 s, σ = 5 s
Standardise
z = (0 − 4)/5 = −0.8
Probability
P = 1 − Φ(−0.8) = 0.7881
Answer
P = 0.7881
15.Confidence intervals
Answer: upper limit = 129.1
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Critical value
df = n − 1 = 20, t = 2.845
Margin of error
E = 2.845·17/√21 = 10.55
Interval
118.5 ± 10.55 → [107.9, 129.1]
Answer
upper limit = 129.1
16.Confidence intervals
Answer: n = 121
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Solve for n
n = (1.96·14/2.5)² = 120.5
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 121
Answer
n = 121
17.Hypothesis tests & confidence intervals
Answer: t = 1.002
Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.
Pooled variance
sp² = (8·10² + 7·11²)/15 = 109.8
sp = 10.48
Standard error
SE = 10.48·√(1/9 + 1/8) = 5.092
Test statistic
t = (105.5 − 100.4)/5.092 = 1.002
Answer
t = 1.002
With df = 15, the two-sided 5% critical value is 2.131; |t| is smaller, so H₀ is not rejected.
18.Hypothesis tests & confidence intervals
Answer: L = 46.73
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
Margin of error
E = 1.96·4/√25 = 1.568
Interval limit
L = 48.3 − 1.568 = 46.73
Answer
L = 46.73
19.Linear regression & correlation
Answer: r = 0.9928
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Centred sums
Sxy = 2756 − 36·418/6 = 248
Sxx = 286 − 36²/6 = 70
Syy = 30012 − 418²/6 = 891.3
Correlation coefficient
r = 248/√(70·891.3) = 0.9928
Answer
r = 0.9928
20.Linear regression & correlation
Answer: x = 1.552
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 15/5 = 3
ȳ = 104/5 = 20.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −29
Regression line
b = −29/10 = −2.9
a = 20.8 − (−2.9)·3 = 29.5
ŷ = 29.5 − 2.9·x
Solve for x
x = (25 − 29.5)/(−2.9) = 1.552
Answer
x = 1.552
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