You invest €1 000 at a nominal rate of 4.8% per year, compounded monthly. Find the value FV of the investment after 6 years.
PV = €1 0004.8% / 12n = 6
FV = €
2.Time value of moneyHard
You borrow €12 000 at 7.2% per year, compounded monthly, and repay €175 at the end of every month. How many months does it take until the loan is repaid (the last payment may be smaller)? Give the number of payments N.
P = €12 000A = €1757.2% / 12
N = payments
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 3: €8 000, €7 000, €8 000. The cost of capital is 6%. Find the net present value NPV.
A project costs €50 000 today. Its cash flow is €5 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 8. The cost of capital is 12%. Find the net present value NPV.
C0 = €50 000CF1 = €5 500g = 3%n = 8r = 12%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 3.5% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 3.5%n = 7
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 4.5% and matures in 3 years. The yield to maturity is 4%. Find the Macaulay duration D in years.
F = €1 000coupon = 4.5%n = 3YTM = 4%
D = years
7.CAPM & WACCMedium
A share with a beta of 1.2 has a required return of 9.8%. The risk-free rate is 2%. What expected market return rm does CAPM imply?
β = 1.2re = 9.8%rf = 2%
rm = %
8.CAPM & WACCHard
A company has 1 000 000 shares trading at €49.50; the book value of its equity is €24 800 000. It just paid a dividend of €2.10 per share, and dividends grow by 4% per year. Its debt has a market value of €39 600 000 and a pre-tax cost of 4.5%. The tax rate is 21%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 22, 34, 24, 37, 18, 16.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is below 40?
μ = 50σ = 5x = 40
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 1, 2, 3, 4, 5p(x): 0.25, 0.25, 0.3, 0.15, 0.05
Var(X) =
12.ProbabilityHard
A test has 10 multiple-choice questions. A student who guesses gets each one right with probability 0.3, independently. What is the probability P of exactly 3 correct answers?
n = 10p = 0.3k = 3
P =
13.Normal distributionMedium
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 178 and 183.6 cm? Give P as a decimal.
μ = 178σ = 7x1 = 178x2 = 183.6
P =
14.Normal distributionHard
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 99% of them lie (the 99th percentile).
μ = 250σ = 5P = 99%
x = g
15.Confidence intervalsMedium
A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 4.1. What is the margin of error E₂ of a 90% interval from the same data?
99%n₁ = 50E₁ = 4.190%
E₂ =
16.Confidence intervalsHard
The population standard deviation is 5. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 0.5?
σ = 5E = 0.590%
n =
17.Hypothesis tests & confidence intervalsMedium
Two teaching methods are compared. Group 1: 15 students, mean score 21, standard deviation 7.5. Group 2: 7 students, mean score 12.6, standard deviation 3.5. 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 = 15x̄₁ = 21s₁ = 7.5n₂ = 7x̄₂ = 12.6s₂ = 3.5
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 1.5? The standard deviation is 12. Find the smallest sample size n.
σ = 12E = 1.595%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (19.5, 74) is made. Find its residual e = y − ŷ.
n = 5(19.5, 74)
e =
3Solutions
1.Time value of money
Answer: FV = 1333 €
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.
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.
Rearrange the annuity formula
i = 0.006
(1 + i)−N = 1 − 12 000·0.006/175 = 0.5886
Solve with logarithms
N = −ln(0.5886)/ln(1.006) = 88.61
Round up to whole payments
88 full payments leave a small balance, which the last, smaller payment clears.
N = 89
Answer
N = 89 payments
3.NPV & IRR
Answer: NPV = 494.1 €
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.
Present value of each cash flow
CF0 = −20 000
CF1/(1 + 0.06)1 = 8 000/1.06 = 7 547.17
CF2/(1 + 0.06)2 = 7 000/1.124 = 6 229.98
CF3/(1 + 0.06)3 = 8 000/1.191 = 6 716.95
Add them up
NPV = €494.10
Answer
NPV = €494.10
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: NPV = −20155 €
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.
Present value of the growing annuity
((1.03)/(1.12))8 = 0.5116
PV = 5 500/(0.12 − 0.03)·(1 − 0.5116) = 29 845.01
NPV
NPV = 29 845.01 − 50 000 = −€20 154.99
Answer
NPV = −€20 154.99
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: P = 1154 €
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.
Per period
C = €30.00, i = 0.0175, N = 14
Present value of the coupons
€30.00·(1 − 0.7844)/0.0175 = €369.66
Present value of the face value
€1 000·0.7844 = €784.36
Price
P = €1 154.03
Answer
P = €1 154.03
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: D = 2.874 years
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.
Present value of each cash flow
t = 1: 45/1.04 = 43.27, t·PV = 43.27
t = 2: 45/1.082 = 41.61, t·PV = 83.21
t = 3: 1045/1.125 = 929.00, t·PV = 2 787.00
Duration
P = 1 013.88
D = 2 913.48/1 013.88 = 2.874
Answer
D = 2.874 years
7.CAPM & WACC
Answer: rm = 8.5 %
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).
Market risk premium
rm − rf = (9.8% − 2%)/1.2 = 6.5%
Market return
rm = 2% + 6.5% = 8.5%
Answer
rm = 8.5%
8.CAPM & WACC
Answer: WACC = 6.253 %
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 = −2
Standard normal table
Φ(−2) = 0.02275
P = 0.02275
Answer
P = 0.02275
11.Probability
Answer: Var(X) = 1.35
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.
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z-scores
z₁ = (178 − 178)/7 = 0
z₂ = (183.6 − 178)/7 = 0.8
Area between
P = Φ(0.8) − Φ(0) = 0.7881 − 0.5 = 0.2881
Answer
P = 0.2881
14.Normal distribution
Answer: x = 261.6 g
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
z for the percentile
Φ(z) = 0.99 → z = 2.326
Back to x
x = 250 + 2.326·5 = 261.63 g
Answer
x = 261.63 g
15.Confidence intervals
Answer: E₂ = 2.618
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.
Ratio of margins
E₂/E₁ = (1.645/2.576)·√(50/50) = 0.6386
New margin
E₂ = 4.1·0.6386 = 2.618
Answer
E₂ = 2.618
16.Confidence intervals
Answer: n = 271
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 = (1.645·5/0.5)² = 270.6
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 271
Answer
n = 271
17.Hypothesis tests & confidence intervals
Answer: t = 2.797
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 = (14·7.5² + 6·3.5²)/20 = 43.05
sp = 6.561
Standard error
SE = 6.561·√(1/15 + 1/7) = 3.003
Test statistic
t = (21 − 12.6)/3.003 = 2.797
Answer
t = 2.797
With df = 20, the two-sided 5% critical value is 2.086; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: n = 246
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).
Solve for n
n = (1.96·12/1.5)² = 245.9
Round up
n = 246
Answer
n = 246
19.Linear regression & correlation
Answer: r = −0.9701
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 = 84/5 = 16.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −32
Syy = Σ(y − ȳ)² = 108.8
Correlation coefficient
r = −32/√(10·108.8) = −0.9701
Answer
r = −0.9701
20.Linear regression & correlation
Answer: e = 2.65
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 = 120/5 = 24
ȳ = 425/5 = 85
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 273
Regression line
b = 273/90 = 3.033
a = 85 − 3.033·24 = 12.2
ŷ = 12.2 + 3.033·x
Predicted value
ŷ = 12.2 + 3.033·19.5 = 71.35
Residual
e = 74 − 71.35 = 2.65
Answer
e = 2.65
The point lies above the line.
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