A property is expected to earn net rent of €8 000 next year. The rent is expected to grow by 2% per year forever. The required return is 6% per year. Find the value PV of the property today.
C1 = €8 000g = 2%r = 6%
PV = €
2.Time value of moneyHard
A mortgage of €250 000 is repaid with equal monthly payments over 25 years at 4.8% per year, compounded monthly. How much is still owed just after the payment at the end of year 14? Find the outstanding balance B.
P = €250 0004.8% / 12N = 300k = 168
B = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 3: €13 000, €20 000, €13 000. The cost of capital is 8%. Find the net present value NPV.
A project costs €200 000 today. Its cash flow is €26 500 at the end of year 1 and grows by 2% per year until the last cash flow at the end of year 9. The cost of capital is 8%. Find the net present value NPV.
C0 = €200 000CF1 = €26 500g = 2%n = 9r = 8%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 14 years. Its yield to maturity changes from 6% to 7.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 5%n = 14y1 = 6%y2 = 7.5%
ΔP = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 pays an annual coupon of 6% and matures in 6 years. The yield to maturity is 7%. Find the Macaulay duration D in years.
F = €1 000coupon = 6%n = 6YTM = 7%
D = years
7.CAPM & WACCMedium
A share with a beta of 1.25 has a required return of 9.13%. The risk-free rate is 1%. What expected market return rm does CAPM imply?
β = 1.25re = 9.13%rf = 1%
rm = %
8.CAPM & WACCHard
A company has 2 000 000 shares at €27.00 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 7% and the tax rate is 25.8%. The share’s beta is 0.9, the risk-free rate 1% and the market return 6%. Find the WACC.
β = 0.9rf = 1%rm = 6%rd = 7%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 48, 30, 25, 56, 20.
n = 5
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 10. What proportion P of values is above 65?
μ = 70σ = 10x = 65
P =
11.ProbabilityMedium
A bag holds 3 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?
3 / 9
P =
12.ProbabilityHard
A game costs €1 to play. You win €3 with probability 0.1 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €1€3: p = 0.1€50: p = 0.05
E(G) = €
13.Normal distributionMedium
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 65.5 and 67 points? Give P as a decimal.
μ = 65σ = 10x1 = 65.5x2 = 67
P =
14.Normal distributionHard
The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. Find the value x below which 95% of them lie (the 95th percentile).
μ = 178σ = 7P = 95%
x = cm
15.Confidence intervalsMedium
The battery life of two phone models is compared. Model 1: 30 phones, mean 97 h, known standard deviation 15 h. Model 2: 50 phones, mean 110.5 h, known standard deviation 13 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.
The population standard deviation is 28. 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 5?
σ = 28E = 595%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 99.44 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 99.44s = 5n = 64
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 80%. The standard deviation is 9. Find the minimum sample size n.
α = 0.01power = 80%δ = 3σ = 9zα = 2.326zβ = 0.842
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 5
r =
20.Linear regression & correlationHard
The table gives 6 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 = 20?
n = 6ŷ = 20
x =
3Solutions
1.Time value of money
Answer: PV = 200000 €
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.
Whatever is still owed must be exactly paid off by the remaining payments.
N − k = 300 − 168 = 132
B = 1432·(1 − (1 + 0.004)−132)/0.004 = €146 685.49
Answer
B = €146 685.49
3.NPV & IRR
Answer: NPV = −20496 €
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 = −60 000
CF1/(1 + 0.08)1 = 13 000/1.08 = 12 037.04
CF2/(1 + 0.08)2 = 20 000/1.166 = 17 146.78
CF3/(1 + 0.08)3 = 13 000/1.26 = 10 319.82
Add them up
NPV = €−20 496.37
Answer
NPV = €−20 496.37
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: NPV = −22381 €
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.
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: ΔP = −119.3 €
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.
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: 60/1.07 = 56.07, t·PV = 56.07
t = 2: 60/1.145 = 52.41, t·PV = 104.81
t = 3: 60/1.225 = 48.98, t·PV = 146.93
t = 4: 60/1.311 = 45.77, t·PV = 183.09
t = 5: 60/1.403 = 42.78, t·PV = 213.90
t = 6: 1060/1.501 = 706.32, t·PV = 4 237.94
Duration
P = 952.33
D = 4 942.75/952.33 = 5.19
Answer
D = 5.19 years
7.CAPM & WACC
Answer: rm = 7.504 %
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 = (9.13% − 1%)/1.25 = 6.504%
Market return
rm = 1% + 6.504% = 7.504%
Answer
rm = 7.504%
8.CAPM & WACC
Answer: WACC = 5.474 %
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 = −0.5
Standard normal table
Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915
Answer
P = 0.6915
11.Probability
Answer: P = 0.04545
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.
First and second draw
P = 3/12 · 2/11 = 6/132 = 0.04545
Answer
P = 0.04545
12.Probability
Answer: E(G) = 1.8 €
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: 2 (p = 0.1), 49 (p = 0.05), −1 (p = 0.85)
Expected value
E(G) = 2·0.1 + 49·0.05 − 1·0.85 = 1.8 €
Answer
E(G) = 1.8 €
A positive value means you win money on average.
13.Normal distribution
Answer: P = 0.05932
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 = (65.5 − 65)/10 = 0.05
z₂ = (67 − 65)/10 = 0.2
Area between
P = Φ(0.2) − Φ(0.05) = 0.5793 − 0.5199 = 0.05932
Answer
P = 0.05932
14.Normal distribution
Answer: x = 189.5 cm
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) = 0.95 → z = 1.645
Back to x
x = 178 + 1.645·7 = 189.51 cm
Answer
x = 189.51 cm
15.Confidence intervals
Answer: upper limit = −8.074 h
Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.
Point estimate
x̄₁ − x̄₂ = 97 − 110.5 = −13.5
Standard error
SE = √(15²/30 + 13²/50) = 3.298
Margin of error
E = 1.645·3.298 = 5.426
Interval
−13.5 ± 5.426 → [−18.93, −8.074]
Answer
upper limit = −8.074 h
The interval does not contain 0, so the means differ significantly.
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 = (1.96·28/5)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 = −0.896
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).
Standard error
SE = 5/√64 = 5/8 = 0.625
Test statistic
t = (99.44 − 100)/0.625 = −0.896
Answer
t = −0.896
t0.025; df = 63 = 1.998
|t| < 1.998, so H₀ is not rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 91
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((2.326 + 0.842)·9/3)² = 90.33
Round up
n = 91
Answer
n = 91
19.Linear regression & correlation
Answer: r = 0.9838
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 = 35/5 = 7
ȳ = 442/5 = 88.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 280
Syy = Σ(y − ȳ)² = 2025
Correlation coefficient
r = 280/√(40·2025) = 0.9838
Answer
r = 0.9838
20.Linear regression & correlation
Answer: x = 1.629
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 = 21/6 = 3.5
ȳ = 86/6 = 14.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −53
Regression line
b = −53/17.5 = −3.029
a = 14.33 − (−3.029)·3.5 = 24.93
ŷ = 24.93 − 3.029·x
Solve for x
x = (20 − 24.93)/(−3.029) = 1.629
Answer
x = 1.629
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