A property is expected to earn net rent of €12 000 next year. The rent is expected to grow by 3% per year forever. The required return is 5% per year. Find the value PV of the property today.
C1 = €12 000g = 3%r = 5%
PV = €
2.Time value of moneyHard
You save €500 at the end of every month for 27 years. The account pays 2.4% per year, compounded monthly. How much have you saved at the end? Find FV.
A = €5002.4% / 12N = 324
FV = €
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 4: €4 000, €8 000, €10 000, €7 000. The cost of capital is 7.000000000000001%. Find the net present value NPV.
A project costs €20 000 and returns €13 500, €12 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
CF0 = −20 000CF1 = 13 500CF2 = 12 500
IRR = %
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 matures in 7 years and pays annual coupons. It sells for €950 when the yield to maturity is 9%. Find the coupon rate c in percent.
F = €1 000n = 7P = €950YTM = 9%
c = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 3% matures in 3 years and sells for €1 045. Find its yield to maturity YTM in percent.
F = €1 000coupon = 3%n = 3P = €1 045
YTM = %
7.CAPM & WACCMedium
A company is financed with ordinary shares worth €5 000 000, preferred shares worth €500 000 and debt worth €6 000 000. The cost of equity is 14%, the cost of preferred stock 6% and the pre-tax cost of debt 4%. The tax rate is 28%. Find the WACC.
A company has 2 500 000 shares at €33.50 each and 5 000 bonds trading at €1 000 each. The pre-tax cost of debt is 4% and the tax rate is 25%. The share’s beta is 1.3, the risk-free rate 1.5% and the market return 5.5%. Find the WACC.
β = 1.3rf = 1.5%rm = 5.5%rd = 4%T = 25%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 43, 22, 38, 45, 25.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 50 and standard deviation 12. Find the z-score z of a score of 26.
μ = 50σ = 12x = 26
z =
11.ProbabilityMedium
A fair die is rolled 7 times. What is the probability P of getting at least one six?
n = 7
P =
12.ProbabilityHard
X has the distribution shown, and Y = 2X + 10. Find the standard deviation σY of Y.
x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 2X + 10
σY =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9783. Find z to two decimals.
A = 0.9783
z =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 65σ = 10P = 97.5%
x = points
15.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 9.2. What is the margin of error E₂ of a 95% interval from the same data?
90%n₁ = 64E₁ = 9.295%
E₂ =
16.Confidence intervalsHard
The population standard deviation is 29. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 5?
σ = 29E = 599%
n =
17.Hypothesis tests & confidence intervalsMedium
11 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 2.75 and the standard deviation of the differences is 2.4. Find the paired test statistic t for H₀: μd = 0.
n = 11d̄ = 2.75sd = 2.4
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 80 against H₁: μ < 80 is done at α = 0.01 with a sample of 64; the standard deviation is known to be 20. In reality the mean is 70.5. Find the power 1 − β of the test in percent.
μ₀ = 80μ = 70.5σ = 20n = 64α = 0.01
1 − β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.
n = 6Σx = 21Σy = 84Σx² = 91Σy² = 1348Σxy = 240
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. Predict ŷ for x = 20.
n = 5x = 20
ŷ =
3Solutions
1.Time value of money
Answer: PV = 600000 €
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.
You paid in €162 000 yourself; the rest is interest.
3.NPV & IRR
Answer: NPV = 4229 €
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 + 0.07)1 = 4 000/1.07 = 3 738.32
CF2/(1 + 0.07)2 = 8 000/1.145 = 6 987.51
CF3/(1 + 0.07)3 = 10 000/1.225 = 8 162.98
CF4/(1 + 0.07)4 = 7 000/1.311 = 5 340.27
Add them up
NPV = €4 229.07
Answer
NPV = €4 229.07
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 19.71 %
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(15%) = 1 190.93
NPV(20%) = −69.44
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 + 0.09)−7 = €547.03
Coupon
C = (950 − 547.03)/5.033 = €80.07
Coupon rate
c = 80.07/1000 = 8.007%
Answer
c = 8.007%
6.Bond pricing & yield
Answer: YTM = 1.456 %
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.
Equation for the yield
€1 045 = €30·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³
Solve numerically
P(1%) = €1 058.82, P(2%) = €1 028.84
YTM = 1.456%
Answer
YTM = 1.456%
7.CAPM & WACC
Answer: WACC = 7.85 %
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).
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 = (26 − 50)/12 = −2
Answer
z = −2
11.Probability
Answer: P = 0.7209
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.
Complement
P(no six) = (5/6)7 = 0.2791
P = 1 − 0.2791 = 0.7209
Answer
P = 0.7209
12.Probability
Answer: σY = 1.673
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.
Spread of X
μ = 1, Var(X) = 0.7, σX = 0.8367
Linear change
σY = |2|·0.8367 = 1.673
Answer
σY = 1.673
The mean does move: E(Y) = 2·1 + 10 = 12.
13.Normal distribution
Answer: z = −2.02
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 − 0.9783 = 0.0217
Read the z-table backwards
Find 0.0217 inside the table and read off z.
z = −2.02
Answer
z = −2.02
14.Normal distribution
Answer: x = 84.6 points
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) = 0.975 → z = 1.96
Back to x
x = 65 + 1.96·10 = 84.6 points
Answer
x = 84.6 points
15.Confidence intervals
Answer: E₂ = 10.96
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 = (1.96/1.645)·√(64/64) = 1.191
New margin
E₂ = 9.2·1.191 = 10.96
Answer
E₂ = 10.96
16.Confidence intervals
Answer: n = 224
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 = (2.576·29/5)² = 223.2
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 224
Answer
n = 224
17.Hypothesis tests & confidence intervals
Answer: t = 3.8
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 of the mean difference
SE = 2.4/√11 = 0.7236
Test statistic
t = 2.75/0.7236 = 3.8
Answer
t = 3.8
With df = 10, the two-sided 5% critical value is 2.228; |t| is larger, so H₀ is rejected.
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 92.98 %
Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).
Critical sample mean
SE = 20/√64 = 2.5
x̄c = 80 − 2.326·2.5 = 74.19
z under the true mean
z = (74.19 − 70.5)/2.5 = 1.474
Power
power = P(Z < 1.474) = 92.98%
1 − β = 92.98%
Answer
1 − β = 92.98%
19.Linear regression & correlation
Answer: r = −0.9843
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 = 240 − 21·84/6 = −54
Sxx = 91 − 21²/6 = 17.5
Syy = 1348 − 84²/6 = 172
Correlation coefficient
r = −54/√(17.5·172) = −0.9843
Answer
r = −0.9843
20.Linear regression & correlation
Answer: ŷ = 77.15
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 = 105/5 = 21
ȳ = 409/5 = 81.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 186
Regression line
b = 186/40 = 4.65
a = 81.8 − 4.65·21 = −15.85
ŷ = −15.85 + 4.65·x
Prediction
ŷ = −15.85 + 4.65·20 = 77.15
Answer
ŷ = 77.15
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