A property is expected to earn net rent of €5 000 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 = €5 000g = 3%r = 5%
PV = €
2.Time value of moneyHard
A loan of €8 000 is repaid with equal monthly payments over 3 years. The interest rate is 4.2% per year, compounded monthly. Find the monthly payment A.
P = €8 0004.2% / 12N = 36
A = €
3.NPV & IRRMedium
An investment of €20 000 gives these cash flows at the end of years 1 to 4: €6 000, €8 000, €4 000, €9 000. The cost of capital is 9%. Find the net present value NPV.
A project costs €10 000 and returns €4 500, €5 000, €3 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.
CF0 = −10 000CF1 = 4 500CF2 = 5 000CF3 = 3 500
IRR = %
5.Bond pricing & yieldMedium
You buy a bond with 4 years to maturity (face value €1 000, annual coupon 8%) when its yield to maturity is 5.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 6.5%. Find your holding period return HPR in percent.
F = €1 000coupon = 8%n = 4y0 = 5.5%y1 = 6.5%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 6% matures in 8 years and sells for €1 100. Find its yield to maturity YTM in percent.
F = €1 000coupon = 6%n = 8P = €1 100
YTM = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €4 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 7% and the tax rate is 28%. Find the WACC.
A company has 2 000 000 shares at €52.00 each and 10 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3% and the tax rate is 28%. The share’s beta is 1.3, the risk-free rate 1% and the market return 7.5%. Find the WACC.
β = 1.3rf = 1%rm = 7.5%rd = 3%T = 28%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 19, 58, 18, 55, 54, 34.
n = 6
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 60 and standard deviation 12. What proportion P of values is below 48?
μ = 60σ = 12x = 48
P =
11.ProbabilityMedium
A fair die is rolled 8 times. What is the probability P of getting at least one six?
n = 8
P =
12.ProbabilityHard
A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.2, independently. What is the probability P of exactly 2 correct answers?
n = 5p = 0.2k = 2
P =
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0901. Find z to two decimals.
A = 0.0901
z =
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 99% of them lie (the 99th percentile).
μ = 178σ = 7P = 99%
x = cm
15.Confidence intervalsMedium
The blood pressure of 7 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 6.4 and standard deviation 7. Find the lower limit of the 90% confidence interval for the mean difference.
n = 7d̄ = 6.4sd = 790%t(6) = 1.943
lower limit = mmHg
16.Confidence intervalsHard
The population standard deviation is 36. 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 2?
σ = 36E = 290%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 36 students has a mean score of 42.25 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 50.
μ₀ = 50x̄ = 42.25s = 15n = 36
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.10 with a sample of 25; the standard deviation is known to be 12.5. In reality the mean is 52.5. Find the power 1 − β of the test in percent.
μ₀ = 50μ = 52.5σ = 12.5n = 25α = 0.10
1 − β = %
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 = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 88?
n = 5ŷ = 88
x =
3Solutions
1.Time value of money
Answer: PV = 250000 €
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.
Monthly rate and number of payments
i = 0.042/12 = 0.0035
N = 12·3 = 36
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 = 0.8818
A = 8 000·0.0035/(1 − 0.8818) = €236.90
Answer
A = €236.90
3.NPV & IRR
Answer: NPV = 1703 €
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 + 0.09)1 = 6 000/1.09 = 5 504.59
CF2/(1 + 0.09)2 = 8 000/1.188 = 6 733.44
CF3/(1 + 0.09)3 = 4 000/1.295 = 3 088.73
CF4/(1 + 0.09)4 = 9 000/1.412 = 6 375.83
Add them up
NPV = €1 702.59
Answer
NPV = €1 702.59
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: IRR = 14.97 %
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%) = 852.74
NPV(15%) = −4.93
Narrow down
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
IRR ≈ 10 + 5·852.74/(852.74 − (−4.93)) = 14.97%
exact: NPV(14.97%) = 0 → IRR = 14.97%
Answer
IRR = 14.97%
5.Bond pricing & yield
Answer: HPR = 2.951 %
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.
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
€1 100 = €60·(1 − (1 + y)⁻⁸)/y + €1 000/(1 + y)⁸
Solve numerically
P(4%) = €1 134.65, P(5%) = €1 064.63
YTM = 4.485%
Answer
YTM = 4.485%
7.CAPM & WACC
Answer: WACC = 8.52 %
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
V = €8 000 000, E/V = 0.5, D/V = 0.5
WACC
WACC = 0.5·12% + 0.5·7%·(1 − 0.28) = 8.52%
Answer
WACC = 8.52%
8.CAPM & WACC
Answer: WACC = 8.781 %
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
Standard normal table
Φ(−1) = 0.1587
P = 0.1587
Answer
P = 0.1587
11.Probability
Answer: P = 0.7674
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)8 = 0.2326
P = 1 − 0.2326 = 0.7674
Answer
P = 0.7674
12.Probability
Answer: P = 0.2048
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.
Binomial formula
P(X = 2) = 10·0.22·0.83 = 10·0.04·0.512 = 0.2048
Answer
P = 0.2048
13.Normal distribution
Answer: z = 1.34
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 − 0.0901 = 0.9099
Read the z-table backwards
Find 0.9099 inside the table and read off z.
z = 1.34
Answer
z = 1.34
14.Normal distribution
Answer: x = 194.3 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) = 0.99 → z = 2.326
Back to x
x = 178 + 2.326·7 = 194.28 cm
Answer
x = 194.28 cm
15.Confidence intervals
Answer: lower limit = 1.259 mmHg
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.
Critical value
df = 6, t = 1.943
Margin of error
E = 1.943·7/√7 = 5.141
Interval
6.4 ± 5.141 → [1.259, 11.54]
Answer
lower limit = 1.259 mmHg
16.Confidence intervals
Answer: n = 877
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 = (1.645·36/2)² = 876.8
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 877
Answer
n = 877
17.Hypothesis tests & confidence intervals
Answer: t = −3.1
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 = 15/√36 = 15/6 = 2.5
Test statistic
t = (42.25 − 50)/2.5 = −3.1
Answer
t = −3.1
t0.025; df = 35 = 2.03
|t| > 2.03, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: 1 − β = 38.9 %
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 = 12.5/√25 = 2.5
x̄c = 50 + 1.282·2.5 = 53.2
z under the true mean
z = (53.2 − 52.5)/2.5 = 0.282
Power
power = P(Z > 0.282) = 38.9%
1 − β = 38.9%
Answer
1 − β = 38.9%
19.Linear regression & correlation
Answer: r = −0.7926
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 = 117/5 = 23.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −14
Syy = Σ(y − ȳ)² = 31.2
Correlation coefficient
r = −14/√(10·31.2) = −0.7926
Answer
r = −0.7926
20.Linear regression & correlation
Answer: x = 5.973
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 = 469/5 = 93.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 226
Regression line
b = 226/40 = 5.65
a = 93.8 − 5.65·7 = 54.25
ŷ = 54.25 + 5.65·x
Solve for x
x = (88 − 54.25)/5.65 = 5.973
Answer
x = 5.973
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