You need €25 000 in 8 years. An account pays 3% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €25 0003% / 12n = 8
PV = €
2.Time value of moneyHard
You deposit €1 000 at the start of every year for 29 years, the first deposit today. The account pays 6% per year. How much is in the account at the end of year 29? Find FV.
A = €1 000i = 6%n = 29
FV = €
3.NPV & IRRMedium
An investment of €75 000 gives these cash flows at the end of years 1 to 4: €20 000, €18 000, €18 000, €18 000. The cost of capital is 8%. Find the net present value NPV.
A project costs €50 000 today. Its cash flow is €10 000 at the end of year 1 and grows by 2% per year until the last cash flow at the end of year 6. The cost of capital is 8%. Find the net present value NPV.
C0 = €50 000CF1 = €10 000g = 2%n = 6r = 8%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 11 years. The yield to maturity is 3.5% per year. Find the price P of the bond.
F = €1 000coupon = 5%YTM = 3.5%n = 11
P = €
6.Bond pricing & yieldHard
A bond trades at €1 010 with a yield to maturity of 7% (annual compounding) and a Macaulay duration of 10.7 years. Use the modified duration to estimate the new price P if the yield rises by 75 basis points.
P0 = €1 010y = 7%D = 10.7Δy = +75 bp
P = €
7.CAPM & WACCMedium
A company has equity worth €8 000 000 and debt worth €5 000 000. Its WACC is 7.06%, its pre-tax cost of debt is 5% and the tax rate is 21%. Find the cost of equity re.
A company has 5 000 000 shares trading at €37.00; the book value of its equity is €92 500 000. It just paid a dividend of €1.60 per share, and dividends grow by 5% per year. Its debt has a market value of €55 500 000 and a pre-tax cost of 4.5%. The tax rate is 25%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 27, 13, 12, 33, 27, 15, 26.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 100 and standard deviation 5. What proportion P of values is above 105?
μ = 100σ = 5x = 105
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 1, 2p(x): 0.3, 0.35, 0.35
Var(X) =
12.ProbabilityHard
A game costs €5 to play. You win €10 with probability 0.15 and €100 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€10: p = 0.15€100: p = 0.01
E(G) = €
13.Normal distributionMedium
The bag weights are normally distributed with mean 250 g and standard deviation 5 g. What fraction lies between 250.25 and 259 g? Give P as a decimal.
μ = 250σ = 5x1 = 250.25x2 = 259
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 20% of them lie (the 20th percentile).
μ = 178σ = 7P = 20%
x = cm
15.Confidence intervalsMedium
A sample of 12 items has mean 67.5 and standard deviation s = 8. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.
n = 12x̄ = 67.5s = 890%t(11) = 1.796
lower limit =
16.Confidence intervalsHard
Two fertilisers are tested. Plot group 1: 12 plots, mean yield 91 kg, standard deviation 8.5 kg. Plot group 2: 5 plots, mean 94 kg, standard deviation 4 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.
Two production lines make the same part. Line 1: a sample of 30 parts has mean length 68 mm, and the line's known standard deviation is 12 mm. Line 2: a sample of 30 parts has mean 75.6 mm, known standard deviation 6 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.05 with a sample of 100; the standard deviation is known to be 30. In reality the mean is 206. Find the probability of a Type II error β in percent.
μ₀ = 200μ = 206σ = 30n = 100α = 0.05
β = %
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 6
r² =
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. For which value x does the line predict ŷ = 62?
n = 5ŷ = 62
x =
3Solutions
1.Time value of money
Answer: PV = 19672 €
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.
Rate and number of periods
i = 0.03/12 = 0.0025
N = 12·8 = 96
Present value
PV = FV/(1 + i)N = 25 000/1.271 = €19 671.59
Answer
PV = €19 671.59
2.Time value of money
Answer: FV = 78058 €
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.
Ordinary annuity factor
((1 + 0.06)29 − 1)/0.06 = 73.64
Annuity due
Each deposit is invested one year longer.
FV = 1 000·73.64·1.06 = €78 058.19
Answer
FV = €78 058.19
3.NPV & IRR
Answer: NPV = −13530 €
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 = −75 000
CF1/(1 + 0.08)1 = 20 000/1.08 = 18 518.52
CF2/(1 + 0.08)2 = 18 000/1.166 = 15 432.10
CF3/(1 + 0.08)3 = 18 000/1.26 = 14 288.98
CF4/(1 + 0.08)4 = 18 000/1.36 = 13 230.54
Add them up
NPV = €−13 529.87
Answer
NPV = €−13 529.87
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: NPV = −1612 €
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: P = 1136 €
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 = €25.00, i = 0.0175, N = 22
Present value of the coupons
€25.00·(1 − 0.6827)/0.0175 = €453.26
Present value of the face value
€1 000·0.6827 = €682.72
Price
P = €1 135.98
Answer
P = €1 135.98
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: P = 934.3 €
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 = 10.7/1.07 = 10
Estimated price change
%ΔP ≈ −10·(0.0075) = −7.5%
P ≈ 1010·(1 − 0.075) = €934.25
Answer
P = €934.25
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: re = 9.004 %
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 = €13 000 000, E/V = 0.6154, D/V = 0.3846
Debt part of the WACC
0.3846·5%·(1 − 0.21) = 1.519%
Cost of equity
re = (7.06% − 1.519%)/0.6154 = 9.004%
Answer
re = 9.004%
8.CAPM & WACC
Answer: WACC = 8.118 %
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) = 0.8413
P = 1 − 0.8413 = 0.1587
Answer
P = 0.1587
11.Probability
Answer: Var(X) = 0.6475
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.
Mean
μ = 0·0.3 + 1·0.35 + 2·0.35 = 1.05
Mean of the squares
E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75
Variance
Var(X) = 1.75 − 1.05² = 0.6475
Answer
Var(X) = 0.6475
12.Probability
Answer: E(G) = −2.5 €
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: 5 (p = 0.15), 95 (p = 0.01), −5 (p = 0.84)
Expected value
E(G) = 5·0.15 + 95·0.01 − 5·0.84 = −2.5 €
Answer
E(G) = −2.5 €
A negative value means you lose money on average.
13.Normal distribution
Answer: P = 0.4441
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 = (250.25 − 250)/5 = 0.05
z₂ = (259 − 250)/5 = 1.8
Area between
P = Φ(1.8) − Φ(0.05) = 0.9641 − 0.5199 = 0.4441
Answer
P = 0.4441
14.Normal distribution
Answer: x = 172.1 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) = 0.2 → z = −0.8416
Back to x
x = 178 + −0.8416·7 = 172.11 cm
Answer
x = 172.11 cm
15.Confidence intervals
Answer: lower limit = 63.35
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 = 11, t = 1.796
Margin of error
E = 1.796·8/√12 = 4.148
Interval
67.5 ± 4.148 → [63.35, 71.65]
Answer
lower limit = 63.35
16.Confidence intervals
Answer: lower limit = −10.06 kg
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.
Pooled standard deviation
sp² = (11·8.5² + 4·4²)/15 = 57.25
sp = 7.566
Margin of error
E = 1.753·7.566·√(1/12 + 1/5) = 7.06
Interval
−3 ± 7.06 → [−10.06, 4.06]
Answer
lower limit = −10.06 kg
17.Hypothesis tests & confidence intervals
Answer: z = −3.103
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.
Standard error of the difference
SE = √(12²/30 + 6²/30) = √6 = 2.449
Test statistic
z = (68 − 75.6)/2.449 = −3.103
Answer
z = −3.103
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: β = 36.13 %
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 = 30/√100 = 3
x̄c = 200 + 1.645·3 = 204.9
z under the true mean
z = (204.9 − 206)/3 = −0.355
Type II error
power = P(Z > −0.355) = 63.87%
β = 100% − 63.87% = 36.13%
Answer
β = 36.13%
19.Linear regression & correlation
Answer: r² = 0.9109
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 = 126/6 = 21
ȳ = 445/6 = 74.17
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 263
Syy = Σ(y − ȳ)² = 1085
Coefficient of determination
r² = 263²/(70·1085) = 0.9109
Answer
r² = 0.9109
About 91.09% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: x = 21.57
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 = 100/5 = 20
ȳ = 289/5 = 57.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 240
Regression line
b = 240/90 = 2.667
a = 57.8 − 2.667·20 = 4.467
ŷ = 4.467 + 2.667·x
Solve for x
x = (62 − 4.467)/2.667 = 21.57
Answer
x = 21.57
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