You invest €1 000 at a nominal rate of 4% per year, compounded monthly. Find the value FV of the investment after 5 years.
PV = €1 0004% / 12n = 5
FV = €
2.Time value of moneyHard
You borrow €12 000 at 3.6% per year, compounded monthly, and repay €275 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 = €2753.6% / 12
N = payments
3.NPV & IRRMedium
A project costs €20 000 today and returns €6 500 at the end of each year for 4 years. The cost of capital is 8%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.
A project costs €25 000 and returns €12 500, €9 500, €9 500, €9 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 12%. Find the modified internal rate of return MIRR in percent.
A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 10 years. The yield to maturity is 5% per year. Find the price P of the bond.
F = €1 000coupon = 6%YTM = 5%n = 10
P = €
6.Bond pricing & yieldHard
A bond trades at €1 060 with a yield to maturity of 4% (annual compounding) and a Macaulay duration of 11 years. Use the modified duration to estimate the new price P if the yield falls by 75 basis points.
P0 = €1 060y = 4%D = 11Δy = −75 bp
P = €
7.CAPM & WACCMedium
A share with a beta of 2 has a required return of 14%. The risk-free rate is 4%. What expected market return rm does CAPM imply?
β = 2re = 14%rf = 4%
rm = %
8.CAPM & WACCHard
A company has 2 500 000 shares at €32.00 each and 5 000 bonds trading at €980 each. The pre-tax cost of debt is 4.5% and the tax rate is 25.8%. The share’s beta is 1.8, the risk-free rate 2.5% and the market return 6.5%. Find the WACC.
β = 1.8rf = 2.5%rm = 6.5%rd = 4.5%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 52, 37, 55, 28, 49, 34, 58.
n = 7
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 12. Find the z-score z of a score of 85.
μ = 70σ = 12x = 85
z =
11.ProbabilityMedium
A fair die is rolled 2 times. What is the probability P of getting at least one six?
n = 2
P =
12.ProbabilityHard
A game costs €5 to play. You win €15 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 = €5€15: p = 0.1€50: p = 0.05
E(G) = €
13.Normal distributionMedium
X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 757.9 g. Round z to two decimals.
μX = 250, σX = 4μY = 500, σY = 9X + Y > 757.9 g
P =
14.Normal distributionHard
The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.
μ = 65 pointsσ = 10 points40%
x = points
15.Confidence intervalsMedium
A 90% confidence interval for a mean, based on 25 observations and a known σ, has a margin of error of 7. What is the margin of error E₂ of a 95% interval from the same data?
90%n₁ = 25E₁ = 795%
E₂ =
16.Confidence intervalsHard
The diameters of 16 machined shafts have a sample standard deviation of 6.6 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 15, 2.5% of the χ² distribution lies below 6.262 and 2.5% lies above 27.488.
n = 16s = 6.6 µmχ² = 6.262, 27.488
σ = µm
17.Hypothesis tests & confidence intervalsMedium
A sample of 64 students has a mean score of 199.1 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 200.
μ₀ = 200x̄ = 199.1s = 5n = 64
t =
18.Hypothesis tests & confidence intervalsHard
How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 15. Find the smallest sample size n.
σ = 15E = 190%
n =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.
The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. A new observation (8, 63) is made. Find its residual e = y − ŷ.
n = 6(8, 63)
e =
3Solutions
1.Time value of money
Answer: FV = 1221 €
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 = 0.003
(1 + i)−N = 1 − 12 000·0.003/275 = 0.8691
Solve with logarithms
N = −ln(0.8691)/ln(1.003) = 46.84
Round up to whole payments
46 full payments leave a small balance, which the last, smaller payment clears.
N = 47
Answer
N = 47 payments
3.NPV & IRR
Answer: DPB = 3.68 years
Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.
Discounted and cumulative cash flows
year 1: 6 500/1.08 = 6 018.52 → Σ = 6 018.52
year 2: 6 500/1.166 = 5 572.70 → Σ = 11 591.22
year 3: 6 500/1.26 = 5 159.91 → Σ = 16 751.13
year 4: 6 500/1.36 = 4 777.69 → Σ = 21 528.82
Fraction of the last year
DPB = 3 + 3 248.87/4 777.69 = 3.68
Answer
DPB = 3.68 years
The discounted payback is always longer than the simple payback (3.077 years).
4.NPV & IRR
Answer: MIRR = 18.69 %
Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.
Terminal value of the inflows
12 500·1.123 = 17 561.60
9 500·1.122 = 11 916.80
9 500·1.121 = 10 640.00
9 500·1.120 = 9 500.00
TV = 49 618.40
MIRR
MIRR = (49 618.40/25 000)1/4 − 1 = 18.69%
Answer
MIRR = 18.69%
5.Bond pricing & yield
Answer: P = 1078 €
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 = 0.025, N = 20
Present value of the coupons
€30.00·(1 − 0.6103)/0.025 = €467.67
Present value of the face value
€1 000·0.6103 = €610.27
Price
P = €1 077.95
Answer
P = €1 077.95
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: P = 1144 €
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 = 11/1.04 = 10.58
Estimated price change
%ΔP ≈ −10.58·(−0.0075) = 7.933%
P ≈ 1060·(1 + 0.07933) = €1 144.09
Answer
P = €1 144.09
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: rm = 9 %
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 = (14% − 4%)/2 = 5%
Market return
rm = 4% + 5% = 9%
Answer
rm = 9%
8.CAPM & WACC
Answer: WACC = 9.333 %
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 = (85 − 70)/12 = 1.25
Answer
z = 1.25
11.Probability
Answer: P = 0.3056
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)2 = 0.6944
P = 1 − 0.6944 = 0.3056
Answer
P = 0.3056
12.Probability
Answer: E(G) = −1 €
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: 10 (p = 0.1), 45 (p = 0.05), −5 (p = 0.85)
Expected value
E(G) = 10·0.1 + 45·0.05 − 5·0.85 = −1 €
Answer
E(G) = −1 €
A negative value means you lose money on average.
13.Normal distribution
Answer: P = 0.2119
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the sum
μ = 750 g, σ = √(4² + 9²) = 9.849 g
Standardise
z = (757.9 − 750)/9.849 = 0.8
Upper tail
P = 1 − Φ(0.8) = 1 − 0.7881 = 0.2119
Answer
P = 0.2119
14.Normal distribution
Answer: x = 67.5 points
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
Φ(z) = 0.6
z from the table
z ≈ 0.25
Back to the original units
x = 65 + 0.25·10 = 67.5 points
Answer
x = 67.5 points
15.Confidence intervals
Answer: E₂ = 8.34
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)·√(25/25) = 1.191
New margin
E₂ = 7·1.191 = 8.34
Answer
E₂ = 8.34
16.Confidence intervals
Answer: σ = 4.875 µm
Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.
Sum of squares
(n − 1)s² = 15·6.6² = 653.4
Limits for σ²
653.4/27.488 = 23.77
653.4/6.262 = 104.3
Limit for σ
σ = √23.77 = 4.875 µm
Answer
σ = 4.875 µm
95% interval for σ: [4.875, 10.21] µm.
17.Hypothesis tests & confidence intervals
Answer: t = −1.44
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 = (199.1 − 200)/0.625 = −1.44
Answer
t = −1.44
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 = 609
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 = (1.645·15/1)² = 608.9
Round up
n = 609
Answer
n = 609
19.Linear regression & correlation
Answer: r = 0.987
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 = 11303 − 110·491/5 = 501
Sxx = 2510 − 110²/5 = 90
Syy = 51079 − 491²/5 = 2863
Correlation coefficient
r = 501/√(90·2863) = 0.987
Answer
r = 0.987
20.Linear regression & correlation
Answer: e = −2.333
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 = 48/6 = 8
ȳ = 392/6 = 65.33
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 316
Regression line
b = 316/70 = 4.514
a = 65.33 − 4.514·8 = 29.22
ŷ = 29.22 + 4.514·x
Predicted value
ŷ = 29.22 + 4.514·8 = 65.33
Residual
e = 63 − 65.33 = −2.333
Answer
e = −2.333
The point lies below the line.
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