You invest €20 000 at 6% per year with continuous compounding. Find the value FV after 4 years.
PV = €20 000r = 6%t = 4
FV = €
2.Time value of moneyHard
You want €30 000 in 15 years. You save the same amount at the end of every month in an account paying 4.8% per year, compounded monthly. Find the monthly deposit A.
FV = €30 0004.8% / 12N = 180
A = €
3.NPV & IRRMedium
An investment of €60 000 gives these cash flows at the end of years 1 to 3: €19 000, €27 000, €15 000. The cost of capital is 9%. Find the net present value NPV.
A project costs €100 000 today. Its cash flow is €22 000 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 5. The cost of capital is 10%. Find the net present value NPV.
C0 = €100 000CF1 = €22 000g = 3%n = 5r = 10%
NPV = €
5.Bond pricing & yieldMedium
A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 4% per year. Find the price P of the bond.
F = €1 000coupon = 7%YTM = 4%n = 7
P = €
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 8% matures in 3 years and sells for €940. Find its yield to maturity YTM in percent.
F = €1 000coupon = 8%n = 3P = €940
YTM = %
7.CAPM & WACCMedium
A company has equity worth €3 000 000 and debt worth €3 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 4% and the tax rate is 21%. Find the WACC.
E = €3 000 000D = €3 000 000re = 9%rd = 4%T = 21%
WACC = %
8.CAPM & WACCHard
A company has 4 000 000 shares at €52.00 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 7% and the tax rate is 28%. The share’s beta is 1.3, the risk-free rate 2.5% and the market return 7.5%. Find the WACC.
β = 1.3rf = 2.5%rm = 7.5%rd = 7%T = 28%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 44, 41, 27, 36, 52.
n = 5
s =
10.Descriptive statisticsHard
Exam scores are normally distributed with mean 70 and standard deviation 8. Find the z-score z of a score of 60.4.
μ = 70σ = 8x = 60.4
z =
11.ProbabilityMedium
A fair die is rolled 3 times. What is the probability P of getting at least one six?
n = 3
P =
12.ProbabilityHard
A game costs €5 to play. You win €15 with probability 0.2 and €250 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.
stake = €5€15: p = 0.2€250: p = 0.01
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area between −z and z under the standard normal curve is 0.3328. Find z to two decimals.
A = 0.3328
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 10% of them lie (the 10th percentile).
μ = 65σ = 10P = 10%
x = points
15.Confidence intervalsMedium
A sample of 21 items has mean 115.5 and standard deviation s = 16.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 21x̄ = 115.5s = 16.590%t(20) = 1.725
E =
16.Confidence intervalsHard
The population standard deviation is 2. 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 0.5?
σ = 2E = 0.590%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 100 students has a mean score of 102.5 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 100.
μ₀ = 100x̄ = 102.5s = 12n = 100
t =
18.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 2.5 units with a power of 95%. The standard deviation is 6. Find the minimum sample size n.
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
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. Predict ŷ for x = 9.
n = 5x = 9
ŷ =
3Solutions
1.Time value of money
Answer: FV = 25425 €
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.
Exponent
r·t = 0.06·4 = 0.24
e0.24 = 1.271
Future value
FV = 20 000·1.271 = €25 424.98
Answer
FV = €25 424.98
2.Time value of money
Answer: A = 114.1 €
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.
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 + 0.09)1 = 19 000/1.09 = 17 431.19
CF2/(1 + 0.09)2 = 27 000/1.188 = 22 725.36
CF3/(1 + 0.09)3 = 15 000/1.295 = 11 582.75
Add them up
NPV = €−8 260.70
Answer
NPV = €−8 260.70
NPV is negative, so the project earns less than the required return.
4.NPV & IRR
Answer: NPV = −11943 €
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 the growing annuity
((1.03)/(1.1))5 = 0.7198
PV = 22 000/(0.1 − 0.03)·(1 − 0.7198) = 88 057.20
NPV
NPV = 88 057.20 − 100 000 = −€11 942.80
Answer
NPV = −€11 942.80
NPV is negative, so the project earns less than the required return.
5.Bond pricing & yield
Answer: P = 1182 €
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 = €35.00, i = 0.02, N = 14
Present value of the coupons
€35.00·(1 − 0.7579)/0.02 = €423.72
Present value of the face value
€1 000·0.7579 = €757.88
Price
P = €1 181.59
Answer
P = €1 181.59
The yield is below the coupon rate, so the bond sells at a premium.
6.Bond pricing & yield
Answer: YTM = 10.43 %
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
€940 = €80·(1 − (1 + y)⁻³)/y + €1 000/(1 + y)³
Solve numerically
P(10%) = €950.26, P(11%) = €926.69
YTM = 10.43%
Answer
YTM = 10.43%
7.CAPM & WACC
Answer: WACC = 6.08 %
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 = €6 000 000, E/V = 0.5, D/V = 0.5
WACC
WACC = 0.5·9% + 0.5·4%·(1 − 0.21) = 6.08%
Answer
WACC = 6.08%
8.CAPM & WACC
Answer: WACC = 8.383 %
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 values
E = 4 000 000·€52.00 = €208 000 000
D = 40 000·€960 = €38 400 000
Cost of equity (CAPM)
re = 2.5% + 1.3·5% = 9%
Weights
V = €246 400 000, E/V = 0.8442, D/V = 0.1558
WACC
WACC = 0.8442·9% + 0.1558·7%·(1 − 0.28) = 8.383%
Answer
WACC = 8.383%
9.Descriptive statistics
Answer: s = 9.301
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.
Mean
x̄ = 200/5 = 40
Squared deviations
Σ(x − x̄)² = 4² + 1² + (−13)² + (−4)² + 12² = 346
Standard deviation
s = √(346/4) = 9.301
Answer
s = 9.301
10.Descriptive statistics
Answer: z = −1.2
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 = (60.4 − 70)/8 = −1.2
Answer
z = −1.2
11.Probability
Answer: P = 0.4213
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)3 = 0.5787
P = 1 − 0.5787 = 0.4213
Answer
P = 0.4213
12.Probability
Answer: E(G) = 0.5 €
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: 10 (p = 0.2), 245 (p = 0.01), −5 (p = 0.79)
Expected value
E(G) = 10·0.2 + 245·0.01 − 5·0.79 = 0.5 €
Answer
E(G) = 0.5 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = 0.43
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 + 0.3328)/2 = 0.6664
Read the z-table backwards
Find 0.6664 inside the table and read off z.
z = 0.43
Answer
z = 0.43
14.Normal distribution
Answer: x = 52.18 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) = 0.1 → z = −1.282
Back to x
x = 65 + −1.282·10 = 52.18 points
Answer
x = 52.18 points
15.Confidence intervals
Answer: E = 6.211
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 = 20, t = 1.725
Margin of error
E = 1.725·16.5/√21 = 6.211
Answer
E = 6.211
16.Confidence intervals
Answer: n = 44
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 = (1.645·2/0.5)² = 43.3
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 44
Answer
n = 44
17.Hypothesis tests & confidence intervals
Answer: t = 2.083
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 = 12/√100 = 12/10 = 1.2
Test statistic
t = (102.5 − 100)/1.2 = 2.083
Answer
t = 2.083
t0.025; df = 99 = 1.984
|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: n = 63
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 = ((1.645 + 1.645)·6/2.5)² = 62.35
Round up
n = 63
Answer
n = 63
19.Linear regression & correlation
Answer: r² = 0.8249
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 = 110/5 = 22
ȳ = 475/5 = 95
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 146
Syy = Σ(y − ȳ)² = 646
Coefficient of determination
r² = 146²/(40·646) = 0.8249
Answer
r² = 0.8249
About 82.49% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 94.8
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 = 30/5 = 6
ȳ = 390/5 = 78
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 224
Regression line
b = 224/40 = 5.6
a = 78 − 5.6·6 = 44.4
ŷ = 44.4 + 5.6·x
Prediction
ŷ = 44.4 + 5.6·9 = 94.8
Answer
ŷ = 94.8
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