You need €40 000 in 6 years. An account pays 3% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €40 0003% / 12n = 6
PV = €
2.Time value of moneyHard
A loan of €200 000 is repaid with equal monthly payments over 30 years at 2.4% per year, compounded monthly. How much interest I is paid in total over the life of the loan?
P = €200 0002.4% / 12N = 360
I = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 3: €18 000, €16 000, €19 000. The cost of capital is 7.000000000000001%. Find the net present value NPV.
Project A costs €30 000 and returns €16 500, €12 000, €6 000 at the end of years 1 to 3. Project B costs €32 000 and returns €8 500, €13 000, €16 500. Find the crossover rate r* at which both projects have the same NPV, in percent.
A bond with a face value of €1 000 pays an annual coupon of 7% and matures in 5 years. Its yield to maturity changes from 6% to 4.5%. By how much ΔP does the bond price change?
F = €1 000coupon = 7%n = 5y1 = 6%y2 = 4.5%
ΔP = €
6.Bond pricing & yieldHard
A bond trades at €900 with a yield to maturity of 5% (annual compounding) and a Macaulay duration of 10.3 years. Use the modified duration to estimate the new price P if the yield falls by 25 basis points.
P0 = €900y = 5%D = 10.3Δy = −25 bp
P = €
7.CAPM & WACCMedium
A company has equity worth €12 000 000 and debt worth €6 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 21%. Find the WACC.
A company has 4 000 000 shares trading at €30.00; the book value of its equity is €36 000 000. It just paid a dividend of €0.60 per share, and dividends grow by 4% per year. Its debt has a market value of €36 000 000 and a pre-tax cost of 5%. The tax rate is 25.8%. Find the WACC, using the dividend growth model for the cost of equity.
Find the sample standard deviation s of the data: 26, 60, 31, 20, 34, 26, 29.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is below 163.6?
μ = 170σ = 8x = 163.6
P =
11.ProbabilityMedium
For the distribution shown, find the variance Var(X) of X.
x: 0, 2, 4, 6p(x): 0.35, 0.2, 0.15, 0.3
Var(X) =
12.ProbabilityHard
A game costs €5 to play. You win €15 with probability 0.2 and €250 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 = 0.2€250: p = 0.05
E(G) = €
13.Normal distributionMedium
Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0778. Find z to two decimals.
A = 0.0778
z =
14.Normal distributionHard
The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
A sample of 21 items has mean 10.5 and standard deviation s = 9.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.
n = 21x̄ = 10.5s = 9.590%t(20) = 1.725
E =
16.Confidence intervalsHard
The population standard deviation is 31. 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 0.5?
σ = 31E = 0.599%
n =
17.Hypothesis tests & confidence intervalsMedium
A sample of 49 students has a mean score of 66.14 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.
μ₀ = 75x̄ = 66.14s = 20n = 49
t =
18.Hypothesis tests & confidence intervalsHard
A sample of 64 measurements has mean 119.4. The population standard deviation is 4. Find the upper limit U of the 99% confidence interval for the mean.
x̄ = 119.4σ = 4n = 6499%
U =
19.Linear regression & correlationMedium
The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.
n = 6
r =
20.Linear regression & correlationHard
The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 3.5.
n = 5x = 3.5
ŷ =
3Solutions
1.Time value of money
Answer: PV = 33418 €
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 = 12·6 = 72
Present value
PV = FV/(1 + i)N = 40 000/1.197 = €33 418.31
Answer
PV = €33 418.31
2.Time value of money
Answer: I = 80758 €
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 payment
i = 0.002, N = 360
A = 200 000·0.002/(1 − (1 + 0.002)−360) = €779.88
Total interest
I = N·A − P = 360·779.9 − 200 000 = €80 757.78
Answer
I = €80 757.78
3.NPV & IRR
Answer: NPV = 1307 €
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 = −45 000
CF1/(1 + 0.07)1 = 18 000/1.07 = 16 822.43
CF2/(1 + 0.07)2 = 16 000/1.145 = 13 975.02
CF3/(1 + 0.07)3 = 19 000/1.225 = 15 509.66
Add them up
NPV = €1 307.11
Answer
NPV = €1 307.11
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: r* = 6.829 %
Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.
Difference in cash flows
ΔCF0 = −32 000 − (−30 000) = −2 000
ΔCF1 = 8 500 − 16 500 = −8 000
ΔCF2 = 13 000 − 12 000 = 1 000
ΔCF3 = 16 500 − 6 000 = 10 500
IRR of the difference
NPVΔ(6%) = 158.83
NPVΔ(7%) = −32.07
r* = 6.829%
Answer
r* = 6.829%
Check: at 6.829% both NPVs are €881.47.
5.Bond pricing & yield
Answer: ΔP = 67.63 €
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.
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 = 10.3/1.05 = 9.81
Estimated price change
%ΔP ≈ −9.81·(−0.0025) = 2.452%
P ≈ 900·(1 + 0.02452) = €922.07
Answer
P = €922.07
Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.
7.CAPM & WACC
Answer: WACC = 7.852 %
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 = −0.8
Standard normal table
Φ(−0.8) = 0.2119
P = 0.2119
Answer
P = 0.2119
11.Probability
Answer: Var(X) = 6.16
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 = 0·0.35 + 2·0.2 + 4·0.15 + 6·0.3 = 2.8
Mean of the squares
E(X²) = 0·0.35 + 4·0.2 + 16·0.15 + 36·0.3 = 14
Variance
Var(X) = 14 − 2.8² = 6.16
Answer
Var(X) = 6.16
12.Probability
Answer: E(G) = 10.5 €
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 = 0.05), −5 (p = 0.75)
Expected value
E(G) = 10·0.2 + 245·0.05 − 5·0.75 = 10.5 €
Answer
E(G) = 10.5 €
A positive value means you win money on average.
13.Normal distribution
Answer: z = 1.42
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 − 0.0778 = 0.9222
Read the z-table backwards
Find 0.9222 inside the table and read off z.
z = 1.42
Answer
z = 1.42
14.Normal distribution
Answer: P = 0.7389
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 difference
μ = 180 − 170 = 10 cm, σ = 15.62 cm
Standardise
z = (0 − 10)/15.62 = −0.64
Probability
P = 1 − Φ(−0.64) = 0.7389
Answer
P = 0.7389
15.Confidence intervals
Answer: E = 3.576
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 = 1.725·9.5/√21 = 3.576
Answer
E = 3.576
16.Confidence intervals
Answer: n = 25508
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 = (2.576·31/0.5)² = 25508
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 25508
Answer
n = 25508
17.Hypothesis tests & confidence intervals
Answer: t = −3.101
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 = 20/√49 = 20/7 = 2.857
Test statistic
t = (66.14 − 75)/2.857 = −3.101
Answer
t = −3.101
t0.025; df = 48 = 2.011
|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: U = 120.7
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).
Margin of error
E = 2.576·4/√64 = 1.288
Interval limit
U = 119.4 + 1.288 = 120.7
Answer
U = 120.7
19.Linear regression & correlation
Answer: r = 0.996
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 = 36/6 = 6
ȳ = 346/6 = 57.67
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 244
Syy = Σ(y − ȳ)² = 857.3
Correlation coefficient
r = 244/√(70·857.3) = 0.996
Answer
r = 0.996
20.Linear regression & correlation
Answer: ŷ = 16.3
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 = 88/5 = 17.6
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −26
Regression line
b = −26/10 = −2.6
a = 17.6 − (−2.6)·3 = 25.4
ŷ = 25.4 − 2.6·x
Prediction
ŷ = 25.4 + (−2.6)·3.5 = 16.3
Answer
ŷ = 16.3
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