You need €20 000 in 7 years. An account pays 4.8% per year, compounded monthly. How much must you deposit today? Find the present value PV.
FV = €20 0004.8% / 12n = 7
PV = €
2.Time value of moneyHard
You deposit €1 000 at the start of every year for 30 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 30? Find FV.
A = €1 000i = 5%n = 30
FV = €
3.NPV & IRRMedium
An investment of €45 000 gives these cash flows at the end of years 1 to 4: €11 000, €22 000, €16 000, €9 000. The cost of capital is 9%. Find the net present value NPV.
A project costs €10 000 and returns €3 000, €5 000, €3 500 at the end of years 1 to 3. Cash flows can be reinvested at the cost of capital of 6%. Find the modified internal rate of return MIRR in percent.
You buy a bond with 12 years to maturity (face value €1 000, annual coupon 4.5%) when its yield to maturity is 5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 4%. Find your holding period return HPR in percent.
F = €1 000coupon = 4.5%n = 12y0 = 5%y1 = 4%
HPR = %
6.Bond pricing & yieldHard
A bond with a face value of €1 000 and an annual coupon of 5% matures in 7 years and sells for €1 060. Find its yield to maturity YTM in percent.
F = €1 000coupon = 5%n = 7P = €1 060
YTM = %
7.CAPM & WACCMedium
A company has equity worth €4 000 000 and debt worth €1 000 000. Its cost of equity is 13%, its pre-tax cost of debt is 3% and the tax rate is 25.8%. Find the WACC.
A company has 2 000 000 shares at €25.50 each and 5 000 bonds trading at €1 050 each. The pre-tax cost of debt is 6% and the tax rate is 20%. The share’s beta is 1.1, the risk-free rate 2.5% and the market return 9%. Find the WACC.
β = 1.1rf = 2.5%rm = 9%rd = 6%T = 20%
WACC = %
9.Descriptive statisticsMedium
Find the sample standard deviation s of the data: 50, 10, 48, 27, 24, 28, 43.
n = 7
s =
10.Descriptive statisticsHard
A quantity is normally distributed with mean 500 and standard deviation 75. What proportion P of values is above 593.8?
μ = 500σ = 75x = 593.8
P =
11.ProbabilityMedium
For the distribution shown, find the standard deviation σ of X.
x: 1, 2, 3, 4, 5p(x): 0.25, 0.1, 0.3, 0.25, 0.1
σ =
12.ProbabilityHard
X has the distribution shown, and Y = 3X + 10. Find the standard deviation σY of Y.
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 502 and 532 h? Give P as a decimal.
μ = 500σ = 40x1 = 502x2 = 532
P =
14.Normal distributionHard
Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
hole: μ = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
The blood pressure of 13 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 1.8 and standard deviation 5.4. Find the lower limit of the 99% confidence interval for the mean difference.
n = 13d̄ = 1.8sd = 5.499%t(12) = 3.055
lower limit = mmHg
16.Confidence intervalsHard
The population standard deviation is 9. 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 4?
σ = 9E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
Two production lines make the same part. Line 1: a sample of 100 parts has mean length 87.5 mm, and the line's known standard deviation is 20 mm. Line 2: a sample of 50 parts has mean 80.2 mm, known standard deviation 14 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).
In a trial, 38 of 150 patients on treatment A recover, and 33 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.
x₁ = 38n₁ = 150x₂ = 33n₂ = 100
z =
19.Linear regression & correlationMedium
The table gives 5 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
20.Linear regression & correlationHard
The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 26.
n = 6x = 26
ŷ =
3Solutions
1.Time value of money
Answer: PV = 14302 €
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 = 0.048/12 = 0.004
N = 12·7 = 84
Present value
PV = FV/(1 + i)N = 20 000/1.398 = €14 302.04
Answer
PV = €14 302.04
2.Time value of money
Answer: FV = 69761 €
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 + 0.05)30 − 1)/0.05 = 66.44
Annuity due
Each deposit is invested one year longer.
FV = 1 000·66.44·1.05 = €69 760.79
Answer
FV = €69 760.79
3.NPV & IRR
Answer: NPV = 2339 €
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 + 0.09)1 = 11 000/1.09 = 10 091.74
CF2/(1 + 0.09)2 = 22 000/1.188 = 18 516.96
CF3/(1 + 0.09)3 = 16 000/1.295 = 12 354.94
CF4/(1 + 0.09)4 = 9 000/1.412 = 6 375.83
Add them up
NPV = €2 339.47
Answer
NPV = €2 339.47
NPV is positive, so the project earns more than the required return.
4.NPV & IRR
Answer: MIRR = 6.768 %
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
3 000·1.062 = 3 370.80
5 000·1.061 = 5 300.00
3 500·1.060 = 3 500.00
TV = 12 170.80
MIRR
MIRR = (12 170.80/10 000)1/3 − 1 = 6.768%
Answer
MIRR = 6.768%
5.Bond pricing & yield
Answer: HPR = 13.93 %
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
€1 060 = €50·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷
Solve numerically
P(4%) = €1 060.02, P(5%) = €1 000.00
YTM = 4%
Answer
YTM = 4%
7.CAPM & WACC
Answer: WACC = 10.85 %
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 = €5 000 000, E/V = 0.8, D/V = 0.2
WACC
WACC = 0.8·13% + 0.2·3%·(1 − 0.258) = 10.85%
Answer
WACC = 10.85%
8.CAPM & WACC
Answer: WACC = 9.197 %
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 = 1.25
Standard normal table
Φ(1.25) = 0.8944
P = 1 − 0.8944 = 0.1056
Answer
P = 0.1056
11.Probability
Answer: σ = 1.314
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.
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.
Spread of X
μ = 3.5, Var(X) = 5.15, σX = 2.269
Linear change
σY = |3|·2.269 = 6.808
Answer
σY = 6.808
The mean does move: E(Y) = 3·3.5 + 10 = 20.5.
13.Normal distribution
Answer: P = 0.2682
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 = (502 − 500)/40 = 0.05
z₂ = (532 − 500)/40 = 0.8
Area between
P = Φ(0.8) − Φ(0.05) = 0.7881 − 0.5199 = 0.2682
Answer
P = 0.2682
14.Normal distribution
Answer: P = 0.9868
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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm
Standardise
z = (0 − 0.04)/0.01803 = −2.22
Probability
P = 1 − Φ(−2.22) = 0.9868
Answer
P = 0.9868
15.Confidence intervals
Answer: lower limit = −2.775 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 = 12, t = 3.055
Margin of error
E = 3.055·5.4/√13 = 4.575
Interval
1.8 ± 4.575 → [−2.775, 6.375]
Answer
lower limit = −2.775 mmHg
16.Confidence intervals
Answer: n = 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.
Solve for n
n = (2.576·9/4)² = 33.59
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 34
Answer
n = 34
17.Hypothesis tests & confidence intervals
Answer: z = 2.594
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 = √(20²/100 + 14²/50) = √7.92 = 2.814
Test statistic
z = (87.5 − 80.2)/2.814 = 2.594
Answer
z = 2.594
|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).
18.Hypothesis tests & confidence intervals
Answer: z = −1.317
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.
Sample proportions
p̂₁ = 38/150 = 0.2533
p̂₂ = 33/100 = 0.33
p̂ = 71/250 = 0.284
Standard error under H₀
SE = √(0.284·0.716·(1/150 + 1/100)) = 0.05822
Test statistic
z = (0.2533 − 0.33)/0.05822 = −1.317
Answer
z = −1.317
|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).
19.Linear regression & correlation
Answer: r² = 0.8547
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 = 109/5 = 21.8
Sums of squares and products
Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −20
Syy = Σ(y − ȳ)² = 46.8
Coefficient of determination
r² = −20²/(10·46.8) = 0.8547
Answer
r² = 0.8547
About 85.47% of the variation in y is explained by the line.
20.Linear regression & correlation
Answer: ŷ = 155.9
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 = 132/6 = 22
ȳ = 792/6 = 132
Sums of squares and products
Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 418
Regression line
b = 418/70 = 5.971
a = 132 − 5.971·22 = 0.6286
ŷ = 0.6286 + 5.971·x
Prediction
ŷ = 0.6286 + 5.971·26 = 155.9
Answer
ŷ = 155.9
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