CFA Level I quantitative methods and fixed income practice: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

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1Formulas and methods

Time value of money

Future value
FV = PV·(1 + i)n
Present value
PV = FV/(1 + i)n
Several periods per year
i = r/m, N = m·n
Annuity, present value
PV = PMT·[1 − (1 + i)−n]/i
Annuity, future value
FV = PMT·[(1 + i)n − 1]/i

Method

  1. Draw a timeline with every cash flow and its timing.
  2. Match the rate to the period: a monthly rate for monthly payments.
  3. Use FV = PV·(1 + i)ⁿ for single amounts and the annuity formulas for equal payments.
  4. Check the sign and the order of magnitude of the answer.

NPV & IRR

Net present value
NPV = −C0 + Σ CFt/(1 + r)t
Equal cash flows
NPV = −C0 + CF·[1 − (1 + r)−n]/r
Internal rate of return
the r at which NPV = 0
Decision
accept when NPV > 0 (or IRR > required return)

Method

  1. List all cash flows with their year; the investment at year 0 is negative.
  2. Discount each cash flow: CFt/(1 + r)t.
  3. Add them up to get the NPV. Positive means the project earns more than r.
  4. For the IRR, find the rate that makes NPV zero by trial or interpolation.

Bond pricing & yield

Bond price
P = C·(1 − (1 + y)−n)/y + F/(1 + y)n
Semi-annual
C/2, y/2 and 2n periods
Current yield
C/P
Price and yield
yield up → price down

Method

  1. Find the coupon per period and the yield per period.
  2. Price = present value of the coupons (an annuity) + present value of the face value.
  3. For semi-annual coupons use half the coupon, half the yield and twice the periods.
  4. Check: if the yield is above the coupon rate, the price is below face value.

CAPM & WACC

CAPM
re = rf + β·(rm − rf)
WACC
E/V·re + D/V·rd·(1 − T)
Firm value
V = E + D

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

2Practice problems

1.Time value of moneyMedium
years0123…20PV = ?€1 000€1 000€1 000€1 000

An annuity pays €1 000 at the end of each year for 20 years. The interest rate is 8% per year. Find the present value PV of the annuity.

A = €1 000i = 8%n = 20
PV = €
2.Time value of moneyHard
months0123…240€150 000A?AAA

A loan of €150 000 is repaid with equal monthly payments over 20 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.

P = €150 0003% / 12N = 240
A = €
3.NPV & IRRMedium
years012345−80 00019 00019 00019 00019 00027 000

A machine costs €80 000 today. It generates €19 000 at the end of each year for 5 years and can then be sold for €8 000. The cost of capital is 6%. Find the net present value NPV.

C0 = €80 000CF = €19 000S = €8 000n = 5r = 6%
NPV = €
4.NPV & IRRHard
years01234567−100 00013 000+5%+5%+5%+5%+5%+5%r = 10%, NPV = ?

A project costs €100 000 today. Its cash flow is €13 000 at the end of year 1 and grows by 5% per year until the last cash flow at the end of year 7. The cost of capital is 10%. Find the net present value NPV.

C0 = €100 000CF1 = €13 000g = 5%n = 7r = 10%
NPV = €
5.Bond pricing & yieldMedium
years0123…15€909C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 15 years and pays annual coupons. It sells for €909 when the yield to maturity is 7%. Find the coupon rate c in percent.

F = €1 000n = 15P = €909YTM = 7%
c = %
6.Bond pricing & yieldHard
half-years0123…16€929€10€10€10 + €1 000

A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 8 years. It sells for €929. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 2% (2×)n = 8P = €929
YTM = %
7.CAPM & WACCMedium
WACC = ?equityEre​ = 13%debtD = 1.5·Erd​ = 7%T = 25%

A company has a debt-to-equity ratio (at market values) of 1.5. Its cost of equity is 13%, its pre-tax cost of debt is 7% and the tax rate is 25%. Find the WACC.

D/E = 1.5re = 13%rd = 7%T = 25%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €24.00D0​ = €1.00g = 4%debt€24 000 000rd​ = 3%T = 28%

A company has 2 000 000 shares trading at €24.00; the book value of its equity is €14 400 000. It just paid a dividend of €1.00 per share, and dividends grow by 4% per year. Its debt has a market value of €24 000 000 and a pre-tax cost of 3%. The tax rate is 28%. Find the WACC, using the dividend growth model for the cost of equity.

N = 2 000 000P0 = €24.00book equity = €14 400 000D0 = €1.00g = 4%D = €24 000 000rd = 3%T = 28%
WACC = %
9.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 41, 58, 45, 29, 42, 34, 46.

n = 7
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 500 and standard deviation 60. Find the z-score z of a score of 428.

μ = 500σ = 60x = 428
z =
11.ProbabilityMedium
xp(x)00.310.3520.130.140.15μ = 1.45

For the distribution shown, find the variance Var(X) of X.

x: 0, 1, 2, 3, 4p(x): 0.3, 0.35, 0.1, 0.1, 0.15
Var(X) =
12.ProbabilityHard
xp(x)10.320.3530.0540.3μ

X has the distribution shown, and Y = 10X + 5. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.3, 0.35, 0.05, 0.3Y = 10X + 5
σY =
13.Normal distributionMedium
70123.25P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 70 and 123.25 points? Give P as a decimal.

μ = 100σ = 15x1 = 70x2 = 123.25
P =
14.Normal distributionHard
380420460500540580620htop 60%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded top 60% is to the right of it. Find the cut-off x.

μ = 500 hσ = 40 h60%
x = h
15.Confidence intervalsMedium
x̄E₁ = 1.295%, n = 50x̄E₂ = ?90%, n = 50

A 95% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 1.2. What is the margin of error E₂ of a 90% interval from the same data?

95%n₁ = 50E₁ = 1.290%
E₂ =
16.Confidence intervalsHard
μ₁ − μ₂, t(12) = 1.7820L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 5 plots, mean yield 24 kg, standard deviation 3.5 kg. Plot group 2: 9 plots, mean 34 kg, standard deviation 5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 5x̄₁ = 24s₁ = 3.5n₂ = 9x̄₂ = 34s₂ = 590%t(12) = 1.782
lower limit = kg
17.Hypothesis tests & confidence intervalsMedium
−2.2012.201t = ?df = 11

12 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 6.1 and the standard deviation of the differences is 6.6. Find the paired test statistic t for H₀: μd = 0.

n = 12d̄ = 6.1sd = 6.6
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 32 of 80 patients on treatment A recover, and 72 of 200 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 32n₁ = 80x₂ = 72n₂ = 200
z =
19.Linear regression & correlationMedium
xy14431641185820632266xy1520405060

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.

n = 5Σx = 90Σy = 271Σx² = 1660Σy² = 15219Σxy = 5014
r =
20.Linear regression & correlationHard
xy274166238291035xy510102030x = 9

The table gives 5 paired observations of x = load in kN and y = extension in mm. 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: PV = 9818 €

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.

Annuity factor

(1 − (1 + i)−n)/i = (1 − 0.2145)/0.08 = 9.818

Present value

PV = 1 000·9.818 = €9 818.15

Answer

PV = €9 818.15
2.Time value of money

Answer: A = 831.9 €

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 rate and number of payments

i = 0.03/12 = 0.0025
N = 12·20 = 240

Payment

The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.

(1 + i)−N = 0.5492
A = 150 000·0.0025/(1 − 0.5492) = €831.90

Answer

A = €831.90
3.NPV & IRR

Answer: NPV = 6013 €

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 yearly cash flows

19 000·4.212 = 80 034.91

Present value of the salvage value

8 000/1.338 = 5 978.07

NPV

NPV = −80 000 + 80 034.91 + 5 978.07 = €6 012.98

Answer

NPV = €6 012.98
4.NPV & IRR

Answer: NPV = −27737 €

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

((1.05)/(1.1))7 = 0.7221
PV = 13 000/(0.1 − 0.05)·(1 − 0.7221) = 72 263.10

NPV

NPV = 72 263.10 − 100 000 = −€27 736.90

Answer

NPV = −€27 736.90

NPV is negative, so the project earns less than the required return.

5.Bond pricing & yield

Answer: c = 6.001 %

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.

Present value of the face value

1000·(1 + 0.07)−15 = €362.45

Coupon

C = (909 − 362.45)/9.108 = €60.01

Coupon rate

c = 60.01/1000 = 6.001%

Answer

c = 6.001%
6.Bond pricing & yield

Answer: YTM = 3.005 %

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 = €10.00, N = 16

Solve numerically

i = 1.503% per half-year

YTM

YTM = 2·1.503% = 3.005%

Answer

YTM = 3.005%
7.CAPM & WACC

Answer: WACC = 8.35 %

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 from D/E

E/V = 1/(1 + 1.5) = 0.4
D/V = 1.5/(1 + 1.5) = 0.6

WACC

WACC = 0.4·13% + 0.6·7%·(1 − 0.25) = 8.35%

Answer

WACC = 8.35%
8.CAPM & WACC

Answer: WACC = 6.276 %

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 equity

D1 = 1·1.04 = €1.04
re = 1.04/24 + 0.04 = 8.333%

Market values

E = 2 000 000·€24.00 = €48 000 000
V = €72 000 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·8.333% + 0.3333·3%·(1 − 0.28) = 6.276%

Answer

WACC = 6.276%
9.Descriptive statistics

Answer: s = 9.263

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 = 295/7 = 42.14

Squared deviations

Σ(x − x̄)² = (−1.143)² + 15.86² + 2.857² + (−13.14)² + (−0.1429)² + (−8.143)² + 3.857² = 514.9

Standard deviation

s = √(514.9/6) = 9.263

Answer

s = 9.263
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 = (428 − 500)/60 = −1.2

Answer

z = −1.2
11.Probability

Answer: Var(X) = 1.947

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 + 2·0.1 + 3·0.1 + 4·0.15 = 1.45

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.1 + 9·0.1 + 16·0.15 = 4.05

Variance

Var(X) = 4.05 − 1.45² = 1.947

Answer

Var(X) = 1.947
12.Probability

Answer: σY = 11.95

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 = 2.35, Var(X) = 1.428, σX = 1.195

Linear change

σY = |10|·1.195 = 11.95

Answer

σY = 11.95

The mean does move: E(Y) = 10·2.35 + 5 = 28.5.

13.Normal distribution

Answer: P = 0.9167

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 = (70 − 100)/15 = −2
z₂ = (123.25 − 100)/15 = 1.55

Area between

P = Φ(1.55) − Φ(−2) = 0.9394 − 0.02275 = 0.9167

Answer

P = 0.9167
14.Normal distribution

Answer: x = 490 h

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) = 0.4

z from the table

z ≈ −0.25

Back to the original units

x = 500 − 0.25·40 = 490 h

Answer

x = 490 h
15.Confidence intervals

Answer: E₂ = 1.007

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 = (1.645/1.96)·√(50/50) = 0.8393

New margin

E₂ = 1.2·0.8393 = 1.007

Answer

E₂ = 1.007
16.Confidence intervals

Answer: lower limit = −14.53 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 = (4·3.5² + 8·5²)/12 = 20.75
sp = 4.555

Margin of error

E = 1.782·4.555·√(1/5 + 1/9) = 4.528

Interval

−10 ± 4.528 → [−14.53, −5.472]

Answer

lower limit = −14.53 kg
17.Hypothesis tests & confidence intervals

Answer: t = 3.202

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 of the mean difference

SE = 6.6/√12 = 1.905

Test statistic

t = 6.1/1.905 = 3.202

Answer

t = 3.202

With df = 11, the two-sided 5% critical value is 2.201; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: z = 0.6258

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 = 32/80 = 0.4
p̂₂ = 72/200 = 0.36
p̂ = 104/280 = 0.3714

Standard error under H₀

SE = √(0.3714·0.6286·(1/80 + 1/200)) = 0.06392

Test statistic

z = (0.4 − 0.36)/0.06392 = 0.6258

Answer

z = 0.6258

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

19.Linear regression & correlation

Answer: r = 0.9333

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 = 5014 − 90·271/5 = 136
Sxx = 1660 − 90²/5 = 40
Syy = 15219 − 271²/5 = 530.8

Correlation coefficient

r = 136/√(40·530.8) = 0.9333

Answer

r = 0.9333
20.Linear regression & correlation

Answer: ŷ = 32.35

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 = 110/5 = 22

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 138

Regression line

b = 138/40 = 3.45
a = 22 − 3.45·6 = 1.3
ŷ = 1.3 + 3.45·x

Prediction

ŷ = 1.3 + 3.45·9 = 32.35

Answer

ŷ = 32.35

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