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
years012345PV = ?€20 000

You need €20 000 in 5 years. An account pays 4% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €20 0004% / 12n = 5
PV = €
2.Time value of moneyHard
months0123…72€500€500€500FV = ?

You save €500 at the end of every month for 6 years. The account pays 4.8% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €5004.8% / 12N = 72
FV = €
3.NPV & IRRMedium
years0123−75 00026 00018 00023 000

An investment of €75 000 gives these cash flows at the end of years 1 to 3: €26 000, €18 000, €23 000. The cost of capital is 8%. Find the net present value NPV.

r = 8%CF0 = −75 000CF1 = 26 000CF2 = 18 000CF3 = 23 000
NPV = €
4.NPV & IRRHard
years01234−10 0003 0003 0005 5003 500

A project costs €10 000 and returns €3 000, €3 000, €5 500, €3 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 8%. Find the modified internal rate of return MIRR in percent.

r = 8%CF0 = −10 000CF1 = 3 000CF2 = 3 000CF3 = 5 500CF4 = 3 500
MIRR = %
5.Bond pricing & yieldMedium
years0123…14P€50€50€50 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 14 years. Its yield to maturity changes from 7% to 7.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 5%n = 14y1 = 7%y2 = 7.5%
ΔP = €
6.Bond pricing & yieldHard
years012345P€40€40€40 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 5 years. The yield to maturity is 7%. Find the Macaulay duration D in years.

F = €1 000coupon = 4%n = 5YTM = 7%
D = years
7.CAPM & WACCMedium
WACC = ?equity€8 000 000re​ = 8%debt€5 000 000rd​ = 4%T = 30%

A company has equity worth €8 000 000 and debt worth €5 000 000. Its cost of equity is 8%, its pre-tax cost of debt is 4% and the tax rate is 30%. Find the WACC.

E = €8 000 000D = €5 000 000re = 8%rd = 4%T = 30%
WACC = %
8.CAPM & WACCHard
rf​ = 3.5%, rm​ − rf​ = 6%no debtequityβU​ = 0.8D/E = 0.5equityre​ = ?debtT = 20%

A company's business has an unlevered (asset) beta of 0.8. It plans a debt-to-equity ratio of 0.5. The tax rate is 20%, the risk-free rate is 3.5% and the market risk premium is 6%. Find the cost of equity re at the new capital structure.

βU = 0.8D/E = 0.5T = 20%rf = 3.5%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
3035404550

Find the sample standard deviation s of the data: 31, 28, 34, 52, 26.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 90.

μ = 70σ = 20x = 90
z =
11.ProbabilityMedium
xp(x)00.110.220.330.3540.05μ = 2.05

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

x: 0, 1, 2, 3, 4p(x): 0.1, 0.2, 0.3, 0.35, 0.05
Var(X) =
12.ProbabilityHard
n = 5, p = 0.7: P(X = 4) = ?k012345

A test has 5 multiple-choice questions. A student who guesses gets each one right with probability 0.7, independently. What is the probability P of exactly 4 correct answers?

n = 5p = 0.7k = 4
P =
13.Normal distributionMedium
100 kg102 kgT = X₁ + … + X₄?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 4 of them exceeds 102 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 4> 102 kg
P =
14.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.02 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 = 25.02 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 40 phones, mean 84.5 h, known standard deviation 8 h. Model 2: 40 phones, mean 74 h, known standard deviation 7 h. Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 84.5 hσ₁ = 8 hn₂ = 40x̄₂ = 74 hσ₂ = 7 h90%
lower limit = h
16.Confidence intervalsHard
p = 0.5, n = ?p̂ − Ep̂ + Ep̂E = 0.0290%

Nothing is known in advance about the proportion of defective products. How many products n must be inspected so that a 90% confidence interval for the proportion has a margin of error of at most 0.02?

p = 0.5E = 0.0290%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 96.7 and a sample standard deviation of 15. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 96.7s = 15n = 100
t =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.05βσ = 21, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 5 units with a power of 80%. The standard deviation is 21. Find the minimum sample size n.

α = 0.05power = 80%δ = 5σ = 21zα = 1.645zβ = 0.842
n =
19.Linear regression & correlationMedium
xy124222319417518xy2420

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 100Σx² = 55Σy² = 2034Σxy = 283
r =
20.Linear regression & correlationHard
xy254126178191026xy5101020ŷ = 18

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. For which value x does the line predict ŷ = 18?

n = 5ŷ = 18
x =

3Solutions

1.Time value of money

Answer: PV = 16380 €

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 = 0.04/12 = 0.003333
N = 12·5 = 60

Present value

PV = FV/(1 + i)N = 20 000/1.221 = €16 380.06

Answer

PV = €16 380.06
2.Time value of money

Answer: FV = 41624 €

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 deposits

i = 0.048/12 = 0.004
N = 12·6 = 72

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 500·(1.333 − 1)/0.004 = €41 623.92

Answer

FV = €41 623.92

You paid in €36 000 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = −17236 €

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 = −75 000
CF1/(1 + 0.08)1 = 26 000/1.08 = 24 074.07
CF2/(1 + 0.08)2 = 18 000/1.166 = 15 432.10
CF3/(1 + 0.08)3 = 23 000/1.26 = 18 258.14

Add them up

NPV = €−17 235.69

Answer

NPV = €−17 235.69

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

4.NPV & IRR

Answer: MIRR = 13.71 %

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

3 000·1.083 = 3 779.14
3 000·1.082 = 3 499.20
5 500·1.081 = 5 940.00
3 500·1.080 = 3 500.00
TV = 16 718.34

MIRR

MIRR = (16 718.34/10 000)1/4 − 1 = 13.71%

Answer

MIRR = 13.71%
5.Bond pricing & yield

Answer: ΔP = −37.32 €

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.

Price at the old yield

C = €50.00, i = 0.07, N = 14
€50.00·(1 − 0.3878)/0.07 + €1 000·0.3878 = €437.27 + €387.82 = €825.09

Price at the new yield

C = €50.00, i = 0.075, N = 14
€50.00·(1 − 0.3633)/0.075 + €1 000·0.3633 = €424.46 + €363.31 = €787.77

Change

ΔP = 787.77 − 825.09 = −€37.32

Answer

ΔP = −€37.32
6.Bond pricing & yield

Answer: D = 4.601 years

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.

Present value of each cash flow

t = 1: 40/1.07 = 37.38, t·PV = 37.38
t = 2: 40/1.145 = 34.94, t·PV = 69.88
t = 3: 40/1.225 = 32.65, t·PV = 97.96
t = 4: 40/1.311 = 30.52, t·PV = 122.06
t = 5: 1040/1.403 = 741.51, t·PV = 3 707.53

Duration

P = 876.99
D = 4 034.81/876.99 = 4.601

Answer

D = 4.601 years
7.CAPM & WACC

Answer: WACC = 6 %

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 = €13 000 000, E/V = 0.6154, D/V = 0.3846

WACC

WACC = 0.6154·8% + 0.3846·4%·(1 − 0.3) = 6%

Answer

WACC = 6%
8.CAPM & WACC

Answer: re = 10.22 %

Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.

Levered beta

βL = 0.8·(1 + 0.8·0.5) = 1.12

Cost of equity (CAPM)

re = 3.5% + 1.12·6% = 10.22%

Answer

re = 10.22%
9.Descriptive statistics

Answer: s = 10.4

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 = 171/5 = 34.2

Squared deviations

Σ(x − x̄)² = (−3.2)² + (−6.2)² + (−0.2)² + 17.8² + (−8.2)² = 432.8

Standard deviation

s = √(432.8/4) = 10.4

Answer

s = 10.4
10.Descriptive statistics

Answer: z = 1

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 = (90 − 70)/20 = 1

Answer

z = 1
11.Probability

Answer: Var(X) = 1.147

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 = 0·0.1 + 1·0.2 + 2·0.3 + 3·0.35 + 4·0.05 = 2.05

Mean of the squares

E(X²) = 0·0.1 + 1·0.2 + 4·0.3 + 9·0.35 + 16·0.05 = 5.35

Variance

Var(X) = 5.35 − 2.05² = 1.147

Answer

Var(X) = 1.147
12.Probability

Answer: P = 0.3601

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.

Binomial formula

P(X = 4) = 5·0.74·0.31 = 5·0.2401·0.3 = 0.3601

Answer

P = 0.3601
13.Normal distribution

Answer: P = 0.02275

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 total

μT = 100 kg, σT = 0.5·√4 = 1 kg

Standardise

z = (102 − 100)/1 = 2

Probability

P = 1 − Φ(2) = 0.02275

Answer

P = 0.02275
14.Normal distribution

Answer: P = 0.8133

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 = 25.02 − 25 = 0.02 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.02)/0.02236 = −0.89

Probability

P = 1 − Φ(−0.89) = 0.8133

Answer

P = 0.8133
15.Confidence intervals

Answer: lower limit = 7.735 h

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.

Point estimate

x̄₁ − x̄₂ = 84.5 − 74 = 10.5

Standard error

SE = √(8²/40 + 7²/40) = 1.681

Margin of error

E = 1.645·1.681 = 2.765

Interval

10.5 ± 2.765 → [7.735, 13.26]

Answer

lower limit = 7.735 h

The interval does not contain 0, so the means differ significantly.

16.Confidence intervals

Answer: n = 1692

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.645²·0.5·0.5/0.02² = 1691

Round up

n = 1692

Answer

n = 1692
17.Hypothesis tests & confidence intervals

Answer: t = −2.2

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 = 15/√100 = 15/10 = 1.5

Test statistic

t = (96.7 − 100)/1.5 = −2.2

Answer

t = −2.2
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 = 110

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 + 0.842)·21/5)² = 109.1

Round up

n = 110

Answer

n = 110
19.Linear regression & correlation

Answer: r = −0.922

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 = 283 − 15·100/5 = −17
Sxx = 55 − 15²/5 = 10
Syy = 2034 − 100²/5 = 34

Correlation coefficient

r = −17/√(10·34) = −0.922

Answer

r = −0.922
20.Linear regression & correlation

Answer: x = 6.898

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 = 79/5 = 15.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 98

Regression line

b = 98/40 = 2.45
a = 15.8 − 2.45·6 = 1.1
ŷ = 1.1 + 2.45·x

Solve for x

x = (18 − 1.1)/2.45 = 6.898

Answer

x = 6.898

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