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
periods0123411 + EAR = ?21% / 4

A credit card charges a nominal interest rate of 21% per year, compounded quarterly. Find the effective annual rate EAR in percent.

r = 21%m = 4
EAR = %
2.Time value of moneyHard
months0123…312€250€250€250FV = ?

You save €250 at the end of every month for 26 years. The account pays 7.2% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €2507.2% / 12N = 312
FV = €
3.NPV & IRRMedium
years01234−75 00031 00024 00029 00016 000

An investment of €75 000 gives these cash flows at the end of years 1 to 4: €31 000, €24 000, €29 000, €16 000. The cost of capital is 9%. Find the net present value NPV.

r = 9%CF0 = −75 000CF1 = 31 000CF2 = 24 000CF3 = 29 000CF4 = 16 000
NPV = €
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €40 000 and returns €22 000, €15 500, €9 500 at the end of years 1 to 3. Project B costs €45 000 and returns €12 000, €17 000, €25 500. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −40 000, 22 000, 15 500, 9 500B: −45 000, 12 000, 17 000, 25 500
r* = %
5.Bond pricing & yieldMedium
years0123456P€20€20€20 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 2% and matures in 6 years. Its yield to maturity changes from 5% to 5.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 2%n = 6y1 = 5%y2 = 5.5%
ΔP = €
6.Bond pricing & yieldHard
years01234P€70€70€70 + €1 000

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

F = €1 000coupon = 7%n = 4YTM = 4%
D = years
7.CAPM & WACCMedium
βr (%)SMLrf​ = 3%β = 1rm​ = ?β = 0.8re​ = 7%

A share with a beta of 0.8 has a required return of 7%. The risk-free rate is 3%. What expected market return rm does CAPM imply?

β = 0.8re = 7%rf = 3%
rm = %
8.CAPM & WACCHard
WACC = ?equity1 000 000 × €44.50β = 1.1rf​ = 1.5%, rm​ = 6.5%debt40 000 × €1 000rd​ = 5%T = 25.8%

A company has 1 000 000 shares at €44.50 each and 40 000 bonds trading at €1 000 each. The pre-tax cost of debt is 5% and the tax rate is 25.8%. The share’s beta is 1.1, the risk-free rate 1.5% and the market return 6.5%. Find the WACC.

β = 1.1rf = 1.5%rm = 6.5%rd = 5%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 24, 52, 28, 16, 16, 33.

n = 6
s =
10.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 170 and standard deviation 12. What proportion P of values is below 158?

μ = 170σ = 12x = 158
P =
11.ProbabilityMedium
n = 15k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 15? The order does not matter.

n = 15k = 2
N =
12.ProbabilityHard
xp(x)00.310.3520.35μ

X has the distribution shown, and Y = 3X + 1. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35Y = 3X + 1
σY =
13.Normal distributionMedium
85112.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 85 and 112.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 85x2 = 112.75
P =
14.Normal distributionHard
μ = 65x = ?80%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 80% of them lie (the 80th percentile).

μ = 65σ = 10P = 80%
x = points
15.Confidence intervalsMedium
n = 15, s = 16.5L = ?Ux̄ = 58.5E99%

A sample of 15 items has mean 58.5 and standard deviation s = 16.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 15x̄ = 58.5s = 16.599%t(14) = 2.977
lower limit =
16.Confidence intervalsHard
σ = 39, n = ?x̄ − Ex̄ + Ex̄E = 499%

The population standard deviation is 39. 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 = 39E = 499%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 40 parts has mean length 133.5 mm, and the line's known standard deviation is 4 mm. Line 2: a sample of 50 parts has mean 138 mm, known standard deviation 17 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 40x̄₁ = 133.5 mmσ₁ = 4 mmn₂ = 50x̄₂ = 138 mmσ₂ = 17 mm
z =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 5α = 0.05βσ = 17, 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 17. Find the minimum sample size n.

α = 0.05power = 80%δ = 5σ = 17zα = 1.645zβ = 0.842
n =
19.Linear regression & correlationMedium
xy185321712481279730100xy2025306080100

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy2554576748871093xy5106080x = 5

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (5, 71) is made. Find its residual e = y − ŷ.

n = 5(5, 71)
e =

3Solutions

1.Time value of money

Answer: EAR = 22.71 %

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 per period

r/m = 0.21/4 = 0.0525

Compound over one year

EAR = (1 + 0.0525)4 − 1 = 1.227 − 1 = 22.71%

Answer

EAR = 22.71%

The EAR is always above the nominal rate when interest compounds more than once a year.

2.Time value of money

Answer: FV = 227709 €

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 = 0.07200000000000001/12 = 0.006
N = 12·26 = 312

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 250·(6.465 − 1)/0.006 = €227 709.24

Answer

FV = €227 709.24

You paid in €78 000 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 7369 €

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 + 0.09)1 = 31 000/1.09 = 28 440.37
CF2/(1 + 0.09)2 = 24 000/1.188 = 20 200.32
CF3/(1 + 0.09)3 = 29 000/1.295 = 22 393.32
CF4/(1 + 0.09)4 = 16 000/1.412 = 11 334.80

Add them up

NPV = €7 368.81

Answer

NPV = €7 368.81

NPV is positive, so the project earns more than the required return.

4.NPV & IRR

Answer: r* = 7.083 %

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 = −45 000 − (−40 000) = −5 000
ΔCF1 = 12 000 − 22 000 = −10 000
ΔCF2 = 17 000 − 15 500 = 1 500
ΔCF3 = 25 500 − 9 500 = 16 000

IRR of the difference

NPVΔ(7%) = 25.13
NPVΔ(8%) = −271.94
r* = 7.083%

Answer

r* = 7.083%

Check: at 7.083% both NPVs are €1 798.95.

5.Bond pricing & yield

Answer: ΔP = −22.57 €

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 = €20.00, i = 0.05, N = 6
€20.00·(1 − 0.7462)/0.05 + €1 000·0.7462 = €101.51 + €746.22 = €847.73

Price at the new yield

C = €20.00, i = 0.055, N = 6
€20.00·(1 − 0.7252)/0.055 + €1 000·0.7252 = €99.91 + €725.25 = €825.16

Change

ΔP = 825.16 − 847.73 = −€22.57

Answer

ΔP = −€22.57
6.Bond pricing & yield

Answer: D = 3.645 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: 70/1.04 = 67.31, t·PV = 67.31
t = 2: 70/1.082 = 64.72, t·PV = 129.44
t = 3: 70/1.125 = 62.23, t·PV = 186.69
t = 4: 1070/1.17 = 914.64, t·PV = 3 658.56

Duration

P = 1 108.90
D = 4 042.00/1 108.90 = 3.645

Answer

D = 3.645 years
7.CAPM & WACC

Answer: rm = 8 %

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 risk premium

rm − rf = (7% − 3%)/0.8 = 5%

Market return

rm = 3% + 5% = 8%

Answer

rm = 8%
8.CAPM & WACC

Answer: WACC = 5.443 %

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 = 1 000 000·€44.50 = €44 500 000
D = 40 000·€1 000 = €40 000 000

Cost of equity (CAPM)

re = 1.5% + 1.1·5% = 7%

Weights

V = €84 500 000, E/V = 0.5266, D/V = 0.4734

WACC

WACC = 0.5266·7% + 0.4734·5%·(1 − 0.258) = 5.443%

Answer

WACC = 5.443%
9.Descriptive statistics

Answer: s = 13.45

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 = 169/6 = 28.17

Squared deviations

Σ(x − x̄)² = (−4.167)² + 23.83² + (−0.1667)² + (−12.17)² + (−12.17)² + 4.833² = 904.8

Standard deviation

s = √(904.8/5) = 13.45

Answer

s = 13.45
10.Descriptive statistics

Answer: P = 0.1587

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

Standard normal table

Φ(−1) = 0.1587
P = 0.1587

Answer

P = 0.1587
11.Probability

Answer: N = 105

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.

Combinations

C(15, 2) = (15·14)/(2·1) = 105

Answer

N = 105
12.Probability

Answer: σY = 2.414

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 = 1.05, Var(X) = 0.6475, σX = 0.8047

Linear change

σY = |3|·0.8047 = 2.414

Answer

σY = 2.414

The mean does move: E(Y) = 3·1.05 + 1 = 4.15.

13.Normal distribution

Answer: P = 0.6437

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 = (85 − 100)/15 = −1
z₂ = (112.75 − 100)/15 = 0.85

Area between

P = Φ(0.85) − Φ(−1) = 0.8023 − 0.1587 = 0.6437

Answer

P = 0.6437
14.Normal distribution

Answer: x = 73.42 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) = 0.8 → z = 0.8416

Back to x

x = 65 + 0.8416·10 = 73.42 points

Answer

x = 73.42 points
15.Confidence intervals

Answer: lower limit = 45.82

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 = 14, t = 2.977

Margin of error

E = 2.977·16.5/√15 = 12.68

Interval

58.5 ± 12.68 → [45.82, 71.18]

Answer

lower limit = 45.82
16.Confidence intervals

Answer: n = 631

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 = (2.576·39/4)² = 630.8

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 631

Answer

n = 631
17.Hypothesis tests & confidence intervals

Answer: z = −1.81

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 = √(4²/40 + 17²/50) = √6.18 = 2.486

Test statistic

z = (133.5 − 138)/2.486 = −1.81

Answer

z = −1.81

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

18.Hypothesis tests & confidence intervals

Answer: n = 72

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)·17/5)² = 71.5

Round up

n = 72

Answer

n = 72
19.Linear regression & correlation

Answer: r = 0.9801

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 = 120/5 = 24
ȳ = 402/5 = 80.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 360
Syy = Σ(y − ȳ)² = 1499

Correlation coefficient

r = 360/√(90·1499) = 0.9801

Answer

r = 0.9801
20.Linear regression & correlation

Answer: e = 3.1

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 = 366/5 = 73.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 212

Regression line

b = 212/40 = 5.3
a = 73.2 − 5.3·6 = 41.4
ŷ = 41.4 + 5.3·x

Predicted value

ŷ = 41.4 + 5.3·5 = 67.9

Residual

e = 71 − 67.9 = 3.1

Answer

e = 3.1

The point lies above the line.

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