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
periods01211 + EAR = ?24% / 2

A credit card charges a nominal interest rate of 24% per year, compounded semi-annually. Find the effective annual rate EAR in percent.

r = 24%m = 2
EAR = %
2.Time value of moneyHard
months01…156…240€250 000AB = ?A5.4% / 12

A mortgage of €250 000 is repaid with equal monthly payments over 20 years at 5.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 13? Find the outstanding balance B.

P = €250 0005.4% / 12N = 240k = 156
B = €
3.NPV & IRRMedium
years0123−45 00010 00011 00016 000

An investment of €45 000 gives these cash flows at the end of years 1 to 3: €10 000, €11 000, €16 000. The cost of capital is 15%. Find the net present value NPV.

r = 15%CF0 = −45 000CF1 = 10 000CF2 = 11 000CF3 = 16 000
NPV = €
4.NPV & IRRHard
years012345−20 0005 0005 0005 0005 0005 000

An investment of €20 000 returns €5 000 at the end of each year for 5 years. Find its internal rate of return IRR in percent.

C0 = €20 000CF = €5 000n = 5
IRR = %
5.Bond pricing & yieldMedium
years0123…13P€40€40€40 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 4% and matures in 13 years. Its yield to maturity changes from 9% to 8.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 4%n = 13y1 = 9%y2 = 8.5%
ΔP = €
6.Bond pricing & yieldHard
years01234567€1 090€60€60 + €1 000

A bond with a face value of €1 000 and an annual coupon of 6% matures in 7 years and sells for €1 090. Find its yield to maturity YTM in percent.

F = €1 000coupon = 6%n = 7P = €1 090
YTM = %
7.CAPM & WACCMedium
WACC = 8.15%equity€5 000 000re​ = ?debt€4 000 000rd​ = 4.5%T = 25.8%

A company has equity worth €5 000 000 and debt worth €4 000 000. Its WACC is 8.15%, its pre-tax cost of debt is 4.5% and the tax rate is 25.8%. Find the cost of equity re.

E = €5 000 000D = €4 000 000WACC = 8.15%rd = 4.5%T = 25.8%
re = %
8.CAPM & WACCHard
rf​ = 1%, rm​ − rf​ = 6%no debtequityβU​ = 0.7D/E = 0.5equityre​ = ?debtT = 25%

A company's business has an unlevered (asset) beta of 0.7. It plans a debt-to-equity ratio of 0.5. The tax rate is 25%, the risk-free rate is 1% and the market risk premium is 6%. Find the cost of equity re at the new capital structure.

βU = 0.7D/E = 0.5T = 25%rf = 1%rm − rf = 6%
re = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 11, 14, 21, 53, 20, 46, 12.

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

A quantity is normally distributed with mean 50 and standard deviation 8. What proportion P of values is above 58?

μ = 50σ = 8x = 58
P =
11.ProbabilityMedium
xp(x)10.120.330.340.3μ = 2.8

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4p(x): 0.1, 0.3, 0.3, 0.3
σ =
12.ProbabilityHard
n = 6, p = 0.25: P(X = 1) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 1 correct answers?

n = 6p = 0.25k = 1
P =
13.Normal distributionMedium
200 mm200.8 mmX + Y?

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 200.8 mm. Round z to two decimals.

μX = 120, σX = 0.3μY = 80, σY = 0.4X + Y > 200.8 mm
P =
14.Normal distributionHard
380420460500540580620hbottom 30%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Drag the cut-off so the shaded bottom 30% is to the left of it. Find the cut-off x.

μ = 500 hσ = 40 h30%
x = h
15.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E90%

The battery life of two phone models is compared. Model 1: 100 phones, mean 40 h, known standard deviation 13 h. Model 2: 80 phones, mean 30.5 h, known standard deviation 10 h. Find the upper limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 100x̄₁ = 40 hσ₁ = 13 hn₂ = 80x̄₂ = 30.5 hσ₂ = 10 h90%
upper limit = h
16.Confidence intervalsHard
σ = 38, n = ?x̄ − Ex̄ + Ex̄E = 290%

The population standard deviation is 38. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 2?

σ = 38E = 290%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 124.7 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 124.7s = 20n = 81
t =
18.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2α = 0.05βσ = 10, power 95%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 95%. The standard deviation is 10. Find the minimum sample size n.

α = 0.05power = 95%δ = 2σ = 10zα = 1.645zβ = 1.645
n =
19.Linear regression & correlationMedium
xy2484486578611072xy510506070

The table gives 5 paired observations of x = hours studied and y = exam score. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 286Σx² = 220Σy² = 16762Σxy = 1838
r =
20.Linear regression & correlationHard
xy153318502163247127783097xy15202530406080100x = 16.5

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. A new observation (16.5, 47) is made. Find its residual e = y − ŷ.

n = 6(16.5, 47)
e =

3Solutions

1.Time value of money

Answer: EAR = 25.44 %

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 = 0.24/2 = 0.12

Compound over one year

EAR = (1 + 0.12)2 − 1 = 1.254 − 1 = 25.44%

Answer

EAR = 25.44%

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

2.Time value of money

Answer: B = 119086 €

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 = 0.0045, N = 240
A = 250 000·0.0045/(1 − (1 + 0.0045)−240) = €1 705.63

Present value of the remaining payments

Whatever is still owed must be exactly paid off by the remaining payments.

N − k = 240 − 156 = 84
B = 1706·(1 − (1 + 0.0045)−84)/0.0045 = €119 086.34

Answer

B = €119 086.34
3.NPV & IRR

Answer: NPV = −17467 €

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 + 0.15)1 = 10 000/1.15 = 8 695.65
CF2/(1 + 0.15)2 = 11 000/1.322 = 8 317.58
CF3/(1 + 0.15)3 = 16 000/1.521 = 10 520.26

Add them up

NPV = €−17 466.51

Answer

NPV = €−17 466.51

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

4.NPV & IRR

Answer: IRR = 7.931 %

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.

Required annuity factor

(1 − (1 + i)−5)/i = 20 000/5 000 = 4

Solve for the rate

The annuity factor falls as the rate rises; bracket the rate and narrow down.

i = 7%: 4.1
i = 8%: 3.993
IRR = 7.931%

Answer

IRR = 7.931%
5.Bond pricing & yield

Answer: ΔP = 28.25 €

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 = €40.00, i = 0.09, N = 13
€40.00·(1 − 0.3262)/0.09 + €1 000·0.3262 = €299.48 + €326.18 = €625.65

Price at the new yield

C = €40.00, i = 0.085, N = 13
€40.00·(1 − 0.3463)/0.085 + €1 000·0.3463 = €307.64 + €346.27 = €653.91

Change

ΔP = 653.91 − 625.65 = €28.25

Answer

ΔP = €28.25
6.Bond pricing & yield

Answer: YTM = 4.474 %

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

€1 090 = €60·(1 − (1 + y)⁻⁷)/y + €1 000/(1 + y)⁷

Solve numerically

P(4%) = €1 120.04, P(5%) = €1 057.86
YTM = 4.474%

Answer

YTM = 4.474%
7.CAPM & WACC

Answer: re = 12 %

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 = €9 000 000, E/V = 0.5556, D/V = 0.4444

Debt part of the WACC

0.4444·4.5%·(1 − 0.258) = 1.484%

Cost of equity

re = (8.15% − 1.484%)/0.5556 = 12%

Answer

re = 12%
8.CAPM & WACC

Answer: re = 6.775 %

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 = 0.7·(1 + 0.75·0.5) = 0.9625

Cost of equity (CAPM)

re = 1% + 0.9625·6% = 6.775%

Answer

re = 6.775%
9.Descriptive statistics

Answer: s = 17.09

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 = 177/7 = 25.29

Squared deviations

Σ(x − x̄)² = (−14.29)² + (−11.29)² + (−4.286)² + 27.71² + (−5.286)² + 20.71² + (−13.29)² = 1751

Standard deviation

s = √(1751/6) = 17.09

Answer

s = 17.09
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) = 0.8413
P = 1 − 0.8413 = 0.1587

Answer

P = 0.1587
11.Probability

Answer: σ = 0.9798

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 = 1·0.1 + 2·0.3 + 3·0.3 + 4·0.3 = 2.8

Mean of the squares

E(X²) = 1·0.1 + 4·0.3 + 9·0.3 + 16·0.3 = 8.8

Variance

Var(X) = 8.8 − 2.8² = 0.96
σ = √0.96 = 0.9798

Answer

σ = 0.9798
12.Probability

Answer: P = 0.356

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 = 1) = 6·0.251·0.755 = 6·0.25·0.2373 = 0.356

Answer

P = 0.356
13.Normal distribution

Answer: P = 0.0548

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (200.8 − 200)/0.5 = 1.6

Upper tail

P = 1 − Φ(1.6) = 1 − 0.9452 = 0.0548

Answer

P = 0.0548
14.Normal distribution

Answer: x = 479 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) = 0.3

z from the table

z ≈ −0.52

Back to the original units

x = 500 − 0.52·40 = 479 h

Answer

x = 479 h
15.Confidence intervals

Answer: upper limit = 12.32 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 = 40 − 30.5 = 9.5

Standard error

SE = √(13²/100 + 10²/80) = 1.715

Margin of error

E = 1.645·1.715 = 2.821

Interval

9.5 ± 2.821 → [6.679, 12.32]

Answer

upper limit = 12.32 h

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

16.Confidence intervals

Answer: n = 977

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 = (1.645·38/2)² = 976.9

Round up

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

n = 977

Answer

n = 977
17.Hypothesis tests & confidence intervals

Answer: t = 2.115

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/√81 = 20/9 = 2.222

Test statistic

t = (124.7 − 120)/2.222 = 2.115

Answer

t = 2.115
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 271

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 + 1.645)·10/2)² = 270.6

Round up

n = 271

Answer

n = 271
19.Linear regression & correlation

Answer: r = 0.9611

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 = 1838 − 30·286/5 = 122
Sxx = 220 − 30²/5 = 40
Syy = 16762 − 286²/5 = 402.8

Correlation coefficient

r = 122/√(40·402.8) = 0.9611

Answer

r = 0.9611
20.Linear regression & correlation

Answer: e = 5.21

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 = 135/6 = 22.5
ȳ = 392/6 = 65.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 618

Regression line

b = 618/157.5 = 3.924
a = 65.33 − 3.924·22.5 = −22.95
ŷ = −22.95 + 3.924·x

Predicted value

ŷ = −22.95 + 3.924·16.5 = 41.79

Residual

e = 47 − 41.79 = 5.21

Answer

e = 5.21

The point lies above the line.

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