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
years0123456PV = ?€20 000

You need €20 000 in 6 years. An account pays 3% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €20 0003% / 4n = 6
PV = €
2.Time value of moneyHard
months0123…204A?AA€10 000

You want €10 000 in 17 years. You save the same amount at the end of every month in an account paying 4.2% per year, compounded monthly. Find the monthly deposit A.

FV = €10 0004.2% / 12N = 204
A = €
3.NPV & IRRMedium
years0123…10€80 000€4 000€4 000€4 000€4 000r = 12%, EAC = ?

A machine costs €80 000 today, lasts 10 years and costs €4 000 per year to run (paid at the end of each year). The cost of capital is 12%. Find its equivalent annual cost EAC.

C0 = €80 000M = €4 000n = 10r = 12%
EAC = €
4.NPV & IRRHard
years0123456−20 0004 5004 5004 5004 5004 5004 500

An investment of €20 000 returns €4 500 at the end of each year for 6 years. Find its internal rate of return IRR in percent.

C0 = €20 000CF = €4 500n = 6
IRR = %
5.Bond pricing & yieldMedium
half-years0123…16€25€25€25 + €1 000P = ?

A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 8 years. The yield to maturity is 3.5% per year. Find the price P of the bond.

F = €1 000coupon = 5%YTM = 3.5%n = 8
P = €
6.Bond pricing & yieldHard
years01234€1 188€80€80€80 + €1 080

A bond with 10 years to maturity (face value €1 000, annual coupon 8%) sells for €1 188. The issuer can call it in 4 years at a call price of €1 080. Find the yield to call YTC in percent.

F = €1 000coupon = 8%P = €1 188call price = €1 080call in 4 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€5 000 000re​ = 12%debt€2 000 000rd​ = 3%T = 30%

A company has equity worth €5 000 000 and debt worth €2 000 000. Its cost of equity is 12%, its pre-tax cost of debt is 3% and the tax rate is 30%. Find the WACC.

E = €5 000 000D = €2 000 000re = 12%rd = 3%T = 30%
WACC = %
8.CAPM & WACCHard
rf​ = 3%, rm​ − rf​ = 4.5%D/E = 0.2equityβL​ = 1.5debtD/E = 1equityre​ = ?debtT = 25%

A company's shares have a beta of 1.5 at its current debt-to-equity ratio of 0.2. It plans to raise its debt-to-equity ratio to 1. The tax rate is 25%, the risk-free rate 3% and the market risk premium 4.5%. Find the cost of equity re after the change.

βL = 1.5D/E1 = 0.2D/E2 = 1T = 25%rf = 3%rm − rf = 4.5%
re = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 52, 18, 58, 27, 38.

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

Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 75.

μ = 70σ = 5x = 75
z =
11.ProbabilityMedium
red: 3blue: 9

A bag holds 3 red and 9 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

3 / 9
P =
12.ProbabilityHard
n = 8, p = 0.6: P(X ≤ 1) = ?k012345678

8 components are tested, and each one fails with probability 0.6, independently. What is the probability P that at most one fails?

n = 8p = 0.6
P =
13.Normal distributionMedium
192 kg198.4 kgT = X₁ + … + X₁₆?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 16 of them is below 198.4 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 16< 198.4 kg
P =
14.Normal distributionHard
0.05 mm0D − d?

Hole diameters D are normal with mean 25.05 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.05 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm
P =
15.Confidence intervalsMedium
n = 15, sd​ = 4.10LUd̄ = 9.2E = ?95%

The blood pressure of 15 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean 9.2 and standard deviation 4.1. Find the margin of error E of the 95% confidence interval for the mean difference.

n = 15d̄ = 9.2sd = 4.195%t(14) = 2.145
E = mmHg
16.Confidence intervalsHard
σ = 19, n = ?x̄ − Ex̄ + Ex̄E = 1.595%

The population standard deviation is 19. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 1.5?

σ = 19E = 1.595%
n =
17.Hypothesis tests & confidence intervalsMedium
−1.52p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 20. A random sample of 64 observations has a mean of 46.2. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.

μ₀ = 50σ = 20n = 64x̄ = 46.2H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.05 with a sample of 36; the standard deviation is known to be 12. In reality the mean is 94.6. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 94.6σ = 12n = 36α = 0.05
1 − β = %
19.Linear regression & correlationMedium
xy25486981310191219xy5105101520

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy124219322417517614xy246152025ŷ = 18

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 18?

n = 6ŷ = 18
x =

3Solutions

1.Time value of money

Answer: PV = 16717 €

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 = 0.03/4 = 0.0075
N = 4·6 = 24

Present value

PV = FV/(1 + i)N = 20 000/1.196 = €16 716.63

Answer

PV = €16 716.63
2.Time value of money

Answer: A = 33.67 €

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 = 0.042/12 = 0.0035
N = 12·17 = 204

Deposit

A = FV·i/((1 + i)N − 1) = 10 000·0.0035/(2.04 − 1) = €33.67

Answer

A = €33.67
3.NPV & IRR

Answer: EAC = 18159 €

Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.

Annuity factor

(1 − 1.12−10)/0.12 = 5.65

Annualised purchase price

80 000/5.65 = 14 158.73

EAC

EAC = 14 158.73 + 4 000 = €18 158.73

Answer

EAC = €18 158.73
4.NPV & IRR

Answer: IRR = 9.312 %

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)−6)/i = 20 000/4 500 = 4.444

Solve for the rate

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

i = 9%: 4.486
i = 10%: 4.355
IRR = 9.312%

Answer

IRR = 9.312%
5.Bond pricing & yield

Answer: P = 1104 €

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 = €25.00, i = 0.0175, N = 16

Present value of the coupons

€25.00·(1 − 0.7576)/0.0175 = €346.26

Present value of the face value

€1 000·0.7576 = €757.62

Price

P = €1 103.88

Answer

P = €1 103.88

The yield is below the coupon rate, so the bond sells at a premium.

6.Bond pricing & yield

Answer: YTC = 4.613 %

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

1188 = 80·(1 − (1 + y)−4)/y + 1080/(1 + y)4

Solve numerically

y = 4%: €1 213.58
y = 5%: €1 172.19
YTC = 4.613%

Answer

YTC = 4.613%

The yield to maturity is 5.505%. Investors in a premium bond expect it to be called, so the lower yield to call is the more realistic one.

7.CAPM & WACC

Answer: WACC = 9.171 %

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 = €7 000 000, E/V = 0.7143, D/V = 0.2857

WACC

WACC = 0.7143·12% + 0.2857·3%·(1 − 0.3) = 9.171%

Answer

WACC = 9.171%
8.CAPM & WACC

Answer: re = 13.27 %

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.

Unlever the beta

βU = 1.5/(1 + 0.75·0.2) = 1.304

Relever at the new D/E

βL = 1.304·(1 + 0.75·1) = 2.283

Cost of equity (CAPM)

re = 3% + 2.283·4.5% = 13.27%

Answer

re = 13.27%
9.Descriptive statistics

Answer: s = 16.7

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 = 193/5 = 38.6

Squared deviations

Σ(x − x̄)² = 13.4² + (−20.6)² + 19.4² + (−11.6)² + (−0.6)² = 1115

Standard deviation

s = √(1115/4) = 16.7

Answer

s = 16.7
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 = (75 − 70)/5 = 1

Answer

z = 1
11.Probability

Answer: P = 0.04545

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.

First and second draw

P = 3/12 · 2/11 = 6/132 = 0.04545

Answer

P = 0.04545
12.Probability

Answer: P = 0.00852

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.

Zero and one failure

P(X = 0) = 0.48 = 0.0006554
P(X = 1) = 8·0.6·0.47 = 0.007864

Add them

P = 0.0006554 + 0.007864 = 0.00852

Answer

P = 0.00852
13.Normal distribution

Answer: P = 0.9772

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 = 192 kg, σT = 0.8·√16 = 3.2 kg

Standardise

z = (198.4 − 192)/3.2 = 2

Probability

P = Φ(2) = 0.9772

Answer

P = 0.9772
14.Normal distribution

Answer: P = 0.9616

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 = 25.05 − 25 = 0.05 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.05)/0.02828 = −1.77

Probability

P = 1 − Φ(−1.77) = 0.9616

Answer

P = 0.9616
15.Confidence intervals

Answer: E = 2.271 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 = 14, t = 2.145

Margin of error

E = 2.145·4.1/√15 = 2.271

Answer

E = 2.271 mmHg
16.Confidence intervals

Answer: n = 617

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 = (1.96·19/1.5)² = 616.4

Round up

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

n = 617

Answer

n = 617
17.Hypothesis tests & confidence intervals

Answer: p = 0.06426

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.

Test statistic

SE = σ/√n = 20/√64 = 2.5
z = (46.2 − 50)/2.5 = −1.52

Tail area

Φ(1.52) = 0.9357
1 − Φ(1.52) = 0.0643

p-value

p = 0.06426

Answer

p = 0.06426

p > 0.05, so H₀ is not rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 85.43 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 12/√36 = 2
x̄c = 100 − 1.645·2 = 96.71

z under the true mean

z = (96.71 − 94.6)/2 = 1.055

Power

power = P(Z < 1.055) = 85.43%
1 − β = 85.43%

Answer

1 − β = 85.43%
19.Linear regression & correlation

Answer: r = 0.9728

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 = 42/6 = 7
ȳ = 73/6 = 12.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 107
Syy = Σ(y − ȳ)² = 172.8

Correlation coefficient

r = 107/√(70·172.8) = 0.9728

Answer

r = 0.9728
20.Linear regression & correlation

Answer: x = 3.978

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 = 21/6 = 3.5
ȳ = 113/6 = 18.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −30.5

Regression line

b = −30.5/17.5 = −1.743
a = 18.83 − (−1.743)·3.5 = 24.93
ŷ = 24.93 − 1.743·x

Solve for x

x = (18 − 24.93)/(−1.743) = 3.978

Answer

x = 3.978

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