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
periods0123…1211 + EAR = ?17.5% / 12

A credit card charges a nominal interest rate of 17.5% per year, compounded monthly. Find the effective annual rate EAR in percent.

r = 17.5%m = 12
EAR = %
2.Time value of moneyHard
months01…36…360€250 000AB = ?A2.4% / 12

A mortgage of €250 000 is repaid with equal monthly payments over 30 years at 2.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 3? Find the outstanding balance B.

P = €250 0002.4% / 12N = 360k = 36
B = €
3.NPV & IRRMedium
years01234−75 00021 00026 00029 00020 000

An investment of €75 000 gives these cash flows at the end of years 1 to 4: €21 000, €26 000, €29 000, €20 000. The cost of capital is 5%. Find the net present value NPV.

r = 5%CF0 = −75 000CF1 = 21 000CF2 = 26 000CF3 = 29 000CF4 = 20 000
NPV = €
4.NPV & IRRHard
years012−50 00032 50023 000

A project costs €50 000 and returns €32 500, €23 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −50 000CF1 = 32 500CF2 = 23 000
IRR = %
5.Bond pricing & yieldMedium
years0123…12P€20€20€20 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 2% and matures in 12 years. Its yield to maturity changes from 5.5% to 7.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 2%n = 12y1 = 5.5%y2 = 7.5%
ΔP = €
6.Bond pricing & yieldHard
years0123€1 310€50€50€50 + €1 050

A bond with 15 years to maturity (face value €1 000, annual coupon 5%) sells for €1 310. The issuer can call it in 3 years at a call price of €1 050. Find the yield to call YTC in percent.

F = €1 000coupon = 5%P = €1 310call price = €1 050call in 3 years
YTC = %
7.CAPM & WACCMedium
βr (%)SMLrf​ = 3.5%β = 1rm​ = ?β = 1.6re​ = 10.7%

A share with a beta of 1.6 has a required return of 10.7%. The risk-free rate is 3.5%. What expected market return rm does CAPM imply?

β = 1.6re = 10.7%rf = 3.5%
rm = %
8.CAPM & WACCHard
rf​ = 3%, rm​ − rf​ = 5.5%D/E = 0.4equityβL​ = 1.4debtD/E = 1.5equityre​ = ?debtT = 21%

A company's shares have a beta of 1.4 at its current debt-to-equity ratio of 0.4. It plans to raise its debt-to-equity ratio to 1.5. The tax rate is 21%, the risk-free rate 3% and the market risk premium 5.5%. Find the cost of equity re after the change.

βL = 1.4D/E1 = 0.4D/E2 = 1.5T = 21%rf = 3%rm − rf = 5.5%
re = %
9.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 46, 41, 51, 55, 20, 57.

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

Exam scores are normally distributed with mean 70 and standard deviation 15. Find the z-score z of a score of 79.

μ = 70σ = 15x = 79
z =
11.ProbabilityMedium
xp(x)100.35200.25300.05400.35μ = 24

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

x: 10, 20, 30, 40p(x): 0.35, 0.25, 0.05, 0.35
Var(X) =
12.ProbabilityHard
€p(x)−10.8930.1190.01

A game costs €1 to play. You win €4 with probability 0.1 and €20 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€4: p = 0.1€20: p = 0.01
E(G) = €
13.Normal distributionMedium
27.241.3P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 27.2 and 41.3 min? Give P as a decimal.

μ = 32σ = 6x1 = 27.2x2 = 41.3
P =
14.Normal distributionHard
35455565758595pointstop 40%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 40% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points40%
x = points
15.Confidence intervalsMedium
n = 7, s = 3LUx̄ = 52E = ?95%

A sample of 7 items has mean 52 and standard deviation s = 3. The population is roughly normal. Find the margin of error E of a 95% confidence interval for the mean.

n = 7x̄ = 52s = 395%t(6) = 2.447
E =
16.Confidence intervalsHard
σ = 37, n = ?x̄ − Ex̄ + Ex̄E = 195%

The population standard deviation is 37. 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?

σ = 37E = 195%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 499 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 499s = 8n = 49
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 50 against H₁: μ > 50 is done at α = 0.10 with a sample of 100; the standard deviation is known to be 5. In reality the mean is 50.6. Find the probability of a Type II error β in percent.

μ₀ = 50μ = 50.6σ = 5n = 100α = 0.10
β = %
19.Linear regression & correlationMedium
xy363472574678790892xy468708090

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy284186238321039xy51010203040x = 3

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 = 3.

n = 5x = 3
ŷ =

3Solutions

1.Time value of money

Answer: EAR = 18.97 %

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 = 0.175/12 = 0.01458

Compound over one year

EAR = (1 + 0.01458)12 − 1 = 1.19 − 1 = 18.97%

Answer

EAR = 18.97%

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

2.Time value of money

Answer: B = 232293 €

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 = 0.002, N = 360
A = 250 000·0.002/(1 − (1 + 0.002)−360) = €974.85

Present value of the remaining payments

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

N − k = 360 − 36 = 324
B = 974.9·(1 − (1 + 0.002)−324)/0.002 = €232 293.17

Answer

B = €232 293.17
3.NPV & IRR

Answer: NPV = 10088 €

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 + 0.05)1 = 21 000/1.05 = 20 000.00
CF2/(1 + 0.05)2 = 26 000/1.103 = 23 582.77
CF3/(1 + 0.05)3 = 29 000/1.158 = 25 051.29
CF4/(1 + 0.05)4 = 20 000/1.216 = 16 454.05

Add them up

NPV = €10 088.11

Answer

NPV = €10 088.11

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

4.NPV & IRR

Answer: IRR = 7.708 %

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.

Bracket the IRR

NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.

NPV(5%) = 1 814.06
NPV(10%) = −1 446.28

Narrow down

Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.

IRR ≈ 5 + 5·1 814.06/(1 814.06 − (−1 446.28)) = 7.782%
exact: NPV(7.708%) = 0 → IRR = 7.708%

Answer

IRR = 7.708%
5.Bond pricing & yield

Answer: ΔP = −123.8 €

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 = 0.055, N = 12
€20.00·(1 − 0.526)/0.055 + €1 000·0.526 = €172.37 + €525.98 = €698.35

Price at the new yield

C = €20.00, i = 0.075, N = 12
€20.00·(1 − 0.4199)/0.075 + €1 000·0.4199 = €154.71 + €419.85 = €574.56

Change

ΔP = 574.56 − 698.35 = −€123.79

Answer

ΔP = −€123.79
6.Bond pricing & yield

Answer: YTC = 50 %

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

1310 = 50·(1 − (1 + y)−3)/y + 1050/(1 + y)3

Solve numerically

y = 50%: €381.48
y = 50%: €381.48
YTC = 50%

Answer

YTC = 50%

The yield to maturity is 2.497%. 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: 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 = (10.7% − 3.5%)/1.6 = 4.5%

Market return

rm = 3.5% + 4.5% = 8%

Answer

rm = 8%
8.CAPM & WACC

Answer: re = 15.78 %

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 = 1.4/(1 + 0.79·0.4) = 1.064

Relever at the new D/E

βL = 1.064·(1 + 0.79·1.5) = 2.324

Cost of equity (CAPM)

re = 3% + 2.324·5.5% = 15.78%

Answer

re = 15.78%
9.Descriptive statistics

Answer: s = 13.58

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 = 270/6 = 45

Squared deviations

Σ(x − x̄)² = 1² + (−4)² + 6² + 10² + (−25)² + 12² = 922

Standard deviation

s = √(922/5) = 13.58

Answer

s = 13.58
10.Descriptive statistics

Answer: z = 0.6

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 = (79 − 70)/15 = 0.6

Answer

z = 0.6
11.Probability

Answer: Var(X) = 164

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 = 10·0.35 + 20·0.25 + 30·0.05 + 40·0.35 = 24

Mean of the squares

E(X²) = 100·0.35 + 400·0.25 + 900·0.05 + 1600·0.35 = 740

Variance

Var(X) = 740 − 24² = 164

Answer

Var(X) = 164
12.Probability

Answer: E(G) = −0.4 €

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.

Net outcomes

G: 3 (p = 0.1), 19 (p = 0.01), −1 (p = 0.89)

Expected value

E(G) = 3·0.1 + 19·0.01 − 1·0.89 = −0.4 €

Answer

E(G) = −0.4 €

A negative value means you lose money on average.

13.Normal distribution

Answer: P = 0.7276

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 = (27.2 − 32)/6 = −0.8
z₂ = (41.3 − 32)/6 = 1.55

Area between

P = Φ(1.55) − Φ(−0.8) = 0.9394 − 0.2119 = 0.7276

Answer

P = 0.7276
14.Normal distribution

Answer: x = 67.5 points

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) = 0.6

z from the table

z ≈ 0.25

Back to the original units

x = 65 + 0.25·10 = 67.5 points

Answer

x = 67.5 points
15.Confidence intervals

Answer: E = 2.775

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 = 6, t = 2.447

Margin of error

E = 2.447·3/√7 = 2.775

Answer

E = 2.775
16.Confidence intervals

Answer: n = 5260

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 = (1.96·37/1)² = 5259

Round up

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

n = 5260

Answer

n = 5260
17.Hypothesis tests & confidence intervals

Answer: t = −0.875

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 = 8/√49 = 8/7 = 1.143

Test statistic

t = (499 − 500)/1.143 = −0.875

Answer

t = −0.875
t0.025; df = 48 = 2.011

|t| < 2.011, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: β = 53.27 %

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 = 5/√100 = 0.5
x̄c = 50 + 1.282·0.5 = 50.64

z under the true mean

z = (50.64 − 50.6)/0.5 = 0.082

Type II error

power = P(Z > 0.082) = 46.73%
β = 100% − 46.73% = 53.27%

Answer

β = 53.27%
19.Linear regression & correlation

Answer: r² = 0.9544

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 = 33/6 = 5.5
ȳ = 469/6 = 78.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 101.5
Syy = Σ(y − ȳ)² = 616.8

Coefficient of determination

r² = 101.5²/(17.5·616.8) = 0.9544

Answer

r² = 0.9544

About 95.44% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 12.6

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 = 120/5 = 24

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 152

Regression line

b = 152/40 = 3.8
a = 24 − 3.8·6 = 1.2
ŷ = 1.2 + 3.8·x

Prediction

ŷ = 1.2 + 3.8·3 = 12.6

Answer

ŷ = 12.6

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