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…36511 + EAR = ?22% / 365

A credit card charges a nominal interest rate of 22% per year, compounded daily. Find the effective annual rate EAR in percent.

r = 22%m = 365
EAR = %
2.Time value of moneyHard
months01…132…300€250 000AB = ?A5.4% / 12

A mortgage of €250 000 is repaid with equal monthly payments over 25 years at 5.4% per year, compounded monthly. How much is still owed just after the payment at the end of year 11? Find the outstanding balance B.

P = €250 0005.4% / 12N = 300k = 132
B = €
3.NPV & IRRMedium
years0123−75 00031 00037 00017 000

An investment of €75 000 gives these cash flows at the end of years 1 to 3: €31 000, €37 000, €17 000. The cost of capital is 9%. Find the net present value NPV.

r = 9%CF0 = −75 000CF1 = 31 000CF2 = 37 000CF3 = 17 000
NPV = €
4.NPV & IRRHard
years0123456−150 00027 500+5%+5%+5%+5%+5%r = 10%, NPV = ?

A project costs €150 000 today. Its cash flow is €27 500 at the end of year 1 and grows by 5% per year until the last cash flow at the end of year 6. The cost of capital is 10%. Find the net present value NPV.

C0 = €150 000CF1 = €27 500g = 5%n = 6r = 10%
NPV = €
5.Bond pricing & yieldMedium
half-years0123…10€15€15€15 + €1 000P = ?

A bond with a face value of €1 000 pays a 3% annual coupon in two half-yearly payments and matures in 5 years. The yield to maturity is 2.5% per year. Find the price P of the bond.

F = €1 000coupon = 3%YTM = 2.5%n = 5
P = €
6.Bond pricing & yieldHard
years0123456€925€50€50 + €1 000

A bond with a face value of €1 000 and an annual coupon of 5% matures in 6 years and sells for €925. Find its yield to maturity YTM in percent.

F = €1 000coupon = 5%n = 6P = €925
YTM = %
7.CAPM & WACCMedium
βr (%)SMLrf​ = 3.5%β = 1rm​ = ?β = 1.25re​ = 9.75%

A share with a beta of 1.25 has a required return of 9.75%. The risk-free rate is 3.5%. What expected market return rm does CAPM imply?

β = 1.25re = 9.75%rf = 3.5%
rm = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €25.00β = 0.8rf​ = 3.5%, rm​ = 10.5%debt20 000 × €980rd​ = 4%T = 30%

A company has 2 500 000 shares at €25.00 each and 20 000 bonds trading at €980 each. The pre-tax cost of debt is 4% and the tax rate is 30%. The share’s beta is 0.8, the risk-free rate 3.5% and the market return 10.5%. Find the WACC.

β = 0.8rf = 3.5%rm = 10.5%rd = 4%T = 30%
WACC = %
9.Descriptive statisticsMedium
20253035404550

Find the sample standard deviation s of the data: 34, 18, 40, 53, 42, 37, 45.

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

A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is above 166?

μ = 170σ = 8x = 166
P =
11.ProbabilityMedium
P(at least one six) = ??1?2?3?4?5P(six) = 1/6 on each roll

A fair die is rolled 5 times. What is the probability P of getting at least one six?

n = 5
P =
12.ProbabilityHard
xp(x)10.220.130.0540.350.35μ

X has the distribution shown, and Y = 2X − 3. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.2, 0.1, 0.05, 0.3, 0.35Y = 2X − 3
σY =
13.Normal distributionMedium
150 kg148.2 kgT = X₁ + … + X₆?

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 6 of them exceeds 148.2 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 6> 148.2 kg
P =
14.Normal distributionHard
235240245250255260265gtop 10%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded top 10% is to the right of it. Find the cut-off x.

μ = 250 gσ = 5 g10%
x = g
15.Confidence intervalsMedium
n = 12, s = 12LUx̄ = 72E = ?99%

A sample of 12 items has mean 72 and standard deviation s = 12. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 12x̄ = 72s = 1299%t(11) = 3.106
E =
16.Confidence intervalsHard
p ≈ 0.8, n = ?p̂ − Ep̂ + Ep̂E = 0.0495%

A pilot study suggests that about 0.8 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.04?

p ≈ 0.8E = 0.0495%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 124.2 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 124.2s = 8n = 16
t =
18.Hypothesis tests & confidence intervalsHard
σ = 20, n = ?x̄ − Ex̄ + Ex̄E = 0.590%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 0.5? The standard deviation is 20. Find the smallest sample size n.

σ = 20E = 0.590%
n =
19.Linear regression & correlationMedium
xy23486108151015xy51051015

The table gives 5 paired observations of x = load in kN and y = extension in mm. Find Pearson's correlation coefficient r.

n = 5
r =
20.Linear regression & correlationHard
xy264761081710221222xy5105101520ŷ = 18

The table gives 6 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 = 6ŷ = 18
x =

3Solutions

1.Time value of money

Answer: EAR = 24.6 %

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 = 0.22/365 = 0.0006027

Compound over one year

EAR = (1 + 0.0006027)365 − 1 = 1.246 − 1 = 24.6%

Answer

EAR = 24.6%

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

2.Time value of money

Answer: B = 178946 €

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 = 300
A = 250 000·0.0045/(1 − (1 + 0.0045)−300) = €1 520.32

Present value of the remaining payments

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

N − k = 300 − 132 = 168
B = 1520·(1 − (1 + 0.0045)−168)/0.0045 = €178 946.34

Answer

B = €178 946.34
3.NPV & IRR

Answer: NPV = −2290 €

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 = 37 000/1.188 = 31 142.16
CF3/(1 + 0.09)3 = 17 000/1.295 = 13 127.12

Add them up

NPV = €−2 290.35

Answer

NPV = €−2 290.35

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

4.NPV & IRR

Answer: NPV = −16047 €

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 the growing annuity

((1.05)/(1.1))6 = 0.7564
PV = 27 500/(0.1 − 0.05)·(1 − 0.7564) = 133 953.02

NPV

NPV = 133 953.02 − 150 000 = −€16 046.98

Answer

NPV = −€16 046.98

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

5.Bond pricing & yield

Answer: P = 1023 €

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 = €15.00, i = 0.0125, N = 10

Present value of the coupons

€15.00·(1 − 0.8832)/0.0125 = €140.18

Present value of the face value

€1 000·0.8832 = €883.18

Price

P = €1 023.36

Answer

P = €1 023.36

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

6.Bond pricing & yield

Answer: YTM = 6.552 %

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

€925 = €50·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶

Solve numerically

P(6%) = €950.83, P(7%) = €904.67
YTM = 6.552%

Answer

YTM = 6.552%
7.CAPM & WACC

Answer: rm = 8.5 %

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 = (9.75% − 3.5%)/1.25 = 5%

Market return

rm = 3.5% + 5% = 8.5%

Answer

rm = 8.5%
8.CAPM & WACC

Answer: WACC = 7.596 %

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 = 2 500 000·€25.00 = €62 500 000
D = 20 000·€980 = €19 600 000

Cost of equity (CAPM)

re = 3.5% + 0.8·7% = 9.1%

Weights

V = €82 100 000, E/V = 0.7613, D/V = 0.2387

WACC

WACC = 0.7613·9.1% + 0.2387·4%·(1 − 0.3) = 7.596%

Answer

WACC = 7.596%
9.Descriptive statistics

Answer: s = 10.88

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 = 269/7 = 38.43

Squared deviations

Σ(x − x̄)² = (−4.429)² + (−20.43)² + 1.571² + 14.57² + 3.571² + (−1.429)² + 6.571² = 709.7

Standard deviation

s = √(709.7/6) = 10.88

Answer

s = 10.88
10.Descriptive statistics

Answer: P = 0.6915

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 = −0.5

Standard normal table

Φ(−0.5) = 0.3085
P = 1 − 0.3085 = 0.6915

Answer

P = 0.6915
11.Probability

Answer: P = 0.5981

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.

Complement

P(no six) = (5/6)5 = 0.4019
P = 1 − 0.4019 = 0.5981

Answer

P = 0.5981
12.Probability

Answer: σY = 3.066

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 = 3.5, Var(X) = 2.35, σX = 1.533

Linear change

σY = |2|·1.533 = 3.066

Answer

σY = 3.066

The mean does move: E(Y) = 2·3.5 − 3 = 4.

13.Normal distribution

Answer: P = 0.9292

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 = 150 kg, σT = 0.5·√6 = 1.225 kg

Standardise

z = (148.2 − 150)/1.225 = −1.47

Probability

P = 1 − Φ(−1.47) = 0.9292

Answer

P = 0.9292
14.Normal distribution

Answer: x = 256.4 g

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) = 0.9

z from the table

z ≈ 1.28

Back to the original units

x = 250 + 1.28·5 = 256.4 g

Answer

x = 256.4 g
15.Confidence intervals

Answer: E = 10.76

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 = 11, t = 3.106

Margin of error

E = 3.106·12/√12 = 10.76

Answer

E = 10.76
16.Confidence intervals

Answer: n = 385

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 = 1.96²·0.8·0.2/0.04² = 384.2

Round up

n = 385

Answer

n = 385
17.Hypothesis tests & confidence intervals

Answer: t = 2.1

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/√16 = 8/4 = 2

Test statistic

t = (124.2 − 120)/2 = 2.1

Answer

t = 2.1
t0.025; df = 15 = 2.131

|t| < 2.131, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 4330

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).

Solve for n

n = (1.645·20/0.5)² = 4330

Round up

n = 4330

Answer

n = 4330
19.Linear regression & correlation

Answer: r = 0.9669

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 = 51/5 = 10.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 62
Syy = Σ(y − ȳ)² = 102.8

Correlation coefficient

r = 62/√(40·102.8) = 0.9669

Answer

r = 0.9669
20.Linear regression & correlation

Answer: x = 9.121

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 = 84/6 = 14

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 132

Regression line

b = 132/70 = 1.886
a = 14 − 1.886·7 = 0.8
ŷ = 0.8 + 1.886·x

Solve for x

x = (18 − 0.8)/1.886 = 9.121

Answer

x = 9.121

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