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
years0123€20 000FV = ?

You invest €20 000 at a nominal rate of 4% per year, compounded monthly. Find the value FV of the investment after 3 years.

PV = €20 0004% / 12n = 3
FV = €
2.Time value of moneyHard
months0123…324€100€100€100FV = ?

You save €100 at the end of every month for 27 years. The account pays 3.6% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €1003.6% / 12N = 324
FV = €
3.NPV & IRRMedium
years01234−30 00014 00011 00010 00013 000

An investment of €30 000 gives these cash flows at the end of years 1 to 4: €14 000, €11 000, €10 000, €13 000. The cost of capital is 6%. Find the net present value NPV.

r = 6%CF0 = −30 000CF1 = 14 000CF2 = 11 000CF3 = 10 000CF4 = 13 000
NPV = €
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €40 000 and returns €21 500, €16 000, €9 500 at the end of years 1 to 3. Project B costs €45 000 and returns €13 000, €18 500, €23 000. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −40 000, 21 500, 16 000, 9 500B: −45 000, 13 000, 18 500, 23 000
r* = %
5.Bond pricing & yieldMedium
half-years0123…14€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 7 years. The yield to maturity is 5.5% per year. Find the price P of the bond.

F = €1 000coupon = 5%YTM = 5.5%n = 7
P = €
6.Bond pricing & yieldHard
years0123456€1 233€60€60€60 + €1 060

A bond with 16 years to maturity (face value €1 000, annual coupon 6%) sells for €1 233. The issuer can call it in 6 years at a call price of €1 060. Find the yield to call YTC in percent.

F = €1 000coupon = 6%P = €1 233call price = €1 060call in 6 years
YTC = %
7.CAPM & WACCMedium
βr (%)SMLrf​ = 4%β = 1rm​ = ?β = 0.6re​ = 6.4%

A share with a beta of 0.6 has a required return of 6.4%. The risk-free rate is 4%. What expected market return rm does CAPM imply?

β = 0.6re = 6.4%rf = 4%
rm = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €18.50β = 1.8rf​ = 2%, rm​ = 8%debt20 000 × €1 050rd​ = 4%T = 21%

A company has 5 000 000 shares at €18.50 each and 20 000 bonds trading at €1 050 each. The pre-tax cost of debt is 4% and the tax rate is 21%. The share’s beta is 1.8, the risk-free rate 2% and the market return 8%. Find the WACC.

β = 1.8rf = 2%rm = 8%rd = 4%T = 21%
WACC = %
9.Descriptive statisticsMedium
1520253035404550

Find the sample standard deviation s of the data: 45, 53, 13, 16, 53, 46.

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

A quantity is normally distributed with mean 50 and standard deviation 10. What proportion P of values is above 38?

μ = 50σ = 10x = 38
P =
11.ProbabilityMedium
n = 7k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 7? The order does not matter.

n = 7k = 2
N =
12.ProbabilityHard
n = 9, p = 0.7: P(X ≤ 1) = ?k0123456789

9 components are tested, and each one fails with probability 0.7, independently. What is the probability P that at most one fails?

n = 9p = 0.7
P =
13.Normal distributionMedium
30.236.8P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 30.2 and 36.8 min? Give P as a decimal.

μ = 32σ = 6x1 = 30.2x2 = 36.8
P =
14.Normal distributionHard
μ = 250x = ?10%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 10% of them lie (the 10th percentile).

μ = 250σ = 5P = 10%
x = g
15.Confidence intervalsMedium
n = 10, sd​ = 7.60LUd̄ = −1E = ?90%

The blood pressure of 10 patients is measured before and after a treatment. The differences d = before − after (mmHg) have mean −1 and standard deviation 7.6. Find the margin of error E of the 90% confidence interval for the mean difference.

n = 10d̄ = −1sd = 7.690%t(9) = 1.833
E = mmHg
16.Confidence intervalsHard
σ = 17, n = ?x̄ − Ex̄ + Ex̄E = 595%

The population standard deviation is 17. 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 5?

σ = 17E = 595%
n =
17.Hypothesis tests & confidence intervalsMedium
1.9p = ?

The mean of a population is claimed to be 120; the population standard deviation is known to be 15. A random sample of 25 observations has a mean of 125.7. Test H₀: μ = 120 against H₁: μ > 120. Find the p-value p.

μ₀ = 120σ = 15n = 25x̄ = 125.7H₁: μ > μ₀
p =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 500 against H₁: μ < 500 is done at α = 0.10 with a sample of 25; the standard deviation is known to be 10. In reality the mean is 497.2. Find the probability of a Type II error β in percent.

μ₀ = 500μ = 497.2σ = 10n = 25α = 0.10
β = %
19.Linear regression & correlationMedium
xy2114156238311039xy51010203040

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
xy1673188420912210824123xy2080100120ŷ = 102

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 102?

n = 5ŷ = 102
x =

3Solutions

1.Time value of money

Answer: FV = 22545 €

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 = r/m = 0.04/12 = 0.003333
N = m·n = 12·3 = 36

Future value

FV = 20 000·(1 + 0.003333)36 = 20 000·1.127 = €22 545.44

Answer

FV = €22 545.44
2.Time value of money

Answer: FV = 54646 €

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 = 0.036000000000000004/12 = 0.003
N = 12·27 = 324

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 100·(2.639 − 1)/0.003 = €54 646.08

Answer

FV = €54 646.08

You paid in €32 400 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = 11691 €

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 = −30 000
CF1/(1 + 0.06)1 = 14 000/1.06 = 13 207.55
CF2/(1 + 0.06)2 = 11 000/1.124 = 9 789.96
CF3/(1 + 0.06)3 = 10 000/1.191 = 8 396.19
CF4/(1 + 0.06)4 = 13 000/1.262 = 10 297.22

Add them up

NPV = €11 690.92

Answer

NPV = €11 690.92

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

4.NPV & IRR

Answer: r* = 7.961 %

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 = 13 000 − 21 500 = −8 500
ΔCF2 = 18 500 − 16 000 = 2 500
ΔCF3 = 23 000 − 9 500 = 13 500

IRR of the difference

NPVΔ(7%) = 259.69
NPVΔ(8%) = −10.29
r* = 7.961%

Answer

r* = 7.961%

Check: at 7.961% both NPVs are €1 191.43.

5.Bond pricing & yield

Answer: P = 971.3 €

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 = 0.0275, N = 14

Present value of the coupons

€25.00·(1 − 0.684)/0.0275 = €287.28

Present value of the face value

€1 000·0.684 = €684.00

Price

P = €971.27

Answer

P = €971.27

The yield is above the coupon rate, so the bond sells at a discount.

6.Bond pricing & yield

Answer: YTC = 2.68 %

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

1233 = 60·(1 − (1 + y)−6)/y + 1060/(1 + y)6

Solve numerically

y = 2%: €1 277.34
y = 3%: €1 212.76
YTC = 2.68%

Answer

YTC = 2.68%

The yield to maturity is 4%. 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 = (6.4% − 4%)/0.6 = 4%

Market return

rm = 4% + 4% = 8%

Answer

rm = 8%
8.CAPM & WACC

Answer: WACC = 11.02 %

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 = 5 000 000·€18.50 = €92 500 000
D = 20 000·€1 050 = €21 000 000

Cost of equity (CAPM)

re = 2% + 1.8·6% = 12.8%

Weights

V = €113 500 000, E/V = 0.815, D/V = 0.185

WACC

WACC = 0.815·12.8% + 0.185·4%·(1 − 0.21) = 11.02%

Answer

WACC = 11.02%
9.Descriptive statistics

Answer: s = 18.28

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 = 226/6 = 37.67

Squared deviations

Σ(x − x̄)² = 7.333² + 15.33² + (−24.67)² + (−21.67)² + 15.33² + 8.333² = 1671

Standard deviation

s = √(1671/5) = 18.28

Answer

s = 18.28
10.Descriptive statistics

Answer: P = 0.8849

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 1 − 0.1151 = 0.8849

Answer

P = 0.8849
11.Probability

Answer: N = 21

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(7, 2) = (7·6)/(2·1) = 21

Answer

N = 21
12.Probability

Answer: P = 0.000433

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) = 0.39 = 1.968 × 10⁻⁵
P(X = 1) = 9·0.7·0.38 = 0.0004133

Add them

P = 1.968 × 10⁻⁵ + 0.0004133 = 0.000433

Answer

P = 0.000433
13.Normal distribution

Answer: P = 0.4061

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 = (30.2 − 32)/6 = −0.3
z₂ = (36.8 − 32)/6 = 0.8

Area between

P = Φ(0.8) − Φ(−0.3) = 0.7881 − 0.3821 = 0.4061

Answer

P = 0.4061
14.Normal distribution

Answer: x = 243.6 g

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) = 0.1 → z = −1.282

Back to x

x = 250 + −1.282·5 = 243.59 g

Answer

x = 243.59 g
15.Confidence intervals

Answer: E = 4.405 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 = 9, t = 1.833

Margin of error

E = 1.833·7.6/√10 = 4.405

Answer

E = 4.405 mmHg
16.Confidence intervals

Answer: n = 45

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 = (1.96·17/5)² = 44.41

Round up

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

n = 45

Answer

n = 45
17.Hypothesis tests & confidence intervals

Answer: p = 0.02872

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 = 15/√25 = 3
z = (125.7 − 120)/3 = 1.9

Tail area

Φ(1.9) = 0.9713
1 − Φ(1.9) = 0.0287

p-value

p = 0.02872

Answer

p = 0.02872

p ≤ 0.05, so H₀ is rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: β = 45.3 %

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 = 10/√25 = 2
x̄c = 500 − 1.282·2 = 497.4

z under the true mean

z = (497.4 − 497.2)/2 = 0.118

Type II error

power = P(Z < 0.118) = 54.7%
β = 100% − 54.7% = 45.3%

Answer

β = 45.3%
19.Linear regression & correlation

Answer: r = 0.9939

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 = 119/5 = 23.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 144
Syy = Σ(y − ȳ)² = 524.8

Correlation coefficient

r = 144/√(40·524.8) = 0.9939

Answer

r = 0.9939
20.Linear regression & correlation

Answer: x = 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 = 100/5 = 20
ȳ = 479/5 = 95.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 248

Regression line

b = 248/40 = 6.2
a = 95.8 − 6.2·20 = −28.2
ŷ = −28.2 + 6.2·x

Solve for x

x = (102 − (−28.2))/6.2 = 21

Answer

x = 21

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