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…8PV = ?€5 000

You need €5 000 in 8 years. An account pays 3.6% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €5 0003.6% / 12n = 8
PV = €
2.Time value of moneyHard
months0123…180A?AA€10 000

You want €10 000 in 15 years. You save the same amount at the end of every month in an account paying 4.8% per year, compounded monthly. Find the monthly deposit A.

FV = €10 0004.8% / 12N = 180
A = €
3.NPV & IRRMedium
years0123−45 00013 00020 00021 000

An investment of €45 000 gives these cash flows at the end of years 1 to 3: €13 000, €20 000, €21 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −45 000CF1 = 13 000CF2 = 20 000CF3 = 21 000
NPV = €
4.NPV & IRRHard
years01234567−40 0009 0009 0009 0009 0009 0009 0009 000

An investment of €40 000 returns €9 000 at the end of each year for 7 years. Find its internal rate of return IRR in percent.

C0 = €40 000CF = €9 000n = 7
IRR = %
5.Bond pricing & yieldMedium
half-years0123…10€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 5 years. The yield to maturity is 9% per year. Find the price P of the bond.

F = €1 000coupon = 5%YTM = 9%n = 5
P = €
6.Bond pricing & yieldHard
years01234€1 080€30€30 + €1 000

A bond with a face value of €1 000 and an annual coupon of 3% matures in 4 years and sells for €1 080. Find its yield to maturity YTM in percent.

F = €1 000coupon = 3%n = 4P = €1 080
YTM = %
7.CAPM & WACCMedium
WACC = ?equityEre​ = 13%debtD = 0.25·Erd​ = 5%T = 21%

A company has a debt-to-equity ratio (at market values) of 0.25. Its cost of equity is 13%, its pre-tax cost of debt is 5% and the tax rate is 21%. Find the WACC.

D/E = 0.25re = 13%rd = 5%T = 21%
WACC = %
8.CAPM & WACCHard
WACC = ?equity5 000 000 × €39.00β = 0.6rf​ = 2%, rm​ = 9%debt10 000 × €1 050rd​ = 3%T = 20%

A company has 5 000 000 shares at €39.00 each and 10 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3% and the tax rate is 20%. The share’s beta is 0.6, the risk-free rate 2% and the market return 9%. Find the WACC.

β = 0.6rf = 2%rm = 9%rd = 3%T = 20%
WACC = %
9.Descriptive statisticsMedium
2025303540

Find the sample standard deviation s of the data: 19, 25, 16, 44, 29.

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 67.5.

μ = 70σ = 5x = 67.5
z =
11.ProbabilityMedium
P(at least one six) = ??1?2P(six) = 1/6 on each roll

A fair die is rolled 2 times. What is the probability P of getting at least one six?

n = 2
P =
12.ProbabilityHard
€p(x)−30.7930.21470.01

A game costs €3 to play. You win €6 with probability 0.2 and €150 with probability 0.01; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€6: p = 0.2€150: p = 0.01
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.9651

Drag z so that the shaded area between −z and z under the standard normal curve is 0.9651. Find z to two decimals.

A = 0.9651
z =
14.Normal distributionHard
μ = 250x = ?90%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 90% of them lie (the 90th percentile).

μ = 250σ = 5P = 90%
x = g
15.Confidence intervalsMedium
n = 30, s = 9.5LU = ?x̄ = 110E99%

A sample of 30 items has mean 110 and standard deviation s = 9.5. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 30x̄ = 110s = 9.599%t(29) = 2.756
upper limit =
16.Confidence intervalsHard
n = 10σ = ?Us = 4.995%

The diameters of 10 machined shafts have a sample standard deviation of 4.9 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 9, 2.5% of the χ² distribution lies below 2.7 and 2.5% lies above 19.023.

n = 10s = 4.9 µmχ² = 2.7, 19.023
σ = µm
17.Hypothesis tests & confidence intervalsMedium
t −2.13102.131

A sample of 16 students has a mean score of 101.5 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 101.5s = 5n = 16
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 100 against H₁: μ < 100 is done at α = 0.10 with a sample of 36; the standard deviation is known to be 3. In reality the mean is 98.85. Find the power 1 − β of the test in percent.

μ₀ = 100μ = 98.85σ = 3n = 36α = 0.10
1 − β = %
19.Linear regression & correlationMedium
xy123224319421516xy2420

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 103Σx² = 55Σy² = 2163Σxy = 292
r =
20.Linear regression & correlationHard
xy25747869189910119xy5106080100120ŷ = 111

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 111?

n = 5ŷ = 111
x =

3Solutions

1.Time value of money

Answer: PV = 3750 €

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 = 0.036000000000000004/12 = 0.003
N = 12·8 = 96

Present value

PV = FV/(1 + i)N = 5 000/1.333 = €3 750.42

Answer

PV = €3 750.42
2.Time value of money

Answer: A = 38.04 €

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 = 0.048/12 = 0.004
N = 12·15 = 180

Deposit

A = FV·i/((1 + i)N − 1) = 10 000·0.004/(2.051 − 1) = €38.04

Answer

A = €38.04
3.NPV & IRR

Answer: NPV = −875.3 €

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 + 0.1)1 = 13 000/1.1 = 11 818.18
CF2/(1 + 0.1)2 = 20 000/1.21 = 16 528.93
CF3/(1 + 0.1)3 = 21 000/1.331 = 15 777.61

Add them up

NPV = €−875.28

Answer

NPV = €−875.28

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

4.NPV & IRR

Answer: IRR = 12.84 %

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)−7)/i = 40 000/9 000 = 4.444

Solve for the rate

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

i = 12%: 4.564
i = 13%: 4.423
IRR = 12.84%

Answer

IRR = 12.84%
5.Bond pricing & yield

Answer: P = 841.7 €

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 = 0.045, N = 10

Present value of the coupons

€25.00·(1 − 0.6439)/0.045 = €197.82

Present value of the face value

€1 000·0.6439 = €643.93

Price

P = €841.75

Answer

P = €841.75

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

6.Bond pricing & yield

Answer: YTM = 0.9522 %

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

€1 080 = €30·(1 − (1 + y)⁻⁴)/y + €1 000/(1 + y)⁴

Solve numerically

P(1%) = €1 078.04, P(1%) = €1 078.04
YTM = 0.9522%

Answer

YTM = 0.9522%
7.CAPM & WACC

Answer: WACC = 11.19 %

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 from D/E

E/V = 1/(1 + 0.25) = 0.8
D/V = 0.25/(1 + 0.25) = 0.2

WACC

WACC = 0.8·13% + 0.2·5%·(1 − 0.21) = 11.19%

Answer

WACC = 11.19%
8.CAPM & WACC

Answer: WACC = 6.006 %

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·€39.00 = €195 000 000
D = 10 000·€1 050 = €10 500 000

Cost of equity (CAPM)

re = 2% + 0.6·7% = 6.2%

Weights

V = €205 500 000, E/V = 0.9489, D/V = 0.05109

WACC

WACC = 0.9489·6.2% + 0.05109·3%·(1 − 0.2) = 6.006%

Answer

WACC = 6.006%
9.Descriptive statistics

Answer: s = 10.97

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 = 133/5 = 26.6

Squared deviations

Σ(x − x̄)² = (−7.6)² + (−1.6)² + (−10.6)² + 17.4² + 2.4² = 481.2

Standard deviation

s = √(481.2/4) = 10.97

Answer

s = 10.97
10.Descriptive statistics

Answer: z = −0.5

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 = (67.5 − 70)/5 = −0.5

Answer

z = −0.5
11.Probability

Answer: P = 0.3056

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)2 = 0.6944
P = 1 − 0.6944 = 0.3056

Answer

P = 0.3056
12.Probability

Answer: E(G) = −0.3 €

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 = 0.2), 147 (p = 0.01), −3 (p = 0.79)

Expected value

E(G) = 3·0.2 + 147·0.01 − 3·0.79 = −0.3 €

Answer

E(G) = −0.3 €

A negative value means you lose money on average.

13.Normal distribution

Answer: z = 2.11

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 of z

Φ(z) = (1 + 0.9651)/2 = 0.9826

Read the z-table backwards

Find 0.9826 inside the table and read off z.

z = 2.11

Answer

z = 2.11
14.Normal distribution

Answer: x = 256.4 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) = 0.9 → z = 1.282

Back to x

x = 250 + 1.282·5 = 256.41 g

Answer

x = 256.41 g
15.Confidence intervals

Answer: upper limit = 114.8

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 = 29, t = 2.756

Margin of error

E = 2.756·9.5/√30 = 4.78

Interval

110 ± 4.78 → [105.2, 114.8]

Answer

upper limit = 114.8
16.Confidence intervals

Answer: σ = 3.37 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 9·4.9² = 216.1

Limits for σ²

216.1/19.023 = 11.36
216.1/2.7 = 80.03

Limit for σ

σ = √11.36 = 3.37 µm

Answer

σ = 3.37 µm

95% interval for σ: [3.37, 8.946] µm.

17.Hypothesis tests & confidence intervals

Answer: t = 1.2

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 = 5/√16 = 5/4 = 1.25

Test statistic

t = (101.5 − 100)/1.25 = 1.2

Answer

t = 1.2
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: 1 − β = 84.57 %

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 = 3/√36 = 0.5
x̄c = 100 − 1.282·0.5 = 99.36

z under the true mean

z = (99.36 − 98.85)/0.5 = 1.018

Power

power = P(Z < 1.018) = 84.57%
1 − β = 84.57%

Answer

1 − β = 84.57%
19.Linear regression & correlation

Answer: r = −0.8375

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 = 292 − 15·103/5 = −17
Sxx = 55 − 15²/5 = 10
Syy = 2163 − 103²/5 = 41.2

Correlation coefficient

r = −17/√(10·41.2) = −0.8375

Answer

r = −0.8375
20.Linear regression & correlation

Answer: x = 9.062

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 = 444/5 = 88.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 290

Regression line

b = 290/40 = 7.25
a = 88.8 − 7.25·6 = 45.3
ŷ = 45.3 + 7.25·x

Solve for x

x = (111 − 45.3)/7.25 = 9.062

Answer

x = 9.062

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