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 = ?€40 000

You need €40 000 in 6 years. An account pays 3% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €40 0003% / 12n = 6
PV = €
2.Time value of moneyHard
months0123…192€200€200€200FV = ?

You save €200 at the end of every month for 16 years. The account pays 4.8% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €2004.8% / 12N = 192
FV = €
3.NPV & IRRMedium
years0123−75 00035 00017 00028 000

An investment of €75 000 gives these cash flows at the end of years 1 to 3: €35 000, €17 000, €28 000. The cost of capital is 8%. Find the net present value NPV.

r = 8%CF0 = −75 000CF1 = 35 000CF2 = 17 000CF3 = 28 000
NPV = €
4.NPV & IRRHard
years0123456−10 0002 5002 5002 5002 5002 5002 500

An investment of €10 000 returns €2 500 at the end of each year for 6 years. Find its internal rate of return IRR in percent.

C0 = €10 000CF = €2 500n = 6
IRR = %
5.Bond pricing & yieldMedium
half-years0123…12€30€30€30 + €1 000P = ?

A bond with a face value of €1 000 pays a 6% annual coupon in two half-yearly payments and matures in 6 years. The yield to maturity is 7% per year. Find the price P of the bond.

F = €1 000coupon = 6%YTM = 7%n = 6
P = €
6.Bond pricing & yieldHard
years0123…10€1 080€50€50 + €1 000

A bond with a face value of €1 000 and an annual coupon of 5% matures in 10 years and sells for €1 080. Find its yield to maturity YTM in percent.

F = €1 000coupon = 5%n = 10P = €1 080
YTM = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€12 000 000re​ = 10%preferred shares€1 500 000rp​ = 8%debt€3 000 000rd​ = 4.5%T = 28%

A company is financed with ordinary shares worth €12 000 000, preferred shares worth €1 500 000 and debt worth €3 000 000. The cost of equity is 10%, the cost of preferred stock 8% and the pre-tax cost of debt 4.5%. The tax rate is 28%. Find the WACC.

E = €12 000 000Ps = €1 500 000D = €3 000 000re = 10%rp = 8%rd = 4.5%T = 28%
WACC = %
8.CAPM & WACCHard
WACC = ?equity€6 000 000re​ = 10%bonds4 000 × €1 080coupon 5%, n = 6T = 28%

A company has equity with a market value of €6 000 000 and a cost of equity of 10%. Its debt consists of 4 000 bonds (face value €1 000, annual coupon 5%, 6 years to maturity) that trade at €1 080 each. The tax rate is 28%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €6 000 000re = 10%bonds: 4 000 × €1 080coupon = 5%n = 6T = 28%
WACC = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 42, 33, 42, 11, 55, 29, 12.

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

Exam scores are normally distributed with mean 60 and standard deviation 10. Find the z-score z of a score of 52.

μ = 60σ = 10x = 52
z =
11.ProbabilityMedium
xp(x)00.2520.340.360.15μ = 2.7

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

x: 0, 2, 4, 6p(x): 0.25, 0.3, 0.3, 0.15
Var(X) =
12.ProbabilityHard
€p(x)−10.840.1590.05

A game costs €1 to play. You win €5 with probability 0.15 and €10 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €1€5: p = 0.15€10: p = 0.05
E(G) = €
13.Normal distributionMedium
32 min30 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 30 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 30 min
P =
14.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
15.Confidence intervalsMedium
n = 15, s = 7LU = ?x̄ = 26.5E90%

A sample of 15 items has mean 26.5 and standard deviation s = 7. The population is roughly normal. Find the upper limit of the 90% confidence interval for the mean.

n = 15x̄ = 26.5s = 790%t(14) = 1.761
upper limit =
16.Confidence intervalsHard
p ≈ 0.15, n = ?p̂ − Ep̂ + Ep̂E = 0.0395%

A pilot study suggests that about 0.15 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 0.03?

p ≈ 0.15E = 0.0395%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.3652.365t = ?df = 7

8 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is −3.25 and the standard deviation of the differences is 10.2. Find the paired test statistic t for H₀: μd = 0.

n = 8d̄ = −3.25sd = 10.2
t =
18.Hypothesis tests & confidence intervalsHard
σ = 5, n = ?x̄ − Ex̄ + Ex̄E = 2.599%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 5. Find the smallest sample size n.

σ = 5E = 2.599%
n =
19.Linear regression & correlationMedium
xy274176258291037xy51010203040

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
xy124221321415516xy24152025x = 2.5

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 5x = 2.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 33418 €

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 = 0.03/12 = 0.0025
N = 12·6 = 72

Present value

PV = FV/(1 + i)N = 40 000/1.197 = €33 418.31

Answer

PV = €33 418.31
2.Time value of money

Answer: FV = 57608 €

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 = 12·16 = 192

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 200·(2.152 − 1)/0.004 = €57 607.58

Answer

FV = €57 607.58

You paid in €38 400 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = −5791 €

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 + 0.08)1 = 35 000/1.08 = 32 407.41
CF2/(1 + 0.08)2 = 17 000/1.166 = 14 574.76
CF3/(1 + 0.08)3 = 28 000/1.26 = 22 227.30

Add them up

NPV = €−5 790.53

Answer

NPV = €−5 790.53

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

4.NPV & IRR

Answer: IRR = 12.98 %

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 = 10 000/2 500 = 4

Solve for the rate

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

i = 12%: 4.111
i = 13%: 3.998
IRR = 12.98%

Answer

IRR = 12.98%
5.Bond pricing & yield

Answer: P = 951.7 €

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 = €30.00, i = 0.035, N = 12

Present value of the coupons

€30.00·(1 − 0.6618)/0.035 = €289.90

Present value of the face value

€1 000·0.6618 = €661.78

Price

P = €951.68

Answer

P = €951.68

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

6.Bond pricing & yield

Answer: YTM = 4.013 %

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 = €50·(1 − (1 + y)⁻¹⁰)/y + €1 000/(1 + y)¹⁰

Solve numerically

P(4%) = €1 081.11, P(5%) = €1 000.00
YTM = 4.013%

Answer

YTM = 4.013%
7.CAPM & WACC

Answer: WACC = 8.589 %

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 = 12 000 000 + 1 500 000 + 3 000 000 = €16 500 000
E/V = 0.7273, Ps/V = 0.09091, D/V = 0.1818

WACC

WACC = 0.7273·10% + 0.09091·8% + 0.1818·4.5%·(1 − 0.28) = 8.589%

Answer

WACC = 8.589%
8.CAPM & WACC

Answer: WACC = 6.868 %

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).

Cost of debt (YTM)

1080 = 50·(1 − (1 + y)−6)/y + 1000/(1 + y)6
y = 3.499%

Market values

D = 4 000·€1 080 = €4 320 000
V = €10 320 000, E/V = 0.5814, D/V = 0.4186

WACC

WACC = 0.5814·10% + 0.4186·3.499%·(1 − 0.28) = 6.868%

Answer

WACC = 6.868%
9.Descriptive statistics

Answer: s = 16.23

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 = 224/7 = 32

Squared deviations

Σ(x − x̄)² = 10² + 1² + 10² + (−21)² + 23² + (−3)² + (−20)² = 1580

Standard deviation

s = √(1580/6) = 16.23

Answer

s = 16.23
10.Descriptive statistics

Answer: z = −0.8

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 = (52 − 60)/10 = −0.8

Answer

z = −0.8
11.Probability

Answer: Var(X) = 4.11

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 = 0·0.25 + 2·0.3 + 4·0.3 + 6·0.15 = 2.7

Mean of the squares

E(X²) = 0·0.25 + 4·0.3 + 16·0.3 + 36·0.15 = 11.4

Variance

Var(X) = 11.4 − 2.7² = 4.11

Answer

Var(X) = 4.11
12.Probability

Answer: E(G) = 0.25 €

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: 4 (p = 0.15), 9 (p = 0.05), −1 (p = 0.8)

Expected value

E(G) = 4·0.15 + 9·0.05 − 1·0.8 = 0.25 €

Answer

E(G) = 0.25 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.7881

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 sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (30 − 32)/2.5 = −0.8

Upper tail

P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
14.Normal distribution

Answer: P = 0.8413

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 = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
15.Confidence intervals

Answer: upper limit = 29.68

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 = 14, t = 1.761

Margin of error

E = 1.761·7/√15 = 3.183

Interval

26.5 ± 3.183 → [23.32, 29.68]

Answer

upper limit = 29.68
16.Confidence intervals

Answer: n = 545

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 = 1.96²·0.15·0.85/0.03² = 544.2

Round up

n = 545

Answer

n = 545
17.Hypothesis tests & confidence intervals

Answer: t = −0.9012

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 of the mean difference

SE = 10.2/√8 = 3.606

Test statistic

t = −3.25/3.606 = −0.9012

Answer

t = −0.9012

With df = 7, the two-sided 5% critical value is 2.365; |t| is smaller, so H₀ is not rejected.

18.Hypothesis tests & confidence intervals

Answer: n = 27

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 = (2.576·5/2.5)² = 26.54

Round up

n = 27

Answer

n = 27
19.Linear regression & correlation

Answer: r = 0.9909

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 = 115/5 = 23

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 144
Syy = Σ(y − ȳ)² = 528

Correlation coefficient

r = 144/√(40·528) = 0.9909

Answer

r = 0.9909
20.Linear regression & correlation

Answer: ŷ = 20.5

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 = 15/5 = 3
ȳ = 97/5 = 19.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −22

Regression line

b = −22/10 = −2.2
a = 19.4 − (−2.2)·3 = 26
ŷ = 26 − 2.2·x

Prediction

ŷ = 26 + (−2.2)·2.5 = 20.5

Answer

ŷ = 20.5

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