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
years0123PV = ?€25 000

You need €25 000 in 3 years. An account pays 4.8% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €25 0004.8% / 12n = 3
PV = €
2.Time value of moneyHard
months0123…36€12 000A?AAA

A loan of €12 000 is repaid with equal monthly payments over 3 years. The interest rate is 4.2% per year, compounded monthly. Find the monthly payment A.

P = €12 0004.2% / 12N = 36
A = €
3.NPV & IRRMedium
years01234567€80 000€4 000€4 000€4 000€4 000€4 000€4 000€4 000r = 8%, EAC = ?

A machine costs €80 000 today, lasts 7 years and costs €4 000 per year to run (paid at the end of each year). The cost of capital is 8%. Find its equivalent annual cost EAC.

C0 = €80 000M = €4 000n = 7r = 8%
EAC = €
4.NPV & IRRHard
years01234−10 0003 5003 0006 0004 500

A project costs €10 000 and returns €3 500, €3 000, €6 000, €4 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 12%. Find the modified internal rate of return MIRR in percent.

r = 12%CF0 = −10 000CF1 = 3 500CF2 = 3 000CF3 = 6 000CF4 = 4 500
MIRR = %
5.Bond pricing & yieldMedium
half-years0123…14€40€40€40 + €1 000P = ?

A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 7 years. The yield to maturity is 5% per year. Find the price P of the bond.

F = €1 000coupon = 8%YTM = 5%n = 7
P = €
6.Bond pricing & yieldHard
years0123456€1 112€70€70€70 + €1 070

A bond with 19 years to maturity (face value €1 000, annual coupon 7%) sells for €1 112. The issuer can call it in 6 years at a call price of €1 070. Find the yield to call YTC in percent.

F = €1 000coupon = 7%P = €1 112call price = €1 070call in 6 years
YTC = %
7.CAPM & WACCMedium
WACC = ?equity€3 000 000re​ = 9%debt€6 000 000rd​ = 4.5%T = 30%

A company has equity worth €3 000 000 and debt worth €6 000 000. Its cost of equity is 9%, its pre-tax cost of debt is 4.5% and the tax rate is 30%. Find the WACC.

E = €3 000 000D = €6 000 000re = 9%rd = 4.5%T = 30%
WACC = %
8.CAPM & WACCHard
WACC = ?equity4 000 000 × €21.50β = 1.1rf​ = 3%, rm​ = 10%debt40 000 × €960rd​ = 3%T = 25.8%

A company has 4 000 000 shares at €21.50 each and 40 000 bonds trading at €960 each. The pre-tax cost of debt is 3% and the tax rate is 25.8%. The share’s beta is 1.1, the risk-free rate 3% and the market return 10%. Find the WACC.

β = 1.1rf = 3%rm = 10%rd = 3%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
26283032343638404244

Find the sample standard deviation s of the data: 41, 37, 37, 25, 37, 26, 45.

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

A quantity is normally distributed with mean 70 and standard deviation 5. What proportion P of values is below 62.5?

μ = 70σ = 5x = 62.5
P =
11.ProbabilityMedium
xp(x)10.1520.230.340.1550.2μ = 3.05

For the distribution shown, find the standard deviation σ of X.

x: 1, 2, 3, 4, 5p(x): 0.15, 0.2, 0.3, 0.15, 0.2
σ =
12.ProbabilityHard
€p(x)−30.890.151470.05

A game costs €3 to play. You win €12 with probability 0.15 and €150 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€12: p = 0.15€150: p = 0.05
E(G) = €
13.Normal distributionMedium
5773P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 57 and 73 points? Give P as a decimal.

μ = 65σ = 10x1 = 57x2 = 73
P =
14.Normal distributionHard
10 cm0X − Y?

The heights X of men and Y of women in a group are independent and normal: X has mean 180 cm and standard deviation 10 cm, Y has mean 170 cm and standard deviation 12 cm. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 180 cm, σX = 10 cmμY = 170 cm, σY = 12 cm
P =
15.Confidence intervalsMedium
μ₁ − μ₂0LU = ?x̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 60 phones, mean 53.5 h, known standard deviation 13 h. Model 2: 80 phones, mean 57 h, known standard deviation 8 h. Find the upper limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 53.5 hσ₁ = 13 hn₂ = 80x̄₂ = 57 hσ₂ = 8 h99%
upper limit = h
16.Confidence intervalsHard
σ = 22, n = ?x̄ − Ex̄ + Ex̄E = 0.599%

The population standard deviation is 22. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 0.5?

σ = 22E = 0.599%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 101.6 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 100.

μ₀ = 100x̄ = 101.6s = 12n = 81
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 53 of 150 patients on treatment A recover, and 41 of 100 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 53n₁ = 150x₂ = 41n₂ = 100
z =
19.Linear regression & correlationMedium
xy274116148191021xy510101520

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
xy12121731741358xy24101520x = 3.5

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 = 3.5.

n = 5x = 3.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 21653 €

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 = 0.048/12 = 0.004
N = 12·3 = 36

Present value

PV = FV/(1 + i)N = 25 000/1.155 = €21 653.41

Answer

PV = €21 653.41
2.Time value of money

Answer: A = 355.4 €

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 payments

i = 0.042/12 = 0.0035
N = 12·3 = 36

Payment

The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.

(1 + i)−N = 0.8818
A = 12 000·0.0035/(1 − 0.8818) = €355.36

Answer

A = €355.36
3.NPV & IRR

Answer: EAC = 19366 €

Equivalent annual cost: spread the present value of all costs over the life as an equal yearly amount (an annuity): EAC = PV·r/(1 − (1 + r)⁻ⁿ). It lets you compare machines with different lives.

Annuity factor

(1 − 1.08−7)/0.08 = 5.206

Annualised purchase price

80 000/5.206 = 15 365.79

EAC

EAC = 15 365.79 + 4 000 = €19 365.79

Answer

EAC = €19 365.79
4.NPV & IRR

Answer: MIRR = 18.77 %

Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.

Terminal value of the inflows

3 500·1.123 = 4 917.25
3 000·1.122 = 3 763.20
6 000·1.121 = 6 720.00
4 500·1.120 = 4 500.00
TV = 19 900.45

MIRR

MIRR = (19 900.45/10 000)1/4 − 1 = 18.77%

Answer

MIRR = 18.77%
5.Bond pricing & yield

Answer: P = 1175 €

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 = €40.00, i = 0.025, N = 14

Present value of the coupons

€40.00·(1 − 0.7077)/0.025 = €467.64

Present value of the face value

€1 000·0.7077 = €707.73

Price

P = €1 175.36

Answer

P = €1 175.36

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

6.Bond pricing & yield

Answer: YTC = 5.75 %

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

1112 = 70·(1 − (1 + y)−6)/y + 1070/(1 + y)6

Solve numerically

y = 5%: €1 153.75
y = 6%: €1 098.52
YTC = 5.75%

Answer

YTC = 5.75%

The yield to maturity is 5.997%. 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: WACC = 5.1 %

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 = €9 000 000, E/V = 0.3333, D/V = 0.6667

WACC

WACC = 0.3333·9% + 0.6667·4.5%·(1 − 0.3) = 5.1%

Answer

WACC = 5.1%
8.CAPM & WACC

Answer: WACC = 8.084 %

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 = 4 000 000·€21.50 = €86 000 000
D = 40 000·€960 = €38 400 000

Cost of equity (CAPM)

re = 3% + 1.1·7% = 10.7%

Weights

V = €124 400 000, E/V = 0.6913, D/V = 0.3087

WACC

WACC = 0.6913·10.7% + 0.3087·3%·(1 − 0.258) = 8.084%

Answer

WACC = 8.084%
9.Descriptive statistics

Answer: s = 7.39

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 = 248/7 = 35.43

Squared deviations

Σ(x − x̄)² = 5.571² + 1.571² + 1.571² + (−10.43)² + 1.571² + (−9.429)² + 9.571² = 327.7

Standard deviation

s = √(327.7/6) = 7.39

Answer

s = 7.39
10.Descriptive statistics

Answer: P = 0.06681

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 = −1.5

Standard normal table

Φ(−1.5) = 0.06681
P = 0.06681

Answer

P = 0.06681
11.Probability

Answer: σ = 1.322

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 = 1·0.15 + 2·0.2 + 3·0.3 + 4·0.15 + 5·0.2 = 3.05

Mean of the squares

E(X²) = 1·0.15 + 4·0.2 + 9·0.3 + 16·0.15 + 25·0.2 = 11.05

Variance

Var(X) = 11.05 − 3.05² = 1.748
σ = √1.748 = 1.322

Answer

σ = 1.322
12.Probability

Answer: E(G) = 6.3 €

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: 9 (p = 0.15), 147 (p = 0.05), −3 (p = 0.8)

Expected value

E(G) = 9·0.15 + 147·0.05 − 3·0.8 = 6.3 €

Answer

E(G) = 6.3 €

A positive value means you win money on average.

13.Normal distribution

Answer: P = 0.5763

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 = (57 − 65)/10 = −0.8
z₂ = (73 − 65)/10 = 0.8

Area between

P = Φ(0.8) − Φ(−0.8) = 0.7881 − 0.2119 = 0.5763

Answer

P = 0.5763
14.Normal distribution

Answer: P = 0.7389

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 = 180 − 170 = 10 cm, σ = 15.62 cm

Standardise

z = (0 − 10)/15.62 = −0.64

Probability

P = 1 − Φ(−0.64) = 0.7389

Answer

P = 0.7389
15.Confidence intervals

Answer: upper limit = 1.399 h

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.

Point estimate

x̄₁ − x̄₂ = 53.5 − 57 = −3.5

Standard error

SE = √(13²/60 + 8²/80) = 1.902

Margin of error

E = 2.576·1.902 = 4.899

Interval

−3.5 ± 4.899 → [−8.399, 1.399]

Answer

upper limit = 1.399 h

The interval contains 0, so the data do not show a significant difference.

16.Confidence intervals

Answer: n = 12847

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 = (2.576·22/0.5)² = 12847

Round up

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

n = 12847

Answer

n = 12847
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 = 12/√81 = 12/9 = 1.333

Test statistic

t = (101.6 − 100)/1.333 = 1.2

Answer

t = 1.2
t0.025; df = 80 = 1.99

|t| < 1.99, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = −0.9062

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.

Sample proportions

p̂₁ = 53/150 = 0.3533
p̂₂ = 41/100 = 0.41
p̂ = 94/250 = 0.376

Standard error under H₀

SE = √(0.376·0.624·(1/150 + 1/100)) = 0.06253

Test statistic

z = (0.3533 − 0.41)/0.06253 = −0.9062

Answer

z = −0.9062

|z| < 1.96, so H₀ is not rejected at the 5% level (two-sided).

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 = 72/5 = 14.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 72
Syy = Σ(y − ȳ)² = 131.2

Correlation coefficient

r = 72/√(40·131.2) = 0.9939

Answer

r = 0.9939
20.Linear regression & correlation

Answer: ŷ = 13.7

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 = 76/5 = 15.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −30

Regression line

b = −30/10 = −3
a = 15.2 − (−3)·3 = 24.2
ŷ = 24.2 − 3·x

Prediction

ŷ = 24.2 + (−3)·3.5 = 13.7

Answer

ŷ = 13.7

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