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
years01234567€3 000FV = ?

You invest €3 000 at a nominal rate of 4.8% per year, compounded quarterly. Find the value FV of the investment after 7 years.

PV = €3 0004.8% / 4n = 7
FV = €
2.Time value of moneyHard
months01…180…360€200 000AB = ?A4.2% / 12

A mortgage of €200 000 is repaid with equal monthly payments over 30 years at 4.2% per year, compounded monthly. How much is still owed just after the payment at the end of year 15? Find the outstanding balance B.

P = €200 0004.2% / 12N = 360k = 180
B = €
3.NPV & IRRMedium
years0123…9−200 00011 000+3%+3%…r = 8%, NPV = ?

A project costs €200 000 today. It generates €11 000 at the end of the first year, and the cash flow then grows by 3% per year forever. The cost of capital is 8%. Find the net present value NPV.

C0 = €200 000CF1 = €11 000g = 3%r = 8%
NPV = €
4.NPV & IRRHard
years01234567−80 00013 500+3%+3%+3%+3%+3%+3%r = 8%, NPV = ?

A project costs €80 000 today. Its cash flow is €13 500 at the end of year 1 and grows by 3% per year until the last cash flow at the end of year 7. The cost of capital is 8%. Find the net present value NPV.

C0 = €80 000CF1 = €13 500g = 3%n = 7r = 8%
NPV = €
5.Bond pricing & yieldMedium
half-years0123…24€35€35€35 + €1 000P = ?

A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 12 years. The yield to maturity is 2.5% per year. Find the price P of the bond.

F = €1 000coupon = 7%YTM = 2.5%n = 12
P = €
6.Bond pricing & yieldHard

A bond trades at €1 085 with a yield to maturity of 6% (annual compounding) and a Macaulay duration of 10.4 years. Use the modified duration to estimate the new price P if the yield falls by 50 basis points.

P0 = €1 085y = 6%D = 10.4Δy = −50 bp
P = €
7.CAPM & WACCMedium
WACC = ?ordinary shares€4 000 000re​ = 11%preferred shares€2 000 000rp​ = 8%debt€6 000 000rd​ = 4.5%T = 30%

A company is financed with ordinary shares worth €4 000 000, preferred shares worth €2 000 000 and debt worth €6 000 000. The cost of equity is 11%, the cost of preferred stock 8% and the pre-tax cost of debt 4.5%. The tax rate is 30%. Find the WACC.

E = €4 000 000Ps = €2 000 000D = €6 000 000re = 11%rp = 8%rd = 4.5%T = 30%
WACC = %
8.CAPM & WACCHard
WACC = ?equity4 000 000 × €51.50β = 1.5rf​ = 3.5%, rm​ = 8%debt5 000 × €1 050rd​ = 3%T = 30%

A company has 4 000 000 shares at €51.50 each and 5 000 bonds trading at €1 050 each. The pre-tax cost of debt is 3% and the tax rate is 30%. The share’s beta is 1.5, the risk-free rate 3.5% and the market return 8%. Find the WACC.

β = 1.5rf = 3.5%rm = 8%rd = 3%T = 30%
WACC = %
9.Descriptive statisticsMedium
303540455055

Find the sample standard deviation s of the data: 49, 38, 26, 56, 39, 45.

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

Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 46.

μ = 50σ = 5x = 46
z =
11.ProbabilityMedium
xp(x)00.3510.320.35μ = 1

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

x: 0, 1, 2p(x): 0.35, 0.3, 0.35
Var(X) =
12.ProbabilityHard
xp(x)10.220.130.3540.35μ

X has the distribution shown, and Y = 3X + 4. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.2, 0.1, 0.35, 0.35Y = 3X + 4
σY =
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.1539

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.1539. Find z to two decimals.

A = 0.1539
z =
14.Normal distributionHard
μ = 65x = ?95%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 95% of them lie (the 95th percentile).

μ = 65σ = 10P = 95%
x = points
15.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 30 phones, mean 79 h, known standard deviation 3 h. Model 2: 80 phones, mean 77 h, known standard deviation 4 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 30x̄₁ = 79 hσ₁ = 3 hn₂ = 80x̄₂ = 77 hσ₂ = 4 h99%
lower limit = h
16.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 199%

The population standard deviation is 31. 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 1?

σ = 31E = 199%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 501.4 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 501.4s = 20n = 100
t =
18.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 75 of 250 patients on treatment A recover, and 19 of 80 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 75n₁ = 250x₂ = 19n₂ = 80
z =
19.Linear regression & correlationMedium
xy27496108141019xy510101520

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
xy252360456562675xy2466070ŷ = 68

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 = 68?

n = 5ŷ = 68
x =

3Solutions

1.Time value of money

Answer: FV = 4190 €

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 = 0.048/4 = 0.012
N = m·n = 4·7 = 28

Future value

FV = 3 000·(1 + 0.012)28 = 3 000·1.397 = €4 189.63

Answer

FV = €4 189.63
2.Time value of money

Answer: B = 130448 €

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 = 0.0035, N = 360
A = 200 000·0.0035/(1 − (1 + 0.0035)−360) = €978.03

Present value of the remaining payments

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

N − k = 360 − 180 = 180
B = 978·(1 − (1 + 0.0035)−180)/0.0035 = €130 448.00

Answer

B = €130 448.00
3.NPV & IRR

Answer: NPV = 20000 €

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.

Value of the cash flows

PV = 11 000/(0.08 − 0.03) = 220 000.00

NPV

NPV = 220 000.00 − 200 000 = €20 000.00

Answer

NPV = €20 000.00

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

4.NPV & IRR

Answer: NPV = −3757 €

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

((1.03)/(1.08))7 = 0.7176
PV = 13 500/(0.08 − 0.03)·(1 − 0.7176) = 76 242.71

NPV

NPV = 76 242.71 − 80 000 = −€3 757.29

Answer

NPV = −€3 757.29

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

5.Bond pricing & yield

Answer: P = 1464 €

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 = €35.00, i = 0.0125, N = 24

Present value of the coupons

€35.00·(1 − 0.7422)/0.0125 = €721.85

Present value of the face value

€1 000·0.7422 = €742.20

Price

P = €1 464.05

Answer

P = €1 464.05

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

6.Bond pricing & yield

Answer: P = 1138 €

Duration: the Macaulay duration is the weighted average time at which a bond pays its cash flows, with the present values of the cash flows as weights: D = Σ t·PVₜ/P. The modified duration D/(1 + y) estimates the percentage price change for a small change in yield: %ΔP ≈ −Dmod·Δy.

Modified duration

Dmod = 10.4/1.06 = 9.811

Estimated price change

%ΔP ≈ −9.811·(−0.005) = 4.906%
P ≈ 1085·(1 + 0.04906) = €1 138.23

Answer

P = €1 138.23

Duration is a linear estimate; for large yield changes convexity makes the true price a little higher.

7.CAPM & WACC

Answer: WACC = 6.575 %

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 = 4 000 000 + 2 000 000 + 6 000 000 = €12 000 000
E/V = 0.3333, Ps/V = 0.1667, D/V = 0.5

WACC

WACC = 0.3333·11% + 0.1667·8% + 0.5·4.5%·(1 − 0.3) = 6.575%

Answer

WACC = 6.575%
8.CAPM & WACC

Answer: WACC = 10.05 %

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·€51.50 = €206 000 000
D = 5 000·€1 050 = €5 250 000

Cost of equity (CAPM)

re = 3.5% + 1.5·4.5% = 10.25%

Weights

V = €211 250 000, E/V = 0.9751, D/V = 0.02485

WACC

WACC = 0.9751·10.25% + 0.02485·3%·(1 − 0.3) = 10.05%

Answer

WACC = 10.05%
9.Descriptive statistics

Answer: s = 10.34

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 = 253/6 = 42.17

Squared deviations

Σ(x − x̄)² = 6.833² + (−4.167)² + (−16.17)² + 13.83² + (−3.167)² + 2.833² = 534.8

Standard deviation

s = √(534.8/5) = 10.34

Answer

s = 10.34
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 = (46 − 50)/5 = −0.8

Answer

z = −0.8
11.Probability

Answer: Var(X) = 0.7

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 = 0·0.35 + 1·0.3 + 2·0.35 = 1

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7

Variance

Var(X) = 1.7 − 1² = 0.7

Answer

Var(X) = 0.7
12.Probability

Answer: σY = 3.324

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 = 2.85, Var(X) = 1.228, σX = 1.108

Linear change

σY = |3|·1.108 = 3.324

Answer

σY = 3.324

The mean does move: E(Y) = 3·2.85 + 4 = 12.55.

13.Normal distribution

Answer: z = 1.02

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 − 0.1539 = 0.8461

Read the z-table backwards

Find 0.8461 inside the table and read off z.

z = 1.02

Answer

z = 1.02
14.Normal distribution

Answer: x = 81.45 points

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) = 0.95 → z = 1.645

Back to x

x = 65 + 1.645·10 = 81.45 points

Answer

x = 81.45 points
15.Confidence intervals

Answer: lower limit = 0.1785 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 = 79 − 77 = 2

Standard error

SE = √(3²/30 + 4²/80) = 0.7071

Margin of error

E = 2.576·0.7071 = 1.822

Interval

2 ± 1.822 → [0.1785, 3.822]

Answer

lower limit = 0.1785 h

The interval does not contain 0, so the means differ significantly.

16.Confidence intervals

Answer: n = 6377

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 = (2.576·31/1)² = 6377

Round up

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

n = 6377

Answer

n = 6377
17.Hypothesis tests & confidence intervals

Answer: t = 0.7

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 = 20/√100 = 20/10 = 2

Test statistic

t = (501.4 − 500)/2 = 0.7

Answer

t = 0.7
t0.025; df = 99 = 1.984

|t| < 1.984, so H₀ is not rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: z = 1.078

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 = 75/250 = 0.3
p̂₂ = 19/80 = 0.2375
p̂ = 94/330 = 0.2848

Standard error under H₀

SE = √(0.2848·0.7152·(1/250 + 1/80)) = 0.05798

Test statistic

z = (0.3 − 0.2375)/0.05798 = 1.078

Answer

z = 1.078

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

19.Linear regression & correlation

Answer: r = 0.9624

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 = 59/5 = 11.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 58
Syy = Σ(y − ȳ)² = 90.8

Correlation coefficient

r = 58/√(40·90.8) = 0.9624

Answer

r = 0.9624
20.Linear regression & correlation

Answer: x = 5.458

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 = 20/5 = 4
ȳ = 305/5 = 61

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 48

Regression line

b = 48/10 = 4.8
a = 61 − 4.8·4 = 41.8
ŷ = 41.8 + 4.8·x

Solve for x

x = (68 − 41.8)/4.8 = 5.458

Answer

x = 5.458

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