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
years012PV = ?€15 000

You need €15 000 in 2 years. An account pays 3% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €15 0003% / 12n = 2
PV = €
2.Time value of moneyHard
years01234567€2 000€2 000€2 000FV = ?

You deposit €2 000 at the start of every year for 7 years, the first deposit today. The account pays 5% per year. How much is in the account at the end of year 7? Find FV.

A = €2 000i = 5%n = 7
FV = €
3.NPV & IRRMedium
years0123−20 0008 0006 0006 000

An investment of €20 000 gives these cash flows at the end of years 1 to 3: €8 000, €6 000, €6 000. The cost of capital is 10%. Find the net present value NPV.

r = 10%CF0 = −20 000CF1 = 8 000CF2 = 6 000CF3 = 6 000
NPV = €
4.NPV & IRRHard
years012−10 0007 0007 500

A project costs €10 000 and returns €7 000, €7 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 7 000CF2 = 7 500
IRR = %
5.Bond pricing & yieldMedium
years0123…8€1 179C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 8 years and pays annual coupons. It sells for €1 179 when the yield to maturity is 2.5%. Find the coupon rate c in percent.

F = €1 000n = 8P = €1 179YTM = 2.5%
c = %
6.Bond pricing & yieldHard
years012345€895€30€30 + €1 000

A bond with a face value of €1 000 and an annual coupon of 3% matures in 5 years and sells for €895. Find its yield to maturity YTM in percent.

F = €1 000coupon = 3%n = 5P = €895
YTM = %
7.CAPM & WACCMedium
βr (%)SMLrf​ = 2%β = 1rm​ = ?β = 0.8re​ = 7.2%

A share with a beta of 0.8 has a required return of 7.2%. The risk-free rate is 2%. What expected market return rm does CAPM imply?

β = 0.8re = 7.2%rf = 2%
rm = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €37.50D0​ = €1.60g = 4%debt€22 500 000rd​ = 4%T = 21%

A company has 2 000 000 shares trading at €37.50; the book value of its equity is €22 500 000. It just paid a dividend of €1.60 per share, and dividends grow by 4% per year. Its debt has a market value of €22 500 000 and a pre-tax cost of 4%. The tax rate is 21%. Find the WACC, using the dividend growth model for the cost of equity.

N = 2 000 000P0 = €37.50book equity = €22 500 000D0 = €1.60g = 4%D = €22 500 000rd = 4%T = 21%
WACC = %
9.Descriptive statisticsMedium
2025303540

Find the sample standard deviation s of the data: 19, 41, 33, 17, 39.

n = 5
s =
10.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 170 and standard deviation 5. Find the z-score z of a score of 165.

μ = 170σ = 5x = 165
z =
11.ProbabilityMedium
red: 8blue: 4

A bag holds 8 red and 4 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

8 / 4
P =
12.ProbabilityHard
xp(x)00.3520.240.260.25μ

X has the distribution shown, and Y = 0.5X + 5. Find the standard deviation σY of Y.

x: 0, 2, 4, 6p(x): 0.35, 0.2, 0.2, 0.25Y = 0.5X + 5
σY =
13.Normal distributionMedium
360 g356.4 gT = X₁ + … + X₉?

The masses of bolts are independent and normal with mean 40 g and standard deviation 1.2 g. Find the probability P that the total mass of 9 of them exceeds 356.4 g. Round z to two decimals.

μ = 40 gσ = 1.2 gn = 9> 356.4 g
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
x̄E₁ = 11.699%, n = 50x̄E₂ = ?90%, n = 50

A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 11.6. What is the margin of error E₂ of a 90% interval from the same data?

99%n₁ = 50E₁ = 11.690%
E₂ =
16.Confidence intervalsHard
n = 30σ = ?Us = 1.195%

The diameters of 30 machined shafts have a sample standard deviation of 1.1 µm. Assuming the diameters are normally distributed, find the lower limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 1.1 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 60 parts has mean length 75.5 mm, and the line's known standard deviation is 12 mm. Line 2: a sample of 80 parts has mean 80.3 mm, known standard deviation 5 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 60x̄₁ = 75.5 mmσ₁ = 12 mmn₂ = 80x̄₂ = 80.3 mmσ₂ = 5 mm
z =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 36 measurements has mean 114.8. The population standard deviation is 12. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 114.8σ = 12n = 3690%
L =
19.Linear regression & correlationMedium
xy2241061481610221224xy5101020

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy244106148191021xy5101020x = 7

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (7, 20) is made. Find its residual e = y − ŷ.

n = 5(7, 20)
e =

3Solutions

1.Time value of money

Answer: PV = 14128 €

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 = 12·2 = 24

Present value

PV = FV/(1 + i)N = 15 000/1.062 = €14 127.53

Answer

PV = €14 127.53
2.Time value of money

Answer: FV = 17098 €

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.

Ordinary annuity factor

((1 + 0.05)7 − 1)/0.05 = 8.142

Annuity due

Each deposit is invested one year longer.

FV = 2 000·8.142·1.05 = €17 098.22

Answer

FV = €17 098.22
3.NPV & IRR

Answer: NPV = −3261 €

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 = −20 000
CF1/(1 + 0.1)1 = 8 000/1.1 = 7 272.73
CF2/(1 + 0.1)2 = 6 000/1.21 = 4 958.68
CF3/(1 + 0.1)3 = 6 000/1.331 = 4 507.89

Add them up

NPV = €−3 260.71

Answer

NPV = €−3 260.71

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

4.NPV & IRR

Answer: IRR = 28.41 %

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.

Bracket the IRR

NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.

NPV(25%) = 400.00
NPV(30%) = −177.51

Narrow down

Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.

IRR ≈ 25 + 5·400.00/(400.00 − (−177.51)) = 28.46%
exact: NPV(28.41%) = 0 → IRR = 28.41%

Answer

IRR = 28.41%
5.Bond pricing & yield

Answer: c = 4.996 %

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.

Present value of the face value

1000·(1 + 0.025)−8 = €820.75

Coupon

C = (1179 − 820.75)/7.17 = €49.96

Coupon rate

c = 49.96/1000 = 4.996%

Answer

c = 4.996%
6.Bond pricing & yield

Answer: YTM = 5.456 %

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

€895 = €30·(1 − (1 + y)⁻⁵)/y + €1 000/(1 + y)⁵

Solve numerically

P(5%) = €913.41, P(6%) = €873.63
YTM = 5.456%

Answer

YTM = 5.456%
7.CAPM & WACC

Answer: rm = 8.5 %

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 = (7.2% − 2%)/0.8 = 6.5%

Market return

rm = 2% + 6.5% = 8.5%

Answer

rm = 8.5%
8.CAPM & WACC

Answer: WACC = 7.219 %

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 equity

D1 = 1.6·1.04 = €1.66
re = 1.664/37.5 + 0.04 = 8.437%

Market values

E = 2 000 000·€37.50 = €75 000 000
V = €97 500 000, E/V = 0.7692, D/V = 0.2308

WACC

WACC = 0.7692·8.437% + 0.2308·4%·(1 − 0.21) = 7.219%

Answer

WACC = 7.219%
9.Descriptive statistics

Answer: s = 11.19

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 = 149/5 = 29.8

Squared deviations

Σ(x − x̄)² = (−10.8)² + 11.2² + 3.2² + (−12.8)² + 9.2² = 500.8

Standard deviation

s = √(500.8/4) = 11.19

Answer

s = 11.19
10.Descriptive statistics

Answer: z = −1

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 = (165 − 170)/5 = −1

Answer

z = −1
11.Probability

Answer: P = 0.4242

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.

First and second draw

P = 8/12 · 7/11 = 56/132 = 0.4242

Answer

P = 0.4242
12.Probability

Answer: σY = 1.195

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 = 2.7, Var(X) = 5.71, σX = 2.39

Linear change

σY = |0.5|·2.39 = 1.195

Answer

σY = 1.195

The mean does move: E(Y) = 0.5·2.7 + 5 = 6.35.

13.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 total

μT = 360 g, σT = 1.2·√9 = 3.6 g

Standardise

z = (356.4 − 360)/3.6 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413
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: E₂ = 7.408

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.

Ratio of margins

E₂/E₁ = (1.645/2.576)·√(50/50) = 0.6386

New margin

E₂ = 11.6·0.6386 = 7.408

Answer

E₂ = 7.408
16.Confidence intervals

Answer: σ = 0.8761 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 = 29·1.1² = 35.09

Limits for σ²

35.09/45.722 = 0.7675
35.09/16.047 = 2.187

Limit for σ

σ = √0.7675 = 0.8761 µm

Answer

σ = 0.8761 µm

95% interval for σ: [0.8761, 1.479] µm.

17.Hypothesis tests & confidence intervals

Answer: z = −2.914

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.

Standard error of the difference

SE = √(12²/60 + 5²/80) = √2.712 = 1.647

Test statistic

z = (75.5 − 80.3)/1.647 = −2.914

Answer

z = −2.914

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: L = 111.5

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).

Margin of error

E = 1.645·12/√36 = 3.29

Interval limit

L = 114.8 − 3.29 = 111.5

Answer

L = 111.5
19.Linear regression & correlation

Answer: r² = 0.9618

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 = 42/6 = 7
ȳ = 88/6 = 14.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 148
Syy = Σ(y − ȳ)² = 325.3

Coefficient of determination

r² = 148²/(70·325.3) = 0.9618

Answer

r² = 0.9618

About 96.18% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = 4.25

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 = 68/5 = 13.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 86

Regression line

b = 86/40 = 2.15
a = 13.6 − 2.15·6 = 0.7
ŷ = 0.7 + 2.15·x

Predicted value

ŷ = 0.7 + 2.15·7 = 15.75

Residual

e = 20 − 15.75 = 4.25

Answer

e = 4.25

The point lies above the line.

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