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…8€10 000FV = ?

You invest €10 000 at 3% per year with continuous compounding. Find the value FV after 8 years.

PV = €10 000r = 3%t = 8
FV = €
2.Time value of moneyHard
months0123…36A?AA€10 000

You want €10 000 in 3 years. You save the same amount at the end of every month in an account paying 3.6% per year, compounded monthly. Find the monthly deposit A.

FV = €10 0003.6% / 12N = 36
A = €
3.NPV & IRRMedium
years0123…8−80 00015 00015 00015 00035 000

A machine costs €80 000 today. It generates €15 000 at the end of each year for 8 years and can then be sold for €20 000. The cost of capital is 8%. Find the net present value NPV.

C0 = €80 000CF = €15 000S = €20 000n = 8r = 8%
NPV = €
4.NPV & IRRHard
years012−40 00024 00025 000

A project costs €40 000 and returns €24 000, €25 000 at the end of years 1 to 2. Find its internal rate of return IRR in percent.

CF0 = −40 000CF1 = 24 000CF2 = 25 000
IRR = %
5.Bond pricing & yieldMedium
years0123…11P€70€70€70 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 7% and matures in 11 years. Its yield to maturity changes from 3.5% to 2.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 7%n = 11y1 = 3.5%y2 = 2.5%
ΔP = €
6.Bond pricing & yieldHard
half-years0123…8€892€10€10€10 + €1 000

A bond with a face value of €1 000 pays a 2% annual coupon in two half-yearly payments and matures in 4 years. It sells for €892. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 2% (2×)n = 4P = €892
YTM = %
7.CAPM & WACCMedium
WACC = ?equityEre​ = 11%debtD = 1.5·Erd​ = 4%T = 25%

A company has a debt-to-equity ratio (at market values) of 1.5. Its cost of equity is 11%, its pre-tax cost of debt is 4% and the tax rate is 25%. Find the WACC.

D/E = 1.5re = 11%rd = 4%T = 25%
WACC = %
8.CAPM & WACCHard
rf​ = 2.5%, rm​ − rf​ = 7%D/E = 0.5equityβL​ = 1.5debtD/E = 1equityre​ = ?debtT = 25%

A company's shares have a beta of 1.5 at its current debt-to-equity ratio of 0.5. It plans to raise its debt-to-equity ratio to 1. The tax rate is 25%, the risk-free rate 2.5% and the market risk premium 7%. Find the cost of equity re after the change.

βL = 1.5D/E1 = 0.5D/E2 = 1T = 25%rf = 2.5%rm − rf = 7%
re = %
9.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 14, 22, 19, 37, 59, 20.

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

Exam scores are normally distributed with mean 100 and standard deviation 12. Find the z-score z of a score of 90.4.

μ = 100σ = 12x = 90.4
z =
11.ProbabilityMedium
xp(x)100.3200.25300.25400.2μ = 23.5

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

x: 10, 20, 30, 40p(x): 0.3, 0.25, 0.25, 0.2
σ =
12.ProbabilityHard
xp(x)00.310.3520.0530.240.1μ

X has the distribution shown, and Y = 5X + 4. Find the standard deviation σY of Y.

x: 0, 1, 2, 3, 4p(x): 0.3, 0.35, 0.05, 0.2, 0.1Y = 5X + 4
σY =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8664

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8664. Find z to two decimals.

A = 0.8664
z =
14.Normal distributionHard
235240245250255260265gbottom 75%

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Drag the cut-off so the shaded bottom 75% is to the left of it. Find the cut-off x.

μ = 250 gσ = 5 g75%
x = g
15.Confidence intervalsMedium
x̄E₁ = 4.295%, n = 100x̄E₂ = ?99%, n = 200

A 95% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 4.2. What is the margin of error E₂ of a 99% interval based on 200 observations?

95%n₁ = 100E₁ = 4.299%n₂ = 200
E₂ =
16.Confidence intervalsHard
σ = 34, n = ?x̄ − Ex̄ + Ex̄E = 0.599%

The population standard deviation is 34. 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 = 34E = 0.599%
n =
17.Hypothesis tests & confidence intervalsMedium
−0.96p = ?

The mean of a population is claimed to be 50; the population standard deviation is known to be 12. A random sample of 144 observations has a mean of 49.04. Test H₀: μ = 50 against H₁: μ < 50. Find the p-value p.

μ₀ = 50σ = 12n = 144x̄ = 49.04H₁: μ < μ₀
p =
18.Hypothesis tests & confidence intervalsHard
z −1.9601.96

A sample of 100 measurements has mean 51.4. The population standard deviation is 12. Find the upper limit U of the 95% confidence interval for the mean.

x̄ = 51.4σ = 12n = 10095%
U =
19.Linear regression & correlationMedium
xy15831780191022110623126xy152080100120

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy24446118111017xy51051015x = 3

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. Predict ŷ for x = 3.

n = 5x = 3
ŷ =

3Solutions

1.Time value of money

Answer: FV = 12712 €

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.

Exponent

r·t = 0.03·8 = 0.24
e0.24 = 1.271

Future value

FV = 10 000·1.271 = €12 712.49

Answer

FV = €12 712.49
2.Time value of money

Answer: A = 263.5 €

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 = 0.036000000000000004/12 = 0.003
N = 12·3 = 36

Deposit

A = FV·i/((1 + i)N − 1) = 10 000·0.003/(1.114 − 1) = €263.46

Answer

A = €263.46
3.NPV & IRR

Answer: NPV = 17005 €

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 yearly cash flows

15 000·5.747 = 86 199.58

Present value of the salvage value

20 000/1.851 = 10 805.38

NPV

NPV = −80 000 + 86 199.58 + 10 805.38 = €17 004.96

Answer

NPV = €17 004.96
4.NPV & IRR

Answer: IRR = 14.56 %

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(10%) = 2 479.34
NPV(15%) = −226.84

Narrow down

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

IRR ≈ 10 + 5·2 479.34/(2 479.34 − (−226.84)) = 14.58%
exact: NPV(14.56%) = 0 → IRR = 14.56%

Answer

IRR = 14.56%
5.Bond pricing & yield

Answer: ΔP = 113.1 €

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.

Price at the old yield

C = €70.00, i = 0.035, N = 11
€70.00·(1 − 0.6849)/0.035 + €1 000·0.6849 = €630.11 + €684.95 = €1 315.05

Price at the new yield

C = €70.00, i = 0.025, N = 11
€70.00·(1 − 0.7621)/0.025 + €1 000·0.7621 = €665.99 + €762.14 = €1 428.14

Change

ΔP = 1 428.14 − 1 315.05 = €113.09

Answer

ΔP = €113.09
6.Bond pricing & yield

Answer: YTM = 5.013 %

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.

Per period

C = €10.00, N = 8

Solve numerically

i = 2.507% per half-year

YTM

YTM = 2·2.507% = 5.013%

Answer

YTM = 5.013%
7.CAPM & WACC

Answer: WACC = 6.2 %

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 + 1.5) = 0.4
D/V = 1.5/(1 + 1.5) = 0.6

WACC

WACC = 0.4·11% + 0.6·4%·(1 − 0.25) = 6.2%

Answer

WACC = 6.2%
8.CAPM & WACC

Answer: re = 15.86 %

Levered beta (Hamada): debt makes equity riskier. βL = βU·(1 + (1 − T)·D/E), where βU is the beta of the business without debt. To move a beta to another capital structure, unlever it first and then relever it at the new D/E.

Unlever the beta

βU = 1.5/(1 + 0.75·0.5) = 1.091

Relever at the new D/E

βL = 1.091·(1 + 0.75·1) = 1.909

Cost of equity (CAPM)

re = 2.5% + 1.909·7% = 15.86%

Answer

re = 15.86%
9.Descriptive statistics

Answer: s = 16.84

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 = 171/6 = 28.5

Squared deviations

Σ(x − x̄)² = (−14.5)² + (−6.5)² + (−9.5)² + 8.5² + 30.5² + (−8.5)² = 1418

Standard deviation

s = √(1418/5) = 16.84

Answer

s = 16.84
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 = (90.4 − 100)/12 = −0.8

Answer

z = −0.8
11.Probability

Answer: σ = 11.08

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 = 10·0.3 + 20·0.25 + 30·0.25 + 40·0.2 = 23.5

Mean of the squares

E(X²) = 100·0.3 + 400·0.25 + 900·0.25 + 1600·0.2 = 675

Variance

Var(X) = 675 − 23.5² = 122.8
σ = √122.8 = 11.08

Answer

σ = 11.08
12.Probability

Answer: σY = 6.796

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 = 1.45, Var(X) = 1.847, σX = 1.359

Linear change

σY = |5|·1.359 = 6.796

Answer

σY = 6.796

The mean does move: E(Y) = 5·1.45 + 4 = 11.25.

13.Normal distribution

Answer: z = 1.5

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 + 0.8664)/2 = 0.9332

Read the z-table backwards

Find 0.9332 inside the table and read off z.

z = 1.50

Answer

z = 1.50
14.Normal distribution

Answer: x = 253.4 g

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

Φ(z) = 0.75

z from the table

z ≈ 0.67

Back to the original units

x = 250 + 0.67·5 = 253.4 g

Answer

x = 253.4 g
15.Confidence intervals

Answer: E₂ = 3.903

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 = (2.576/1.96)·√(100/200) = 0.9293

New margin

E₂ = 4.2·0.9293 = 3.903

Answer

E₂ = 3.903
16.Confidence intervals

Answer: n = 30684

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 = (2.576·34/0.5)² = 30684

Round up

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

n = 30684

Answer

n = 30684
17.Hypothesis tests & confidence intervals

Answer: p = 0.1685

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.

Test statistic

SE = σ/√n = 12/√144 = 1
z = (49.04 − 50)/1 = −0.96

Tail area

Φ(0.96) = 0.8315
1 − Φ(0.96) = 0.1685

p-value

p = 0.1685

Answer

p = 0.1685

p > 0.05, so H₀ is not rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: U = 53.75

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 = 1.96·12/√100 = 2.352

Interval limit

U = 51.4 + 2.352 = 53.75

Answer

U = 53.75
19.Linear regression & correlation

Answer: r² = 0.894

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 = 95/5 = 19
ȳ = 497/5 = 99.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 224
Syy = Σ(y − ȳ)² = 1403

Coefficient of determination

r² = 224²/(40·1403) = 0.894

Answer

r² = 0.894

About 89.4% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 4.45

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 = 47/5 = 9.4

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 66

Regression line

b = 66/40 = 1.65
a = 9.4 − 1.65·6 = −0.5
ŷ = −0.5 + 1.65·x

Prediction

ŷ = −0.5 + 1.65·3 = 4.45

Answer

ŷ = 4.45

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