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…9€10 000FV = ?

You invest €10 000 at 8% per year with continuous compounding. Find the value FV after 9 years.

PV = €10 000r = 8%t = 9
FV = €
2.Time value of moneyHard
years0123…24€6 000€6 000€6 000FV = ?

You deposit €6 000 at the start of every year for 24 years, the first deposit today. The account pays 8% per year. How much is in the account at the end of year 24? Find FV.

A = €6 000i = 8%n = 24
FV = €
3.NPV & IRRMedium
years012345−40 0008 0008 0008 0008 00014 000

A machine costs €40 000 today. It generates €8 000 at the end of each year for 5 years and can then be sold for €6 000. The cost of capital is 6%. Find the net present value NPV.

C0 = €40 000CF = €8 000S = €6 000n = 5r = 6%
NPV = €
4.NPV & IRRHard
years0123−10 0006 5006 5005 000

A project costs €10 000 and returns €6 500, €6 500, €5 000 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 6 500CF2 = 6 500CF3 = 5 000
IRR = %
5.Bond pricing & yieldMedium
years0123…11€1 278C = ?C = ?C = ? + €1 000

A bond with a face value of €1 000 matures in 11 years and pays annual coupons. It sells for €1 278 when the yield to maturity is 3%. Find the coupon rate c in percent.

F = €1 000n = 11P = €1 278YTM = 3%
c = %
6.Bond pricing & yieldHard
years0123€1 106€50€50€50 + €1 050

A bond with 14 years to maturity (face value €1 000, annual coupon 5%) sells for €1 106. The issuer can call it in 3 years at a call price of €1 050. Find the yield to call YTC in percent.

F = €1 000coupon = 5%P = €1 106call price = €1 050call in 3 years
YTC = %
7.CAPM & WACCMedium
WACC = ?ordinary shares€6 000 000re​ = 13%preferred shares€2 000 000rp​ = 8%debt€5 000 000rd​ = 5%T = 30%

A company is financed with ordinary shares worth €6 000 000, preferred shares worth €2 000 000 and debt worth €5 000 000. The cost of equity is 13%, the cost of preferred stock 8% and the pre-tax cost of debt 5%. The tax rate is 30%. Find the WACC.

E = €6 000 000Ps = €2 000 000D = €5 000 000re = 13%rp = 8%rd = 5%T = 30%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €49.50D0​ = €2.30g = 2%debt€59 400 000rd​ = 4.5%T = 21%

A company has 2 000 000 shares trading at €49.50; the book value of its equity is €39 600 000. It just paid a dividend of €2.30 per share, and dividends grow by 2% per year. Its debt has a market value of €59 400 000 and a pre-tax cost of 4.5%. The tax rate is 21%. Find the WACC, using the dividend growth model for the cost of equity.

N = 2 000 000P0 = €49.50book equity = €39 600 000D0 = €2.30g = 2%D = €59 400 000rd = 4.5%T = 21%
WACC = %
9.Descriptive statisticsMedium
202530354045

Find the sample standard deviation s of the data: 35, 33, 20, 45, 27, 49.

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

Exam scores are normally distributed with mean 100 and standard deviation 20. Find the z-score z of a score of 60.

μ = 100σ = 20x = 60
z =
11.ProbabilityMedium
xp(x)10.320.130.0540.2550.3μ = 3.15

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

x: 1, 2, 3, 4, 5p(x): 0.3, 0.1, 0.05, 0.25, 0.3
σ =
12.ProbabilityHard
xp(x)00.3510.320.35μ

X has the distribution shown, and Y = 0.5X + 10. Find the standard deviation σY of Y.

x: 0, 1, 2p(x): 0.35, 0.3, 0.35Y = 0.5X + 10
σY =
13.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.8064

Drag z so that the shaded area between −z and z under the standard normal curve is 0.8064. Find z to two decimals.

A = 0.8064
z =
14.Normal distributionHard
0.04 mm0D − d?

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.015 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.04 mm, σ = 0.015 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
15.Confidence intervalsMedium
n = 20, s = 2.5L = ?Ux̄ = 54E90%

A sample of 20 items has mean 54 and standard deviation s = 2.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 20x̄ = 54s = 2.590%t(19) = 1.729
lower limit =
16.Confidence intervalsHard
σ = 23, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 23. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 4?

σ = 23E = 495%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.382.38p = ?

The mean of a population is claimed to be 80; the population standard deviation is known to be 2.5. A random sample of 25 observations has a mean of 78.81. Test H₀: μ = 80 against H₁: μ ≠ 80. Find the p-value p.

μ₀ = 80σ = 2.5n = 25x̄ = 78.81H₁: μ ≠ μ₀
p =
18.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 25 measurements has mean 498.2. The population standard deviation is 15. Find the lower limit L of the 90% confidence interval for the mean.

x̄ = 498.2σ = 15n = 2590%
L =
19.Linear regression & correlationMedium
xy152254351462560xy245560

The table gives 5 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
20.Linear regression & correlationHard
xy127222321422518617xy2462025x = 5.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (5.5, 15) is made. Find its residual e = y − ŷ.

n = 6(5.5, 15)
e =

3Solutions

1.Time value of money

Answer: FV = 20544 €

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 = 0.08·9 = 0.72
e0.72 = 2.054

Future value

FV = 10 000·2.054 = €20 544.33

Answer

FV = €20 544.33
2.Time value of money

Answer: FV = 432636 €

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 + 0.08)24 − 1)/0.08 = 66.76

Annuity due

Each deposit is invested one year longer.

FV = 6 000·66.76·1.08 = €432 635.64

Answer

FV = €432 635.64
3.NPV & IRR

Answer: NPV = −1818 €

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

8 000·4.212 = 33 698.91

Present value of the salvage value

6 000/1.338 = 4 483.55

NPV

NPV = −40 000 + 33 698.91 + 4 483.55 = −€1 817.54

Answer

NPV = −€1 817.54
4.NPV & IRR

Answer: IRR = 38.21 %

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(35%) = 413.55
NPV(40%) = −218.66

Narrow down

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

IRR ≈ 35 + 5·413.55/(413.55 − (−218.66)) = 38.27%
exact: NPV(38.21%) = 0 → IRR = 38.21%

Answer

IRR = 38.21%
5.Bond pricing & yield

Answer: c = 6.005 %

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 + 0.03)−11 = €722.42

Coupon

C = (1278 − 722.42)/9.253 = €60.05

Coupon rate

c = 60.05/1000 = 6.005%

Answer

c = 6.005%
6.Bond pricing & yield

Answer: YTC = 2.881 %

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

1106 = 50·(1 − (1 + y)−3)/y + 1050/(1 + y)3

Solve numerically

y = 2%: €1 133.63
y = 3%: €1 102.33
YTC = 2.881%

Answer

YTC = 2.881%

The yield to maturity is 3.997%. 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 = 8.577 %

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 = 6 000 000 + 2 000 000 + 5 000 000 = €13 000 000
E/V = 0.4615, Ps/V = 0.1538, D/V = 0.3846

WACC

WACC = 0.4615·13% + 0.1538·8% + 0.3846·5%·(1 − 0.3) = 8.577%

Answer

WACC = 8.577%
8.CAPM & WACC

Answer: WACC = 5.545 %

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 = 2.3·1.02 = €2.35
re = 2.346/49.5 + 0.02 = 6.739%

Market values

E = 2 000 000·€49.50 = €99 000 000
V = €158 400 000, E/V = 0.625, D/V = 0.375

WACC

WACC = 0.625·6.739% + 0.375·4.5%·(1 − 0.21) = 5.545%

Answer

WACC = 5.545%
9.Descriptive statistics

Answer: s = 10.85

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 = 209/6 = 34.83

Squared deviations

Σ(x − x̄)² = 0.1667² + (−1.833)² + (−14.83)² + 10.17² + (−7.833)² + 14.17² = 588.8

Standard deviation

s = √(588.8/5) = 10.85

Answer

s = 10.85
10.Descriptive statistics

Answer: z = −2

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 = (60 − 100)/20 = −2

Answer

z = −2
11.Probability

Answer: σ = 1.652

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 = 1·0.3 + 2·0.1 + 3·0.05 + 4·0.25 + 5·0.3 = 3.15

Mean of the squares

E(X²) = 1·0.3 + 4·0.1 + 9·0.05 + 16·0.25 + 25·0.3 = 12.65

Variance

Var(X) = 12.65 − 3.15² = 2.728
σ = √2.728 = 1.652

Answer

σ = 1.652
12.Probability

Answer: σY = 0.4183

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, Var(X) = 0.7, σX = 0.8367

Linear change

σY = |0.5|·0.8367 = 0.4183

Answer

σY = 0.4183

The mean does move: E(Y) = 0.5·1 + 10 = 10.5.

13.Normal distribution

Answer: z = 1.3

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 + 0.8064)/2 = 0.9032

Read the z-table backwards

Find 0.9032 inside the table and read off z.

z = 1.30

Answer

z = 1.30
14.Normal distribution

Answer: P = 0.9868

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 = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.04)/0.01803 = −2.22

Probability

P = 1 − Φ(−2.22) = 0.9868

Answer

P = 0.9868
15.Confidence intervals

Answer: lower limit = 53.03

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 = 19, t = 1.729

Margin of error

E = 1.729·2.5/√20 = 0.9665

Interval

54 ± 0.9665 → [53.03, 54.97]

Answer

lower limit = 53.03
16.Confidence intervals

Answer: n = 128

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 = (1.96·23/4)² = 127

Round up

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

n = 128

Answer

n = 128
17.Hypothesis tests & confidence intervals

Answer: p = 0.01731

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 = 2.5/√25 = 0.5
z = (78.81 − 80)/0.5 = −2.38

Tail area

Φ(2.38) = 0.9913
1 − Φ(2.38) = 0.0087

p-value

p = 2·0.0087 = 0.01731

Answer

p = 0.01731

p ≤ 0.05, so H₀ is rejected at the 5% level.

18.Hypothesis tests & confidence intervals

Answer: L = 493.3

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 = 1.645·15/√25 = 4.935

Interval limit

L = 498.2 − 4.935 = 493.3

Answer

L = 493.3
19.Linear regression & correlation

Answer: r² = 0.595

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 = 279/5 = 55.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 24
Syy = Σ(y − ȳ)² = 96.8

Coefficient of determination

r² = 24²/(10·96.8) = 0.595

Answer

r² = 0.595

About 59.5% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: e = −2.681

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 = 21/6 = 3.5
ȳ = 127/6 = 21.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −30.5

Regression line

b = −30.5/17.5 = −1.743
a = 21.17 − (−1.743)·3.5 = 27.27
ŷ = 27.27 − 1.743·x

Predicted value

ŷ = 27.27 + (−1.743)·5.5 = 17.68

Residual

e = 15 − 17.68 = −2.681

Answer

e = −2.681

The point lies below the line.

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