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
years0123456PV = ?€40 000

You need €40 000 in 6 years. An account pays 3% per year, compounded monthly. How much must you deposit today? Find the present value PV.

FV = €40 0003% / 12n = 6
PV = €
2.Time value of moneyHard
months0123…360€200 000AAAA2.4% / 12, I = ?

A loan of €200 000 is repaid with equal monthly payments over 30 years at 2.4% per year, compounded monthly. How much interest I is paid in total over the life of the loan?

P = €200 0002.4% / 12N = 360
I = €
3.NPV & IRRMedium
years0123−45 00018 00016 00019 000

An investment of €45 000 gives these cash flows at the end of years 1 to 3: €18 000, €16 000, €19 000. The cost of capital is 7.000000000000001%. Find the net present value NPV.

r = 7.000000000000001%CF0 = −45 000CF1 = 18 000CF2 = 16 000CF3 = 19 000
NPV = €
4.NPV & IRRHard
r (%)NPVABr* = ?0

Project A costs €30 000 and returns €16 500, €12 000, €6 000 at the end of years 1 to 3. Project B costs €32 000 and returns €8 500, €13 000, €16 500. Find the crossover rate r* at which both projects have the same NPV, in percent.

A: −30 000, 16 500, 12 000, 6 000B: −32 000, 8 500, 13 000, 16 500
r* = %
5.Bond pricing & yieldMedium
years012345P€70€70€70 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 7% and matures in 5 years. Its yield to maturity changes from 6% to 4.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 7%n = 5y1 = 6%y2 = 4.5%
ΔP = €
6.Bond pricing & yieldHard

A bond trades at €900 with a yield to maturity of 5% (annual compounding) and a Macaulay duration of 10.3 years. Use the modified duration to estimate the new price P if the yield falls by 25 basis points.

P0 = €900y = 5%D = 10.3Δy = −25 bp
P = €
7.CAPM & WACCMedium
WACC = ?equity€12 000 000re​ = 10%debt€6 000 000rd​ = 4.5%T = 21%

A company has equity worth €12 000 000 and debt worth €6 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 4.5% and the tax rate is 21%. Find the WACC.

E = €12 000 000D = €6 000 000re = 10%rd = 4.5%T = 21%
WACC = %
8.CAPM & WACCHard
WACC = ?equity4 000 000 × €30.00D0​ = €0.60g = 4%debt€36 000 000rd​ = 5%T = 25.8%

A company has 4 000 000 shares trading at €30.00; the book value of its equity is €36 000 000. It just paid a dividend of €0.60 per share, and dividends grow by 4% per year. Its debt has a market value of €36 000 000 and a pre-tax cost of 5%. The tax rate is 25.8%. Find the WACC, using the dividend growth model for the cost of equity.

N = 4 000 000P0 = €30.00book equity = €36 000 000D0 = €0.60g = 4%D = €36 000 000rd = 5%T = 25.8%
WACC = %
9.Descriptive statisticsMedium
202530354045505560

Find the sample standard deviation s of the data: 26, 60, 31, 20, 34, 26, 29.

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

A quantity is normally distributed with mean 170 and standard deviation 8. What proportion P of values is below 163.6?

μ = 170σ = 8x = 163.6
P =
11.ProbabilityMedium
xp(x)00.3520.240.1560.3μ = 2.8

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

x: 0, 2, 4, 6p(x): 0.35, 0.2, 0.15, 0.3
Var(X) =
12.ProbabilityHard
€p(x)−50.75100.22450.05

A game costs €5 to play. You win €15 with probability 0.2 and €250 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€15: p = 0.2€250: p = 0.05
E(G) = €
13.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.0778

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.0778. Find z to two decimals.

A = 0.0778
z =
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
n = 21, s = 9.5LUx̄ = 10.5E = ?90%

A sample of 21 items has mean 10.5 and standard deviation s = 9.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 21x̄ = 10.5s = 9.590%t(20) = 1.725
E =
16.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 0.599%

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 0.5?

σ = 31E = 0.599%
n =
17.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 66.14 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 75.

μ₀ = 75x̄ = 66.14s = 20n = 49
t =
18.Hypothesis tests & confidence intervalsHard
z −2.57602.576

A sample of 64 measurements has mean 119.4. The population standard deviation is 4. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 119.4σ = 4n = 6499%
U =
19.Linear regression & correlationMedium
xy1423465537619681176xy510406080

The table gives 6 paired observations of x = hours studied and y = exam score. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy123220317416512xy241520x = 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 = 33418 €

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 = 12·6 = 72

Present value

PV = FV/(1 + i)N = 40 000/1.197 = €33 418.31

Answer

PV = €33 418.31
2.Time value of money

Answer: I = 80758 €

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 = 0.002, N = 360
A = 200 000·0.002/(1 − (1 + 0.002)−360) = €779.88

Total interest

I = N·A − P = 360·779.9 − 200 000 = €80 757.78

Answer

I = €80 757.78
3.NPV & IRR

Answer: NPV = 1307 €

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 = −45 000
CF1/(1 + 0.07)1 = 18 000/1.07 = 16 822.43
CF2/(1 + 0.07)2 = 16 000/1.145 = 13 975.02
CF3/(1 + 0.07)3 = 19 000/1.225 = 15 509.66

Add them up

NPV = €1 307.11

Answer

NPV = €1 307.11

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

4.NPV & IRR

Answer: r* = 6.829 %

Crossover rate: the discount rate at which two projects have the same NPV. It is the IRR of the difference in their cash flows (B − A). Below it the project with the later, larger cash flows has the higher NPV; above it the other one wins.

Difference in cash flows

ΔCF0 = −32 000 − (−30 000) = −2 000
ΔCF1 = 8 500 − 16 500 = −8 000
ΔCF2 = 13 000 − 12 000 = 1 000
ΔCF3 = 16 500 − 6 000 = 10 500

IRR of the difference

NPVΔ(6%) = 158.83
NPVΔ(7%) = −32.07
r* = 6.829%

Answer

r* = 6.829%

Check: at 6.829% both NPVs are €881.47.

5.Bond pricing & yield

Answer: ΔP = 67.63 €

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 = 0.06, N = 5
€70.00·(1 − 0.7473)/0.06 + €1 000·0.7473 = €294.87 + €747.26 = €1 042.12

Price at the new yield

C = €70.00, i = 0.045, N = 5
€70.00·(1 − 0.8025)/0.045 + €1 000·0.8025 = €307.30 + €802.45 = €1 109.75

Change

ΔP = 1 109.75 − 1 042.12 = €67.63

Answer

ΔP = €67.63
6.Bond pricing & yield

Answer: P = 922.1 €

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 = 10.3/1.05 = 9.81

Estimated price change

%ΔP ≈ −9.81·(−0.0025) = 2.452%
P ≈ 900·(1 + 0.02452) = €922.07

Answer

P = €922.07

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

7.CAPM & WACC

Answer: WACC = 7.852 %

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 = €18 000 000, E/V = 0.6667, D/V = 0.3333

WACC

WACC = 0.6667·10% + 0.3333·4.5%·(1 − 0.21) = 7.852%

Answer

WACC = 7.852%
8.CAPM & WACC

Answer: WACC = 5.533 %

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 = 0.6·1.04 = €0.62
re = 0.624/30 + 0.04 = 6.08%

Market values

E = 4 000 000·€30.00 = €120 000 000
V = €156 000 000, E/V = 0.7692, D/V = 0.2308

WACC

WACC = 0.7692·6.08% + 0.2308·5%·(1 − 0.258) = 5.533%

Answer

WACC = 5.533%
9.Descriptive statistics

Answer: s = 13

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 = 226/7 = 32.29

Squared deviations

Σ(x − x̄)² = (−6.286)² + 27.71² + (−1.286)² + (−12.29)² + 1.714² + (−6.286)² + (−3.286)² = 1013

Standard deviation

s = √(1013/6) = 13

Answer

s = 13
10.Descriptive statistics

Answer: P = 0.2119

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 = −0.8

Standard normal table

Φ(−0.8) = 0.2119
P = 0.2119

Answer

P = 0.2119
11.Probability

Answer: Var(X) = 6.16

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 + 2·0.2 + 4·0.15 + 6·0.3 = 2.8

Mean of the squares

E(X²) = 0·0.35 + 4·0.2 + 16·0.15 + 36·0.3 = 14

Variance

Var(X) = 14 − 2.8² = 6.16

Answer

Var(X) = 6.16
12.Probability

Answer: E(G) = 10.5 €

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: 10 (p = 0.2), 245 (p = 0.05), −5 (p = 0.75)

Expected value

E(G) = 10·0.2 + 245·0.05 − 5·0.75 = 10.5 €

Answer

E(G) = 10.5 €

A positive value means you win money on average.

13.Normal distribution

Answer: z = 1.42

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 − 0.0778 = 0.9222

Read the z-table backwards

Find 0.9222 inside the table and read off z.

z = 1.42

Answer

z = 1.42
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: E = 3.576

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 = 20, t = 1.725

Margin of error

E = 1.725·9.5/√21 = 3.576

Answer

E = 3.576
16.Confidence intervals

Answer: n = 25508

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 = (2.576·31/0.5)² = 25508

Round up

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

n = 25508

Answer

n = 25508
17.Hypothesis tests & confidence intervals

Answer: t = −3.101

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/√49 = 20/7 = 2.857

Test statistic

t = (66.14 − 75)/2.857 = −3.101

Answer

t = −3.101
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: U = 120.7

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).

Margin of error

E = 2.576·4/√64 = 1.288

Interval limit

U = 119.4 + 1.288 = 120.7

Answer

U = 120.7
19.Linear regression & correlation

Answer: r = 0.996

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 = 36/6 = 6
ȳ = 346/6 = 57.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 244
Syy = Σ(y − ȳ)² = 857.3

Correlation coefficient

r = 244/√(70·857.3) = 0.996

Answer

r = 0.996
20.Linear regression & correlation

Answer: ŷ = 16.3

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 = 88/5 = 17.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −26

Regression line

b = −26/10 = −2.6
a = 17.6 − (−2.6)·3 = 25.4
ŷ = 25.4 − 2.6·x

Prediction

ŷ = 25.4 + (−2.6)·3.5 = 16.3

Answer

ŷ = 16.3

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