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…10PV = ?€20 000

You need €20 000 in 10 years. An account pays 3.6% per year, compounded quarterly. How much must you deposit today? Find the present value PV.

FV = €20 0003.6% / 4n = 10
PV = €
2.Time value of moneyHard
months0123…300€150 000A?AAA

A loan of €150 000 is repaid with equal monthly payments over 25 years. The interest rate is 3% per year, compounded monthly. Find the monthly payment A.

P = €150 0003% / 12N = 300
A = €
3.NPV & IRRMedium
years01234−30 00011 00011 00011 00011 000

A project costs €30 000 today and returns €11 000 at the end of each year for 4 years. The cost of capital is 12%. Assume cash flows arrive evenly through each year. Find the discounted payback period DPB in years.

r = 12%CF0 = −30 000CF1 = 11 000CF2 = 11 000CF3 = 11 000CF4 = 11 000
DPB = years
4.NPV & IRRHard
years0123−40 00016 00021 00017 500

A project costs €40 000 and returns €16 000, €21 000, €17 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −40 000CF1 = 16 000CF2 = 21 000CF3 = 17 500
IRR = %
5.Bond pricing & yieldMedium
years0123…11P€20€20€20 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 2% and matures in 11 years. Its yield to maturity changes from 5.5% to 6.5%. By how much ΔP does the bond price change?

F = €1 000coupon = 2%n = 11y1 = 5.5%y2 = 6.5%
ΔP = €
6.Bond pricing & yieldHard
years0123456P€50€50€50 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 5% and matures in 6 years. The yield to maturity is 7%. Find the Macaulay duration D in years.

F = €1 000coupon = 5%n = 6YTM = 7%
D = years
7.CAPM & WACCMedium
βr (%)SMLrf​ = 3.5%β = 1rm​ = ?β = 1.6re​ = 13.1%

A share with a beta of 1.6 has a required return of 13.1%. The risk-free rate is 3.5%. What expected market return rm does CAPM imply?

β = 1.6re = 13.1%rf = 3.5%
rm = %
8.CAPM & WACCHard
WACC = ?equity2 500 000 × €10.50β = 0.9rf​ = 2.5%, rm​ = 7.5%debt10 000 × €1 000rd​ = 5%T = 20%

A company has 2 500 000 shares at €10.50 each and 10 000 bonds trading at €1 000 each. The pre-tax cost of debt is 5% and the tax rate is 20%. The share’s beta is 0.9, the risk-free rate 2.5% and the market return 7.5%. Find the WACC.

β = 0.9rf = 2.5%rm = 7.5%rd = 5%T = 20%
WACC = %
9.Descriptive statisticsMedium
25303540455055

Find the sample standard deviation s of the data: 43, 33, 57, 26, 37, 21, 52.

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

A quantity is normally distributed with mean 100 and standard deviation 8. What proportion P of values is below 84?

μ = 100σ = 8x = 84
P =
11.ProbabilityMedium
xp(x)00.1510.3520.230.3μ = 1.65

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

x: 0, 1, 2, 3p(x): 0.15, 0.35, 0.2, 0.3
Var(X) =
12.ProbabilityHard
€p(x)−20.8540.1180.05

A game costs €2 to play. You win €6 with probability 0.1 and €20 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €2€6: p = 0.1€20: p = 0.05
E(G) = €
13.Normal distributionMedium
5780P = ?

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. What fraction lies between 57 and 80 points? Give P as a decimal.

μ = 65σ = 10x1 = 57x2 = 80
P =
14.Normal distributionHard
μ = 100x = ?80%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 80% of them lie (the 80th percentile).

μ = 100σ = 15P = 80%
x = points
15.Confidence intervalsMedium
n = 10, s = 1.5LUx̄ = 106E = ?99%

A sample of 10 items has mean 106 and standard deviation s = 1.5. The population is roughly normal. Find the margin of error E of a 99% confidence interval for the mean.

n = 10x̄ = 106s = 1.599%t(9) = 3.25
E =
16.Confidence intervalsHard
σ = 5, n = ?x̄ − Ex̄ + Ex̄E = 1.590%

The population standard deviation is 5. How large a sample n do you need so that a 90% confidence interval for the mean has a margin of error of at most 1.5?

σ = 5E = 1.590%
n =
17.Hypothesis tests & confidence intervalsMedium
t −1.9901.99

A sample of 81 students has a mean score of 117.1 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 117.1s = 12n = 81
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 80 against H₁: μ > 80 is done at α = 0.05 with a sample of 25; the standard deviation is known to be 15. In reality the mean is 86.6. Find the power 1 − β of the test in percent.

μ₀ = 80μ = 86.6σ = 15n = 25α = 0.05
1 − β = %
19.Linear regression & correlationMedium
xy178919108211222312825146xy2025100120140

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 105Σy = 593Σx² = 2245Σy² = 72169Σxy = 12721
r =
20.Linear regression & correlationHard
xy151369586796911111124xy51050100ŷ = 75

The table gives 6 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 = 75?

n = 6ŷ = 75
x =

3Solutions

1.Time value of money

Answer: PV = 13976 €

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 = 0.036000000000000004/4 = 0.009
N = 4·10 = 40

Present value

PV = FV/(1 + i)N = 20 000/1.431 = €13 976.01

Answer

PV = €13 976.01
2.Time value of money

Answer: A = 711.3 €

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 payments

i = 0.03/12 = 0.0025
N = 12·25 = 300

Payment

The present value of N equal payments A is A·(1 − (1 + i)⁻ᴺ)/i. Set it equal to the loan and solve for A.

(1 + i)−N = 0.4728
A = 150 000·0.0025/(1 − 0.4728) = €711.32

Answer

A = €711.32
3.NPV & IRR

Answer: DPB = 3.512 years

Payback period: the time until the cumulative cash flows repay the initial investment. Cash is assumed to come in evenly during each year, so the last part is a fraction of a year. The simple payback ignores the time value of money; the discounted payback period uses the discounted cash flows instead.

Discounted and cumulative cash flows

year 1: 11 000/1.12 = 9 821.43 → Σ = 9 821.43
year 2: 11 000/1.254 = 8 769.13 → Σ = 18 590.56
year 3: 11 000/1.405 = 7 829.58 → Σ = 26 420.14
year 4: 11 000/1.574 = 6 990.70 → Σ = 33 410.84

Fraction of the last year

DPB = 3 + 3 579.86/6 990.70 = 3.512

Answer

DPB = 3.512 years

The discounted payback is always longer than the simple payback (2.727 years).

4.NPV & IRR

Answer: IRR = 16.91 %

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(15%) = 1 298.59
NPV(20%) = −1 956.02

Narrow down

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

IRR ≈ 15 + 5·1 298.59/(1 298.59 − (−1 956.02)) = 17%
exact: NPV(16.91%) = 0 → IRR = 16.91%

Answer

IRR = 16.91%
5.Bond pricing & yield

Answer: ΔP = −62.77 €

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 = €20.00, i = 0.055, N = 11
€20.00·(1 − 0.5549)/0.055 + €1 000·0.5549 = €161.85 + €554.91 = €716.76

Price at the new yield

C = €20.00, i = 0.065, N = 11
€20.00·(1 − 0.5002)/0.065 + €1 000·0.5002 = €153.78 + €500.21 = €653.99

Change

ΔP = 653.99 − 716.76 = −€62.77

Answer

ΔP = −€62.77
6.Bond pricing & yield

Answer: D = 5.29 years

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.

Present value of each cash flow

t = 1: 50/1.07 = 46.73, t·PV = 46.73
t = 2: 50/1.145 = 43.67, t·PV = 87.34
t = 3: 50/1.225 = 40.81, t·PV = 122.44
t = 4: 50/1.311 = 38.14, t·PV = 152.58
t = 5: 50/1.403 = 35.65, t·PV = 178.25
t = 6: 1050/1.501 = 699.66, t·PV = 4 197.96

Duration

P = 904.67
D = 4 785.30/904.67 = 5.29

Answer

D = 5.29 years
7.CAPM & WACC

Answer: rm = 9.5 %

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 = (13.1% − 3.5%)/1.6 = 6%

Market return

rm = 3.5% + 6% = 9.5%

Answer

rm = 9.5%
8.CAPM & WACC

Answer: WACC = 6.172 %

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 = 2 500 000·€10.50 = €26 250 000
D = 10 000·€1 000 = €10 000 000

Cost of equity (CAPM)

re = 2.5% + 0.9·5% = 7%

Weights

V = €36 250 000, E/V = 0.7241, D/V = 0.2759

WACC

WACC = 0.7241·7% + 0.2759·5%·(1 − 0.2) = 6.172%

Answer

WACC = 6.172%
9.Descriptive statistics

Answer: s = 13.16

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 = 269/7 = 38.43

Squared deviations

Σ(x − x̄)² = 4.571² + (−5.429)² + 18.57² + (−12.43)² + (−1.429)² + (−17.43)² + 13.57² = 1040

Standard deviation

s = √(1040/6) = 13.16

Answer

s = 13.16
10.Descriptive statistics

Answer: P = 0.02275

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 0.02275

Answer

P = 0.02275
11.Probability

Answer: Var(X) = 1.127

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 = 0·0.15 + 1·0.35 + 2·0.2 + 3·0.3 = 1.65

Mean of the squares

E(X²) = 0·0.15 + 1·0.35 + 4·0.2 + 9·0.3 = 3.85

Variance

Var(X) = 3.85 − 1.65² = 1.127

Answer

Var(X) = 1.127
12.Probability

Answer: E(G) = −0.4 €

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: 4 (p = 0.1), 18 (p = 0.05), −2 (p = 0.85)

Expected value

E(G) = 4·0.1 + 18·0.05 − 2·0.85 = −0.4 €

Answer

E(G) = −0.4 €

A negative value means you lose money on average.

13.Normal distribution

Answer: P = 0.7213

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-scores

z₁ = (57 − 65)/10 = −0.8
z₂ = (80 − 65)/10 = 1.5

Area between

P = Φ(1.5) − Φ(−0.8) = 0.9332 − 0.2119 = 0.7213

Answer

P = 0.7213
14.Normal distribution

Answer: x = 112.6 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) = 0.8 → z = 0.8416

Back to x

x = 100 + 0.8416·15 = 112.62 points

Answer

x = 112.62 points
15.Confidence intervals

Answer: E = 1.542

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 = 9, t = 3.25

Margin of error

E = 3.25·1.5/√10 = 1.542

Answer

E = 1.542
16.Confidence intervals

Answer: n = 31

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 = (1.645·5/1.5)² = 30.07

Round up

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

n = 31

Answer

n = 31
17.Hypothesis tests & confidence intervals

Answer: t = −2.175

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 = 12/√81 = 12/9 = 1.333

Test statistic

t = (117.1 − 120)/1.333 = −2.175

Answer

t = −2.175
t0.025; df = 80 = 1.99

|t| > 1.99, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 71.06 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 15/√25 = 3
x̄c = 80 + 1.645·3 = 84.94

z under the true mean

z = (84.94 − 86.6)/3 = −0.555

Power

power = P(Z > −0.555) = 71.06%
1 − β = 71.06%

Answer

1 − β = 71.06%
19.Linear regression & correlation

Answer: r = 0.9881

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.

Centred sums

Sxy = 12721 − 105·593/5 = 268
Sxx = 2245 − 105²/5 = 40
Syy = 72169 − 593²/5 = 1839

Correlation coefficient

r = 268/√(40·1839) = 0.9881

Answer

r = 0.9881
20.Linear regression & correlation

Answer: x = 3.974

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 = 537/6 = 89.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 501

Regression line

b = 501/70 = 7.157
a = 89.5 − 7.157·6 = 46.56
ŷ = 46.56 + 7.157·x

Solve for x

x = (75 − 46.56)/7.157 = 3.974

Answer

x = 3.974

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