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…9PV = ?€5 000+1%+1%…r = 7%

A property is expected to earn net rent of €5 000 next year. The rent is expected to grow by 1% per year forever. The required return is 7% per year. Find the value PV of the property today.

C1 = €5 000g = 1%r = 7%
PV = €
2.Time value of moneyHard
months0123…60€25 000AAAA3% / 12, I = ?

A loan of €25 000 is repaid with equal monthly payments over 5 years at 3% per year, compounded monthly. How much interest I is paid in total over the life of the loan?

P = €25 0003% / 12N = 60
I = €
3.NPV & IRRMedium
years0123…9−200 00023 000+1%+1%…r = 10%, NPV = ?

A project costs €200 000 today. It generates €23 000 at the end of the first year, and the cash flow then grows by 1% per year forever. The cost of capital is 10%. Find the net present value NPV.

C0 = €200 000CF1 = €23 000g = 1%r = 10%
NPV = €
4.NPV & IRRHard
years01234−50 00024 00021 00019 00017 500

A project costs €50 000 and returns €24 000, €21 000, €19 000, €17 500 at the end of years 1 to 4. Cash flows can be reinvested at the cost of capital of 12%. Find the modified internal rate of return MIRR in percent.

r = 12%CF0 = −50 000CF1 = 24 000CF2 = 21 000CF3 = 19 000CF4 = 17 500
MIRR = %
5.Bond pricing & yieldMedium
years0123…8P€45€45€45 + €1 000

A bond with a face value of €1 000 pays an annual coupon of 4.5% and matures in 8 years. Its yield to maturity changes from 7% to 5%. By how much ΔP does the bond price change?

F = €1 000coupon = 4.5%n = 8y1 = 7%y2 = 5%
ΔP = €
6.Bond pricing & yieldHard
half-years0123…20€740€25€25€25 + €1 000

A bond with a face value of €1 000 pays a 5% annual coupon in two half-yearly payments and matures in 10 years. It sells for €740. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 5% (2×)n = 10P = €740
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€8 000 000re​ = 10%debt€3 000 000rd​ = 5%T = 20%

A company has equity worth €8 000 000 and debt worth €3 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 5% and the tax rate is 20%. Find the WACC.

E = €8 000 000D = €3 000 000re = 10%rd = 5%T = 20%
WACC = %
8.CAPM & WACCHard
WACC = ?equity2 000 000 × €44.50β = 1.5rf​ = 1.5%, rm​ = 8%debt20 000 × €1 000rd​ = 7%T = 30%

A company has 2 000 000 shares at €44.50 each and 20 000 bonds trading at €1 000 each. The pre-tax cost of debt is 7% and the tax rate is 30%. The share’s beta is 1.5, the risk-free rate 1.5% and the market return 8%. Find the WACC.

β = 1.5rf = 1.5%rm = 8%rd = 7%T = 30%
WACC = %
9.Descriptive statisticsMedium
42444648505254565860

Find the sample standard deviation s of the data: 56, 47, 59, 60, 48, 46, 42.

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

Exam scores are normally distributed with mean 100 and standard deviation 5. Find the z-score z of a score of 103.

μ = 100σ = 5x = 103
z =
11.ProbabilityMedium
P(at least one six) = ??1?2?3P(six) = 1/6 on each roll

A fair die is rolled 3 times. What is the probability P of getting at least one six?

n = 3
P =
12.ProbabilityHard
xp(x)00.2510.2520.3530.15μ

X has the distribution shown, and Y = 10X + 4. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.25, 0.25, 0.35, 0.15Y = 10X + 4
σY =
13.Normal distributionMedium
2341.3P = ?

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. What fraction lies between 23 and 41.3 min? Give P as a decimal.

μ = 32σ = 6x1 = 23x2 = 41.3
P =
14.Normal distributionHard
35455565758595pointstop 15%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Drag the cut-off so the shaded top 15% is to the right of it. Find the cut-off x.

μ = 65 pointsσ = 10 points15%
x = points
15.Confidence intervalsMedium
x̄E₁ = 5.199%, n = 50x̄E₂ = ?95%, n = 50

A 99% confidence interval for a mean, based on 50 observations and a known σ, has a margin of error of 5.1. What is the margin of error E₂ of a 95% interval from the same data?

99%n₁ = 50E₁ = 5.195%
E₂ =
16.Confidence intervalsHard
μ₁ − μ₂, t(20) = 1.7250L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 9 plots, mean yield 93.5 kg, standard deviation 5 kg. Plot group 2: 13 plots, mean 97 kg, standard deviation 6.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 9x̄₁ = 93.5s₁ = 5n₂ = 13x̄₂ = 97s₂ = 6.590%t(20) = 1.725
lower limit = kg
17.Hypothesis tests & confidence intervalsMedium
t −1.98401.984

A sample of 100 students has a mean score of 502.9 and a sample standard deviation of 10. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 502.9s = 10n = 100
t =
18.Hypothesis tests & confidence intervalsHard
μ₀x̄c

A one-sided z test of H₀: μ = 200 against H₁: μ > 200 is done at α = 0.10 with a sample of 36; the standard deviation is known to be 6. In reality the mean is 201.2. Find the power 1 − β of the test in percent.

μ₀ = 200μ = 201.2σ = 6n = 36α = 0.10
1 − β = %
19.Linear regression & correlationMedium
xy1523585667759811183xy51050607080

The table gives 6 paired observations of x = hours studied and y = exam score. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy126221321416516xy242025x = 2.5

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 = 2.5.

n = 5x = 2.5
ŷ =

3Solutions

1.Time value of money

Answer: PV = 83333 €

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.

Growing perpetuity

PV = C1/(r − g) = 5 000/(0.07 − 0.01) = 5 000/0.06 = €83 333.33

Answer

PV = €83 333.33
2.Time value of money

Answer: I = 1953 €

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 = 0.0025, N = 60
A = 25 000·0.0025/(1 − (1 + 0.0025)−60) = €449.22

Total interest

I = N·A − P = 60·449.2 − 25 000 = €1 953.04

Answer

I = €1 953.04
3.NPV & IRR

Answer: NPV = 55556 €

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.

Value of the cash flows

PV = 23 000/(0.1 − 0.01) = 255 555.56

NPV

NPV = 255 555.56 − 200 000 = €55 555.56

Answer

NPV = €55 555.56

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

4.NPV & IRR

Answer: MIRR = 18.57 %

Modified IRR: compound every inflow forward to the end of the project at the reinvestment rate (usually the cost of capital) to get the terminal value TV. MIRR is the rate that turns the initial investment into TV: MIRR = (TV/C₀)^(1/n) − 1.

Terminal value of the inflows

24 000·1.123 = 33 718.27
21 000·1.122 = 26 342.40
19 000·1.121 = 21 280.00
17 500·1.120 = 17 500.00
TV = 98 840.67

MIRR

MIRR = (98 840.67/50 000)1/4 − 1 = 18.57%

Answer

MIRR = 18.57%
5.Bond pricing & yield

Answer: ΔP = 117 €

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 = €45.00, i = 0.07, N = 8
€45.00·(1 − 0.582)/0.07 + €1 000·0.582 = €268.71 + €582.01 = €850.72

Price at the new yield

C = €45.00, i = 0.05, N = 8
€45.00·(1 − 0.6768)/0.05 + €1 000·0.6768 = €290.84 + €676.84 = €967.68

Change

ΔP = 967.68 − 850.72 = €116.97

Answer

ΔP = €116.97
6.Bond pricing & yield

Answer: YTM = 8.997 %

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 = €25.00, N = 20

Solve numerically

i = 4.499% per half-year

YTM

YTM = 2·4.499% = 8.997%

Answer

YTM = 8.997%
7.CAPM & WACC

Answer: WACC = 8.364 %

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 = €11 000 000, E/V = 0.7273, D/V = 0.2727

WACC

WACC = 0.7273·10% + 0.2727·5%·(1 − 0.2) = 8.364%

Answer

WACC = 8.364%
8.CAPM & WACC

Answer: WACC = 10.08 %

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 = 2 000 000·€44.50 = €89 000 000
D = 20 000·€1 000 = €20 000 000

Cost of equity (CAPM)

re = 1.5% + 1.5·6.5% = 11.25%

Weights

V = €109 000 000, E/V = 0.8165, D/V = 0.1835

WACC

WACC = 0.8165·11.25% + 0.1835·7%·(1 − 0.3) = 10.08%

Answer

WACC = 10.08%
9.Descriptive statistics

Answer: s = 7.081

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 = 358/7 = 51.14

Squared deviations

Σ(x − x̄)² = 4.857² + (−4.143)² + 7.857² + 8.857² + (−3.143)² + (−5.143)² + (−9.143)² = 300.9

Standard deviation

s = √(300.9/6) = 7.081

Answer

s = 7.081
10.Descriptive statistics

Answer: z = 0.6

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 = (103 − 100)/5 = 0.6

Answer

z = 0.6
11.Probability

Answer: P = 0.4213

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.

Complement

P(no six) = (5/6)3 = 0.5787
P = 1 − 0.5787 = 0.4213

Answer

P = 0.4213
12.Probability

Answer: σY = 10.2

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 = 1.4, Var(X) = 1.04, σX = 1.02

Linear change

σY = |10|·1.02 = 10.2

Answer

σY = 10.2

The mean does move: E(Y) = 10·1.4 + 4 = 18.

13.Normal distribution

Answer: P = 0.8726

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 = (23 − 32)/6 = −1.5
z₂ = (41.3 − 32)/6 = 1.55

Area between

P = Φ(1.55) − Φ(−1.5) = 0.9394 − 0.06681 = 0.8726

Answer

P = 0.8726
14.Normal distribution

Answer: x = 75.4 points

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) = 0.85

z from the table

z ≈ 1.04

Back to the original units

x = 65 + 1.04·10 = 75.4 points

Answer

x = 75.4 points
15.Confidence intervals

Answer: E₂ = 3.88

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 = (1.96/2.576)·√(50/50) = 0.7609

New margin

E₂ = 5.1·0.7609 = 3.88

Answer

E₂ = 3.88
16.Confidence intervals

Answer: lower limit = −7.947 kg

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Pooled standard deviation

sp² = (8·5² + 12·6.5²)/20 = 35.35
sp = 5.946

Margin of error

E = 1.725·5.946·√(1/9 + 1/13) = 4.447

Interval

−3.5 ± 4.447 → [−7.947, 0.9474]

Answer

lower limit = −7.947 kg
17.Hypothesis tests & confidence intervals

Answer: t = 2.9

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 = 10/√100 = 10/10 = 1

Test statistic

t = (502.9 − 500)/1 = 2.9

Answer

t = 2.9
t0.025; df = 99 = 1.984

|t| > 1.984, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: 1 − β = 46.73 %

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 = 6/√36 = 1
x̄c = 200 + 1.282·1 = 201.3

z under the true mean

z = (201.3 − 201.2)/1 = 0.082

Power

power = P(Z > 0.082) = 46.73%
1 − β = 46.73%

Answer

1 − β = 46.73%
19.Linear regression & correlation

Answer: r² = 0.9757

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 = 415/6 = 69.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 233
Syy = Σ(y − ȳ)² = 794.8

Coefficient of determination

r² = 233²/(70·794.8) = 0.9757

Answer

r² = 0.9757

About 97.57% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 21.25

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 = 15/5 = 3
ȳ = 100/5 = 20

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = −25

Regression line

b = −25/10 = −2.5
a = 20 − (−2.5)·3 = 27.5
ŷ = 27.5 − 2.5·x

Prediction

ŷ = 27.5 + (−2.5)·2.5 = 21.25

Answer

ŷ = 21.25

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