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
periods0123411 + EAR = ?19.5% / 4

A credit card charges a nominal interest rate of 19.5% per year, compounded quarterly. Find the effective annual rate EAR in percent.

r = 19.5%m = 4
EAR = %
2.Time value of moneyHard
years012345€1 000€1 000€1 000FV = ?

You deposit €1 000 at the start of every year for 5 years, the first deposit today. The account pays 6% per year. How much is in the account at the end of year 5? Find FV.

A = €1 000i = 6%n = 5
FV = €
3.NPV & IRRMedium
years0123…9−250 00020 000+1%+1%…r = 11%, NPV = ?

A project costs €250 000 today. It generates €20 000 at the end of the first year, and the cash flow then grows by 1% per year forever. The cost of capital is 11%. Find the net present value NPV.

C0 = €250 000CF1 = €20 000g = 1%r = 11%
NPV = €
4.NPV & IRRHard
years0123−10 0004 0005 5003 500

A project costs €10 000 and returns €4 000, €5 500, €3 500 at the end of years 1 to 3. Find its internal rate of return IRR in percent.

CF0 = −10 000CF1 = 4 000CF2 = 5 500CF3 = 3 500
IRR = %
5.Bond pricing & yieldMedium
half-years0123…8€40€40€40 + €1 000P = ?

A bond with a face value of €1 000 pays an 8% annual coupon in two half-yearly payments and matures in 4 years. The yield to maturity is 6% per year. Find the price P of the bond.

F = €1 000coupon = 8%YTM = 6%n = 4
P = €
6.Bond pricing & yieldHard
half-years0123456€974€35€35€35 + €1 000

A bond with a face value of €1 000 pays a 7% annual coupon in two half-yearly payments and matures in 3 years. It sells for €974. Find its yield to maturity YTM as an annual rate (twice the half-yearly yield), in percent.

F = €1 000coupon = 7% (2×)n = 3P = €974
YTM = %
7.CAPM & WACCMedium
WACC = 12.1%equity€8 000 000re​ = ?debt€2 000 000rd​ = 6%T = 25%

A company has equity worth €8 000 000 and debt worth €2 000 000. Its WACC is 12.1%, its pre-tax cost of debt is 6% and the tax rate is 25%. Find the cost of equity re.

E = €8 000 000D = €2 000 000WACC = 12.1%rd = 6%T = 25%
re = %
8.CAPM & WACCHard
WACC = ?equity4 000 000 × €16.00β = 1.3rf​ = 2%, rm​ = 8%debt5 000 × €980rd​ = 6%T = 25%

A company has 4 000 000 shares at €16.00 each and 5 000 bonds trading at €980 each. The pre-tax cost of debt is 6% and the tax rate is 25%. The share’s beta is 1.3, the risk-free rate 2% and the market return 8%. Find the WACC.

β = 1.3rf = 2%rm = 8%rd = 6%T = 25%
WACC = %
9.Descriptive statisticsMedium
10152025303540

Find the sample standard deviation s of the data: 22, 42, 42, 10, 13, 23.

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

A quantity is normally distributed with mean 50 and standard deviation 5. What proportion P of values is above 60?

μ = 50σ = 5x = 60
P =
11.ProbabilityMedium
xp(x)10.3520.1530.240.150.2μ = 2.65

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

x: 1, 2, 3, 4, 5p(x): 0.35, 0.15, 0.2, 0.1, 0.2
σ =
12.ProbabilityHard
n = 11, p = 0.2: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.2, independently. What is the probability P that at most one fails?

n = 11p = 0.2
P =
13.Normal distributionMedium
120 kg125.1 kgT = X₁ + … + X₁₀?

The masses of boxes are independent and normal with mean 12 kg and standard deviation 0.8 kg. Find the probability P that the total mass of 10 of them is below 125.1 kg. Round z to two decimals.

μ = 12 kgσ = 0.8 kgn = 10< 125.1 kg
P =
14.Normal distributionHard
μ = 100x = ?90%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Find the value x below which 90% of them lie (the 90th percentile).

μ = 100σ = 15P = 90%
x = points
15.Confidence intervalsMedium
n = 13, s = 7LU = ?x̄ = 117.5E99%

A sample of 13 items has mean 117.5 and standard deviation s = 7. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 13x̄ = 117.5s = 799%t(12) = 3.055
upper limit =
16.Confidence intervalsHard
σ = 26, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

The population standard deviation is 26. 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 2.5?

σ = 26E = 2.595%
n =
17.Hypothesis tests & confidence intervalsMedium
−2.2282.228t = ?df = 10

Two teaching methods are compared. Group 1: 6 students, mean score 42, standard deviation 6.5. Group 2: 6 students, mean score 53.7, standard deviation 10. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 6x̄₁ = 42s₁ = 6.5n₂ = 6x̄₂ = 53.7s₂ = 10
t =
18.Hypothesis tests & confidence intervalsHard
σ = 4, n = ?x̄ − Ex̄ + Ex̄E = 299%

How large must a sample be so that the 99% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 4. Find the smallest sample size n.

σ = 4E = 299%
n =
19.Linear regression & correlationMedium
xy157117701985211032310825117xy15202580100120

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Find Pearson's correlation coefficient r.

n = 6
r =
20.Linear regression & correlationHard
xy2534646678791080xy510607080x = 9

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 9.

n = 5x = 9
ŷ =

3Solutions

1.Time value of money

Answer: EAR = 20.97 %

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 per period

r/m = 0.195/4 = 0.04875

Compound over one year

EAR = (1 + 0.04875)4 − 1 = 1.21 − 1 = 20.97%

Answer

EAR = 20.97%

The EAR is always above the nominal rate when interest compounds more than once a year.

2.Time value of money

Answer: FV = 5975 €

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 + 0.06)5 − 1)/0.06 = 5.637

Annuity due

Each deposit is invested one year longer.

FV = 1 000·5.637·1.06 = €5 975.32

Answer

FV = €5 975.32
3.NPV & IRR

Answer: NPV = −50000 €

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 = 20 000/(0.11 − 0.01) = 200 000.00

NPV

NPV = 200 000.00 − 250 000 = −€50 000.00

Answer

NPV = −€50 000.00

NPV is negative, so the project earns less than the required return.

4.NPV & IRR

Answer: IRR = 14.62 %

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%) = 811.42
NPV(15%) = −61.64

Narrow down

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

IRR ≈ 10 + 5·811.42/(811.42 − (−61.64)) = 14.65%
exact: NPV(14.62%) = 0 → IRR = 14.62%

Answer

IRR = 14.62%
5.Bond pricing & yield

Answer: P = 1070 €

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 = €40.00, i = 0.03, N = 8

Present value of the coupons

€40.00·(1 − 0.7894)/0.03 = €280.79

Present value of the face value

€1 000·0.7894 = €789.41

Price

P = €1 070.20

Answer

P = €1 070.20

The yield is below the coupon rate, so the bond sells at a premium.

6.Bond pricing & yield

Answer: YTM = 7.992 %

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 = €35.00, N = 6

Solve numerically

i = 3.996% per half-year

YTM

YTM = 2·3.996% = 7.992%

Answer

YTM = 7.992%
7.CAPM & WACC

Answer: re = 14 %

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 = €10 000 000, E/V = 0.8, D/V = 0.2

Debt part of the WACC

0.2·6%·(1 − 0.25) = 0.9%

Cost of equity

re = (12.1% − 0.9%)/0.8 = 14%

Answer

re = 14%
8.CAPM & WACC

Answer: WACC = 9.423 %

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 = 4 000 000·€16.00 = €64 000 000
D = 5 000·€980 = €4 900 000

Cost of equity (CAPM)

re = 2% + 1.3·6% = 9.8%

Weights

V = €68 900 000, E/V = 0.9289, D/V = 0.07112

WACC

WACC = 0.9289·9.8% + 0.07112·6%·(1 − 0.25) = 9.423%

Answer

WACC = 9.423%
9.Descriptive statistics

Answer: s = 13.85

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 = 152/6 = 25.33

Squared deviations

Σ(x − x̄)² = (−3.333)² + 16.67² + 16.67² + (−15.33)² + (−12.33)² + (−2.333)² = 959.3

Standard deviation

s = √(959.3/5) = 13.85

Answer

s = 13.85
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) = 0.9772
P = 1 − 0.9772 = 0.02275

Answer

P = 0.02275
11.Probability

Answer: σ = 1.526

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 = 1·0.35 + 2·0.15 + 3·0.2 + 4·0.1 + 5·0.2 = 2.65

Mean of the squares

E(X²) = 1·0.35 + 4·0.15 + 9·0.2 + 16·0.1 + 25·0.2 = 9.35

Variance

Var(X) = 9.35 − 2.65² = 2.327
σ = √2.327 = 1.526

Answer

σ = 1.526
12.Probability

Answer: P = 0.3221

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.

Zero and one failure

P(X = 0) = 0.811 = 0.0859
P(X = 1) = 11·0.2·0.810 = 0.2362

Add them

P = 0.0859 + 0.2362 = 0.3221

Answer

P = 0.3221
13.Normal distribution

Answer: P = 0.9783

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 total

μT = 120 kg, σT = 0.8·√10 = 2.53 kg

Standardise

z = (125.1 − 120)/2.53 = 2.02

Probability

P = Φ(2.02) = 0.9783

Answer

P = 0.9783
14.Normal distribution

Answer: x = 119.2 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) = 0.9 → z = 1.282

Back to x

x = 100 + 1.282·15 = 119.22 points

Answer

x = 119.22 points
15.Confidence intervals

Answer: upper limit = 123.4

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 = 12, t = 3.055

Margin of error

E = 3.055·7/√13 = 5.931

Interval

117.5 ± 5.931 → [111.6, 123.4]

Answer

upper limit = 123.4
16.Confidence intervals

Answer: n = 416

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 = (1.96·26/2.5)² = 415.5

Round up

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

n = 416

Answer

n = 416
17.Hypothesis tests & confidence intervals

Answer: t = −2.403

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.

Pooled variance

sp² = (5·6.5² + 5·10²)/10 = 71.13
sp = 8.434

Standard error

SE = 8.434·√(1/6 + 1/6) = 4.869

Test statistic

t = (42 − 53.7)/4.869 = −2.403

Answer

t = −2.403

With df = 10, the two-sided 5% critical value is 2.228; |t| is larger, so H₀ is rejected.

18.Hypothesis tests & confidence intervals

Answer: n = 27

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).

Solve for n

n = (2.576·4/2)² = 26.54

Round up

n = 27

Answer

n = 27
19.Linear regression & correlation

Answer: r = 0.9735

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 = 120/6 = 20
ȳ = 554/6 = 92.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 362
Syy = Σ(y − ȳ)² = 1975

Correlation coefficient

r = 362/√(70·1975) = 0.9735

Answer

r = 0.9735
20.Linear regression & correlation

Answer: ŷ = 78.95

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 = 343/5 = 68.6

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 138

Regression line

b = 138/40 = 3.45
a = 68.6 − 3.45·6 = 47.9
ŷ = 47.9 + 3.45·x

Prediction

ŷ = 47.9 + 3.45·9 = 78.95

Answer

ŷ = 78.95

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