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
years0123456€1 000FV = ?

You invest €1 000 at a nominal rate of 2% per year, compounded monthly. Find the value FV of the investment after 6 years.

PV = €1 0002% / 12n = 6
FV = €
2.Time value of moneyHard
months0123…288€100€100€100FV = ?

You save €100 at the end of every month for 24 years. The account pays 5.4% per year, compounded monthly. How much have you saved at the end? Find FV.

A = €1005.4% / 12N = 288
FV = €
3.NPV & IRRMedium
years0123−20 0006 0006 0006 000

An investment of €20 000 gives these cash flows at the end of years 1 to 3: €6 000, €6 000, €6 000. The cost of capital is 6%. Find the net present value NPV.

r = 6%CF0 = −20 000CF1 = 6 000CF2 = 6 000CF3 = 6 000
NPV = €
4.NPV & IRRHard
years0123…9−80 00012 00012 00012 00012 000

An investment of €80 000 returns €12 000 at the end of each year for 9 years. Find its internal rate of return IRR in percent.

C0 = €80 000CF = €12 000n = 9
IRR = %
5.Bond pricing & yieldMedium

You buy a bond with 7 years to maturity (face value €1 000, annual coupon 4%) when its yield to maturity is 5.5%. One year later, just after receiving the first coupon, you sell it when the yield to maturity is 4.5%. Find your holding period return HPR in percent.

F = €1 000coupon = 4%n = 7y0 = 5.5%y1 = 4.5%
HPR = %
6.Bond pricing & yieldHard
years0123456€925€30€30 + €1 000

A bond with a face value of €1 000 and an annual coupon of 3% matures in 6 years and sells for €925. Find its yield to maturity YTM in percent.

F = €1 000coupon = 3%n = 6P = €925
YTM = %
7.CAPM & WACCMedium
WACC = ?equity€8 000 000re​ = 10%debt€6 000 000rd​ = 6%T = 25.8%

A company has equity worth €8 000 000 and debt worth €6 000 000. Its cost of equity is 10%, its pre-tax cost of debt is 6% and the tax rate is 25.8%. Find the WACC.

E = €8 000 000D = €6 000 000re = 10%rd = 6%T = 25.8%
WACC = %
8.CAPM & WACCHard
WACC = ?equity€15 000 000re​ = 10%bonds2 000 × €878coupon 5%, n = 12T = 20%

A company has equity with a market value of €15 000 000 and a cost of equity of 10%. Its debt consists of 2 000 bonds (face value €1 000, annual coupon 5%, 12 years to maturity) that trade at €878 each. The tax rate is 20%. Find the WACC, using the yield to maturity of the bonds as the pre-tax cost of debt.

E = €15 000 000re = 10%bonds: 2 000 × €878coupon = 5%n = 12T = 20%
WACC = %
9.Descriptive statisticsMedium
152025303540

Find the sample standard deviation s of the data: 44, 44, 34, 14, 41, 20.

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

A quantity is normally distributed with mean 60 and standard deviation 20. What proportion P of values is below 36?

μ = 60σ = 20x = 36
P =
11.ProbabilityMedium
xp(x)10.3520.330.0540.3μ = 2.3

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

x: 1, 2, 3, 4p(x): 0.35, 0.3, 0.05, 0.3
Var(X) =
12.ProbabilityHard
€p(x)−20.7520.2180.05

A game costs €2 to play. You win €4 with probability 0.2 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€4: p = 0.2€20: p = 0.05
E(G) = €
13.Normal distributionMedium
88118.75P = ?

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 88 and 118.75 points? Give P as a decimal.

μ = 100σ = 15x1 = 88x2 = 118.75
P =
14.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
15.Confidence intervalsMedium
n = 8, s = 4.5L = ?Ux̄ = 99E90%

A sample of 8 items has mean 99 and standard deviation s = 4.5. The population is roughly normal. Find the lower limit of the 90% confidence interval for the mean.

n = 8x̄ = 99s = 4.590%t(7) = 1.895
lower limit =
16.Confidence intervalsHard
n = 30Lσ = ?s = 9.795%

The diameters of 30 machined shafts have a sample standard deviation of 9.7 µm. Assuming the diameters are normally distributed, find the upper limit σ of the 95% confidence interval for the population standard deviation. With df = 29, 2.5% of the χ² distribution lies below 16.047 and 2.5% lies above 45.722.

n = 30s = 9.7 µmχ² = 16.047, 45.722
σ = µm
17.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 60 parts has mean length 143 mm, and the line's known standard deviation is 8 mm. Line 2: a sample of 30 parts has mean 139.7 mm, known standard deviation 3 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 60x̄₁ = 143 mmσ₁ = 8 mmn₂ = 30x̄₂ = 139.7 mmσ₂ = 3 mm
z =
18.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 190%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 1? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 190%
n =
19.Linear regression & correlationMedium
xy244761181410181225xy5101020

The table gives 6 paired observations of x = load in kN and y = extension in mm. Find the coefficient of determination r² of the least-squares line.

n = 6
r² =
20.Linear regression & correlationHard
xy1293415507609631174xy5104060x = 6

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 6.

n = 6x = 6
ŷ =

3Solutions

1.Time value of money

Answer: FV = 1127 €

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 = r/m = 0.02/12 = 0.001667
N = m·n = 12·6 = 72

Future value

FV = 1 000·(1 + 0.001667)72 = 1 000·1.127 = €1 127.38

Answer

FV = €1 127.38
2.Time value of money

Answer: FV = 58756 €

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 deposits

i = 0.054000000000000006/12 = 0.0045
N = 12·24 = 288

Future value of the deposits

FV = A·((1 + i)N − 1)/i = 100·(3.644 − 1)/0.0045 = €58 756.43

Answer

FV = €58 756.43

You paid in €28 800 yourself; the rest is interest.

3.NPV & IRR

Answer: NPV = −3962 €

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 = −20 000
CF1/(1 + 0.06)1 = 6 000/1.06 = 5 660.38
CF2/(1 + 0.06)2 = 6 000/1.124 = 5 339.98
CF3/(1 + 0.06)3 = 6 000/1.191 = 5 037.72

Add them up

NPV = €−3 961.93

Answer

NPV = €−3 961.93

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

4.NPV & IRR

Answer: IRR = 6.463 %

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.

Required annuity factor

(1 − (1 + i)−9)/i = 80 000/12 000 = 6.667

Solve for the rate

The annuity factor falls as the rate rises; bracket the rate and narrow down.

i = 6%: 6.802
i = 7%: 6.515
IRR = 6.463%

Answer

IRR = 6.463%
5.Bond pricing & yield

Answer: HPR = 10.87 %

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.

Purchase price

C = €40.00, i = 0.055, N = 7
€40.00·(1 − 0.6874)/0.055 + €1 000·0.6874 = €227.32 + €687.44 = €914.76

Selling price

C = €40.00, i = 0.045, N = 6
€40.00·(1 − 0.7679)/0.045 + €1 000·0.7679 = €206.31 + €767.90 = €974.21

Holding period return

HPR = (40 + 974.21 − 914.76)/914.76 = 10.87%

Answer

HPR = 10.87%
6.Bond pricing & yield

Answer: YTM = 4.452 %

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

€925 = €30·(1 − (1 + y)⁻⁶)/y + €1 000/(1 + y)⁶

Solve numerically

P(4%) = €947.58, P(5%) = €898.49
YTM = 4.452%

Answer

YTM = 4.452%
7.CAPM & WACC

Answer: WACC = 7.622 %

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 = €14 000 000, E/V = 0.5714, D/V = 0.4286

WACC

WACC = 0.5714·10% + 0.4286·6%·(1 − 0.258) = 7.622%

Answer

WACC = 7.622%
8.CAPM & WACC

Answer: WACC = 9.497 %

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 debt (YTM)

878 = 50·(1 − (1 + y)−12)/y + 1000/(1 + y)12
y = 6.495%

Market values

D = 2 000·€878 = €1 756 000
V = €16 756 000, E/V = 0.8952, D/V = 0.1048

WACC

WACC = 0.8952·10% + 0.1048·6.495%·(1 − 0.2) = 9.497%

Answer

WACC = 9.497%
9.Descriptive statistics

Answer: s = 12.94

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 = 197/6 = 32.83

Squared deviations

Σ(x − x̄)² = 11.17² + 11.17² + 1.167² + (−18.83)² + 8.167² + (−12.83)² = 836.8

Standard deviation

s = √(836.8/5) = 12.94

Answer

s = 12.94
10.Descriptive statistics

Answer: P = 0.1151

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 = −1.2

Standard normal table

Φ(−1.2) = 0.1151
P = 0.1151

Answer

P = 0.1151
11.Probability

Answer: Var(X) = 1.51

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 + 2·0.3 + 3·0.05 + 4·0.3 = 2.3

Mean of the squares

E(X²) = 1·0.35 + 4·0.3 + 9·0.05 + 16·0.3 = 6.8

Variance

Var(X) = 6.8 − 2.3² = 1.51

Answer

Var(X) = 1.51
12.Probability

Answer: E(G) = −0.2 €

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.

Net outcomes

G: 2 (p = 0.2), 18 (p = 0.05), −2 (p = 0.75)

Expected value

E(G) = 2·0.2 + 18·0.05 − 2·0.75 = −0.2 €

Answer

E(G) = −0.2 €

A negative value means you lose money on average.

13.Normal distribution

Answer: P = 0.6825

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 = (88 − 100)/15 = −0.8
z₂ = (118.75 − 100)/15 = 1.25

Area between

P = Φ(1.25) − Φ(−0.8) = 0.8944 − 0.2119 = 0.6825

Answer

P = 0.6825
14.Normal distribution

Answer: P = 0.7881

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 = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
15.Confidence intervals

Answer: lower limit = 95.99

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 = 7, t = 1.895

Margin of error

E = 1.895·4.5/√8 = 3.015

Interval

99 ± 3.015 → [95.99, 102]

Answer

lower limit = 95.99
16.Confidence intervals

Answer: σ = 13.04 µm

Interval for a variance: (n − 1)s²/σ² has a χ² distribution with n − 1 degrees of freedom. The 95% interval for σ² runs from (n − 1)s² divided by the upper χ² value (2.5% above it) to (n − 1)s² divided by the lower χ² value (2.5% below it); the interval for σ is the square root of both limits. It is not symmetric around s.

Sum of squares

(n − 1)s² = 29·9.7² = 2729

Limits for σ²

2729/45.722 = 59.68
2729/16.047 = 170

Limit for σ

σ = √170 = 13.04 µm

Answer

σ = 13.04 µm

95% interval for σ: [7.725, 13.04] µm.

17.Hypothesis tests & confidence intervals

Answer: z = 2.823

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.

Standard error of the difference

SE = √(8²/60 + 3²/30) = √1.367 = 1.169

Test statistic

z = (143 − 139.7)/1.169 = 2.823

Answer

z = 2.823

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

18.Hypothesis tests & confidence intervals

Answer: n = 390

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 = (1.645·12/1)² = 389.7

Round up

n = 390

Answer

n = 390
19.Linear regression & correlation

Answer: r² = 0.9766

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 = 42/6 = 7
ȳ = 79/6 = 13.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 141
Syy = Σ(y − ȳ)² = 290.8

Coefficient of determination

r² = 141²/(70·290.8) = 0.9766

Answer

r² = 0.9766

About 97.66% of the variation in y is explained by the line.

20.Linear regression & correlation

Answer: ŷ = 52.83

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 = 317/6 = 52.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 301

Regression line

b = 301/70 = 4.3
a = 52.83 − 4.3·6 = 27.03
ŷ = 27.03 + 4.3·x

Prediction

ŷ = 27.03 + 4.3·6 = 52.83

Answer

ŷ = 52.83

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