Solved sample questions

Solution

1. 'At least one' is the union, so inclusion-exclusion applies: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
2. Substituting the given proportions gives $0.48+0.31-0.17$.
3. That is $0.62$. The subtraction is essential because the 17% with both were counted once in each of the first two figures.
4. As a check, the proportion with neither is $1-0.62=0.38$, which is consistent with the numbers given.

Trap. Adding 0.48 and 0.31 to get 0.79 forgets that the overlap has been double-counted.

Solution

1. 'Exactly one' is the union minus the intersection: the union counts everything that happens, and removing the overlap leaves the parts where only one occurred.
2. $P(A\cup B)=0.5+0.6-0.2=0.9$.
3. $P(\text{exactly one})=0.9-0.2=0.7$.
4. Equivalently $\left[P(A)-P(A\cap B)\right]+\left[P(B)-P(A\cap B)\right]=0.3+0.4=0.7$, the same computation written as two disjoint pieces.

Trap. Reporting the union (0.9), which includes the outcomes where BOTH events occur.

Solution

1. The group is unordered - sending analysts A, B, C, D is the same as sending D, C, B, A - so this is a combination.
2. $\left(\genfrac{}{}{0ex}{}{11}{4}\right)=\frac{11!}{4!7!}$.
3. $=\frac{11\times 10\times 9\times 8}{4\times 3\times 2\times 1}=\frac{7920}{24}=330$.
4. The ordered count $P(11,4)=7920$ is exactly $4!=24$ times too large, which is why it appears as a distractor.

Trap. Using the permutation count 7,920, which treats two identical groups in different orders as different selections.

Solution

1. Each position is chosen independently, so the multiplication rule applies directly.
2. The letters contribute $26^3=17,576$.
3. The digits contribute $10^2=100$.
4. $17,576\times 100=1,757,600$ passwords.

Trap. Using permutations without repetition when the problem explicitly allows repeats.

Solution

1. Let $A$ be 'above 10,000' and $B$ be 'commercial'. The 18% is a JOINT probability out of all claims.
2. $P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{0.18}{0.30}$.
3. $=0.60$.
4. The conditional probability is larger than the joint one because the denominator has shrunk to the commercial claims only.

Trap. Reading the 18% as already conditional on being commercial.

Solution

1. The conditioning event is 'owns a home', so $P(\text{home})=0.40$ is the denominator.
2. $P(\text{boat}\mid \text{home})=\frac{P(\text{both})}{P(\text{home})}=\frac{0.15}{0.40}$.
3. $=0.375$.
4. Since $P(\text{boat})=0.25\ne 0.375$, home and boat ownership are NOT independent in this book - owning a home raises the chance of owning a boat.

Solution

1. This is the law of total probability across a three-part partition.
2. $P(\text{claim})=0.5(0.02)+0.3(0.05)+0.2(0.11)$.
3. $=0.010+0.015+0.022$.
4. $=0.047$. This figure is the denominator of every Bayes calculation on this book, so it is worth computing once and reusing.

Trap. Averaging 2%, 5% and 11% without weights, which gives 6% and assumes equal class sizes.

Solution

1. Weight each supplier's failure rate by its share of parts.
2. $0.25(0.02)+0.35(0.04)+0.40(0.05)$.
3. $=0.005+0.014+0.020$.
4. $=0.039$, between the smallest and largest rates as any weighted average must be.

Solution

1. Any interval probability is a difference of CDF values: $P(a<X\le b)=F(b)-F(a)$.
2. $F(4)=1-5^{-3}=1-0.008=0.992$.
3. $F(1)=1-2^{-3}=1-0.125=0.875$.
4. $P=0.992-0.875=0.117$, so about 11.7% of losses fall in that band.
5. Because the variable is continuous, using strict or weak inequalities makes no difference to the answer.

Trap. Integrating the density from scratch when the CDF is already given - slower and an easy place to lose a constant.

Solution

1. Each of the 20 integers has probability $1/20$.
2. 'Greater than 14' means the values 15 through 20 - six of them.
3. $P(X>14)=6/20$.
4. $=0.30$. Counting carefully matters: 'at least 14' would give seven values and 0.35.

Trap. Including 14 itself and answering 0.35.

Solution

1. Fixed number of independent trials with a constant success probability makes this binomial with $n=10$, $p=0.25$.
2. $P(X=3)=\left(\genfrac{}{}{0ex}{}{10}{3}\right)(0.25{)}^3(0.75{)}^7$.
3. $=120\times 0.015625\times 0.133484$.
4. $=0.2503$.

Solution

1. $E[X]=\sum xp(x)$ over the support.
2. $=0(0.25)+1(0.35)+2(0.28)+3(0.12)$.
3. $=0+0.35+0.56+0.36$.
4. $=1.27$ claims per policyholder per year.
5. The mean is not one of the attainable values, which is normal: an expectation is a weighted centre of mass, not an outcome.
6. It sits above 1 because the upper values, though less likely, pull the average up.

Solution

1. For an exponential with mean $\theta$, the survival function is $S(x)=e^{-x/\theta }$ - the single most useful form of the distribution.
2. Here $\theta =800$, so $S(1200)=e^{-1200/800}=e^{-1.5}$.
3. $=0.223130$.
4. The rate parameter is $\lambda =1/800$; mixing up rate and mean is the classic slip, and it would give $e^{-960000}$, an obviously absurd answer.

Trap. Using λ = 800 instead of λ = 1/800.

Solution

1. The exponential is memoryless: $P(X>s+t\mid X>s)=P(X>t)$.
2. Here $s=300$ and $s+t=800$, so $t=500$.
3. $P(X>500)=e^{-500/500}=e^{-1}$.
4. $=0.367879$. Computing $S(800)/S(300)=e^{-1.6}/e^{-0.6}$ gives the same thing - which IS the memoryless property.

Trap. Reporting the unconditional S(800) = 0.2019 and ignoring the information already given.

Solution

1. The computational formula is $\mathrm{Var}(X)=E[X^2]-(E[X]{)}^2$.
2. $(E[X]{)}^2=16$.
3. $\mathrm{Var}(X)=25-16$.
4. $=9$, so the standard deviation is 3. Never subtract $E[X]$ from $E[X^2]$ directly - the square matters.

Trap. Reporting the standard deviation 3, or subtracting 4 from 25 to get 21.

Solution

1. For a non-negative variable, $E[X]={\int }_0^{\infty }S(x)dx$ - usually much faster than integrating $xf(x)$.
2. ${\int }_0^{\infty }(1+x{)}^{-2}dx={\left[-(1+x{)}^{-1}\right]}_0^{\infty }$.
3. $=0-(-1)=1$.
4. The survival-function route avoids differentiating $S$ to get the density at all.

Trap. Differentiating S to get f and then integrating xf(x) - correct but twice the work.

Solution

1. The normal MGF is $\exp \left(\mu t+\frac{{\sigma }^2{t}^2}{2}\right)$, so match coefficients.
2. The $t$ coefficient gives $\mu =3$.
3. The $t^2$ coefficient gives $\frac{{\sigma }^2}{2}=8$, so ${\sigma }^2=16$.
4. The variance is 16. Recognising the shape is far faster than differentiating twice.

Trap. Reading the t² coefficient as the variance and answering 8.

Solution

1. This MGF is a weighted sum of $e^{tx}$ terms, so it describes a discrete variable taking values 0, 1 and 3 with probabilities 0.2, 0.3 and 0.5.
2. Read the distribution straight off the exponents and coefficients.
3. $E[X]=0(0.2)+1(0.3)+3(0.5)$.
4. $=1.8$. Differentiating and setting $t=0$ gives the same, but reading it off is faster and less error-prone.

Trap. Differentiating carelessly and dropping the factor of 3 from the e^{3t} term.

Solution

1. Sketch the region: the triangle below the line $y=1-x$ inside the unit square.
2. $P={\int }_0^1{\int }_0^{1-x}(x+y)dydx$.
3. The inner integral is $x(1-x)+\frac{(1-x{)}^2}{2}$.
4. Integrating over $x$ from 0 to 1 gives $\frac{1}{2}-\frac{1}{3}+\frac{1}{6}=\frac{1}{3}=0.3333$.

Trap. Using constant limits 0 to 1 for y, which integrates over the whole square rather than the triangle.

Solution

1. Use the conditional variance decomposition: $\mathrm{Var}(S)=E\left[\mathrm{Var}(S\mid N)\right]+\mathrm{Var}\left(E[S\mid N]\right)$.
2. $E\left[\mathrm{Var}(S\mid N)\right]=E[10,000N]=10,000(4)=40,000$.
3. $\mathrm{Var}\left(E[S\mid N]\right)=\mathrm{Var}(300N)=90,000\mathrm{Var}(N)=90,000(4)=360,000$.
4. Total: $40,000+360,000=400,000$.

Trap. Using only one of the two terms - the 'variance of the mean' piece is the one usually forgotten.

Solution

1. Work through the CDF: $Y>3⟺-2\ln U>3⟺\ln U<-1.5$.
2. The inequality FLIPPED when dividing by $-2$ - the step most often missed.
3. $\ln U<-1.5⟺U<e^{-1.5}$, and for a standard uniform $P(U<c)=c$.
4. $P(Y>3)=e^{-1.5}=0.2231$. In fact $Y$ is exponential with mean 2, which is the standard way to simulate exponentials.

Trap. Failing to reverse the inequality when dividing by a negative number.

Solution

1. A linear transform of a normal is normal: $X$ has mean 20 and standard deviation 5.
2. Standardise: $z=\frac{28-20}{5}=1.6$.
3. $P(X>28)=1-\Phi (1.6)=1-0.945201$.
4. $=0.054799$, so about one observation in eighteen exceeds 28.
5. Recognising that a linear transform of a normal stays normal is what makes this a two-line problem rather than a change-of-variable exercise.

Solution

1. $E[(X-d{)}_{+}]={\int }_d^{\infty }S(x)dx$, and for an exponential this is $\theta e^{-d/\theta }$.
2. $=1,000e^{-250/1000}=1,000e^{-0.25}$.
3. $=1,000(0.778801)$.
4. $=778.80$. This averages over ALL losses, including the small ones that pay nothing.

Trap. Answering 750 by subtracting the deductible from the mean, which ignores losses below the deductible.

Solution

1. For the two-parameter Pareto, $E[X]=\frac{\theta }{\alpha -1}=\frac{1000}{2}=500$.
2. The variance exists whenever $\alpha >2$, and here $\alpha =3$.
3. $\mathrm{Var}(X)=\frac{\alpha {\theta }^2}{(\alpha -1{)}^2(\alpha -2)}=\frac{3\times 10^6}{4}=750,000$.
4. So the standard deviation is about 866, well above the mean - heavy-tailed, but with both moments finite.

Trap. Assuming any Pareto has infinite variance; it depends entirely on whether α exceeds 2.

Solution

1. $E[S]=100(50)=5,000$ and $\mathrm{Var}(S)=100(144)=14,400$, so ${\sigma }_S=120$.
2. Variances add; standard deviations do not - that is the whole content of the $\sqrt{n}$ scaling.
3. $z=\frac{5200-5000}{120}=1.6667$.
4. $P(S>5200)\approx 1-\Phi (1.6667)=1-0.952210=0.047790$.

Trap. Using n·σ = 1,200 for the standard deviation of the sum instead of √n·σ = 120.

Solution

1. The sum of independent Poissons is Poisson with mean $50\times 2=100$, and its variance is also 100.
2. $\sigma =10$.
3. $z=\frac{115-100}{10}=1.5$ (ignoring the continuity correction for a first pass).
4. Applying the half-unit continuity correction at 115.5 gives $z=1.55$ and $P\approx 1-\Phi (1.55)=0.0606$.

Trap. Using √(50) × 2 for the standard deviation rather than √(50 × 2).

Solution

1. The relationship between the two rates is $d=\frac{i}{1+i}$, which rearranges to $i=\frac{d}{1-d}$.
2. $i=\frac{0.06}{1-0.06}=\frac{0.06}{0.94}$.
3. $=0.063830$, or 6.3830%.
4. The interest rate must EXCEED the discount rate, because discount is interest paid in advance - a useful check on the direction of the algebra.

Trap. Computing d(1+d) = 6.36% or treating the two rates as equal.

Solution

1. Solve $(1+i{)}^{12}=2$.
2. $1+i=2^{1/12}=1.059463$.
3. $i=5.9463\%$.
4. The 'rule of 72' estimate $72/12=6\%$ is close, which is exactly why that approximation survives - but it is an approximation.
5. The rule is most accurate near 8%; it drifts noticeably at very low or very high rates.

Trap. Using the rule of 72 as though it were exact.

Solution

1. Payments at the END of each year make this an annuity-immediate: $PV=2,500a_{\overline{12}|0.06}$.
2. $a_{\overline{12}|0.06}=\frac{1-1.06^{-12}}{0.06}=8.383844$.
3. $PV=2,500\times 8.383844$.
4. $=20,959.61$. The annuity-due version would be 6% larger, at 22,217.19.

Trap. Using the due factor when the first payment is a year away.

Solution

1. An annuity-due pays each instalment exactly ONE period earlier than the corresponding annuity-immediate.
2. Shifting every payment one period earlier multiplies its present value by $(1+i)$ - once, not $n$ times.
3. So ${\ddot{a}}_{\overline{n}|}=a_{\overline{n}|}(1+i)$, equivalently $\frac{1-v^n}{d}$.
4. The related identity ${\ddot{a}}_{\overline{n}|}=1+a_{\overline{n-1}|}$ says the same thing from the other direction: a payment now plus a shorter immediate annuity.

Trap. Raising (1+i) to the power n, which shifts the whole annuity n periods rather than one.

Solution

1. The payments are $1,000\times (1,2,\dots ,10)$, so this is $1,000(Ia{)}_{\overline{10}|0.06}$.
2. $(Ia{)}_{\overline{n}|}=\frac{{\ddot{a}}_{\overline{n}|}-n{v}^n}{i}$ - note the DUE factor in the numerator.
3. ${\ddot{a}}_{\overline{10}|}=7.801692$ and $10v^{10}=5.583948$.
4. $(Ia{)}_{\overline{10}|}=\frac{2.217744}{0.06}=36.9624$, so the present value is 36,962.41.

Trap. Using a rather than ä in the numerator, which is the signature error on increasing annuities.

Solution

1. The payments are $100\times (1,2,\dots ,8)$, so this is $100(Ia{)}_{\overline{8}|0.05}$.
2. $a_{\overline{8}|0.05}=6.463213$, so ${\ddot{a}}_{\overline{8}|}=6.786373$.
3. $8v^8=8(0.676839)=5.414715$.
4. $(Ia{)}_{\overline{8}|}=\frac{6.786373-5.414715}{0.05}=27.43316$, so $PV=2,743.32$.
5. The average payment is 450 across eight years, an undiscounted total of 3,600 - discounting removes about a quarter of it.

Solution

1. The loan equals the present value of the payments: $L=P{a}_{\overline{n}|i}$.
2. $a_{\overline{20}|0.055}=11.950382$.
3. $P=\frac{250,000}{11.950382}$.
4. $=20,919.83$. Total payments come to 418,397 against a 250,000 loan, so interest is about 168,000.

Solution

1. The first payment's interest is charged on the full original balance.
2. $I_1=L\times i=40,000\times 0.08$.
3. $=3,200$.
4. The payment is $40,000/a_{\overline{10}|0.08}=5,961.16$, so the principal portion is $2,761.16$.

Trap. Applying the general formula when the first payment's interest is simply L·i - much faster.

Solution

1. Use the basic formula $P=Fr{a}_{\overline{n}|i}+C{v}^n$, with the coupon rate applied to FACE and the yield discounting to REDEMPTION.
2. $Fr=1,000(0.08)=80$, $a_{\overline{10}|0.06}=7.360087$, $v^{10}=0.558395$.
3. $P=80(7.360087)+1,000(0.558395)=588.81+558.40$.
4. $=1,147.20$ - a premium bond, because the coupon rate exceeds the yield.

Trap. Discounting the coupons at the coupon rate rather than the yield.

Solution

1. The ISSUER chooses when to call, and will choose whichever date is worst for the investor.
2. A premium bond's book value falls towards redemption over time, so the investor loses most if the bond is called EARLY.
3. Pricing to the earliest call therefore guarantees at least the assumed yield whatever the issuer does.
4. For a DISCOUNT bond the reasoning reverses: the book value rises, so the worst case is the latest call date.

Trap. Pricing a callable premium bond to maturity and overpaying for a bond that will be called.

Solution

1. Discount each inflow at 9% and subtract the outlay, which is NOT discounted because it occurs at time 0.
2. $18,000(0.917431)=16,513.76$; $22,000(0.841680)=18,516.96$; $25,000(0.772183)=19,304.58$.
3. Total inflows: $54,335.31$.
4. $NPV=54,335.31-50,000=4,335.31$, so the project clears the 9% hurdle.

Trap. Discounting the initial outlay by one period, which overstates the NPV by about 4,100.

Solution

1. The time-weighted return chains sub-period growth factors, so the size and timing of deposits cancel out.
2. That isolates the performance of the underlying investments, which is what a manager controls.
3. The dollar-weighted return (an IRR on the fund's cash flows) does depend on timing, which is what the INVESTOR controls.
4. A manager who performs well while investors add money at the wrong moments will show a good time-weighted and a poor dollar-weighted return.

Solution

1. Each cash flow is discounted at ITS OWN spot rate - one rate per date, not one rate per bond.
2. $1,000/1.04=961.54$; $1,000/1.046^2=913.98$; $1,000/1.051^3=861.37$.
3. Total: $961.54+913.98+861.37$.
4. $=2,736.89$. Using a flat 4.6% for all three would give 2,743.79 - close, and wrong.

Trap. Applying the three-year spot rate to all three cash flows.

Solution

1. Chaining: $(1+s_3{)}^3=(1+s_2{)}^2(1+f_{2,3})$.
2. $(1.05{)}^2(1.07)=1.1025\times 1.07=1.179675$.
3. $1+s_3=1.179675^{1/3}=1.056625$.
4. $s_3=5.6625\%$ - between the two-year spot and the forward, as a geometric mean must be.

Trap. Averaging 5% and 7% arithmetically to get 6%, which ignores that the 5% applies to two years and the 7% to one.

Solution

1. Macaulay duration is the present-value-weighted average time to the cash flows.
2. Cash flows are 60 at times 1 to 4 and 1,060 at time 5; at a 6% yield the price is exactly 1,000.
3. $D=\frac{\sum t{C}_t{v}^t}{1,000}$ with $v=1/1.06$.
4. $=4.4651$ years, less than the 5-year term because four coupons arrive first.

Trap. Weighting by the cash flows themselves rather than by their present values.

Solution

1. Macaulay convexity is $\frac{\sum t^2{C}_t{v}^t}{P}$ - the second PV-weighted moment of the payment times.
2. Working that sum over the five cash flows at a 6% yield and dividing by the price of 1,000 gives 21.2863.
3. The MODIFIED convexity $\frac{\sum t(t+1)C_t{v}^{t+2}}{P}=22.9187$ is a different quantity.
4. Only the modified version belongs in the price-change estimate; mixing them is the standard error here.

Trap. Using Macaulay convexity in the second-order price approximation.

Solution

1. Write the surplus $S(i)=P{V}_A(i)-P{V}_L(i)$. Matching present values makes $S=0$ at the current rate.
2. Matching durations makes $S^{\prime }(i)=0$, so the current rate is a stationary point of the surplus.
3. Requiring asset convexity STRICTLY greater makes $S^{\prime \prime }(i)>0$, so that stationary point is a minimum.
4. A minimum at zero means any small rate move in either direction produces a surplus, which is exactly what immunisation promises.

Trap. Matching convexity as well, which leaves the surplus flat to second order and provides no protection.

Solution

1. The derivation is a second-order Taylor expansion of the surplus around the CURRENT rate, assuming the whole curve moves by the same amount.
2. That makes it valid for SMALL PARALLEL shifts only.
3. A large shift takes the expansion outside its useful range, and a twist in the curve breaks the parallel assumption entirely.
4. Durations also drift as time passes, so an immunised portfolio must be rebalanced rather than left alone.

Trap. Treating immunisation as permanent protection rather than a local, and perishable, condition.

Solution

1. The swap rate is $R=\frac{1-P(n)}{{\sum }_{t}P(t)}$, where $P(t)$ are the zero-coupon prices.
2. $P(1)=0.961538$, $P(2)=0.913980$, $P(3)=0.861374$, summing to $2.736893$.
3. $R=\frac{1-0.861374}{2.736893}=\frac{0.138626}{2.736893}$.
4. $=5.0651\%$, just below the 3-year spot rate - a swap rate is a PV-weighted average of the forwards, so it lags the longest spot on a rising curve.

Trap. Averaging the spot rates rather than working with the discount factors.

Solution

1. The borrower already PAYS floating on its debt.
2. Entering a swap to RECEIVE floating offsets that exposure, and the swap's fixed leg becomes the borrower's net obligation.
3. So the position needed is pay fixed, receive floating - a payer swap.
4. The combination synthesises a fixed-rate loan without renegotiating the original debt.

Trap. Choosing the receiver swap, which doubles the floating exposure rather than removing it.