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1. #1Exam PSet theory and probability axiomsCore

   A motor insurer finds that 48% of its policyholders have a telematics device, 31% have a named second driver, and 17% have both. What proportion have at least one of the two?

   1. A$0.1700$
   2. B$0.6200$
   3. C$0.6500$
   4. D$0.7900$
   5. E$0.9600$

   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.

2. #2Exam PConditional probability and independenceExam level

   An insurer knows $P(\text{claim})=0.12$ and $P(\text{claim}\cap \text{young driver})=0.045$. What proportion of claimants are young drivers?

   1. A$0.0054$
   2. B$0.1200$
   3. C$0.2667$
   4. D$0.3750$
   5. E$0.4500$

   Solution

   1. The question asks for $P(\text{young}\mid \text{claim})$: the conditioning event is 'claim', so it is the denominator.
   2. $P(\text{young}\mid \text{claim})=\frac{0.045}{0.12}$.
   3. $=0.375$.
   4. Reading the question carefully matters here: 'what proportion of claimants' fixes the denominator, and getting that backwards gives 0.375 versus its reciprocal-style alternative.

   Trap. Conditioning the wrong way round and computing P(claim | young) with the data given.

3. #3Exam PRandom variables and distribution functionsExam level

   A device's lifetime in years has survival function $S(t)=e^{-t/4}$ for $t>0$. What is the probability it survives more than 6 years?

   1. A$0.2231$
   2. B$0.2865$
   3. C$0.4724$
   4. D$0.6065$
   5. E$0.7769$

   Solution

   1. The survival function IS the answer: $P(T>6)=S(6)$.
   2. $S(6)=e^{-6/4}=e^{-1.5}$.
   3. $=0.223130$.
   4. This is an exponential lifetime with mean 4 years, so surviving beyond 1.5 mean lifetimes has probability $e^{-1.5}$.

   Trap. Computing 1 − S(6) = 0.7769, which is the probability of failing before 6 years.

4. #4Exam PDiscrete distributionsExam level

   In 8 independent trials with success probability 0.35, what is the probability of at most 2 successes?

   1. A$0.1373$
   2. B$0.2786$
   3. C$0.3428$
   4. D$0.4279$
   5. E$0.5721$

   Solution

   1. Sum the binomial mass function over $k=0,1,2$ with $n=8$, $p=0.35$.
   2. $P(X=0)=(0.65{)}^8=0.031864$.
   3. $P(X=1)=8(0.35)(0.65{)}^7=0.137276$; $P(X=2)=28(0.35{)}^2(0.65{)}^6=0.258712$.
   4. The total is $0.031864+0.137276+0.258712=0.427852$.
   5. So a little over two fifths of samples contain at most two successes, against a mean of 2.8.
5. #5Exam PContinuous distributionsExam level

   Losses follow an exponential with mean 1,000. What is the probability a loss lies between 500 and 1,500?

   1. A$0.2498$
   2. B$0.3834$
   3. C$0.4712$
   4. D$0.6065$
   5. E$0.7769$

   Solution

   1. $P(500<X<1500)=S(500)-S(1500)$ — easiest through the survival function for an exponential.
   2. $S(500)=e^{-0.5}=0.606531$.
   3. $S(1500)=e^{-1.5}=0.223130$.
   4. $0.606531-0.223130=0.383401$.
   5. Roughly two fifths of losses land in this middle band, with the rest split between small and very large claims.
6. #6Exam PMoment generating functionsExam level

   For the MGF $M(t)=0.2+0.3e^t+0.5e^{3t}$, find the variance.

   1. A$1.5600$
   2. B$1.8000$
   3. C$3.2400$
   4. D$4.8000$
   5. E$5.4000$

   Solution

   1. The distribution puts mass 0.2, 0.3 and 0.5 on the values 0, 1 and 3.
   2. $E[X]=1.8$ from before, so $(E[X]{)}^2=3.24$.
   3. $E[X^2]=0(0.2)+1(0.3)+9(0.5)=4.8$.
   4. $\mathrm{Var}(X)=4.8-3.24=1.56$.

   Trap. Reporting E[X²] = 4.8 as the variance.

7. #7Exam PTransformations and order statisticsExam level

   $X$ is exponential with mean 5 and $Y=\min (X,8)$. Find $E[Y]$.

   1. A$3.9906$
   2. B$4.0000$
   3. C$5.0000$
   4. D$6.5000$
   5. E$8.0000$

   Solution

   1. $Y$ is the limited variable, and $E[X\wedge u]={\int }_0^{u}S(x)dx$ for a non-negative variable.
   2. For an exponential, ${\int }_0^u{e}^{-x/\theta }dx=\theta \left(1-e^{-u/\theta }\right)$.
   3. $=5\left(1-e^{-8/5}\right)=5(1-0.201897)$.
   4. $=5(0.798103)=3.990515$, comfortably below the unlimited mean of 5 as any cap must be.

   Trap. Reporting the unlimited mean 5, or capping the mean at 8 and answering 5 anyway.

8. #8Exam FMMeasurement of interestExam level

   At what effective annual rate does money double in 12 years?

   1. A$5.9463\%$
   2. B$6.0000\%$
   3. C$6.1163\%$
   4. D$8.3333\%$
   5. E$12.0000\%$

   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.

9. #9Exam FMLevel annuitiesExam level

   A 10-year annuity pays 1 at the end of each half-year, at a nominal 6% convertible semiannually. Find the present value.

   1. A$7.3601$
   2. B$10.0000$
   3. C$14.8775$
   4. D$16.0000$
   5. E$20.0000$

   Solution

   1. Work in half-years: the rate is 3% per period and there are 20 periods.
   2. $a_{\overline{20}|0.03}=\frac{1-1.03^{-20}}{0.03}$.
   3. $1.03^{-20}=0.553676$, so the factor is $\frac{0.446324}{0.03}$.
   4. $=14.877475$. Treating the payments as annual at 6% would give 7.36, less than half the correct value.

   Trap. Using 10 periods at 6% rather than 20 periods at 3%.

10. #10Exam FMLoan amortisation and sinking fundsExam level

    A 100,000 loan charges 7% annually and is repaid in 12 years by the sinking-fund method, with the fund earning 5%. Find the total annual outlay.

    1. A$7,000.00$
    2. B$12,590.20$
    3. C$13,282.54$
    4. D$14,000.00$
    5. E$15,000.00$

    Solution

    1. The lender receives interest only: $100,000\times 0.07=7,000$ a year.
    2. The sinking fund must accumulate to 100,000: $D=\frac{100,000}{s_{\overline{12}|0.05}}=\frac{100,000}{15.917127}=6,282.54$.
    3. Total outlay: $7,000+6,282.54=13,282.54$.
    4. Straight amortisation at 7% would cost $100,000/a_{\overline{12}|0.07}=12,590.20$ — cheaper, because the fund earns less than the loan costs.

    Trap. Using a rather than s for the fund, which discounts instead of accumulating.

11. #11Exam FMBond pricing, premium and discountExam level

    A 1,000 par 12-year bond with 4% annual coupons is bought to yield 6%. Find the discount (the amount by which C exceeds P).

    1. A$120.00$
    2. B$167.68$
    3. C$200.00$
    4. D$240.00$
    5. E$300.00$

    Solution

    1. $P-C=(Fr-Ci)a_{\overline{n}|}$, so the discount is $(Ci-Fr)a_{\overline{n}|}$.
    2. $Ci-Fr=60-40=20$ per year.
    3. $a_{\overline{12}|0.06}=8.383844$.
    4. Discount $=20(8.383844)=167.68$, so the bond costs about 832 and accumulates up to 1,000 over its life.

    Trap. Reporting the price rather than the discount, or getting the sign backwards.

12. #12Exam FMDuration and convexityExam level

    Adding convexity, estimate the same bond's price at a 6.5% yield. Its modified convexity is 22.9187.

    1. A$978.94$
    2. B$979.22$
    3. C$980.00$
    4. D$981.50$
    5. E$985.00$

    Solution

    1. The second-order estimate is $P\left(1-D_{Mod}\Delta i+\frac{1}{2}{C}_{Mod}(\Delta i{)}^2\right)$.
    2. The convexity correction is $\frac{1}{2}(22.9187)(0.005{)}^2=0.000286$.
    3. $P_{\text{est}}=1,000(1-0.021062+0.000286)$.
    4. $=979.22$, within 0.07 of the true price — a threefold improvement on duration alone.

    Trap. Forgetting the factor of one half on the convexity term.

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