Bank of solved practice problems and exam simulator

Correct answer: A. Let \(J\) be the event young driver and \(S\) the event claim. By total probability, \(P(S)=0.30(0.18)+0.70(0.07)=0.054+0.049=0.103\). By Bayes, \(P(J\mid S)=\dfrac{0.054}{0.103}=0.5243\). The options 0.18 and 0.30 confuse the inverse conditional probability, or the prior, with the posterior.

Correct answer: A. By inclusion and exclusion, \(P(A\cup B)=0.40+0.25-0.12=0.53\). The proportion with neither is \(1-0.53=0.47\). Subtracting without removing the intersection (0.35) counts it twice; using \(0.60\times0.75=0.45\) assumes independence, which the problem does not state.

Correct answer: D. \(P(N\ge 3)=1-[p_0+p_1+p_2]=1-e^{-2.5}\left(1+2.5+\tfrac{2.5^2}{2}\right)=1-0.082085\times6.625=0.4562\). The most common mistake is to calculate \(P(N>3)\) or to stop at \(P(N=3)=0.2138\).

Correct answer: E. For the exponential, \(E[(X-d)_+]=\int_d^\infty S(x)\,dx=\theta e^{-d/\theta}=8000\,e^{-0.25}=6,230.41\). By memorylessness, the expected payment per payment (conditional on \(X>d\)) would be 8,000; subtracting the deductible from the mean (6,000) is wrong because it ignores losses below the deductible.

Correct answer: B. \(E[X]=\int_0^1 4x(1-x)^3dx=\tfrac15\) and \(E[X^2]=\int_0^1 4x^2(1-x)^3dx=\tfrac{1}{15}\). So \(\operatorname{Var}(X)=\tfrac1{15}-\tfrac1{25}=\tfrac{2}{75}=0.0267\). It is a Beta(1, 4). The option 0.0667 is the second moment, not the variance.

Correct answer: E. Marginals: \(f_X(x)=2x\) and \(f_Y(y)=2(1-y)\), so \(E[X]=\tfrac23\) and \(E[Y]=\tfrac13\). Also \(E[XY]=\int_0^1\!\int_0^x 2xy\,dy\,dx=\int_0^1 x^3dx=\tfrac14\). Therefore \(\operatorname{Cov}(X,Y)=\tfrac14-\tfrac29=\tfrac1{36}\approx0.0278\). The covariance is positive because \(Y\) is bounded by \(X\).

Correct answer: E. The sum of two independent exponentials with the same mean is Gamma(\(\alpha=2,\theta=1000\)). Its survival function is \(P(S>s)=e^{-s/\theta}\left(1+\tfrac{s}{\theta}\right)\), so \(P(S>3000)=e^{-3}(1+3)=4e^{-3}=0.1991\). Using \(e^{-3}\) treats the sum as if it were a single exponential.

Correct answer: C. The total \(S\) has mean \(400(12000)\) and standard deviation \(\sqrt{400}(30000)=600{,}000\). We need \(400\,r=1.645\times600{,}000\), that is, \(r=1.645\times\dfrac{30000}{\sqrt{400}}=2,467.50\). Dividing by 400 instead of \(\sqrt{400}\) seriously understates the loading.

Correct answer: B. \(i=\left(1+\tfrac{0.105}{12}\right)^{12}-1=11.020%\). The other options correspond to continuous, semiannual or quarterly compounding, or to confusing the nominal rate with the effective rate.

Correct answer: E. The monthly rate is 1%. \(PV=5000\,a_{\overline{60}|1\%}=5000\cdot\dfrac{1-1.01^{-60}}{0.01}=5000(44.955038)=$224,775.19\). Multiplying by 1.01 would give the annuity due; using \(1.12^{1/12}-1\) treats the 12% as an effective annual rate.

Correct answer: C. Payment: \(R=\dfrac{1{,}200{,}000}{a_{\overline{20}|11\%}}=$150,690.76\). By the prospective method, the balance is the present value of the 15 remaining payments: \(R\,a_{\overline{15}|11\%}=$1,083,597.63\). Subtracting five payments from the loan ignores that the early payments are almost entirely interest.

Correct answer: D. With 20 semiannual periods, a coupon of 4 and a rate of 4.5%: \(P=4\,a_{\overline{20}|4.5\%}+100\,v^{20}=93.4960\). It trades below par because the coupon rate is lower than the yield. Using 9% as an effective annual rate (4.403% semiannual) gives a different price.

Correct answer: A. Present values: \(6/1.07=5.6075\), \(6/1.07^2=5.2406\), \(106/1.07^3=86.5276\); price 97.3757. \(D_{Mac}=\dfrac{1(5.6075)+2(5.2406)+3(86.5276)}{97.3757}=2.8310\) years. Dividing by 1.07 gives the modified duration, 2.6458.

Correct answer: C. \(PV_L=1{,}000{,}000\,v^6=$746,215.40\). The duration of each zero-coupon bond is its term, so the weight of the 10-year bond satisfies \(3(1-w)+10w=6\), that is, \(w=3/7\). The fund invests \($319,806.60\) in the 10-year bond, whose face value is \($319,806.60\times1.05^{10}=$520,931.25\).

Correct answer: C. \((Ia)_{\overline{10}|}=\dfrac{\ddot a_{\overline{10}|}-10v^{10}}{i}=\dfrac{7.801692-10(0.558395)}{0.06}=36.962408\). The present value is \(1000\times36.962408=$36,962.41\). Using the average payment (5,500) on a level annuity understates the value because the large payments are further away.

Correct answer: E. By no arbitrage, \((1.09)^2=(1.08)(1+f_{1,2})\), so \(f_{1,2}=\dfrac{1.1881}{1.08}-1=10.009%\). The linear approximation \(2(9\%)-8\%=10\%\) is close but not exact.

Correct answer: E. It is a deferred probability: \({}_{2|2}q_{60}=\dfrac{l_{62}-l_{64}}{l_{60}}=\dfrac{2{,}880}{100{,}000}=0.0288\). Dividing by \(l_{62}\) gives the conditional probability for someone who has already reached 62.

Correct answer: C. Under De Moivre, the future lifetime of \((x)\) is uniform on \((0,\omega-x)\), so \(\mathring e_x=\dfrac{\omega-x}{2}=\dfrac{60}{2}=30\). The curtate expectation would be \(e_{45}=29.5\).

Correct answer: C. \(\bar A_x=\dfrac{\mu}{\mu+\delta}=\dfrac{0.02}{0.07}=0.285714\) and \(\bar a_x=\dfrac{1}{\mu+\delta}=14.2857\). So \(\bar P=\dfrac{\bar A_x}{\bar a_x}=\mu=0.02\) per peso insured, that is, $20,000 per year. With a constant force of mortality, the premium does not depend on interest.

Correct answer: E. \(A^1_{60:\overline{2}|}=v\,q_{60}+v^2\,{}_{1|}q_{60}=\dfrac{0.011}{1.05}+\dfrac{0.0122}{1.05^2}=0.021542\). Multiplied by 500,000 it gives $10,770.98. Using \(q_{61}\) instead of \({}_{1|}q_{60}\) omits the probability of surviving the first year.

Correct answer: A. \({}_5E_{60}=v^5\,\dfrac{l_{65}}{l_{60}}=\dfrac{0.931}{1.06^5}=0.695697\). The premium is $695,697.36. Leaving out the discounting or the survival probability are the most frequent mistakes.

Correct answer: E. \(A_{60:\overline2|}=v\,q_{60}+v^2p_{60}=0.907528\) and \(\ddot a_{60:\overline2|}=1+v\,p_{60}=1.941905\), so \(P=$46,733.93\). At the end of year 1 the contract will pay 100,000 at the end of year 2 whether or not death occurs, and one premium remains to be collected: \({}_1V=100{,}000\,v-P=$48,504.17\).

Correct answer: B. \(({}_tV+P)(1+i)=q\,b+p\,{}_{t+1}V\). So \({}_{t+1}V=\dfrac{15{,}000(1.05)-0.012(200{,}000)}{0.988}=\dfrac{15{,}750-2{,}400}{0.988}=$13,512.15\). Not dividing by \(p_{x+t}\) ignores that the reserve is required only for survivors.

Correct answer: B. The identity \(A_x=1-d\,\ddot a_x\) with \(d=\dfrac{0.04}{1.04}=0.038462\) gives \(\ddot a_x=\dfrac{0.68}{0.038462}=17.68\). Using \(i\) instead of \(d\) gives 17.00; using \(\delta\) corresponds to the continuous version.

Correct answer: E. The payment is \(Y=\min\{(X-5000)_+,\,30000\}\), which reaches the maximum when \(X\ge35{,}000\). \(E[Y]=\int_{5000}^{35000}\dfrac{x-5000}{50000}dx+30000\,P(X>35000)=9{,}000+9{,}000=$18,000.00\). Without the limit it would be 20,250; 20,000 is the expected payment per payment.

Correct answer: B. In a compound Poisson, \(\operatorname{Var}(S)=\lambda\,E[X^2]=50(15000^2+20000^2)=3.125\times10^{10}\). The standard deviation is \(\sqrt{3.125\times10^{10}}=$176,776.70\). Using only the severity variance forgets that the second moment includes the square of the mean.

Correct answer: B. For the exponential, \(LER(d)=\dfrac{E[X\wedge d]}{E[X]}=1-e^{-d/\theta}\). With inflation, \(\theta'=11{,}000\) and \(LER=1-e^{-1000/11000}=0.0869\), lower than the original 0.0952: a fixed deductible loses effectiveness under inflation.

Correct answer: D. \(a=\dfrac{\beta}{1+\beta}=0.6\), \(b=(r-1)a=0.6\) and \(p_0=(1+\beta)^{-2}=0.16\). Then \(p_1=(a+b)p_0=0.1920\) and \(p_2=(a+b/2)p_1=0.1728\). Sum: 0.5248. A Poisson with the same mean (3) would give 0.4232, lower because it has less dispersion.

Correct answer: A. VaR: \(\left(\tfrac{\theta}{x+\theta}\right)^3=0.01\Rightarrow x=20000(0.01^{-1/3}-1)=$72,831.78\). For the Pareto, the mean excess loss is \(e(x)=\dfrac{x+\theta}{\alpha-1}\), so \(TVaR=VaR+\dfrac{VaR+\theta}{2}=$119,247.67\).

Correct answer: A. Per claim, the reinsurer pays \(\min\{(X-1{,}000{,}000)_+,\,2{,}000{,}000\}\): 0, 800,000 and 2,000,000. Total: $2,800,000. Without the limit it would be 4,300,000; 4,100,000 is what the ceding company retains.

Correct answer: D. By convolution: \(P(S=0)=0.5\), \(P(S=1)=0.18\), \(P(S=2)=0.3(0.4)+0.2(0.6^2)=0.192\), \(P(S=3)=0.2(2)(0.6)(0.4)=0.096\) and \(P(S=4)=0.032\). So \(E[(S-2)_+]=1(0.096)+2(0.032)=0.16\) million, that is, $160,000. Check: \(E[S]=0.7\times1.4=0.98\).

Correct answer: C. The deductible is applied first and then the coinsurance: \(0.80\times(85{,}000-10{,}000)=60{,}000\). Applying the coinsurance before the deductible gives 58,000, a sequence that contradicts the contract's definition.

Correct answer: E. \(n_0=\left(\dfrac{z}{k}\right)^2=\left(\dfrac{1.645}{0.05}\right)^2=1,082.41\). The 90% probability is two-sided, which is why the 95th percentile (1.645) is used. Using 1.96 corresponds to 95% two-sided.

Correct answer: A. \(Z=\sqrt{400/1,082.41}=0.6079\). The estimate is \(Z(1250)+(1-Z)(1100)=$1,191.19\). Using \(n/n_0\) without the square root gives a smaller factor and understates the group's own experience.

Correct answer: D. Overall mean \(\mu=0.2\). EPV \(=E[\lambda]=0.2\) (Poisson), VHM \(=\operatorname{Var}(\lambda)=0.01\), \(k=20\) and \(Z=\dfrac{5}{5+20}=0.2\). Estimate: \(0.2(3/5)+0.8(0.2)=0.2800\).

Correct answer: C. \(L(\theta)=\prod_{i=1}^4\tfrac1\theta e^{-x_i/\theta}\cdot\left(e^{-10000/\theta}\right)^2\). Maximizing, \(\hat\theta=\dfrac{\text{sum of all amounts, censored included}}{\text{number of uncensored observations}}=\dfrac{38{,}500}{4}=$9,625.00\). Dividing by 6 treats the censored claims as exact losses.

Correct answer: C. Mean 7; deviations \(-3,0,-2,2,3\); sum of squares 26. The unbiased estimator divides by \(n-1\): \(s^2=26/4=6.5\). Dividing by 5 (5.2) gives the maximum likelihood estimator, which is biased downward.

Correct answer: A. The standard error is \(6000/\sqrt{100}=600\). Upper limit: \(18{,}400+1.96(600)=19{,}576\). Using the standard deviation of a single claim instead of the standard error produces an absurdly wide interval.

Correct answer: D. For the gamma, \(E[X]=\alpha\theta\) and \(\operatorname{Var}(X)=\alpha\theta^2\). Dividing, \(\hat\theta=\dfrac{10{,}000{,}000}{5{,}000}=2{,}000\) and \(\hat\alpha=2.5\).

Correct answer: A. \(P(N=1\mid\text{good})=0.1e^{-0.1}=0.090484\) and \(P(N=1\mid\text{bad})=0.4e^{-0.4}=0.268128\). Posterior: \(\dfrac{0.75(0.090484)}{0.75(0.090484)+0.25(0.268128)}=0.5031\). Leaving out the factor \(e^{-\lambda}\) gives 0.4286.

Correct answer: B. \(f_{12\to24}=\dfrac{6200+6750+7600}{4000+4500+5000}=\dfrac{20{,}550}{13{,}500}=1.5222\). The simple average of the individual factors (1.55, 1.50, 1.52) is 1.5233; including 2025 in the denominator is a mistake because that year has no 24-month value yet.

Correct answer: B. Factors: \(f_{12\to24}=1.522222\), \(f_{24\to36}=\dfrac{14{,}930}{12{,}950}=1.152896\), \(f_{36\to48}=\dfrac{7{,}400}{7{,}100}=1.042254\). Cumulative factor 1.829117; ultimate \(5{,}600\times1.829117=10,243.1\).

Correct answer: D. Reserves by year: 2023, \(7{,}830(1.042254-1)=330.8\); 2024, \(7{,}600(1.201610-1)=1,532.2\); 2025, 4,643.1; 2022, 0 (no tail). Total 6,506.1. In practice, since the triangle is on a paid basis, this reserve includes reported but unpaid claims as well as incurred but not reported claims.

Correct answer: A. Year 2024 is at 24 months: it needs \(f_{24\to36}\times f_{36\to48}=1.201610\). Ultimate 9,132.2 and reserve \(9,132.2-7{,}600=1,532.2\). Applying only one factor leaves out development from 36 to 48 months.

Correct answer: A. Expected ultimate \(12{,}000\times0.78=9{,}360\). Percentage undeveloped \(1-1/1.829117=0.453288\). BF reserve \(=9{,}360\times0.453288=4,242.8\). The method gives less weight to the volatile early data than chain-ladder does (4,643.1).

Correct answer: D. From August 1 to December 31, 31+30+31+30+31 = 153 days elapse; 212 remain. Unearned premium: \(36{,}500\times212/365=21{,}200\). Using whole months would give 21,291.67, an approximation the exact days method avoids.

Correct answer: D. Used-up premiums: \(9{,}800/1\), \(10{,}500/1.0423\), \(11{,}200/1.2016\), \(12{,}000/1.8291\), which add up to 35,755.7. Current paid: 7,400 + 7,830 + 7,600 + 5,600 = 28,430. \(ELR_{CC}=28{,}430/35,755.7=79.51%\). Dividing paid by the full premium ignores that recent years have not yet developed.

Correct answer: E. \(BE=\dfrac{500}{1.08}+\dfrac{300}{1.085^2}+\dfrac{200}{1.09^3}=872.24\). Each cash flow is discounted at the spot rate for its term; using a single flat rate biases the result according to the slope of the curve.

Correct answer: E. The premium is 12 days of salary per year of service (art. 162, sec. I) and it is payable on voluntary resignation because the employee has more than 15 years of service (sec. III). The salary base is capped at twice the minimum wage (art. 486): \(\min(900,600)=600\). Premium: \(12\times600\times18=129{,}600\).

Basis: LFT, arts. 162 and 486

Correct answer: D. Projected salary at 65: \(40{,}000(1.04)^{15}=$72,037.74\). Total projected benefit: \(25\times S_{65}\); the accrued fraction is \(10/25\). DBO \(=10\times S_{65}\times1.09^{-15}=$197,771.00\). Using the current salary corresponds to the accumulated benefit obligation without projection, which NIF D-3 does not use for this calculation.

Correct answer: C. Net interest is calculated by applying the discount rate to the net liability: \(0.095\times(5{,}000{,}000-3{,}200{,}000)=171{,}000\). The 10.5% expected return does not enter profit or loss; the difference between the actual return and the return recognized at the discount rate is a remeasurement.

Correct answer: A. Article 171 of the 1973 Law sets the cesantía amount as a percentage of the old-age pension: 75% at 60, 80% at 61, 85% at 62, 90% at 63 and 95% at 64. At 62: \(0.85\times12{,}000=10{,}200\). Family allowances and welfare supplements are left out.

Basis: LSS 1973, art. 171

Correct answer: B. The IMSS states that a member under the 1973 Law must be at least 60 when applying for the pension and have at least 500 weeks of contributions. In addition, the third transitory article of the current LSS allows those who contributed before July 1, 1997 to choose between the 1973 and 1997 regimes when they retire. The figures of 750 or 1,000 weeks correspond to the individual account regime.

Basis: IMSS; current LSS, third transitory article

Correct answer: B. The deferred annuity is \({}_{10|}\ddot a_{55}={}_{10}E_{55}\,\ddot a_{65}=0.52\times14.2=7.384\). Present value: \(120{,}000\times7.384=886{,}080\). Subtracting 1 from the annuity turns it into an annuity in arrears, contrary to what the plan states.

Correct answer: D. Expected year-end DBO: \((10{,}000{,}000+600{,}000)(1.09)-800{,}000=10,754,000\). The actual DBO is lower, so there is an actuarial gain of \(10,754,000-10{,}350{,}000=404,000\). Under NIF D-3, remeasurements of the post-employment plan are presented in OCI and recycled to profit or loss based on the average remaining working life (or, as an option, taken directly to profit or loss); under IAS 19 they stay in OCI without recycling.

Correct answer: D. \(\Delta DBO\approx-D_{mod}\,\Delta i\,DBO=-12(0.01)(20{,}000{,}000)=-2{,}400{,}000\), so the DBO falls to 17,600,000. Dividing again by \(1+i\) is appropriate only if the given duration is a Macaulay duration. Because of convexity, the actual decrease is somewhat smaller.

Correct answer: B. Article 235, section III, of the LISF provides that the RCS shall consider unexpected losses at a 99.5% confidence level over a one-year horizon, except for risks whose nature calls for periods suited to their characteristics.

Basis: LISF, art. 235, sec. III

Correct answer: B. Article 233 of the LISF provides that companies must calculate the RCS monthly using the standard formula of article 236 or an internal model under article 237.

Basis: LISF, art. 233

Correct answer: D. Article 218, section I, item e), of the LISF provides that the amount of technical reserves shall equal the sum of the best estimate and a risk margin, which must be calculated separately.

Basis: LISF, art. 218, sec. I, item e)

Correct answer: C. Provision 5.4.3 of the CUSF sets at 10% the net cost of capital rate that insurance companies must use for the risk margin. Solvency II uses 6%, hence the most common distractor.

Basis: CUSF, Provision 5.4.3

Correct answer: C. Article 217, section II, item c), of the LISF includes within the reserves for outstanding obligations the reserve for incurred but not reported claims, as well as for the adjustment expenses allocated to claims.

Basis: LISF, art. 217, sec. II, item c)

Correct answer: A. The CUSF sets a maximum duration of six hours (31.1.7), allows books, calculators and laptops (31.1.8, sec. III), requires at least 70% correct answers (31.1.9), gives results at the end of the exam (31.1.10) and grants a letter valid for two years, renewable (31.1.11). The eight hours correspond to the first part of the exam for independent actuaries (31.2.17).

Basis: CUSF, Provisions 31.1.7 to 31.1.11

Correct answer: B. Article 245 requires the test to be performed at least annually to assess the adequacy of eligible own funds under prospective scenarios. Article 246 makes the board of directors responsible and requires the signature of an actuary with a cédula profesional and current certification from the professional college of the specialty or accreditation before the CNSF, registered with it.

Basis: LISF, arts. 245 and 246

Correct answer: C. Article 218 of the current LSS grants the right to continue voluntarily to insured persons with at least 52 weekly contributions credited in the mandatory regime in the last five years, upon deregistration, enrolling with their last salary or a higher one and paying monthly in advance.

Basis: LSS, art. 218