Exam FAM — Life Insurance Practice Flashcards

Thirty exam-realistic multiple-choice problems on SOA Exam FAM life insurance — whole life, term, pure endowment, and endowment APVs from explicit $q_x$ and $i$ values; second moments and variances via the rule of moments at double force of interest; the $A_x=vq_x+vp_xA_{x+1}$ recursion; UDD and constant-force relationships; and increasing/decreasing insurances — each with a fully worked solution and distractors tied to common candidate errors.

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Whole life
A discrete whole life insurance of $1$ is issued at $i=0.06$. Mortality is $q_x=0.10$, $q_{x+1}=0.15$, $q_{x+2}=1.00$ (the life cannot survive past age $x+3$). Calculate $A_x$. (A) $0.250000$ (B) $0.808493$ (C) $0.856798$ (D) $0.908000$ (E) $0.937500$
**Answer: (C).** With $v=\dfrac{1}{1.06}\approx 0.943396$, build the survival probabilities ${}_0p_x=1$, ${}_1p_x=0.90$, ${}_2p_x=0.90(0.85)=0.765$. Death-year terms $v^{k+1}\,{}_kp_x\,q_{x+k}$: $k=0:\;0.943396(1)(0.10)=0.094340$ $k=1:\;0.943396^{2}(0.90)(0.15)=0.889996(0.135)=0.120150$ $k=2:\;0.943396^{3}(0.765)(1.00)=0.839619(0.765)=0.642309$ $A_x = 0.094340+0.120150+0.642309 = 0.856798$. Forgetting to discount the certain final payment (using ${}_2p_x=0.765$ undiscounted plus the first two terms) overstates the value; ignoring survival probabilities entirely gives the inflated distractor (D).
Whole life
A life has a **constant** annual mortality rate $q=0.05$ at all ages, with $i=0.08$. Calculate the actuarial present value $A_x$ of a discrete unit whole life insurance. (A) $0.050000$ (B) $0.384615$ (C) $0.400000$ (D) $0.578704$ (E) $0.625000$
**Answer: (B).** With $q_{x+k}=q$ for all $k$, the death terms form a geometric series and collapse to the closed form $A_x = \dfrac{q}{q+i}$. $A_x = \dfrac{0.05}{0.05+0.08} = \dfrac{0.05}{0.13} = 0.384615$. Using $\dfrac{q}{q+i}$ but with $i$ replaced by the discount $d=\dfrac{0.08}{1.08}\approx 0.074074$ gives $\dfrac{0.05}{0.124074}\approx 0.403$ near distractor (C); the raw $q=0.05$ (distractor A) forgets discounting of the survival branch entirely.
Whole life
An insurer issues a $\$200{,}000$ whole life policy at $i=0.05$. Mortality is $q_x=0.08$, $q_{x+1}=0.12$, $q_{x+2}=0.20$, $q_{x+3}=1.00$ (the life cannot survive past age $x+4$). Calculate the single-premium APV. (A) $\$152{,}000$ (B) $\$169{,}809$ (C) $\$176{,}000$ (D) $\$192{,}000$ (E) $\$236{,}101$
**Answer: (B).** $v=\dfrac{1}{1.05}\approx 0.952381$. Survival: ${}_1p_x=0.92$, ${}_2p_x=0.92(0.88)=0.8096$, ${}_3p_x=0.8096(0.80)=0.64768$. Death terms $v^{k+1}\,{}_kp_x\,q_{x+k}$: $k=0:\;0.952381(0.08)=0.076190$ $k=1:\;0.907029(0.92)(0.12)=0.100136$ $k=2:\;0.863838(0.8096)(0.20)=0.139873$ $k=3:\;0.822702(0.64768)(1.00)=0.532848$ $A_x = 0.076190+0.100136+0.139873+0.532848 = 0.849047$. APV $= 200{,}000(0.849047) = \$169{,}809$. Ignoring the survival probabilities (weighting each $q$ only by $v^{k+1}$) inflates the value to $\approx\$236{,}101$ (distractor E); pricing the guaranteed final payment undiscounted overstates it further.
Moments & variance
A discrete whole life insurance has $A_x = 0.856798$ at $i=0.06$, for the life with $q_x=0.10$, $q_{x+1}=0.15$, $q_{x+2}=1.00$. Using the rule of moments, calculate the variance of the present-value random variable $Z$. (A) $0.001122$ (B) $0.013500$ (C) $0.121577$ (D) $0.734103$ (E) $0.735225$
**Answer: (A).** The second moment doubles the force of interest: use $j=(1.06)^2-1=0.1236$, so $v_j=\dfrac{1}{1.1236}\approx 0.889996$. ${}^2A_x = \sum v_j^{k+1}\,{}_kp_x\,q_{x+k}$: $k=0:\;0.889996(0.10)=0.089000$ $k=1:\;0.889996^{2}(0.90)(0.15)=0.792092(0.135)=0.106932$ $k=2:\;0.889996^{3}(0.765)=0.704958(0.765)=0.539293$ ${}^2A_x = 0.089000+0.106932+0.539293 = 0.735225$. $Var(Z) = {}^2A_x - A_x^2 = 0.735225 - 0.856798^2 = 0.735225 - 0.734103 = 0.001122$. Reporting ${}^2A_x$ itself (distractor E) or $A_x^2$ (distractor D) skips the subtraction; using the **single** force of interest for the second moment understates the dispersion.
Moments & variance
A unit whole life insurance has a **constant** annual mortality rate $q=0.04$ at all ages and $i=0.06$. Calculate $Var(Z)$. (A) $0.084499$ (B) $0.160000$ (C) $0.244499$ (D) $0.400000$ (E) $0.404499$
**Answer: (A).** Closed form: $A_x = \dfrac{q}{q+i} = \dfrac{0.04}{0.10} = 0.40$. Double the force of interest with $j=(1.06)^2-1 = 0.1236$: ${}^2A_x = \dfrac{q}{q+j} = \dfrac{0.04}{0.04+0.1236} = \dfrac{0.04}{0.1636} = 0.244499$. $Var(Z) = {}^2A_x - A_x^2 = 0.244499 - 0.40^2 = 0.244499 - 0.16 = 0.084499$. Reporting ${}^2A_x$ alone gives distractor (C); forgetting to double the force and using $\dfrac{q}{q+i}=0.40$ for the second moment leaves $0.40-0.16=0.24$, also wrong.
Moments & variance
A $\$250{,}000$ benefit is paid at the end of the year of death. The unit whole life insurance has $A_x=0.30$ and ${}^2A_x=0.11$. Calculate the standard deviation of the present value of the payout. (A) $\$5{,}000$ (B) $\$35{,}355$ (C) $\$50{,}000$ (D) $\$75{,}000$ (E) $\$82{,}916$
**Answer: (B).** Unit variance via the rule of moments: $Var(Z) = {}^2A_x - A_x^2 = 0.11 - 0.30^2 = 0.11 - 0.09 = 0.02$. For a benefit of $b=250{,}000$ the PV is $bZ$, so $Var(bZ)=b^2\,Var(Z)$ and the standard deviation is $250{,}000\sqrt{0.02} = 250{,}000(0.141421) = \$35{,}355$. Standard deviation scales **linearly** with the benefit; scaling the variance by $b$ (not $b^2$) or taking $b\cdot Var(Z)$ without the square root produces the larger distractors.
Moments & variance
At $i=0.10$, a unit $3$-year term insurance has $q_x=0.05$, $q_{x+1}=0.10$, $q_{x+2}=0.20$. The first moment is $A^{1}_{x:\overline{3|}}=0.252442$. Using the rule of moments, calculate $Var(Z)$. (A) $0.063727$ (B) $0.139006$ (C) $0.202733$ (D) $0.252442$ (E) $0.455175$
**Answer: (B).** Second moment at doubled interest $j=(1.1)^2-1=0.21$, $v_j=\dfrac{1}{1.21}\approx 0.826446$; survival ${}_1p_x=0.95$, ${}_2p_x=0.855$. $k=0:\;0.826446(0.05)=0.041322$ $k=1:\;0.826446^{2}(0.95)(0.10)=0.683013(0.095)=0.064886$ $k=2:\;0.826446^{3}(0.855)(0.20)=0.564474(0.171)=0.096525$ ${}^2A^{1}_{x:\overline{3|}} = 0.041322+0.064886+0.096525 = 0.202733$. $Var(Z) = {}^2A^{1}_{x:\overline{3|}} - (A^{1}_{x:\overline{3|}})^2 = 0.202733 - 0.252442^2 = 0.202733 - 0.063727 = 0.139006$. Reporting ${}^2A$ alone gives (C); $A^2$ alone gives (A).
Moments & variance
At $i=0.10$ with $q_x=0.05$, $q_{x+1}=0.10$, $q_{x+2}=0.20$, a $3$-year **endowment** insurance has $A_{x:\overline{3|}}=0.766342$. Its second moment (doubled force) is ${}^2A_{x:\overline{3|}}=0.588833$. Calculate $Var(Z)$. (A) $0.001553$ (B) $0.004671$ (C) $0.139006$ (D) $0.177509$ (E) $0.587280$
**Answer: (A).** $Var(Z) = {}^2A_{x:\overline{3|}} - (A_{x:\overline{3|}})^2 = 0.588833 - 0.766342^2 = 0.588833 - 0.587280 = 0.001553$. The variance is tiny compared with the pure $3$-year term ($Var\approx 0.139$, distractor C): the guaranteed survival payment bounds $Z$ between $v^3$ and $v$, removing almost all dispersion. Reporting $A^2$ alone gives distractor (E).