Exam FAM — Life Annuities Practice Flashcards

**Answer: (C).** Use the fundamental relation $\ddot a_x = \dfrac{1 - A_x}{d}$ with $d = \dfrac{i}{1+i} = \dfrac{0.05}{1.05} \approx 0.047619$. $\ddot a_x = \dfrac{1 - 0.28}{0.047619} = \dfrac{0.72}{0.047619} \approx 15.12$. Using $d = i = 0.05$ instead of $\dfrac{i}{1+i}$ gives $\dfrac{0.72}{0.05} = 14.40$ (distractor B). That premium-rate slip is the most common error here.

**Answer: (B).** The recursion is $\ddot a_x = 1 + v\,p_x\,\ddot a_{x+1}$ with $v = \dfrac{1}{1.05} \approx 0.952381$. $\ddot a_x = 1 + 0.952381(0.99)(13.00) = 1 + 0.952381(12.87) = 1 + 12.25714 \approx 13.26$. Forgetting the leading $+1$ (the certain time-$0$ payment) gives $12.26$ (distractor A); forgetting to discount by $v$ gives $1 + 0.99(13.00) = 13.87$ (distractor D).

**Answer: (C).** An annuity-due paying $B$ per year has EPV $B\,\ddot a_x$. Equate to the premium: $B\,\ddot a_x = 100{,}000 \;\Rightarrow\; B = \dfrac{100{,}000}{\ddot a_x} = \dfrac{100{,}000}{16.00} = \$6{,}250$ per year. Using the annuity-immediate $a_x = \ddot a_x - 1 = 15.00$ in the denominator gives $\dfrac{100{,}000}{15} \approx \$6{,}667$ (distractor D) — wrong because the contract is a due annuity.

**Answer: (D).** Survival: $p_x = 0.96$, $p_{x+1} = 0.95$, $p_{x+2} = 0.94$, so ${}_1p_x = 0.96$, ${}_2p_x = 0.96(0.95) = 0.912$, ${}_3p_x = 0.912(0.94) = 0.85728$. Temporary piece: $\ddot a_{x:\overline{3|}} = 1 + v(0.96) + v^2(0.912)$ with $v = 0.952381$, $v^2 = 0.907029$ $= 1 + 0.914286 + 0.827210 = 2.741496$. Deferred tail: ${}_3E_x\,\ddot a_{x+3} = v^3(0.85728)(12.00)$; $v^3 = 0.863838$, so ${}_3E_x = 0.740551$ and the tail is $0.740551(12.00) = 8.886608$. $\ddot a_x = 2.741496 + 8.886608 \approx 11.6281$. Reporting only the temporary piece ($2.74$, distractor A) or only the tail ($8.89$, distractor B) forgets the other half.

**Answer: (B).** The recursion is $\ddot a_x = 1 + v\,p_x\,\ddot a_{x+1}$ with $p_x = 1 - q_x = 0.985$ and $v = \dfrac{1}{1.04} \approx 0.961538$. $\ddot a_x = 1 + 0.961538(0.985)(12.40) = 1 + 0.961538(12.214) = 1 + 11.74423 \approx 12.7442$. Using $q_x$ in place of $p_x$ ($1 + v\,q_x\,\ddot a_{x+1} = 1 + 0.961538(0.015)(12.40) \approx 1.1788$, distractor A) is the classic survival/death mix-up. Forgetting to discount gives $1 + 0.985(12.40) = 13.214$ (distractor D).

**Answer: (A).** $\ddot a_{x:\overline{3|}} = \sum_{k=0}^{2} v^{k}\,{}_kp_x$ with ${}_0p_x = 1$, ${}_1p_x = 0.97$, ${}_2p_x = 0.97(0.96) = 0.9312$, and $v = 0.952381$, $v^2 = 0.907029$. $= 1 + 0.952381(0.97) + 0.907029(0.9312) = 1 + 0.923810 + 0.844626 \approx 2.7684$. Ignoring survival ($1 + v + v^2 = 2.8594$, distractor C) overstates the value; a 3-year temporary annuity-due makes its last payment at time $2$, so there is no time-$3$ term.

**Answer: (B).** First the pure endowment ${}_{10}E_x = v^{10}\,{}_{10}p_x = (1.05)^{-10}(0.85)$. Since $(1.05)^{-10} \approx 0.613913$, ${}_{10}E_x \approx 0.613913(0.85) = 0.521826$. ${}_{10|}\ddot a_x = {}_{10}E_x\,\ddot a_{x+10} = 0.521826(11.40) \approx 5.9488$. Forgetting the survival factor (using $v^{10}\ddot a_{x+10} = 0.613913 \times 11.40 \approx 6.9986$) gives distractor C; forgetting to discount (using ${}_{10}p_x\,\ddot a_{x+10} = 0.85 \times 11.40 = 9.69$) gives distractor D.

**Answer: (A).** The whole life annuity splits into temporary plus deferred pieces: $\ddot a_x = \ddot a_{x:\overline{10|}} + {}_{10|}\ddot a_x \;\Rightarrow\; {}_{10|}\ddot a_x = \ddot a_x - \ddot a_{x:\overline{10|}} = 15.00 - 7.80 = 7.20$. The temporary piece covers payments at times $0$–$9$; the deferred piece covers times $10, 11, \dots$. Adding the two values ($22.80$, distractor E) double-counts the whole life.