Exam FAM — Survival Models & Life Tables Practice Flashcards

**Answer: (B).** Use ${}_tp_x=\exp\!\bigl(-\int_0^{t}\mu_{x+s}\,ds\bigr)$. Writing $\mu_{x}=0.002+0.0001x$, the cleanest route is to integrate in $x$ from $40$ to $50$: $\displaystyle\int_{40}^{50}(0.002+0.0001x)\,dx=0.002(10)+0.0001\cdot\dfrac{50^2-40^2}{2}=0.02+0.0001(450)=0.02+0.045=0.065.$ ${}_{10}p_{40}=e^{-0.065}\approx 0.9371$. Forgetting the $\tfrac12$ in $\int x\,dx$ (using $0.0001\cdot 900=0.09$, total $0.11$) gives $e^{-0.11}\approx 0.8958$ (A); dropping the $Bc^x$… no — dropping the quadratic piece and keeping only $A t=0.02$ gives $e^{-0.02}\approx 0.9802$ (D).

**Answer: (D).** Under a constant force, $T_x\sim\text{Exponential}(\mu)$ with ${}_tp_x=e^{-\mu t}$. The median $m$ solves ${}_mp_x=0.5$: $e^{-0.02m}=0.5\ \Rightarrow\ -0.02m=\ln 0.5\ \Rightarrow\ m=\dfrac{\ln 2}{0.02}=\dfrac{0.693147}{0.02}\approx 34.66.$ The **mean** future lifetime $\overset{\circ}{e}_x=\tfrac{1}{\mu}=\tfrac{1}{0.02}=50$ is the distractor (E) — for a right-skewed exponential the median is below the mean. Using $\tfrac{0.5}{\mu}=25$ (B is $20$, C is $25$) confuses the median with half the mean.

**Answer: (B).** For a beta–de Moivre form $S_0(x)=\bigl(1-\tfrac{x}{\omega}\bigr)^{\alpha}$, the force is $\mu_x=-\dfrac{d}{dx}\ln S_0(x)=\dfrac{\alpha}{\omega-x}$. Here $\alpha=\tfrac15$ and $\omega=110$, so $\mu_{45}=\dfrac{1/5}{110-45}=\dfrac{0.2}{65}\approx 0.003077.$ Forgetting the exponent $\alpha$ and using the ordinary de Moivre $\dfrac{1}{\omega-x}=\dfrac{1}{65}\approx 0.015385$ gives (D); using $\dfrac{1}{\alpha(\omega-x)}=\dfrac{1}{13}\approx 0.076923$ inverts the role of $\alpha$ (E).

**Answer: (C).** First read off the force: $\ln S_0(x)=-0.00002x^2$, so $\mu_x=-\dfrac{d}{dx}\ln S_0(x)=0.00004x$. ${}_{10}p_{50}=\dfrac{S_0(60)}{S_0(50)}=\exp\!\Bigl(-\!\int_{50}^{60}\mu_x\,dx\Bigr)$, and $\displaystyle\int_{50}^{60}0.00004x\,dx=0.00004\cdot\dfrac{60^2-50^2}{2}=0.00004\cdot\dfrac{1100}{2}=0.022.$ ${}_{10}p_{50}=e^{-0.022}\approx 0.9782$. Equivalently $\dfrac{S_0(60)}{S_0(50)}=e^{-0.00002(3600-2500)}=e^{-0.022}$ — the same number directly from the survival function. Reading off $\mu_{50}=0.00004(50)=0.002$ and forgetting to multiply by the $10$ years gives the exponent $0.002$ and $e^{-0.002}\approx 0.9980$ (E). Dropping the $\tfrac12$ in $\int x\,dx$ (using $0.044$) gives $e^{-0.044}\approx 0.9569$, which lands between options — a sign the $\tfrac12$ was missed.

**Answer: (C).** This is a de Moivre force ($\mu_{x+s}=\tfrac{1}{(\omega-x)-s}$ with $\omega-x=60$). Integrate: $\displaystyle\int_0^{10}\dfrac{ds}{60-s}=\Bigl[-\ln(60-s)\Bigr]_0^{10}=\ln 60-\ln 50=\ln\dfrac{60}{50}.$ ${}_{10}p_{40}=\exp\!\Bigl(-\ln\tfrac{60}{50}\Bigr)=\dfrac{50}{60}\approx 0.8333.$ This matches the de Moivre survival shortcut ${}_{t}p_x=\dfrac{(\omega-x)-t}{\omega-x}=\dfrac{60-10}{60}=\dfrac{50}{60}$. Reading the limiting age as $\omega-x=50$ (so $\tfrac{40}{50}=0.80$) gives the distractor (B).

**Answer: (C).** The force $\mu_x=\dfrac{1}{\omega-x}$ with $\omega=120$ is the de Moivre (uniform) law, whose survival function is $S_0(x)=1-\dfrac{x}{\omega}$. $S_0(48)=1-\dfrac{48}{120}=1-0.4=0.60.$ Directly: $S_0(48)=\exp\!\bigl(-\int_0^{48}\tfrac{dx}{120-x}\bigr)=\exp\!\bigl(\ln\tfrac{120-48}{120}\bigr)=\tfrac{72}{120}=0.60$. Reporting $\dfrac{x}{\omega}=0.40$ — the death probability $F_0(48)$ — instead of the survival probability gives the distractor (A).

**Answer: (B).** Survival probabilities multiply across single years: ${}_3p_{70}=p_{70}\,p_{71}\,p_{72}$, with $p_x=1-q_x$. $p_{70}=0.97,\quad p_{71}=0.96,\quad p_{72}=0.95.$ ${}_3p_{70}=0.97(0.96)(0.95)=0.8846.$ Adding the death rates and subtracting once — $1-(0.03+0.04+0.05)=0.88$ — ignores the compounding and is the distractor (A). Using only the first year, $p_{70}=0.97$, gives (E).

**Answer: (A).** The deferred death probability is ${}_{2|}q_{70}={}_2p_{70}\cdot q_{72}$. ${}_2p_{70}=p_{70}\,p_{71}=0.97(0.96)=0.9312.$ ${}_{2|}q_{70}=0.9312(0.05)=0.04656\approx 0.0466.$ Check: ${}_{2|}q_{70}={}_2p_{70}-{}_3p_{70}=0.9312-0.8846=0.0466$. ✓ Using the raw $q_{72}=0.05$ without first surviving two years (omitting the ${}_2p_{70}$ factor) gives the distractor (B).