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Exam FAM — Survival Models & Life Tables Flashcards

The survival-model foundations of Exam FAM: the survival function $S_0(x)$ and lifetime CDF $F_0(x)$, the force of mortality $\mu_x$ and its link to ${}_tp_x$ via $\exp(-\int\mu)$, the future-lifetime random variable $T_x$ with its survival/death/deferred probabilities, complete and curtate expectations $\overset{\circ}{e}_x$ and $e_x$, the life-table functions $\ell_x$ and $d_x$ with their recursions, the UDD and constant-force fractional-age assumptions, and the standard mortality laws (de Moivre, constant force, Gompertz, Makeham) — with fully worked computations of ${}_tp_x$, $\mu_x$, and expectations from explicit forces, $q_x$ values, and small $\ell_x$ tables.

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  1. Survival function & force
    Define the **survival function** $S_0(x)$ and the **lifetime CDF** $F_0(x)$ for a newborn life, and give the relations among $S_0$, $F_0$, and the density $f_0$.
    Let $X$ be the age at death of a newborn (life aged $0$). $F_0(x)=\Pr[X\le x]$ is the CDF, and $S_0(x)=\Pr[X>x]=1-F_0(x)$ is the **survival function** — the probability of surviving past age $x$. Valid $S_0$ requires $S_0(0)=1$, $S_0(\infty)=0$, and $S_0$ non-increasing. The density is $f_0(x)=F_0'(x)=-S_0'(x)$.
  2. Survival function & force
    Define the **force of mortality** $\mu_x$ and state its relation to $S_0(x)$.
    $\mu_x$ is the instantaneous death rate at age $x$: $\mu_x=\dfrac{f_0(x)}{S_0(x)}=-\dfrac{S_0'(x)}{S_0(x)}=-\dfrac{d}{dx}\ln S_0(x)$. Equivalently the density factors as $f_0(x)=S_0(x)\,\mu_x$. The force is a *rate* (can exceed $1$), not a probability.
  3. Survival function & force
    Recover the survival function from the force of mortality. State $S_0(x)$ in terms of $\mu$.
    Integrating $\mu_x=-\dfrac{d}{dx}\ln S_0(x)$ from $0$ to $x$: $S_0(x)=\exp\!\Bigl(-\displaystyle\int_0^{x}\mu_s\,ds\Bigr)$. More generally, the $t$-year survival probability of a life aged $x$ is ${}_tp_x=\exp\!\Bigl(-\displaystyle\int_0^{t}\mu_{x+s}\,ds\Bigr)$.
  4. Survival function & force
    Given $S_0(x)=\bigl(1-\tfrac{x}{120}\bigr)^{1/6}$ for $0\le x\le 120$, find $\mu_{60}$ and the density $f_0(60)$.
    $\ln S_0(x)=\tfrac16\ln\!\bigl(1-\tfrac{x}{120}\bigr)$, so $\mu_x=-\dfrac{d}{dx}\ln S_0(x)=\dfrac{1/6}{120-x}$. At $x=60$: $\mu_{60}=\dfrac{1/6}{60}=\dfrac{1}{360}\approx 0.0027778$. $S_0(60)=(1-0.5)^{1/6}=0.5^{1/6}\approx 0.890899$. Density $f_0(60)=S_0(60)\,\mu_{60}\approx 0.890899(0.0027778)\approx 0.0024747$.
  5. Survival function & force
    For $S_0(x)=\bigl(1-\tfrac{x}{120}\bigr)^{1/6}$, compute $F_0(60)$ and ${}_{10}p_{40}$.
    $S_0(60)=0.5^{1/6}\approx 0.890899$, so $F_0(60)=1-S_0(60)\approx 0.109101$. For the survival probability, ${}_{10}p_{40}=\dfrac{S_0(50)}{S_0(40)}$. $S_0(50)=(1-\tfrac{50}{120})^{1/6}=(0.583333)^{1/6}\approx 0.914084$ and $S_0(40)=(1-\tfrac13)^{1/6}=(0.666667)^{1/6}\approx 0.934655$. ${}_{10}p_{40}=\dfrac{0.914084}{0.934655}\approx 0.978$.
  6. Future lifetime
    Define the **future-lifetime random variable** $T_x$ and the symbols ${}_tp_x$ and ${}_tq_x$.
    $T_x$ is the remaining lifetime of a life **currently aged** $x$ (so $T_x=X-x\mid X>x$). ${}_tp_x=\Pr[T_x>t]=\dfrac{S_0(x+t)}{S_0(x)}$ — probability $(x)$ survives $t$ more years. ${}_tq_x=\Pr[T_x\le t]=1-{}_tp_x$ — probability $(x)$ dies within $t$ years. When $t=1$ the leading $1$ is dropped: $p_x$, $q_x$.
  7. Future lifetime
    State the **deferred mortality probability** ${}_{u|t}q_x$ in words and as a formula, and give the two standard decompositions.
    ${}_{u|t}q_x=\Pr[(x)$ survives $u$ years, then dies in the next $t]$. $ {}_{u|t}q_x={}_up_x\cdot{}_tq_{x+u}={}_up_x-{}_{u+t}p_x={}_{u+t}q_x-{}_uq_x.$ With $t=1$ it is written ${}_{u|}q_x={}_up_x\,q_{x+u}$.
  8. Future lifetime
    Express ${}_tp_x$ as a product over single years (the **chain rule** for survival).
    Survival probabilities multiply across consecutive years: ${}_np_x=p_x\cdot p_{x+1}\cdots p_{x+n-1}=\displaystyle\prod_{k=0}^{n-1}p_{x+k}$. More generally ${}_{s+t}p_x={}_sp_x\cdot{}_tp_{x+s}$. This factorization underlies every life-table recursion.
  9. Future lifetime
    Given the one-year rates $q_{70}=0.020$, $q_{71}=0.025$, $q_{72}=0.030$, find ${}_3p_{70}$ and ${}_{2|}q_{70}$.
    $p_{70}=0.980$, $p_{71}=0.975$, $p_{72}=0.970$. ${}_3p_{70}=0.980(0.975)(0.970)\approx 0.926835$. ${}_2p_{70}=0.980(0.975)=0.955500$. Deferred death: ${}_{2|}q_{70}={}_2p_{70}\cdot q_{72}=0.955500(0.030)\approx 0.028665$. Check: ${}_{2|}q_{70}={}_2p_{70}-{}_3p_{70}=0.955500-0.926835\approx 0.028665$. ✓
  10. Future lifetime
    Express the density and CDF of $T_x$ in terms of ${}_tp_x$ and $\mu_{x+t}$.
    CDF: $\Pr[T_x\le t]={}_tq_x$, with $\dfrac{d}{dt}\,{}_tq_x={}_tp_x\,\mu_{x+t}$. Density of $T_x$: $f_{T_x}(t)={}_tp_x\,\mu_{x+t}$ — "survive $t$ years, then die at the instant $x+t$." Force form: $\mu_{x+t}=-\dfrac{d}{dt}\ln{}_tp_x=\dfrac{f_{T_x}(t)}{{}_tp_x}$.
  11. Future lifetime
    Define the **complete expectation of life** $\overset{\circ}{e}_x$ and give its integral formula.
    $\overset{\circ}{e}_x=E[T_x]$, the expected *exact* future lifetime of $(x)$. Using $f_{T_x}(t)={}_tp_x\mu_{x+t}$ and integration by parts, the clean form is $\overset{\circ}{e}_x=\displaystyle\int_0^{\infty}{}_tp_x\,dt$. The temporary version is $\overset{\circ}{e}_{x:\overline{n|}}=\displaystyle\int_0^{n}{}_tp_x\,dt$.
  12. Curtate lifetime
    Define the **curtate expectation of life** $e_x$ and relate it to $\overset{\circ}{e}_x$ under UDD.
    $e_x=E[K_x]$, the expected number of *complete* future years lived, where $K_x=\lfloor T_x\rfloor$. $e_x=\displaystyle\sum_{k=1}^{\infty}{}_kp_x$ (temporary: $e_{x:\overline{n|}}=\sum_{k=1}^{n}{}_kp_x$). Under a **uniform distribution of deaths** across each year, $\overset{\circ}{e}_x\approx e_x+\tfrac12$ — the extra half-year is the average fraction lived in the year of death.
  13. Curtate lifetime
    Define the **curtate future lifetime** $K_x$ and give its probability mass function.
    $K_x=\lfloor T_x\rfloor$ is the number of *whole* years $(x)$ survives. Its pmf is $\Pr[K_x=k]={}_kp_x\,q_{x+k}={}_{k|}q_x$ for $k=0,1,2,\dots$ — survive $k$ years then die in year $k+1$. Tail: $\Pr[K_x\ge k]={}_kp_x$, which is why $e_x=\sum_{k=1}^{\infty}{}_kp_x$.
  14. Curtate lifetime
    Give the **recursion** for the curtate expectation $e_x$ and use it: $p_{60}=0.99$, $p_{61}=0.985$, $p_{62}=0.98$, and $e_{63}=18.00$. Find $e_{60}$.
    Recursion: $e_x=p_x\,(1+e_{x+1})$. $e_{62}=p_{62}(1+e_{63})=0.98(19.00)=18.6200$. $e_{61}=p_{61}(1+e_{62})=0.985(19.6200)=19.3257$. $e_{60}=p_{60}(1+e_{61})=0.99(20.3257)\approx 20.1224$.
  15. Curtate lifetime
    Compute the **temporary curtate expectation** $e_{50:\overline{3|}}$ from the life table $\ell_{50}=1000,\ \ell_{51}=990,\ \ell_{52}=975,\ \ell_{53}=955$.
    $e_{50:\overline{3|}}=\displaystyle\sum_{k=1}^{3}{}_kp_{50}=\dfrac{\ell_{51}+\ell_{52}+\ell_{53}}{\ell_{50}}$. ${}_1p_{50}=\tfrac{990}{1000}=0.990$, ${}_2p_{50}=\tfrac{975}{1000}=0.975$, ${}_3p_{50}=\tfrac{955}{1000}=0.955$. $e_{50:\overline{3|}}=0.990+0.975+0.955=2.920$ years.
  16. Curtate lifetime
    Compute the curtate expectation $e_{90}$ from the table $\ell_{90}=1000,\ \ell_{91}=920,\ \ell_{92}=790,\ \ell_{93}=600,\ \ell_{94}=350,\ \ell_{95}=0$.
    $e_{90}=\displaystyle\sum_{k=1}^{\infty}{}_kp_{90}=\dfrac{\ell_{91}+\ell_{92}+\ell_{93}+\ell_{94}}{\ell_{90}}$ (terms vanish once $\ell=0$). $=\dfrac{920+790+600+350}{1000}=\dfrac{2660}{1000}=2.66$ complete years. Under UDD the complete expectation is $\overset{\circ}{e}_{90}\approx e_{90}+\tfrac12=3.16$ years.
  17. Life-table functions
    Define the life-table functions $\ell_x$ and $d_x$ and give the survival/death probabilities they produce.
    $\ell_x$ = expected number alive at exact age $x$ from a radix cohort $\ell_0$; $d_x=\ell_x-\ell_{x+1}$ = expected deaths between $x$ and $x+1$. Probabilities: ${}_np_x=\dfrac{\ell_{x+n}}{\ell_x},\qquad {}_nq_x=\dfrac{\ell_x-\ell_{x+n}}{\ell_x},\qquad q_x=\dfrac{d_x}{\ell_x}.$ Deferred: ${}_{n|}q_x=\dfrac{d_{x+n}}{\ell_x}=\dfrac{\ell_{x+n}-\ell_{x+n+1}}{\ell_x}$.
  18. Life-table functions
    State the **recursions** that build a life table from one-year rates $q_x$.
    From the radix $\ell_0$: $\ell_{x+1}=\ell_x\,p_x=\ell_x(1-q_x)=\ell_x-d_x$, with $d_x=\ell_x\,q_x$. Iteratively, $\ell_{x+n}=\ell_x\displaystyle\prod_{k=0}^{n-1}(1-q_{x+k})$. The table also gives the integral of survivors $T_x=\int_0^\infty \ell_{x+t}\,dt$ used for $\overset{\circ}{e}_x=T_x/\ell_x$.
  19. Life-table functions
    Build three rows of a life table: $\ell_{70}=1000$, $q_{70}=0.020$, $q_{71}=0.025$, $q_{72}=0.030$. Find $\ell_{71},\ell_{72},\ell_{73}$ and $d_{70}$.
    $\ell_{71}=\ell_{70}(1-q_{70})=1000(0.980)=980.00$. $\ell_{72}=\ell_{71}(1-q_{71})=980(0.975)=955.50$. $\ell_{73}=\ell_{72}(1-q_{72})=955.50(0.970)\approx 926.835$. Deaths in the first year: $d_{70}=\ell_{70}-\ell_{71}=1000-980=20.00$ (also $=\ell_{70}q_{70}=1000(0.02)$). ✓
  20. Life-table functions
    From the table $\ell_{50}=1000,\ \ell_{51}=990,\ \ell_{52}=975,\ \ell_{53}=955$, find $d_{50},d_{51},d_{52}$, then ${}_3q_{50}$ and ${}_{2|}q_{50}$.
    $d_{50}=1000-990=10$, $d_{51}=990-975=15$, $d_{52}=975-955=20$. ${}_3q_{50}=\dfrac{\ell_{50}-\ell_{53}}{\ell_{50}}=\dfrac{1000-955}{1000}=0.045$. Deferred death: ${}_{2|}q_{50}=\dfrac{d_{52}}{\ell_{50}}=\dfrac{20}{1000}=0.020$ (check: ${}_2p_{50}\,q_{52}=0.975\cdot\tfrac{20}{975}=0.020$). ✓
  21. Life-table functions
    A cohort has $\ell_{90}=1000,\ \ell_{91}=920,\ \ell_{92}=790,\ \ell_{93}=600$. Find ${}_2p_{90}$, ${}_2q_{90}$, and ${}_{1|2}q_{90}$.
    ${}_2p_{90}=\dfrac{\ell_{92}}{\ell_{90}}=\dfrac{790}{1000}=0.790$. ${}_2q_{90}=1-{}_2p_{90}=0.210$ (also $\tfrac{\ell_{90}-\ell_{92}}{\ell_{90}}=\tfrac{210}{1000}$). ${}_{1|2}q_{90}=\dfrac{\ell_{91}-\ell_{93}}{\ell_{90}}=\dfrac{920-600}{1000}=0.320$ (check: ${}_1p_{90}\cdot{}_2q_{91}=0.920\cdot\tfrac{920-600}{920}=0.320$). ✓
  22. Fractional ages
    State the **two fractional-age assumptions** (UDD and constant force) and what each holds fixed within a year of age.
    Both interpolate between integer-age table values $\ell_x$ for $0\le s<1$. **Uniform Distribution of Deaths (UDD):** deaths spread evenly across the year, so $\ell_{x+s}$ is **linear**: $\ell_{x+s}=\ell_x-s\,d_x$. **Constant Force (CF):** the force $\mu_{x+s}$ is **constant** over the year, so $\ell_{x+s}$ is **exponential**: $\ell_{x+s}=\ell_x\,(p_x)^{s}$.
  23. Fractional ages
    Under **UDD**, state ${}_sq_x$, ${}_sp_x$, and $\mu_{x+s}$ for $0\le s<1$.
    Deaths are linear in $s$: ${}_sq_x=s\,q_x,\qquad {}_sp_x=1-s\,q_x.$ Force of mortality (rises through the year): $\mu_{x+s}=\dfrac{q_x}{1-s\,q_x}.$ Also the density of death within the year is constant: ${}_sp_x\,\mu_{x+s}=q_x$.
  24. Fractional ages
    Under **constant force**, state ${}_sp_x$, ${}_sq_x$, and $\mu_{x+s}$ for $0\le s<1$.
    $\mu_{x+s}=\mu$ is constant within the year, with $\mu=-\ln p_x$. ${}_sp_x=(p_x)^{s}=e^{-\mu s},\qquad {}_sq_x=1-(p_x)^{s}.$ Unlike UDD, the force does not depend on $s$ inside the year — it jumps only at integer ages.
  25. Fractional ages
    Given $q_x=0.04$, use **UDD** to find ${}_{0.5}q_x$, ${}_{0.3}p_x$, and $\mu_{x+0.5}$.
    ${}_{0.5}q_x=0.5\,q_x=0.5(0.04)=0.020$. ${}_{0.3}p_x=1-0.3\,q_x=1-0.3(0.04)=0.988$. $\mu_{x+0.5}=\dfrac{q_x}{1-0.5\,q_x}=\dfrac{0.04}{1-0.02}=\dfrac{0.04}{0.98}\approx 0.040816$.
  26. Fractional ages
    Given $p_x=0.96$, use the **constant-force** assumption to find ${}_{0.5}p_x$, ${}_{0.25}q_x$, and the force $\mu$.
    $\mu=-\ln p_x=-\ln 0.96\approx 0.040822$. ${}_{0.5}p_x=(0.96)^{0.5}\approx 0.979796$. ${}_{0.25}q_x=1-(0.96)^{0.25}\approx 1-0.989846=0.010154$. Compare: under UDD, $\mu_{x+0.5}=\tfrac{0.04}{1-0.02}\approx 0.040816$ — very close, since the methods agree to first order in $q_x$.
  27. Fractional ages
    With $\ell_{50}=1000$ and $\ell_{51}=990$, find the interpolated $\ell_{50.4}$ under **UDD** versus **constant force**.
    $q_{50}=\tfrac{10}{1000}=0.010$, $p_{50}=0.990$. **UDD (linear):** $\ell_{50.4}=\ell_{50}-0.4\,d_{50}=1000-0.4(10)=996.00$. **Constant force (exponential):** $\ell_{50.4}=\ell_{50}(p_{50})^{0.4}=1000(0.990)^{0.4}=1000(0.995988)\approx 995.99$. The two differ only slightly (here by about $0.01$) because $q_{50}$ is small.
  28. Fractional ages
    Why does UDD make $\mu_{x+s}$ *increase* across the year while constant force keeps it flat?
    Under UDD the death *density* ${}_sp_x\mu_{x+s}=q_x$ is constant, but survivors ${}_sp_x=1-sq_x$ shrink as $s\to 1$, so $\mu_{x+s}=\dfrac{q_x}{1-sq_x}$ must rise to keep the death count steady against a thinning population. Under constant force $\mu$ is held fixed by assumption, so survival decays purely exponentially and the death density ${}_sp_x\mu$ *falls* across the year.
  29. Mortality laws
    State the **de Moivre (uniform)** mortality law with limiting age $\omega$: give $S_0(x)$, $\mu_x$, ${}_tp_x$, and $\overset{\circ}{e}_x$.
    Lifetimes are uniform on $[0,\omega]$: $S_0(x)=1-\dfrac{x}{\omega},\qquad \mu_x=\dfrac{1}{\omega-x},\qquad 0\le x<\omega.$ ${}_tp_x=\dfrac{\omega-x-t}{\omega-x}=1-\dfrac{t}{\omega-x}$ for $0\le t\le\omega-x$. Future lifetime is uniform on $[0,\omega-x]$, so $\overset{\circ}{e}_x=\dfrac{\omega-x}{2}$ and $\operatorname{Var}(T_x)=\dfrac{(\omega-x)^2}{12}$.
  30. Mortality laws
    Under de Moivre with $\omega=100$, compute $\mu_{40}$, ${}_{10}p_{40}$, and $\overset{\circ}{e}_{40}$.
    $\mu_{40}=\dfrac{1}{\omega-40}=\dfrac{1}{60}\approx 0.016667$. ${}_{10}p_{40}=\dfrac{\omega-40-10}{\omega-40}=\dfrac{50}{60}\approx 0.833333$. $\overset{\circ}{e}_{40}=\dfrac{\omega-40}{2}=\dfrac{60}{2}=30$ years (uniform on $[0,60]$).
  31. Mortality laws
    State the **generalized (beta) de Moivre** law $S_0(x)=\bigl(1-\tfrac{x}{\omega}\bigr)^{\alpha}$: give $\mu_x$, ${}_tp_x$, and $\overset{\circ}{e}_x$.
    $\mu_x=\dfrac{\alpha}{\omega-x}$, and ${}_tp_x=\left(\dfrac{\omega-x-t}{\omega-x}\right)^{\alpha}$. The complete expectation is $\overset{\circ}{e}_x=\dfrac{\omega-x}{\alpha+1}$ (ordinary de Moivre is the case $\alpha=1$, giving $\tfrac{\omega-x}{2}$).
  32. Mortality laws
    For the beta law $S_0(x)=\bigl(1-\tfrac{x}{100}\bigr)^{1/2}$, find $\mu_{36}$, ${}_{19}p_{36}$, and $\overset{\circ}{e}_{36}$.
    Here $\omega=100$, $\alpha=\tfrac12$. $\mu_{36}=\dfrac{\alpha}{\omega-36}=\dfrac{0.5}{64}\approx 0.0078125$. ${}_{19}p_{36}=\left(\dfrac{100-36-19}{100-36}\right)^{1/2}=\left(\dfrac{45}{64}\right)^{1/2}\approx 0.838525$. $\overset{\circ}{e}_{36}=\dfrac{\omega-36}{\alpha+1}=\dfrac{64}{1.5}\approx 42.6667$ years.
  33. Mortality laws
    State the **constant-force (exponential)** mortality law: $\mu_x=\mu$. Give ${}_tp_x$, $\overset{\circ}{e}_x$, and the median future lifetime.
    If $\mu_x=\mu$ for all ages, then $T_x\sim\text{Exponential}(\mu)$ (memoryless): ${}_tp_x=e^{-\mu t},\qquad \overset{\circ}{e}_x=\dfrac{1}{\mu},\qquad \operatorname{Var}(T_x)=\dfrac{1}{\mu^2}.$ Median future lifetime solves ${}_tp_x=0.5$: $t=\dfrac{\ln 2}{\mu}$. Future lifetime is independent of current age $x$.
  34. Mortality laws
    Under a constant force $\mu=0.04$, compute ${}_{10}p_x$, $\overset{\circ}{e}_x$, the median future lifetime, and the temporary $\overset{\circ}{e}_{x:\overline{20|}}$.
    ${}_{10}p_x=e^{-0.04(10)}=e^{-0.4}\approx 0.670320$. $\overset{\circ}{e}_x=\dfrac{1}{\mu}=\dfrac{1}{0.04}=25$ years. Median: $t=\dfrac{\ln 2}{0.04}\approx \dfrac{0.693147}{0.04}\approx 17.3287$ years. Temporary: $\overset{\circ}{e}_{x:\overline{20|}}=\displaystyle\int_0^{20}e^{-\mu t}\,dt=\dfrac{1-e^{-\mu\cdot20}}{\mu}=\dfrac{1-e^{-0.8}}{0.04}\approx \dfrac{0.550671}{0.04}\approx 13.7668$ years.
  35. Mortality laws
    State **Gompertz's law** $\mu_x=Bc^{x}$ and derive ${}_tp_x$.
    Gompertz: force grows geometrically, $\mu_x=Bc^{x}$ with $B>0,\ c>1$. $\displaystyle\int_0^{t}\mu_{x+s}\,ds=\int_0^t Bc^{x+s}\,ds=\dfrac{Bc^{x}(c^{t}-1)}{\ln c}$. Hence ${}_tp_x=\exp\!\left(-\dfrac{Bc^{x}(c^{t}-1)}{\ln c}\right)$.
  36. Mortality laws
    For Gompertz with $B=0.00005$, $c=1.09$, find $\mu_{60}$ and ${}_{10}p_{50}$.
    $\mu_{60}=Bc^{60}=0.00005(1.09)^{60}$. Since $1.09^{60}\approx 176.031$, $\mu_{60}\approx 0.0088016$. For survival: $\displaystyle\int_0^{10}\mu_{50+s}\,ds=\dfrac{Bc^{50}(c^{10}-1)}{\ln c}$. $c^{50}=1.09^{50}\approx 74.3575$, $c^{10}-1=1.09^{10}-1\approx 1.367364$, $\ln 1.09\approx 0.086178$. Integral $=\dfrac{0.00005(74.3575)(1.367364)}{0.086178}\approx 0.058991$. ${}_{10}p_{50}=e^{-0.058991}\approx 0.942716$.
  37. Mortality laws
    State **Makeham's law** $\mu_x=A+Bc^{x}$, explain the role of $A$, and derive ${}_tp_x$.
    Makeham adds an **age-independent** background hazard $A$ (accidents, infection) to the Gompertz senescent term $Bc^{x}$: $\mu_x=A+Bc^{x}$, with $A\ge -B$, $B>0$, $c>1$. $\displaystyle\int_0^{t}\mu_{x+s}\,ds=A\,t+\dfrac{Bc^{x}(c^{t}-1)}{\ln c}$. Thus ${}_tp_x=\exp\!\left(-A\,t-\dfrac{Bc^{x}(c^{t}-1)}{\ln c}\right)=e^{-At}\cdot\bigl({}_tp_x^{\text{Gomp}}\bigr)$.
  38. Mortality laws
    For Makeham with $A=0.002$, $B=0.0003$, $c=1.08$, find $\mu_{50}$ and ${}_{10}p_{50}$.
    $\mu_{50}=A+Bc^{50}=0.002+0.0003(1.08)^{50}$. Since $1.08^{50}\approx 46.9016$, $\mu_{50}\approx 0.002+0.014070=0.016070$. Integral $0\to10$: $A(10)+\dfrac{Bc^{50}(c^{10}-1)}{\ln c}$. $c^{10}-1=1.08^{10}-1\approx 1.158925$, $\ln 1.08\approx 0.076961$. Gompertz part $=\dfrac{0.0003(46.9016)(1.158925)}{0.076961}\approx 0.211882$; plus $A\cdot10=0.020$. Total $\approx 0.231882$, so ${}_{10}p_{50}=e^{-0.231882}\approx 0.793040$.
  39. Survival function & force
    How do you read off $\mu_x$ from a survival function given as $S_0(x)=e^{-(0.0001x+0.00003x^2)}$? Find $\mu_{50}$ and ${}_{10}p_{50}$.
    $\ln S_0(x)=-(0.0001x+0.00003x^2)$, so $\mu_x=-\dfrac{d}{dx}\ln S_0(x)=0.0001+0.00006\,x$. $\mu_{50}=0.0001+0.00006(50)=0.0031$. Survival: ${}_{10}p_{50}=\dfrac{S_0(60)}{S_0(50)}=\exp\!\Bigl(-\!\int_{50}^{60}\mu_x\,dx\Bigr)$. $\int_{50}^{60}\mu_x\,dx=0.0001(10)+0.00003(60^2-50^2)=0.001+0.00003(1100)=0.034$. ${}_{10}p_{50}=e^{-0.034}\approx 0.966572$.
  40. Future lifetime
    Compute $\overset{\circ}{e}_x$ under the de Moivre law $\omega-x=50$ by integrating ${}_tp_x$, and confirm the shortcut.
    Here ${}_tp_x=1-\dfrac{t}{50}$ for $0\le t\le 50$. $\overset{\circ}{e}_x=\displaystyle\int_0^{50}\!\Bigl(1-\dfrac{t}{50}\Bigr)dt=\Bigl[t-\dfrac{t^2}{100}\Bigr]_0^{50}=50-\dfrac{2500}{100}=50-25=25$. Shortcut: $\overset{\circ}{e}_x=\dfrac{\omega-x}{2}=\dfrac{50}{2}=25$. ✓
  41. Survival function & force
    A life has a constant force $\mu_{x+s}=0.05$ for all $s$. Find ${}_{20}p_{x}$, ${}_{20}q_x$, and $\Pr[T_x>20]$ in one step.
    With constant force, ${}_tp_x=e^{-0.05t}$. ${}_{20}p_x=e^{-0.05(20)}=e^{-1}\approx 0.367879$. ${}_{20}q_x=1-{}_{20}p_x\approx 0.632121$. $\Pr[T_x>20]={}_{20}p_x\approx 0.367879$ — the survival function evaluated at $20$.
  42. Curtate lifetime
    Distinguish $\overset{\circ}{e}_x$ (complete) from $e_x$ (curtate), and explain why $\overset{\circ}{e}_x>e_x$.
    $\overset{\circ}{e}_x=E[T_x]$ counts the *exact* time lived, including the partial final year; $e_x=E[K_x]$ with $K_x=\lfloor T_x\rfloor$ counts only *whole* completed years and discards the final fraction. Since curtate truncates downward, $e_x\le\overset{\circ}{e}_x$ always, and under UDD the discarded piece averages a half-year, giving $\overset{\circ}{e}_x\approx e_x+\tfrac12$.
  43. Future lifetime
    Given $\overset{\circ}{e}_{x:\overline{n|}}=\int_0^n {}_tp_x\,dt$, compute the temporary complete expectation $\overset{\circ}{e}_{x:\overline{20|}}$ under de Moivre with $\omega-x=50$.
    ${}_tp_x=1-\dfrac{t}{50}$, so $\overset{\circ}{e}_{x:\overline{20|}}=\displaystyle\int_0^{20}\!\Bigl(1-\dfrac{t}{50}\Bigr)dt=\Bigl[t-\dfrac{t^2}{100}\Bigr]_0^{20}=20-\dfrac{400}{100}=20-4=16$ years. This is the expected time lived in the next $20$ years, with deaths after $t=20$ contributing the full $20$.
  44. Mortality laws
    Under de Moivre with $\omega=110$, find $\mu_{30}$, ${}_{20}p_{30}$, and the probability a life aged $30$ dies between ages $50$ and $60$.
    $\mu_{30}=\dfrac{1}{110-30}=\dfrac{1}{80}=0.0125$. ${}_{20}p_{30}=\dfrac{110-30-20}{110-30}=\dfrac{60}{80}=0.750$. Death between $50$ and $60$ is ${}_{20|10}q_{30}={}_{20}p_{30}-{}_{30}p_{30}$. ${}_{30}p_{30}=\dfrac{110-60}{80}=\dfrac{50}{80}=0.625$. So ${}_{20|10}q_{30}=0.750-0.625=0.125$.