{
  "deckName": "Exam FAM — Survival Models & Life Tables",
  "examCode": "Exam FAM",
  "cards": [
    {
      "front": "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 $f_0$.",
      "back": "Let $X$ be the age at death of a newborn (life aged $0$).\n$F_0(x)=\\Pr[X\\le x]$ is the CDF, and $S_0(x)=\\Pr[X>x]=1-F_0(x)$ is the **survival function** — the probability of surviving past age $x$.\nValid $S_0$ requires $S_0(0)=1$, $S_0(\\infty)=0$, and $S_0$ non-increasing.\nThe density is $f_0(x)=F_0'(x)=-S_0'(x)$.",
      "tag": "Survival function & force"
    },
    {
      "front": "Define the **force of mortality** $\\mu_x$ and state its relation to $S_0(x)$.",
      "back": "$\\mu_x$ is the instantaneous death rate at age $x$:\n$\\mu_x=\\dfrac{f_0(x)}{S_0(x)}=-\\dfrac{S_0'(x)}{S_0(x)}=-\\dfrac{d}{dx}\\ln S_0(x)$.\nEquivalently the density factors as $f_0(x)=S_0(x)\\,\\mu_x$. The force is a *rate* (can exceed $1$), not a probability.",
      "tag": "Survival function & force"
    },
    {
      "front": "Recover the survival function from the force of mortality. State $S_0(x)$ in terms of $\\mu$.",
      "back": "Integrating $\\mu_x=-\\dfrac{d}{dx}\\ln S_0(x)$ from $0$ to $x$:\n$S_0(x)=\\exp\\!\\Bigl(-\\displaystyle\\int_0^{x}\\mu_s\\,ds\\Bigr)$.\nMore generally, the $t$-year survival probability of a life aged $x$ is\n${}_tp_x=\\exp\\!\\Bigl(-\\displaystyle\\int_0^{t}\\mu_{x+s}\\,ds\\Bigr)$.",
      "tag": "Survival function & force"
    },
    {
      "front": "Given $S_0(x)=\\bigl(1-\\tfrac{x}{120}\\bigr)^{1/6}$ for $0\\le x\\le 120$, find $\\mu_{60}$ and the density $f_0(60)$.",
      "back": "$\\ln S_0(x)=\\tfrac16\\ln\\!\\bigl(1-\\tfrac{x}{120}\\bigr)$, so\n$\\mu_x=-\\dfrac{d}{dx}\\ln S_0(x)=\\dfrac{1/6}{120-x}$.\nAt $x=60$: $\\mu_{60}=\\dfrac{1/6}{60}=\\dfrac{1}{360}\\approx 0.0027778$.\n$S_0(60)=(1-0.5)^{1/6}=0.5^{1/6}\\approx 0.890899$.\nDensity $f_0(60)=S_0(60)\\,\\mu_{60}\\approx 0.890899(0.0027778)\\approx 0.0024747$.",
      "tag": "Survival function & force"
    },
    {
      "front": "For $S_0(x)=\\bigl(1-\\tfrac{x}{120}\\bigr)^{1/6}$, compute $F_0(60)$ and ${}_{10}p_{40}$.",
      "back": "$S_0(60)=0.5^{1/6}\\approx 0.890899$, so $F_0(60)=1-S_0(60)\\approx 0.109101$.\nFor the survival probability, ${}_{10}p_{40}=\\dfrac{S_0(50)}{S_0(40)}$.\n$S_0(50)=(1-\\tfrac{50}{120})^{1/6}=(0.583333)^{1/6}\\approx 0.914084$ and $S_0(40)=(1-\\tfrac13)^{1/6}=(0.666667)^{1/6}\\approx 0.934655$.\n${}_{10}p_{40}=\\dfrac{0.914084}{0.934655}\\approx 0.978$.",
      "tag": "Survival function & force"
    },
    {
      "front": "Define the **future-lifetime random variable** $T_x$ and the symbols ${}_tp_x$ and ${}_tq_x$.",
      "back": "$T_x$ is the remaining lifetime of a life **currently aged** $x$ (so $T_x=X-x\\mid X>x$).\n${}_tp_x=\\Pr[T_x>t]=\\dfrac{S_0(x+t)}{S_0(x)}$ — probability $(x)$ survives $t$ more years.\n${}_tq_x=\\Pr[T_x\\le t]=1-{}_tp_x$ — probability $(x)$ dies within $t$ years.\nWhen $t=1$ the leading $1$ is dropped: $p_x$, $q_x$.",
      "tag": "Future lifetime"
    },
    {
      "front": "State the **deferred mortality probability** ${}_{u|t}q_x$ in words and as a formula, and give the two standard decompositions.",
      "back": "${}_{u|t}q_x=\\Pr[(x)$ survives $u$ years, then dies in the next $t]$.\n$ {}_{u|t}q_x={}_up_x\\cdot{}_tq_{x+u}={}_up_x-{}_{u+t}p_x={}_{u+t}q_x-{}_uq_x.$\nWith $t=1$ it is written ${}_{u|}q_x={}_up_x\\,q_{x+u}$.",
      "tag": "Future lifetime"
    },
    {
      "front": "Express ${}_tp_x$ as a product over single years (the **chain rule** for survival).",
      "back": "Survival probabilities multiply across consecutive years:\n${}_np_x=p_x\\cdot p_{x+1}\\cdots p_{x+n-1}=\\displaystyle\\prod_{k=0}^{n-1}p_{x+k}$.\nMore generally ${}_{s+t}p_x={}_sp_x\\cdot{}_tp_{x+s}$. This factorization underlies every life-table recursion.",
      "tag": "Future lifetime"
    },
    {
      "front": "Given the one-year rates $q_{70}=0.020$, $q_{71}=0.025$, $q_{72}=0.030$, find ${}_3p_{70}$ and ${}_{2|}q_{70}$.",
      "back": "$p_{70}=0.980$, $p_{71}=0.975$, $p_{72}=0.970$.\n${}_3p_{70}=0.980(0.975)(0.970)\\approx 0.926835$.\n${}_2p_{70}=0.980(0.975)=0.955500$.\nDeferred death: ${}_{2|}q_{70}={}_2p_{70}\\cdot q_{72}=0.955500(0.030)\\approx 0.028665$.\nCheck: ${}_{2|}q_{70}={}_2p_{70}-{}_3p_{70}=0.955500-0.926835\\approx 0.028665$. ✓",
      "tag": "Future lifetime"
    },
    {
      "front": "Express the density and CDF of $T_x$ in terms of ${}_tp_x$ and $\\mu_{x+t}$.",
      "back": "CDF: $\\Pr[T_x\\le t]={}_tq_x$, with $\\dfrac{d}{dt}\\,{}_tq_x={}_tp_x\\,\\mu_{x+t}$.\nDensity of $T_x$: $f_{T_x}(t)={}_tp_x\\,\\mu_{x+t}$ — \"survive $t$ years, then die at the instant $x+t$.\"\nForce form: $\\mu_{x+t}=-\\dfrac{d}{dt}\\ln{}_tp_x=\\dfrac{f_{T_x}(t)}{{}_tp_x}$.",
      "tag": "Future lifetime"
    },
    {
      "front": "Define the **complete expectation of life** $\\overset{\\circ}{e}_x$ and give its integral formula.",
      "back": "$\\overset{\\circ}{e}_x=E[T_x]$, the expected *exact* future lifetime of $(x)$.\nUsing $f_{T_x}(t)={}_tp_x\\mu_{x+t}$ and integration by parts, the clean form is\n$\\overset{\\circ}{e}_x=\\displaystyle\\int_0^{\\infty}{}_tp_x\\,dt$.\nThe temporary version is $\\overset{\\circ}{e}_{x:\\overline{n|}}=\\displaystyle\\int_0^{n}{}_tp_x\\,dt$.",
      "tag": "Future lifetime"
    },
    {
      "front": "Define the **curtate expectation of life** $e_x$ and relate it to $\\overset{\\circ}{e}_x$ under UDD.",
      "back": "$e_x=E[K_x]$, the expected number of *complete* future years lived, where $K_x=\\lfloor T_x\\rfloor$.\n$e_x=\\displaystyle\\sum_{k=1}^{\\infty}{}_kp_x$ (temporary: $e_{x:\\overline{n|}}=\\sum_{k=1}^{n}{}_kp_x$).\nUnder 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.",
      "tag": "Curtate lifetime"
    },
    {
      "front": "Define the **curtate future lifetime** $K_x$ and give its probability mass function.",
      "back": "$K_x=\\lfloor T_x\\rfloor$ is the number of *whole* years $(x)$ survives.\nIts 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 $k+1$.\nTail: $\\Pr[K_x\\ge k]={}_kp_x$, which is why $e_x=\\sum_{k=1}^{\\infty}{}_kp_x$.",
      "tag": "Curtate lifetime"
    },
    {
      "front": "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 $e_{60}$.",
      "back": "Recursion: $e_x=p_x\\,(1+e_{x+1})$.\n$e_{62}=p_{62}(1+e_{63})=0.98(19.00)=18.6200$.\n$e_{61}=p_{61}(1+e_{62})=0.985(19.6200)=19.3257$.\n$e_{60}=p_{60}(1+e_{61})=0.99(20.3257)\\approx 20.1224$.",
      "tag": "Curtate lifetime"
    },
    {
      "front": "Compute the **temporary curtate expectation** $e_{50:\\overline{3|}}$ from the life table $\\ell_{50}=1000,\\ \\ell_{51}=990,\\ \\ell_{52}=975,\\ \\ell_{53}=955$.",
      "back": "$e_{50:\\overline{3|}}=\\displaystyle\\sum_{k=1}^{3}{}_kp_{50}=\\dfrac{\\ell_{51}+\\ell_{52}+\\ell_{53}}{\\ell_{50}}$.\n${}_1p_{50}=\\tfrac{990}{1000}=0.990$, ${}_2p_{50}=\\tfrac{975}{1000}=0.975$, ${}_3p_{50}=\\tfrac{955}{1000}=0.955$.\n$e_{50:\\overline{3|}}=0.990+0.975+0.955=2.920$ years.",
      "tag": "Curtate lifetime"
    },
    {
      "front": "Compute the curtate expectation $e_{90}$ from the table $\\ell_{90}=1000,\\ \\ell_{91}=920,\\ \\ell_{92}=790,\\ \\ell_{93}=600,\\ \\ell_{94}=350,\\ \\ell_{95}=0$.",
      "back": "$e_{90}=\\displaystyle\\sum_{k=1}^{\\infty}{}_kp_{90}=\\dfrac{\\ell_{91}+\\ell_{92}+\\ell_{93}+\\ell_{94}}{\\ell_{90}}$ (terms vanish once $\\ell=0$).\n$=\\dfrac{920+790+600+350}{1000}=\\dfrac{2660}{1000}=2.66$ complete years.\nUnder UDD the complete expectation is $\\overset{\\circ}{e}_{90}\\approx e_{90}+\\tfrac12=3.16$ years.",
      "tag": "Curtate lifetime"
    },
    {
      "front": "Define the life-table functions $\\ell_x$ and $d_x$ and give the survival/death probabilities they produce.",
      "back": "$\\ell_x$ = 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 $x+1$.\nProbabilities:\n${}_np_x=\\dfrac{\\ell_{x+n}}{\\ell_x},\\qquad {}_nq_x=\\dfrac{\\ell_x-\\ell_{x+n}}{\\ell_x},\\qquad q_x=\\dfrac{d_x}{\\ell_x}.$\nDeferred: ${}_{n|}q_x=\\dfrac{d_{x+n}}{\\ell_x}=\\dfrac{\\ell_{x+n}-\\ell_{x+n+1}}{\\ell_x}$.",
      "tag": "Life-table functions"
    },
    {
      "front": "State the **recursions** that build a life table from one-year rates $q_x$.",
      "back": "From the radix $\\ell_0$:\n$\\ell_{x+1}=\\ell_x\\,p_x=\\ell_x(1-q_x)=\\ell_x-d_x$, with $d_x=\\ell_x\\,q_x$.\nIteratively, $\\ell_{x+n}=\\ell_x\\displaystyle\\prod_{k=0}^{n-1}(1-q_{x+k})$.\nThe table also gives the integral of survivors $T_x=\\int_0^\\infty \\ell_{x+t}\\,dt$ used for $\\overset{\\circ}{e}_x=T_x/\\ell_x$.",
      "tag": "Life-table functions"
    },
    {
      "front": "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 $d_{70}$.",
      "back": "$\\ell_{71}=\\ell_{70}(1-q_{70})=1000(0.980)=980.00$.\n$\\ell_{72}=\\ell_{71}(1-q_{71})=980(0.975)=955.50$.\n$\\ell_{73}=\\ell_{72}(1-q_{72})=955.50(0.970)\\approx 926.835$.\nDeaths in the first year: $d_{70}=\\ell_{70}-\\ell_{71}=1000-980=20.00$ (also $=\\ell_{70}q_{70}=1000(0.02)$). ✓",
      "tag": "Life-table functions"
    },
    {
      "front": "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 ${}_{2|}q_{50}$.",
      "back": "$d_{50}=1000-990=10$, $d_{51}=990-975=15$, $d_{52}=975-955=20$.\n${}_3q_{50}=\\dfrac{\\ell_{50}-\\ell_{53}}{\\ell_{50}}=\\dfrac{1000-955}{1000}=0.045$.\nDeferred death: ${}_{2|}q_{50}=\\dfrac{d_{52}}{\\ell_{50}}=\\dfrac{20}{1000}=0.020$\n(check: ${}_2p_{50}\\,q_{52}=0.975\\cdot\\tfrac{20}{975}=0.020$). ✓",
      "tag": "Life-table functions"
    },
    {
      "front": "A cohort has $\\ell_{90}=1000,\\ \\ell_{91}=920,\\ \\ell_{92}=790,\\ \\ell_{93}=600$. Find ${}_2p_{90}$, ${}_2q_{90}$, and ${}_{1|2}q_{90}$.",
      "back": "${}_2p_{90}=\\dfrac{\\ell_{92}}{\\ell_{90}}=\\dfrac{790}{1000}=0.790$.\n${}_2q_{90}=1-{}_2p_{90}=0.210$ (also $\\tfrac{\\ell_{90}-\\ell_{92}}{\\ell_{90}}=\\tfrac{210}{1000}$).\n${}_{1|2}q_{90}=\\dfrac{\\ell_{91}-\\ell_{93}}{\\ell_{90}}=\\dfrac{920-600}{1000}=0.320$\n(check: ${}_1p_{90}\\cdot{}_2q_{91}=0.920\\cdot\\tfrac{920-600}{920}=0.320$). ✓",
      "tag": "Life-table functions"
    },
    {
      "front": "State the **two fractional-age assumptions** (UDD and constant force) and what each holds fixed within a year of age.",
      "back": "Both interpolate between integer-age table values $\\ell_x$ for $0\\le s<1$.\n**Uniform Distribution of Deaths (UDD):** deaths spread evenly across the year, so $\\ell_{x+s}$ is **linear**: $\\ell_{x+s}=\\ell_x-s\\,d_x$.\n**Constant Force (CF):** the force $\\mu_{x+s}$ is **constant** over the year, so $\\ell_{x+s}$ is **exponential**: $\\ell_{x+s}=\\ell_x\\,(p_x)^{s}$.",
      "tag": "Fractional ages"
    },
    {
      "front": "Under **UDD**, state ${}_sq_x$, ${}_sp_x$, and $\\mu_{x+s}$ for $0\\le s<1$.",
      "back": "Deaths are linear in $s$:\n${}_sq_x=s\\,q_x,\\qquad {}_sp_x=1-s\\,q_x.$\nForce of mortality (rises through the year):\n$\\mu_{x+s}=\\dfrac{q_x}{1-s\\,q_x}.$\nAlso the density of death within the year is constant: ${}_sp_x\\,\\mu_{x+s}=q_x$.",
      "tag": "Fractional ages"
    },
    {
      "front": "Under **constant force**, state ${}_sp_x$, ${}_sq_x$, and $\\mu_{x+s}$ for $0\\le s<1$.",
      "back": "$\\mu_{x+s}=\\mu$ is constant within the year, with $\\mu=-\\ln p_x$.\n${}_sp_x=(p_x)^{s}=e^{-\\mu s},\\qquad {}_sq_x=1-(p_x)^{s}.$\nUnlike UDD, the force does not depend on $s$ inside the year — it jumps only at integer ages.",
      "tag": "Fractional ages"
    },
    {
      "front": "Given $q_x=0.04$, use **UDD** to find ${}_{0.5}q_x$, ${}_{0.3}p_x$, and $\\mu_{x+0.5}$.",
      "back": "${}_{0.5}q_x=0.5\\,q_x=0.5(0.04)=0.020$.\n${}_{0.3}p_x=1-0.3\\,q_x=1-0.3(0.04)=0.988$.\n$\\mu_{x+0.5}=\\dfrac{q_x}{1-0.5\\,q_x}=\\dfrac{0.04}{1-0.02}=\\dfrac{0.04}{0.98}\\approx 0.040816$.",
      "tag": "Fractional ages"
    },
    {
      "front": "Given $p_x=0.96$, use the **constant-force** assumption to find ${}_{0.5}p_x$, ${}_{0.25}q_x$, and the force $\\mu$.",
      "back": "$\\mu=-\\ln p_x=-\\ln 0.96\\approx 0.040822$.\n${}_{0.5}p_x=(0.96)^{0.5}\\approx 0.979796$.\n${}_{0.25}q_x=1-(0.96)^{0.25}\\approx 1-0.989846=0.010154$.\nCompare: 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 $q_x$.",
      "tag": "Fractional ages"
    },
    {
      "front": "With $\\ell_{50}=1000$ and $\\ell_{51}=990$, find the interpolated $\\ell_{50.4}$ under **UDD** versus **constant force**.",
      "back": "$q_{50}=\\tfrac{10}{1000}=0.010$, $p_{50}=0.990$.\n**UDD (linear):** $\\ell_{50.4}=\\ell_{50}-0.4\\,d_{50}=1000-0.4(10)=996.00$.\n**Constant force (exponential):** $\\ell_{50.4}=\\ell_{50}(p_{50})^{0.4}=1000(0.990)^{0.4}=1000(0.995988)\\approx 995.99$.\nThe two differ only slightly (here by about $0.01$) because $q_{50}$ is small.",
      "tag": "Fractional ages"
    },
    {
      "front": "Why does UDD make $\\mu_{x+s}$ *increase* across the year while constant force keeps it flat?",
      "back": "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.\nUnder constant force $\\mu$ is held fixed by assumption, so survival decays purely exponentially and the death density ${}_sp_x\\mu$ *falls* across the year.",
      "tag": "Fractional ages"
    },
    {
      "front": "State the **de Moivre (uniform)** mortality law with limiting age $\\omega$: give $S_0(x)$, $\\mu_x$, ${}_tp_x$, and $\\overset{\\circ}{e}_x$.",
      "back": "Lifetimes are uniform on $[0,\\omega]$:\n$S_0(x)=1-\\dfrac{x}{\\omega},\\qquad \\mu_x=\\dfrac{1}{\\omega-x},\\qquad 0\\le x<\\omega.$\n${}_tp_x=\\dfrac{\\omega-x-t}{\\omega-x}=1-\\dfrac{t}{\\omega-x}$ for $0\\le t\\le\\omega-x$.\nFuture lifetime is uniform on $[0,\\omega-x]$, so $\\overset{\\circ}{e}_x=\\dfrac{\\omega-x}{2}$ and $\\operatorname{Var}(T_x)=\\dfrac{(\\omega-x)^2}{12}$.",
      "tag": "Mortality laws"
    },
    {
      "front": "Under de Moivre with $\\omega=100$, compute $\\mu_{40}$, ${}_{10}p_{40}$, and $\\overset{\\circ}{e}_{40}$.",
      "back": "$\\mu_{40}=\\dfrac{1}{\\omega-40}=\\dfrac{1}{60}\\approx 0.016667$.\n${}_{10}p_{40}=\\dfrac{\\omega-40-10}{\\omega-40}=\\dfrac{50}{60}\\approx 0.833333$.\n$\\overset{\\circ}{e}_{40}=\\dfrac{\\omega-40}{2}=\\dfrac{60}{2}=30$ years (uniform on $[0,60]$).",
      "tag": "Mortality laws"
    },
    {
      "front": "State the **generalized (beta) de Moivre** law $S_0(x)=\\bigl(1-\\tfrac{x}{\\omega}\\bigr)^{\\alpha}$: give $\\mu_x$, ${}_tp_x$, and $\\overset{\\circ}{e}_x$.",
      "back": "$\\mu_x=\\dfrac{\\alpha}{\\omega-x}$, and\n${}_tp_x=\\left(\\dfrac{\\omega-x-t}{\\omega-x}\\right)^{\\alpha}$.\nThe complete expectation is $\\overset{\\circ}{e}_x=\\dfrac{\\omega-x}{\\alpha+1}$ (ordinary de Moivre is the case $\\alpha=1$, giving $\\tfrac{\\omega-x}{2}$).",
      "tag": "Mortality laws"
    },
    {
      "front": "For the beta law $S_0(x)=\\bigl(1-\\tfrac{x}{100}\\bigr)^{1/2}$, find $\\mu_{36}$, ${}_{19}p_{36}$, and $\\overset{\\circ}{e}_{36}$.",
      "back": "Here $\\omega=100$, $\\alpha=\\tfrac12$.\n$\\mu_{36}=\\dfrac{\\alpha}{\\omega-36}=\\dfrac{0.5}{64}\\approx 0.0078125$.\n${}_{19}p_{36}=\\left(\\dfrac{100-36-19}{100-36}\\right)^{1/2}=\\left(\\dfrac{45}{64}\\right)^{1/2}\\approx 0.838525$.\n$\\overset{\\circ}{e}_{36}=\\dfrac{\\omega-36}{\\alpha+1}=\\dfrac{64}{1.5}\\approx 42.6667$ years.",
      "tag": "Mortality laws"
    },
    {
      "front": "State the **constant-force (exponential)** mortality law: $\\mu_x=\\mu$. Give ${}_tp_x$, $\\overset{\\circ}{e}_x$, and the median future lifetime.",
      "back": "If $\\mu_x=\\mu$ for all ages, then $T_x\\sim\\text{Exponential}(\\mu)$ (memoryless):\n${}_tp_x=e^{-\\mu t},\\qquad \\overset{\\circ}{e}_x=\\dfrac{1}{\\mu},\\qquad \\operatorname{Var}(T_x)=\\dfrac{1}{\\mu^2}.$\nMedian future lifetime solves ${}_tp_x=0.5$: $t=\\dfrac{\\ln 2}{\\mu}$.\nFuture lifetime is independent of current age $x$.",
      "tag": "Mortality laws"
    },
    {
      "front": "Under a constant force $\\mu=0.04$, compute ${}_{10}p_x$, $\\overset{\\circ}{e}_x$, the median future lifetime, and the temporary $\\overset{\\circ}{e}_{x:\\overline{20|}}$.",
      "back": "${}_{10}p_x=e^{-0.04(10)}=e^{-0.4}\\approx 0.670320$.\n$\\overset{\\circ}{e}_x=\\dfrac{1}{\\mu}=\\dfrac{1}{0.04}=25$ years.\nMedian: $t=\\dfrac{\\ln 2}{0.04}\\approx \\dfrac{0.693147}{0.04}\\approx 17.3287$ years.\nTemporary: $\\overset{\\circ}{e}_{x:\\overline{20|}}=\\displaystyle\\int_0^{20}e^{-\\mu t}\\,dt=\\dfrac{1-e^{-\\mu\\cdot20}}{\\mu}=\\dfrac{1-e^{-0.8}}{0.04}\\approx \\dfrac{0.550671}{0.04}\\approx 13.7668$ years.",
      "tag": "Mortality laws"
    },
    {
      "front": "State **Gompertz's law** $\\mu_x=Bc^{x}$ and derive ${}_tp_x$.",
      "back": "Gompertz: force grows geometrically, $\\mu_x=Bc^{x}$ with $B>0,\\ c>1$.\n$\\displaystyle\\int_0^{t}\\mu_{x+s}\\,ds=\\int_0^t Bc^{x+s}\\,ds=\\dfrac{Bc^{x}(c^{t}-1)}{\\ln c}$.\nHence ${}_tp_x=\\exp\\!\\left(-\\dfrac{Bc^{x}(c^{t}-1)}{\\ln c}\\right)$.",
      "tag": "Mortality laws"
    },
    {
      "front": "For Gompertz with $B=0.00005$, $c=1.09$, find $\\mu_{60}$ and ${}_{10}p_{50}$.",
      "back": "$\\mu_{60}=Bc^{60}=0.00005(1.09)^{60}$. Since $1.09^{60}\\approx 176.031$, $\\mu_{60}\\approx 0.0088016$.\nFor survival: $\\displaystyle\\int_0^{10}\\mu_{50+s}\\,ds=\\dfrac{Bc^{50}(c^{10}-1)}{\\ln c}$.\n$c^{50}=1.09^{50}\\approx 74.3575$, $c^{10}-1=1.09^{10}-1\\approx 1.367364$, $\\ln 1.09\\approx 0.086178$.\nIntegral $=\\dfrac{0.00005(74.3575)(1.367364)}{0.086178}\\approx 0.058991$.\n${}_{10}p_{50}=e^{-0.058991}\\approx 0.942716$.",
      "tag": "Mortality laws"
    },
    {
      "front": "State **Makeham's law** $\\mu_x=A+Bc^{x}$, explain the role of $A$, and derive ${}_tp_x$.",
      "back": "Makeham adds an **age-independent** background hazard $A$ (accidents, infection) to the Gompertz senescent term $Bc^{x}$: $\\mu_x=A+Bc^{x}$, with $A\\ge -B$, $B>0$, $c>1$.\n$\\displaystyle\\int_0^{t}\\mu_{x+s}\\,ds=A\\,t+\\dfrac{Bc^{x}(c^{t}-1)}{\\ln c}$.\nThus ${}_tp_x=\\exp\\!\\left(-A\\,t-\\dfrac{Bc^{x}(c^{t}-1)}{\\ln c}\\right)=e^{-At}\\cdot\\bigl({}_tp_x^{\\text{Gomp}}\\bigr)$.",
      "tag": "Mortality laws"
    },
    {
      "front": "For Makeham with $A=0.002$, $B=0.0003$, $c=1.08$, find $\\mu_{50}$ and ${}_{10}p_{50}$.",
      "back": "$\\mu_{50}=A+Bc^{50}=0.002+0.0003(1.08)^{50}$. Since $1.08^{50}\\approx 46.9016$, $\\mu_{50}\\approx 0.002+0.014070=0.016070$.\nIntegral $0\\to10$: $A(10)+\\dfrac{Bc^{50}(c^{10}-1)}{\\ln c}$.\n$c^{10}-1=1.08^{10}-1\\approx 1.158925$, $\\ln 1.08\\approx 0.076961$.\nGompertz part $=\\dfrac{0.0003(46.9016)(1.158925)}{0.076961}\\approx 0.211882$; plus $A\\cdot10=0.020$.\nTotal $\\approx 0.231882$, so ${}_{10}p_{50}=e^{-0.231882}\\approx 0.793040$.",
      "tag": "Mortality laws"
    },
    {
      "front": "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 ${}_{10}p_{50}$.",
      "back": "$\\ln S_0(x)=-(0.0001x+0.00003x^2)$, so $\\mu_x=-\\dfrac{d}{dx}\\ln S_0(x)=0.0001+0.00006\\,x$.\n$\\mu_{50}=0.0001+0.00006(50)=0.0031$.\nSurvival: ${}_{10}p_{50}=\\dfrac{S_0(60)}{S_0(50)}=\\exp\\!\\Bigl(-\\!\\int_{50}^{60}\\mu_x\\,dx\\Bigr)$.\n$\\int_{50}^{60}\\mu_x\\,dx=0.0001(10)+0.00003(60^2-50^2)=0.001+0.00003(1100)=0.034$.\n${}_{10}p_{50}=e^{-0.034}\\approx 0.966572$.",
      "tag": "Survival function & force"
    },
    {
      "front": "Compute $\\overset{\\circ}{e}_x$ under the de Moivre law $\\omega-x=50$ by integrating ${}_tp_x$, and confirm the shortcut.",
      "back": "Here ${}_tp_x=1-\\dfrac{t}{50}$ for $0\\le t\\le 50$.\n$\\overset{\\circ}{e}_x=\\displaystyle\\int_0^{50}\\!\\Bigl(1-\\dfrac{t}{50}\\Bigr)dt=\\Bigl[t-\\dfrac{t^2}{100}\\Bigr]_0^{50}=50-\\dfrac{2500}{100}=50-25=25$.\nShortcut: $\\overset{\\circ}{e}_x=\\dfrac{\\omega-x}{2}=\\dfrac{50}{2}=25$. ✓",
      "tag": "Future lifetime"
    },
    {
      "front": "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.",
      "back": "With constant force, ${}_tp_x=e^{-0.05t}$.\n${}_{20}p_x=e^{-0.05(20)}=e^{-1}\\approx 0.367879$.\n${}_{20}q_x=1-{}_{20}p_x\\approx 0.632121$.\n$\\Pr[T_x>20]={}_{20}p_x\\approx 0.367879$ — the survival function evaluated at $20$.",
      "tag": "Survival function & force"
    },
    {
      "front": "Distinguish $\\overset{\\circ}{e}_x$ (complete) from $e_x$ (curtate), and explain why $\\overset{\\circ}{e}_x>e_x$.",
      "back": "$\\overset{\\circ}{e}_x=E[T_x]$ 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.\nSince curtate truncates downward, $e_x\\le\\overset{\\circ}{e}_x$ always, and under UDD the discarded piece averages a half-year, giving $\\overset{\\circ}{e}_x\\approx e_x+\\tfrac12$.",
      "tag": "Curtate lifetime"
    },
    {
      "front": "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 $\\omega-x=50$.",
      "back": "${}_tp_x=1-\\dfrac{t}{50}$, so\n$\\overset{\\circ}{e}_{x:\\overline{20|}}=\\displaystyle\\int_0^{20}\\!\\Bigl(1-\\dfrac{t}{50}\\Bigr)dt=\\Bigl[t-\\dfrac{t^2}{100}\\Bigr]_0^{20}=20-\\dfrac{400}{100}=20-4=16$ years.\nThis is the expected time lived in the next $20$ years, with deaths after $t=20$ contributing the full $20$.",
      "tag": "Future lifetime"
    },
    {
      "front": "Under de Moivre with $\\omega=110$, find $\\mu_{30}$, ${}_{20}p_{30}$, and the probability a life aged $30$ dies between ages $50$ and $60$.",
      "back": "$\\mu_{30}=\\dfrac{1}{110-30}=\\dfrac{1}{80}=0.0125$.\n${}_{20}p_{30}=\\dfrac{110-30-20}{110-30}=\\dfrac{60}{80}=0.750$.\nDeath between $50$ and $60$ is ${}_{20|10}q_{30}={}_{20}p_{30}-{}_{30}p_{30}$.\n${}_{30}p_{30}=\\dfrac{110-60}{80}=\\dfrac{50}{80}=0.625$.\nSo ${}_{20|10}q_{30}=0.750-0.625=0.125$.",
      "tag": "Mortality laws"
    }
  ]
}