{
  "deckName": "Exam FAM — Life Insurance",
  "examCode": "Exam FAM",
  "cards": [
    {
      "front": "Define the present-value random variable $Z$ for a discrete whole life insurance of $1$, and write its actuarial present value (EPV) $A_x$.",
      "back": "Let $K_x$ be the curtate future lifetime (complete years lived). The benefit of $1$ is paid at the **end of the year of death**, so the PV is $Z=v^{K_x+1}$.\nThe EPV is\n$A_x=E[Z]=\\sum_{k=0}^{\\infty} v^{k+1}\\,{}_kp_x\\,q_{x+k}$,\nwhere $v=\\frac{1}{1+i}$, ${}_kp_x$ is the probability of surviving $k$ years, and $q_{x+k}$ is the probability of dying in the following year.",
      "tag": "Whole life"
    },
    {
      "front": "Write the EPV of a discrete $n$-year **term** insurance $A^{1}_{x:\\overline{n|}}$ and say what the PV variable looks like.",
      "back": "The unit benefit is paid at the end of the year of death **only if death occurs within $n$ years**:\n$Z=\\begin{cases} v^{K_x+1}, & K_x \\le n-1\\\\ 0, & K_x \\ge n\\end{cases}$\nso\n$A^{1}_{x:\\overline{n|}}=\\sum_{k=0}^{n-1} v^{k+1}\\,{}_kp_x\\,q_{x+k}$.\nThe little $1$ over the $x$ marks that the benefit is contingent on **death** (the life status) before the term expires.",
      "tag": "Term & endowment"
    },
    {
      "front": "Write the EPV of an $n$-year **pure endowment** ${}_nE_x$ and describe the benefit.",
      "back": "A pure endowment pays $1$ at time $n$ **only if the life survives** to age $x+n$; nothing is paid on death.\n${}_nE_x = v^{n}\\,{}_np_x$.\nIt is also written $A_{x:\\overline{n|}}^{\\;\\;1}$ (the $1$ over the $n$ marks survival of the term as the contingency) and acts as the discount factor for survival benefits.",
      "tag": "Pure endowment & deferred"
    },
    {
      "front": "Write the EPV of a discrete $n$-year **endowment** insurance $A_{x:\\overline{n|}}$ and decompose it.",
      "back": "An endowment insurance pays $1$ at the end of the year of death **if death is within $n$ years**, or at time $n$ **if the life survives**. Either way $1$ is paid by time $n$.\n$A_{x:\\overline{n|}} = A^{1}_{x:\\overline{n|}} + {}_nE_x$,\nthe sum of the $n$-year term insurance and the $n$-year pure endowment. Its PV variable is $Z=v^{\\min(K_x+1,\\;n)}$.",
      "tag": "Term & endowment"
    },
    {
      "front": "Write the EPV of an $n$-year **deferred whole life** insurance ${}_{n|}A_x$ and give its product form.",
      "back": "Coverage starts after $n$ years: the unit is paid at the end of the year of death **only if death occurs after age $x+n$**.\n${}_{n|}A_x = \\sum_{k=n}^{\\infty} v^{k+1}\\,{}_kp_x\\,q_{x+k} = {}_nE_x\\,A_{x+n} = v^{n}\\,{}_np_x\\,A_{x+n}$.\nYou survive $n$ years (factor ${}_nE_x$) and then hold an ordinary whole life on $(x+n)$.",
      "tag": "Pure endowment & deferred"
    },
    {
      "front": "Give the basic identity linking term, deferred, and whole life insurance.",
      "back": "Whole life splits into the term piece plus the deferred piece:\n$A_x = A^{1}_{x:\\overline{n|}} + {}_{n|}A_x = A^{1}_{x:\\overline{n|}} + {}_nE_x\\,A_{x+n}$.\nDeath in the first $n$ years is covered by the term insurance; death after $n$ years is covered by the deferred insurance. This is the basis for many exam decompositions.",
      "tag": "Pure endowment & deferred"
    },
    {
      "front": "Write the EPV of a **continuous** whole life insurance $\\bar A_x$, where the benefit is paid at the exact moment of death.",
      "back": "With $T_x$ the (continuous) future lifetime, the PV is $Z=v^{T_x}=e^{-\\delta T_x}$, where $\\delta=\\ln(1+i)$ is the force of interest.\n$\\bar A_x = E[Z]=\\int_0^{\\infty} v^{t}\\,{}_tp_x\\,\\mu_{x+t}\\,dt = \\int_0^{\\infty} e^{-\\delta t}\\,{}_tp_x\\,\\mu_{x+t}\\,dt$,\nwhere ${}_tp_x\\,\\mu_{x+t}$ is the density of the time of death.",
      "tag": "Continuous/mthly & relations"
    },
    {
      "front": "Write the EPV of an $m$-thly whole life insurance $A_x^{(m)}$ and explain the benefit timing.",
      "back": "The benefit is paid at the **end of the $\\tfrac1m$-year period of death** (e.g. end of the month for $m=12$). Letting $K_x^{(m)}$ be the curtate lifetime in $\\tfrac1m$-year units,\n$A_x^{(m)} = \\sum_{j=0}^{\\infty} v^{(j+1)/m}\\,{}_{j/m}p_x\\,{}_{1/m}q_{x+j/m}$.\nAs $m\\to\\infty$, $A_x^{(m)}\\to\\bar A_x$. Faster payment means a larger EPV: $A_x \\le A_x^{(m)} \\le \\bar A_x$.",
      "tag": "Continuous/mthly & relations"
    },
    {
      "front": "State the **rule of moments**: how do you get the second moment ${}^2A_x$ of a life-insurance PV?",
      "back": "Because $Z^2=(v^{K+1})^2=(v^2)^{K+1}$, squaring the PV is the same as **doubling the force of interest**. The second moment ${}^2A_x$ is $A_x$ recomputed with $v$ replaced by $v^2$ — i.e. at force $2\\delta$, equivalently interest rate $(1+i)^2-1$.\nSame holds for term, endowment, and deferred forms: ${}^2A^{1}_{x:\\overline{n|}}$, ${}^2A_{x:\\overline{n|}}$, etc., each evaluated at the doubled force.",
      "tag": "Moments & variance"
    },
    {
      "front": "Give the variance of the PV of a unit life insurance, and of a benefit of $b$.",
      "back": "For a unit benefit, $Var(Z) = E[Z^2]-(E[Z])^2 = {}^2A_x - (A_x)^2$, where ${}^2A_x$ uses the doubled force of interest.\nFor a benefit of $b$, the PV is $bZ$, so $Var(bZ)=b^2\\,Var(Z)=b^2\\big({}^2A_x-A_x^2\\big)$ and the standard deviation scales by $b$.\nThe same template works with term/endowment $A$'s.",
      "tag": "Moments & variance"
    },
    {
      "front": "State the backward **recursion** for discrete whole life insurance $A_x$.",
      "back": "$A_x = v\\,q_x + v\\,p_x\\,A_{x+1}$.\nIntuition: in the first year either the life dies (prob $q_x$, paying $1$ discounted one year, $v$) or survives (prob $p_x$) and then holds a whole life on $(x+1)$ worth $A_{x+1}$, discounted one year.\nTo move forward instead, rearrange: $A_{x+1}=\\dfrac{A_x - v q_x}{v\\,p_x}$.",
      "tag": "Recursions"
    },
    {
      "front": "State the recursions for $n$-year **term** and **endowment** insurance.",
      "back": "Term: $A^{1}_{x:\\overline{n|}} = v\\,q_x + v\\,p_x\\,A^{1}_{x+1:\\overline{n-1|}}$, with $A^{1}_{x:\\overline{0|}}=0$.\nEndowment: $A_{x:\\overline{n|}} = v\\,q_x + v\\,p_x\\,A_{x+1:\\overline{n-1|}}$, with $A_{x:\\overline{0|}}=1$ (the survival payment is due immediately at term end).\nBoth peel off the first year: pay on death now, otherwise discount one year and hold a contract one term shorter on $(x+1)$.",
      "tag": "Recursions"
    },
    {
      "front": "State the relationship between whole life insurance $A_x$ and the annuity-due $\\ddot a_x$.",
      "back": "$A_x = 1 - d\\,\\ddot a_x$, where $d=\\dfrac{i}{1+i}=iv$ is the effective discount rate. Equivalently $\\ddot a_x = \\dfrac{1-A_x}{d}$.\nThe endowment version is $A_{x:\\overline{n|}} = 1 - d\\,\\ddot a_{x:\\overline{n|}}$.\nThe continuous analogue is $\\bar A_x = 1 - \\delta\\,\\bar a_x$.",
      "tag": "Continuous/mthly & relations"
    },
    {
      "front": "Under the **UDD** assumption, relate the continuous insurance $\\bar A_x$ to the discrete $A_x$.",
      "back": "Under a uniform distribution of deaths within each year of age,\n$\\bar A_x = \\dfrac{i}{\\delta}\\,A_x$,\nwhere $\\delta=\\ln(1+i)$. The factor $\\dfrac{i}{\\delta}>1$ inflates the discrete EPV because paying the benefit on average mid-year (rather than year-end) advances each payment by about half a year. The same factor applies to term and deferred death benefits: $\\bar A^{1}_{x:\\overline{n|}}=\\dfrac{i}{\\delta}A^{1}_{x:\\overline{n|}}$ (the pure-endowment piece of an endowment is unchanged).",
      "tag": "Continuous/mthly & relations"
    },
    {
      "front": "Under **UDD**, relate the $m$-thly insurance $A_x^{(m)}$ to the annual $A_x$.",
      "back": "$A_x^{(m)} = \\dfrac{i}{i^{(m)}}\\,A_x$,\nwhere $i^{(m)} = m\\big[(1+i)^{1/m}-1\\big]$ is the nominal rate compounded $m$ times a year. The factor $\\dfrac{i}{i^{(m)}}$ (between $1$ and $\\tfrac{i}{\\delta}$) advances each death benefit to the end of its $\\tfrac1m$-year subinterval. As $m\\to\\infty$, $i^{(m)}\\to\\delta$ and $A_x^{(m)}\\to\\bar A_x$.",
      "tag": "Continuous/mthly & relations"
    },
    {
      "front": "Under a **constant force of mortality** $\\mu$ and force of interest $\\delta$, give closed forms for $\\bar A_x$ and ${}^2\\bar A_x$, and hence $Var(Z)$.",
      "back": "With ${}_tp_x=e^{-\\mu t}$ and density $\\mu e^{-\\mu t}$,\n$\\bar A_x = \\int_0^\\infty e^{-\\delta t}\\,e^{-\\mu t}\\mu\\,dt = \\dfrac{\\mu}{\\mu+\\delta}$.\nDoubling the force of interest gives ${}^2\\bar A_x = \\dfrac{\\mu}{\\mu+2\\delta}$.\nSo $Var(Z) = \\dfrac{\\mu}{\\mu+2\\delta} - \\left(\\dfrac{\\mu}{\\mu+\\delta}\\right)^2$.",
      "tag": "Continuous/mthly & relations"
    },
    {
      "front": "Under a **constant** annual mortality rate $q$ (so $q_{x+k}=q$ for all $k$), give the closed form for discrete whole life $A_x$.",
      "back": "Then ${}_kp_x\\,q_{x+k}=(1-q)^k q$, a geometric distribution, so\n$A_x = \\sum_{k=0}^\\infty v^{k+1}(1-q)^k q = \\dfrac{vq}{1-v(1-q)} = \\dfrac{q}{q+i}$.\nThe second moment uses interest $(1+i)^2-1$ in place of $i$: ${}^2A_x = \\dfrac{q}{q+\\big[(1+i)^2-1\\big]}$, and $Var(Z)={}^2A_x - A_x^2$.",
      "tag": "Whole life"
    },
    {
      "front": "Define the **increasing** whole life insurance $(IA)_x$ and the increasing $n$-year term $(IA)^{1}_{x:\\overline{n|}}$.",
      "back": "The benefit paid at the end of the year of death equals $K_x+1$ (i.e. $1$ for death in year 1, $2$ in year 2, …). The PV is $Z=(K_x+1)v^{K_x+1}$.\n$(IA)_x = \\sum_{k=0}^{\\infty}(k+1)\\,v^{k+1}\\,{}_kp_x\\,q_{x+k}$,\nand the $n$-year term version truncates the sum at $k=n-1$.",
      "tag": "Whole life"
    },
    {
      "front": "Define the **decreasing** $n$-year term insurance $(DA)^{1}_{x:\\overline{n|}}$ and give the identity with the increasing term.",
      "back": "The benefit paid at the end of the year of death in year $k+1$ is $n-k$ (i.e. $n$ for death in year 1, decreasing to $1$ in year $n$):\n$(DA)^{1}_{x:\\overline{n|}} = \\sum_{k=0}^{n-1}(n-k)\\,v^{k+1}\\,{}_kp_x\\,q_{x+k}$.\nUseful identity: $(IA)^{1}_{x:\\overline{n|}} + (DA)^{1}_{x:\\overline{n|}} = (n+1)\\,A^{1}_{x:\\overline{n|}}$, since the two benefit patterns sum to the constant $n+1$.",
      "tag": "Whole life"
    },
    {
      "front": "Given $i=0.06$ and the mortality $q_x=0.10,\\;q_{x+1}=0.15,\\;q_{x+2}=1.00$ (so the life cannot survive past age $x+3$), compute the whole life insurance $A_x$.",
      "back": "$v=\\frac{1}{1.06}\\approx 0.943396$. Survival: ${}_0p_x=1$, ${}_1p_x=0.90$, ${}_2p_x=0.90(0.85)=0.765$.\nDeath-year terms $v^{k+1}\\,{}_kp_x\\,q_{x+k}$:\n$k=0:\\;0.943396(1)(0.10)=0.094340$\n$k=1:\\;0.943396^{2}(0.90)(0.15)=0.889996(0.135)=0.120150$\n$k=2:\\;0.943396^{3}(0.765)(1.00)=0.839619(0.765)=0.642309$\n$A_x = 0.094340+0.120150+0.642309 = \\boxed{0.856798}$.",
      "tag": "Whole life"
    },
    {
      "front": "For the same life ($i=0.06$; $q_x=0.10,\\,q_{x+1}=0.15,\\,q_{x+2}=1.00$, with $A_x=0.856798$), find $Var(Z)$ for the unit whole life insurance using the rule of moments.",
      "back": "Double the force of interest: use rate $j=(1.06)^2-1=0.1236$, so $v_j=\\frac{1}{1.1236}\\approx 0.889996$.\n${}^2A_x = \\sum v_j^{k+1}\\,{}_kp_x\\,q_{x+k}$:\n$k=0:\\;0.889996(0.10)=0.089000$\n$k=1:\\;0.889996^{2}(0.90)(0.15)=0.792092(0.135)=0.106932$\n$k=2:\\;0.889996^{3}(0.765)=0.704958(0.765)=0.539293$\n${}^2A_x = 0.089000+0.106932+0.539293 = 0.735225$.\n$Var(Z) = 0.735225 - 0.856798^2 = 0.735225 - 0.734103 = \\boxed{0.001122}$.",
      "tag": "Moments & variance"
    },
    {
      "front": "An insurer issues a $\\$100{,}000$ whole life policy. With $i=0.06$ and $q_x=0.08,\\,q_{x+1}=0.12,\\,q_{x+2}=0.20,\\,q_{x+3}=1.00$, find the single-premium APV.",
      "back": "$v=0.943396$. Survival: ${}_1p_x=0.92$, ${}_2p_x=0.92(0.88)=0.8096$, ${}_3p_x=0.8096(0.80)=0.64768$.\nDeath terms:\n$k=0:\\;0.943396(0.08)=0.075472$\n$k=1:\\;0.889996(0.92)(0.12)=0.098255$\n$k=2:\\;0.839619(0.8096)(0.20)=0.135950$\n$k=3:\\;0.792094(0.64768)(1.00)=0.513025$\n$A_x=0.075472+0.098255+0.135950+0.513025=0.822702$.\nAPV $=100{,}000(0.822702)=\\boxed{\\$82{,}270.20}$.",
      "tag": "Whole life"
    },
    {
      "front": "With $i=0.05$ and $q_x=0.04,\\;q_{x+1}=0.06,\\;q_{x+2}=0.09$, compute the $3$-year term insurance $A^{1}_{x:\\overline{3|}}$.",
      "back": "$v=\\frac{1}{1.05}\\approx 0.952381$. Survival: ${}_1p_x=0.96$, ${}_2p_x=0.96(0.94)=0.9024$.\nDeath terms $v^{k+1}\\,{}_kp_x\\,q_{x+k}$:\n$k=0:\\;0.952381(0.04)=0.038095$\n$k=1:\\;0.907029(0.96)(0.06)=0.052245$\n$k=2:\\;0.863838(0.9024)(0.09)=0.070157$\n$A^{1}_{x:\\overline{3|}} = 0.038095+0.052245+0.070157 = \\boxed{0.160498}$.",
      "tag": "Term & endowment"
    },
    {
      "front": "For the same data ($i=0.05$; $q_x=0.04,\\,q_{x+1}=0.06,\\,q_{x+2}=0.09$; $A^{1}_{x:\\overline{3|}}=0.160498$), find the $3$-year pure endowment ${}_3E_x$ and the $3$-year endowment insurance $A_{x:\\overline{3|}}$.",
      "back": "${}_3p_x = 0.96(0.94)(0.91)=0.821184$ and $v^3=0.952381^3\\approx 0.863838$.\n${}_3E_x = v^3\\,{}_3p_x = 0.863838(0.821184) = \\boxed{0.709370}$.\nEndowment $=$ term $+$ pure endowment:\n$A_{x:\\overline{3|}} = A^{1}_{x:\\overline{3|}} + {}_3E_x = 0.160498 + 0.709370 = \\boxed{0.869868}$.",
      "tag": "Term & endowment"
    },
    {
      "front": "A $\\$50{,}000$ $2$-year term insurance is sold at $i=0.05$ with $q_x=0.03$ and $q_{x+1}=0.05$. Find the APV.",
      "back": "$v=0.952381$. ${}_1p_x=0.97$.\nDeath terms:\n$k=0:\\;v\\,q_x = 0.952381(0.03)=0.028571$\n$k=1:\\;v^2\\,{}_1p_x\\,q_{x+1}=0.907029(0.97)(0.05)=0.043991$\n$A^{1}_{x:\\overline{2|}} = 0.028571+0.043991 = 0.072562$.\nAPV $= 50{,}000(0.072562) = \\boxed{\\$3{,}628.12}$.",
      "tag": "Term & endowment"
    },
    {
      "front": "Compute a $5$-year pure endowment ${}_5E_x$ at $i=0.04$ when the yearly survival probabilities are $p_x=0.97,\\,p_{x+1}=0.96,\\,p_{x+2}=0.95,\\,p_{x+3}=0.93,\\,p_{x+4}=0.90$.",
      "back": "Survival to $5$ years: ${}_5p_x = 0.97(0.96)(0.95)(0.93)(0.90)$.\n$0.97(0.96)=0.9312$; $\\times 0.95 = 0.88464$; $\\times 0.93 = 0.822715$; $\\times 0.90 = 0.740444$.\nDiscount: $v^5 = 1.04^{-5} \\approx 0.821927$.\n${}_5E_x = v^5\\,{}_5p_x = 0.821927(0.740444) = \\boxed{0.608592}$.",
      "tag": "Pure endowment & deferred"
    },
    {
      "front": "Find a $2$-year **deferred** whole life insurance ${}_{2|}A_x$ at $i=0.06$, given $p_x=0.95,\\,p_{x+1}=0.93$ and a constant force at and beyond age $x+2$ producing $A_{x+2}=0.454545$.",
      "back": "Pure-endowment factor: ${}_2p_x = 0.95(0.93)=0.8835$ and $v^2 = 1.06^{-2}\\approx 0.889996$, so\n${}_2E_x = v^2\\,{}_2p_x = 0.889996(0.8835)=0.786312$.\nDeferred insurance $=$ survive $2$ years, then hold a whole life on $(x+2)$:\n${}_{2|}A_x = {}_2E_x\\,A_{x+2} = 0.786312(0.454545) = \\boxed{0.357415}$.",
      "tag": "Pure endowment & deferred"
    },
    {
      "front": "Use the recursion $A_x = v q_x + v p_x A_{x+1}$ to find $A_x$ at $i=0.05$ when $q_x=0.03$ and $A_{x+1}=0.40$.",
      "back": "$v = \\frac{1}{1.05}\\approx 0.952381$, $p_x = 0.97$.\n$A_x = v q_x + v p_x A_{x+1}$\n$= 0.952381(0.03) + 0.952381(0.97)(0.40)$\n$= 0.028571 + 0.952381(0.388)$\n$= 0.028571 + 0.369524 = \\boxed{0.398095}$.",
      "tag": "Recursions"
    },
    {
      "front": "Use the endowment recursion to build $A_{x:\\overline{2|}}$ at $i=0.05$ from $q_x=0.04$ and $q_{x+1}=0.07$, and verify it directly.",
      "back": "$v=0.952381$, $p_x=0.96$. A $1$-year endowment always pays $1$ at year-end (death or survival), so $A_{x+1:\\overline{1|}} = v = 0.952381$.\nRecursion: $A_{x:\\overline{2|}} = v q_x + v p_x A_{x+1:\\overline{1|}} = 0.952381(0.04) + 0.952381(0.96)(0.952381)$\n$= 0.038095 + 0.870748 = \\boxed{0.908844}$.\nDirect check: $v q_x + v^2 p_x q_{x+1} + v^2 p_x p_{x+1} = 0.038095 + 0.907029(0.96)(0.07) + 0.907029(0.96)(0.93) = 0.038095 + 0.060952 + 0.809797 = 0.908844$. \\checkmark",
      "tag": "Recursions"
    },
    {
      "front": "A life has **constant** annual mortality $q=0.04$ at all ages and $i=0.06$. Find $A_x$, ${}^2A_x$, and $Var(Z)$ for a unit whole life insurance.",
      "back": "Constant-$q$ closed form: $A_x = \\dfrac{q}{q+i} = \\dfrac{0.04}{0.04+0.06} = \\boxed{0.40}$.\nSecond moment doubles the force of interest; use $j=(1.06)^2-1 = 0.1236$:\n${}^2A_x = \\dfrac{q}{q+j} = \\dfrac{0.04}{0.04+0.1236} = \\dfrac{0.04}{0.1636} = 0.244499$.\n$Var(Z) = {}^2A_x - A_x^2 = 0.244499 - 0.40^2 = 0.244499 - 0.16 = \\boxed{0.084499}$.",
      "tag": "Moments & variance"
    },
    {
      "front": "A $1$-unit whole life insurance has a **constant force** of mortality $\\mu=0.02$ and force of interest $\\delta=0.05$. Find $\\bar A_x$ and the variance of the PV.",
      "back": "$\\bar A_x = \\dfrac{\\mu}{\\mu+\\delta} = \\dfrac{0.02}{0.02+0.05} = \\dfrac{0.02}{0.07} = \\boxed{0.285714}$.\nSecond moment doubles $\\delta$: ${}^2\\bar A_x = \\dfrac{\\mu}{\\mu+2\\delta} = \\dfrac{0.02}{0.02+0.10} = \\dfrac{0.02}{0.12} = 0.166667$.\n$Var(Z) = 0.166667 - 0.285714^2 = 0.166667 - 0.081633 = \\boxed{0.085034}$.",
      "tag": "Moments & variance"
    },
    {
      "front": "At $i=0.10$, a unit $3$-year **term** insurance has $q_x=0.05,\\,q_{x+1}=0.10,\\,q_{x+2}=0.20$. Find $A^{1}_{x:\\overline{3|}}$, then its variance.",
      "back": "$v=\\frac{1}{1.1}\\approx 0.909091$; ${}_1p_x=0.95$, ${}_2p_x=0.95(0.90)=0.855$.\nFirst moment:\n$k=0:\\;0.909091(0.05)=0.045455$\n$k=1:\\;0.826446(0.95)(0.10)=0.078512$\n$k=2:\\;0.751315(0.855)(0.20)=0.128475$\n$A^{1}_{x:\\overline{3|}} = 0.252442$.\nSecond moment at $j=(1.1)^2-1=0.21$, $v_j=\\frac{1}{1.21}\\approx 0.826446$:\n$0.826446(0.05)=0.041322$; $0.683013(0.95)(0.10)=0.064886$; $0.564474(0.855)(0.20)=0.096525$; sum $={}^2A^{1}_{x:\\overline{3|}}=0.202733$.\n$Var(Z)=0.202733-0.252442^2 = 0.202733-0.063727 = \\boxed{0.139006}$.",
      "tag": "Moments & variance"
    },
    {
      "front": "For the same $i=0.10$ data ($q_x=0.05,\\,q_{x+1}=0.10,\\,q_{x+2}=0.20$), find the $3$-year **endowment** insurance $A_{x:\\overline{3|}}$ and its variance.",
      "back": "Add the pure endowment to the term piece ($A^{1}_{x:\\overline{3|}}=0.252442$). ${}_3p_x=0.95(0.90)(0.80)=0.684$, $v^3=0.751315$.\n${}_3E_x = 0.751315(0.684)=0.513900$, so $A_{x:\\overline{3|}}=0.252442+0.513900=0.766342$.\nSecond moment: term part $0.202733$ (from before) plus endowment part at doubled interest $v_j^3\\,{}_3p_x=0.564474(0.684)=0.386100$.\n${}^2A_{x:\\overline{3|}}=0.202733+0.386100=0.588833$.\n$Var(Z)=0.588833-0.766342^2 = 0.588833-0.587280 = \\boxed{0.001553}$.\nNote the variance is far smaller than the pure term case — the guaranteed survival payment removes most of the dispersion.",
      "tag": "Moments & variance"
    },
    {
      "front": "Given $A_x = 0.35$ and $i=0.06$, find the continuous insurance $\\bar A_x$ under UDD.",
      "back": "Force of interest $\\delta = \\ln(1.06) \\approx 0.058269$.\nUnder UDD, $\\bar A_x = \\dfrac{i}{\\delta}\\,A_x = \\dfrac{0.06}{0.058269}\\,(0.35)$.\n$\\dfrac{0.06}{0.058269} \\approx 1.029709$, so\n$\\bar A_x \\approx 1.029709(0.35) = \\boxed{0.360398}$.\nThe continuous EPV exceeds the discrete one because the benefit is paid on average half a year sooner.",
      "tag": "Continuous/mthly & relations"
    },
    {
      "front": "Given $A_x = 0.35$ and $i=0.06$, find the **monthly** ($m=12$) insurance $A_x^{(12)}$ under UDD.",
      "back": "Nominal rate: $i^{(12)} = 12\\big[(1.06)^{1/12}-1\\big]$. Since $(1.06)^{1/12}\\approx 1.004868$, $i^{(12)}\\approx 12(0.004868)=0.058411$.\nUnder UDD, $A_x^{(12)} = \\dfrac{i}{i^{(12)}}\\,A_x = \\dfrac{0.06}{0.058411}(0.35)$.\n$\\dfrac{0.06}{0.058411}\\approx 1.027211$, so $A_x^{(12)} \\approx 1.027211(0.35) = \\boxed{0.359524}$.\nThis sits between $A_x=0.35$ and $\\bar A_x\\approx 0.360398$, as expected.",
      "tag": "Continuous/mthly & relations"
    },
    {
      "front": "Given $\\ddot a_x = 14.20$ and $i=0.06$, find $A_x$ using the annuity-insurance relationship.",
      "back": "Effective discount rate $d = \\dfrac{i}{1+i} = \\dfrac{0.06}{1.06} \\approx 0.056604$.\n$A_x = 1 - d\\,\\ddot a_x = 1 - 0.056604(14.20) = 1 - 0.803774 = \\boxed{0.196226}$.\nCheck: a large annuity value (long expected lifetime) goes with a small insurance value — death is far off and heavily discounted.",
      "tag": "Continuous/mthly & relations"
    },
    {
      "front": "At $i=0.06$ with $q_x=0.10,\\,q_{x+1}=0.15,\\,q_{x+2}=0.20$, compute the **increasing** $3$-year term $(IA)^{1}_{x:\\overline{3|}}$.",
      "back": "$v=0.943396$; ${}_1p_x=0.90$, ${}_2p_x=0.90(0.85)=0.765$. Benefit in year $k+1$ is $k+1$:\n$k=0:\\;1\\cdot 0.943396(0.10)=0.094340$\n$k=1:\\;2\\cdot 0.889996(0.90)(0.15)=2(0.120149)=0.240299$\n$k=2:\\;3\\cdot 0.839619(0.765)(0.20)=3(0.128462)=0.385385$\n$(IA)^{1}_{x:\\overline{3|}} = 0.094340+0.240299+0.385385 = \\boxed{0.720024}$.",
      "tag": "Whole life"
    },
    {
      "front": "For the same data ($i=0.06$; $q_x=0.10,\\,q_{x+1}=0.15,\\,q_{x+2}=0.20$), find the **decreasing** $3$-year term $(DA)^{1}_{x:\\overline{3|}}$ and verify with the $(IA)+(DA)$ identity.",
      "back": "Benefit in year $k+1$ is $n-k = 3,2,1$:\n$k=0:\\;3\\cdot 0.943396(0.10)=3(0.094340)=0.283019$\n$k=1:\\;2\\cdot 0.889996(0.90)(0.15)=0.240299$\n$k=2:\\;1\\cdot 0.839619(0.765)(0.20)=0.128462$\n$(DA)^{1}_{x:\\overline{3|}} = 0.283019+0.240299+0.128462 = \\boxed{0.651780}$.\nCheck: the level term is $A^{1}_{x:\\overline{3|}} = 0.094340+0.120149+0.128462 = 0.342951$, and $(IA)+(DA)=0.720024+0.651780=1.371804 = 4(0.342951)=(n+1)A^{1}$. \\checkmark",
      "tag": "Whole life"
    },
    {
      "front": "Why does an $n$-year **endowment** insurance always have a far smaller variance than the corresponding $n$-year **term** insurance (same $i$, same mortality)?",
      "back": "The endowment guarantees a payment by time $n$ whether the life dies or survives, so the PV variable $Z=v^{\\min(K+1,n)}$ is bounded between $v^n$ and $v$ — a narrow range. The term insurance pays $v^{K+1}$ on death but **$0$ on survival**, so $Z$ jumps from possibly large values down to $0$, a much wider spread.\nNumerically (the $i=0.10$, $q=0.05/0.10/0.20$ data): term $Var\\approx 0.139$ vs endowment $Var\\approx 0.0016$ — about $90\\times$ smaller.",
      "tag": "Moments & variance"
    },
    {
      "front": "Distinguish $A^{1}_{x:\\overline{n|}}$, $A_{x:\\overline{n|}}^{\\;\\;1}$, and $A_{x:\\overline{n|}}$ by where the \"$1$\" sits.",
      "back": "$A^{1}_{x:\\overline{n|}}$ — the $1$ is over **$x$**: pays on **death** within $n$ years ($n$-year **term** insurance).\n$A_{x:\\overline{n|}}^{\\;\\;1}$ — the $1$ is over the **$n$**: pays on **survival** to time $n$ ($n$-year **pure endowment**, $={}_nE_x$).\n$A_{x:\\overline{n|}}$ — **no** $1$: pays on whichever comes first, death or survival ($n$-year **endowment** insurance $=$ sum of the other two).",
      "tag": "Term & endowment"
    },
    {
      "front": "Order $A_x$, $A_x^{(m)}$, and $\\bar A_x$ for a fixed life and interest rate, and explain the ordering.",
      "back": "$A_x \\;\\le\\; A_x^{(m)} \\;\\le\\; \\bar A_x$.\nAll three EPV the same death benefit; they differ only in **when** the benefit is paid. The discrete $A_x$ waits until year-end; the $m$-thly version pays at the end of the (shorter) $\\tfrac1m$-year subinterval; the continuous version pays at the instant of death. Earlier payment means less discounting and a larger present value, hence the ordering. As $m\\to\\infty$, $A_x^{(m)}\\to\\bar A_x$.",
      "tag": "Continuous/mthly & relations"
    },
    {
      "front": "A life has **constant force** $\\mu=0.03$ and $\\delta=0.06$. Find the EPV of a **continuous** $10$-year term insurance $\\bar A^{1}_{x:\\overline{10|}}$.",
      "back": "Under constant force, $\\bar A^{1}_{x:\\overline{n|}} = \\dfrac{\\mu}{\\mu+\\delta}\\big(1 - e^{-(\\mu+\\delta)n}\\big)$ (integrate $e^{-(\\mu+\\delta)t}\\mu$ from $0$ to $n$).\n$\\dfrac{\\mu}{\\mu+\\delta} = \\dfrac{0.03}{0.09} = 0.333333$. With $n=10$: $(\\mu+\\delta)n = 0.9$, $e^{-0.9}\\approx 0.406570$.\n$\\bar A^{1}_{x:\\overline{10|}} = 0.333333(1 - 0.406570) = 0.333333(0.593430) = \\boxed{0.197810}$.",
      "tag": "Continuous/mthly & relations"
    },
    {
      "front": "A $\\$250{,}000$ benefit is paid on death; the unit insurance has $A_x=0.30$ and ${}^2A_x=0.11$. Find the standard deviation of the present value of the payout.",
      "back": "Unit variance via rule of moments: $Var(Z) = {}^2A_x - A_x^2 = 0.11 - 0.30^2 = 0.11 - 0.09 = 0.02$.\nThe payout PV is $250{,}000\\,Z$, so its variance is $250{,}000^2(0.02)$ and its standard deviation is\n$250{,}000\\sqrt{0.02} = 250{,}000(0.141421) = \\boxed{\\$35{,}355.34}$.\nStandard deviation scales linearly with the benefit; variance scales with its square.",
      "tag": "Moments & variance"
    }
  ]
}