{
  "deckName": "Exam FAM — Life Annuities",
  "examCode": "Exam FAM",
  "cards": [
    {
      "front": "Define the random variable $Y$ for a whole life annuity-due and give its EPV $\\ddot a_x$ as a sum.",
      "back": "$Y=\\ddot a_{\\overline{K+1|}}$ pays $1$ at the start of each year while $(x)$ survives, where $K=K_x$ is the curtate future lifetime.\nIts EPV is\n$\\ddot a_x = E[Y] = \\sum_{k=0}^{\\infty} v^{k}\\,{}_kp_x$,\nthe present value of \\$1 paid at time $k$ weighted by the probability ${}_kp_x$ that $(x)$ is alive then. (The first payment at $k=0$ is certain, contributing $1$.)",
      "tag": "Whole life annuity"
    },
    {
      "front": "State the relation between the whole life annuity-immediate $a_x$ and the annuity-due $\\ddot a_x$.",
      "back": "$a_x = \\ddot a_x - 1$.\nThe annuity-immediate pays \\$1 at the **end** of each year of survival, i.e. $a_x=\\sum_{k=1}^{\\infty} v^{k}\\,{}_kp_x$. It is the annuity-due minus the certain time-$0$ payment of $1$, since both share the payments at times $1,2,\\dots$",
      "tag": "Due vs immediate"
    },
    {
      "front": "Given $\\ddot a_x = 14.20$, find the annuity-immediate $a_x$.",
      "back": "$a_x = \\ddot a_x - 1 = 14.20 - 1 = 13.20$.\nThe difference of exactly $1$ is the time-$0$ payment that the due annuity makes but the immediate annuity does not.",
      "tag": "Due vs immediate"
    },
    {
      "front": "State the fundamental insurance–annuity relation linking $\\ddot a_x$ and $A_x$, and solve it both ways.",
      "back": "$\\ddot a_x = \\dfrac{1 - A_x}{d}$, equivalently $A_x = 1 - d\\,\\ddot a_x$,\nwhere $d=\\dfrac{i}{1+i}=1-v$ is the effective annual discount rate.\nThe identity comes from $1 = d\\,\\ddot a_{\\overline{K+1|}} + v^{K+1}$ taken in expectation: $1 = d\\,\\ddot a_x + A_x$.",
      "tag": "Insurance-annuity relations"
    },
    {
      "front": "Given $A_x = 0.30$ and $i = 0.05$, compute $\\ddot a_x$.",
      "back": "First $d = \\dfrac{i}{1+i} = \\dfrac{0.05}{1.05} \\approx 0.047619$.\n$\\ddot a_x = \\dfrac{1 - A_x}{d} = \\dfrac{1 - 0.30}{0.047619} = \\dfrac{0.70}{0.047619} \\approx 14.7000$.\nThen $a_x = \\ddot a_x - 1 \\approx 13.7000$.",
      "tag": "Insurance-annuity relations"
    },
    {
      "front": "Given $\\ddot a_x = 12.50$ and $i = 0.06$, find $A_x$.",
      "back": "$d = \\dfrac{0.06}{1.06} \\approx 0.056604$.\n$A_x = 1 - d\\,\\ddot a_x = 1 - 0.056604(12.50) = 1 - 0.707547 \\approx 0.29245$.",
      "tag": "Insurance-annuity relations"
    },
    {
      "front": "State the continuous insurance–annuity relation for $\\bar a_x$ and $\\bar A_x$.",
      "back": "$\\bar a_x = \\dfrac{1 - \\bar A_x}{\\delta}$, equivalently $\\bar A_x = 1 - \\delta\\,\\bar a_x$,\nwhere $\\delta = \\ln(1+i)$ is the force of interest and $\\bar a_x = \\int_0^{\\infty} v^{t}\\,{}_tp_x\\,dt$ values a continuously paid \\$1-per-year annuity.\nIt mirrors the discrete case with $\\delta$ in place of $d$ and $\\bar A_x$ in place of $A_x$.",
      "tag": "Insurance-annuity relations"
    },
    {
      "front": "Given $\\bar A_x = 0.40$ and $\\delta = 0.05$, find $\\bar a_x$.",
      "back": "$\\bar a_x = \\dfrac{1 - \\bar A_x}{\\delta} = \\dfrac{1 - 0.40}{0.05} = \\dfrac{0.60}{0.05} = 12.0000$.",
      "tag": "Insurance-annuity relations"
    },
    {
      "front": "State the backward recursion for the whole life annuity-due $\\ddot a_x$.",
      "back": "$\\ddot a_x = 1 + v\\,p_x\\,\\ddot a_{x+1}$.\nThe \\$1 at the start of the year is certain; with probability $p_x$ the life survives to age $x+1$, at which point the remaining annuity is worth $\\ddot a_{x+1}$, discounted one year by $v$.\nThis is the workhorse for building $\\ddot a_x$ down a life table from an endpoint value.",
      "tag": "Whole life annuity"
    },
    {
      "front": "A life table gives $\\ddot a_{x+1} = 13.00$, $p_x = 0.99$, and $i = 0.05$. Find $\\ddot a_x$.",
      "back": "$v = \\dfrac{1}{1.05} \\approx 0.952381$.\n$\\ddot a_x = 1 + v\\,p_x\\,\\ddot a_{x+1} = 1 + 0.952381(0.99)(13.00)$\n$= 1 + 0.952381(12.87) = 1 + 12.25714 \\approx 13.2571$.",
      "tag": "Whole life annuity"
    },
    {
      "front": "Define the $n$-year temporary life annuity-due $\\ddot a_{x:\\overline{n|}}$ as a sum.",
      "back": "$\\ddot a_{x:\\overline{n|}} = \\sum_{k=0}^{n-1} v^{k}\\,{}_kp_x$,\npaying \\$1 at the start of each year for at most $n$ years, only while $(x)$ survives. There is no payment at time $n$ — the last possible payment is at the start of year $n$, i.e. time $n-1$.",
      "tag": "Temporary & deferred"
    },
    {
      "front": "Compute the 3-year temporary annuity-due $\\ddot a_{x:\\overline{3|}}$ given $i = 0.05$, $p_x = 0.97$, $p_{x+1} = 0.96$.",
      "back": "Need ${}_0p_x = 1$, ${}_1p_x = 0.97$, ${}_2p_x = 0.97(0.96) = 0.9312$, and $v=0.952381$.\n$\\ddot a_{x:\\overline{3|}} = v^{0}(1) + v^{1}(0.97) + v^{2}(0.9312)$\n$= 1 + 0.952381(0.97) + 0.907029(0.9312)$\n$= 1 + 0.923810 + 0.844626 \\approx 2.76844$.",
      "tag": "Temporary & deferred"
    },
    {
      "front": "Relate the $n$-year temporary annuity-due to the $n$-year endowment insurance: $\\ddot a_{x:\\overline{n|}} = ?$",
      "back": "$\\ddot a_{x:\\overline{n|}} = \\dfrac{1 - A_{x:\\overline{n|}}}{d}$, equivalently $A_{x:\\overline{n|}} = 1 - d\\,\\ddot a_{x:\\overline{n|}}$,\nwhere $A_{x:\\overline{n|}}$ is the $n$-year **endowment** insurance (death benefit during the term plus pure-endowment survival benefit). Same form as the whole life relation, with the term replacing the whole life on both sides.",
      "tag": "Insurance-annuity relations"
    },
    {
      "front": "Given $A_{x:\\overline{20|}} = 0.45$ and $i = 0.06$, find $\\ddot a_{x:\\overline{20|}}$.",
      "back": "$d = \\dfrac{0.06}{1.06} \\approx 0.056604$.\n$\\ddot a_{x:\\overline{20|}} = \\dfrac{1 - A_{x:\\overline{20|}}}{d} = \\dfrac{1 - 0.45}{0.056604} = \\dfrac{0.55}{0.056604} \\approx 9.71667$.",
      "tag": "Insurance-annuity relations"
    },
    {
      "front": "Define the pure endowment ${}_nE_x$ and state how it splits a whole life annuity-due into temporary + deferred pieces.",
      "back": "${}_nE_x = v^{n}\\,{}_np_x$ is the EPV of \\$1 paid at time $n$ only if $(x)$ survives to $x+n$ — the $n$-year discount factor.\nIt links the pieces by\n$\\ddot a_x = \\ddot a_{x:\\overline{n|}} + {}_{n|}\\ddot a_x$, where the deferred annuity is ${}_{n|}\\ddot a_x = {}_nE_x\\,\\ddot a_{x+n}$.",
      "tag": "Temporary & deferred"
    },
    {
      "front": "Define the $n$-year deferred whole life annuity-due ${}_{n|}\\ddot a_x$ and give its EPV.",
      "back": "${}_{n|}\\ddot a_x$ pays \\$1 at the start of each year **beginning at age $x+n$**, while $(x)$ survives.\nIts EPV is ${}_{n|}\\ddot a_x = {}_nE_x\\,\\ddot a_{x+n} = v^{n}\\,{}_np_x\\,\\ddot a_{x+n} = \\sum_{k=n}^{\\infty} v^{k}\\,{}_kp_x$.\nEquivalently ${}_{n|}\\ddot a_x = \\ddot a_x - \\ddot a_{x:\\overline{n|}}$.",
      "tag": "Temporary & deferred"
    },
    {
      "front": "Given ${}_{10}p_x = 0.85$, $i = 0.05$, and $\\ddot a_{x+10} = 11.40$, find the 10-year deferred annuity ${}_{10|}\\ddot a_x$.",
      "back": "First the pure endowment: ${}_{10}E_x = v^{10}\\,{}_{10}p_x = (1.05)^{-10}(0.85)$.\n$(1.05)^{-10} \\approx 0.613913$, so ${}_{10}E_x \\approx 0.613913(0.85) = 0.521826$.\n${}_{10|}\\ddot a_x = {}_{10}E_x\\,\\ddot a_{x+10} = 0.521826(11.40) \\approx 5.94882$.",
      "tag": "Temporary & deferred"
    },
    {
      "front": "Given $\\ddot a_x = 15.00$ and $\\ddot a_{x:\\overline{10|}} = 7.80$, find the 10-year deferred annuity ${}_{10|}\\ddot a_x$.",
      "back": "$ {}_{10|}\\ddot a_x = \\ddot a_x - \\ddot a_{x:\\overline{10|}} = 15.00 - 7.80 = 7.20$.\nThe whole life annuity is exactly the temporary piece (years $0$–$9$) plus the deferred piece (years $10,11,\\dots$).",
      "tag": "Temporary & deferred"
    },
    {
      "front": "Define an $n$-year **guaranteed (certain-and-life)** annuity-due $\\ddot a_{\\overline{x:\\overline{n|}}}$ and give its EPV.",
      "back": "It pays \\$1 at the start of each year for $n$ years **guaranteed** (paid even if $(x)$ dies), then continues for life thereafter only while $(x)$ survives.\nEPV: $\\ddot a_{\\overline{x:\\overline{n|}}} = \\ddot a_{\\overline{n|}} + {}_{n|}\\ddot a_x = \\ddot a_{\\overline{n|}} + {}_nE_x\\,\\ddot a_{x+n}$,\nwhere $\\ddot a_{\\overline{n|}} = \\dfrac{1-v^{n}}{d}$ is the ordinary annuity-certain-due for the first $n$ guaranteed payments.",
      "tag": "Temporary & deferred"
    },
    {
      "front": "Compute the 10-year certain-and-life annuity-due given $i = 0.05$, ${}_{10}p_x = 0.85$, and $\\ddot a_{x+10} = 11.40$.",
      "back": "Annuity-certain-due: $\\ddot a_{\\overline{10|}} = \\dfrac{1-v^{10}}{d}$ with $v^{10}\\approx0.613913$, $d=\\dfrac{0.05}{1.05}\\approx0.047619$.\n$\\ddot a_{\\overline{10|}} = \\dfrac{1-0.613913}{0.047619} = \\dfrac{0.386087}{0.047619} \\approx 8.10783$.\nDeferred life piece: ${}_{10}E_x\\,\\ddot a_{x+10} = 0.613913(0.85)(11.40) \\approx 5.94882$.\nTotal $= 8.10783 + 5.94882 \\approx 14.0567$.",
      "tag": "Temporary & deferred"
    },
    {
      "front": "State the variance of the present value $Y$ of a whole life annuity-due in terms of insurance moments.",
      "back": "Since $Y = \\dfrac{1 - v^{K+1}}{d} = \\dfrac{1 - Z}{d}$ where $Z=v^{K+1}$ is the whole life insurance PV,\n$\\operatorname{Var}(Y) = \\dfrac{\\operatorname{Var}(Z)}{d^{2}} = \\dfrac{{}^2A_x - (A_x)^{2}}{d^{2}}$,\nwhere ${}^2A_x$ is the second moment of $Z$ (the insurance evaluated at double the force of interest, i.e. at rate $i'=2i+i^{2}$).",
      "tag": "Variance"
    },
    {
      "front": "Given $A_x = 0.25$, ${}^2A_x = 0.09$, and $i = 0.05$, find $\\operatorname{Var}(Y)$ for the whole life annuity-due.",
      "back": "$d = \\dfrac{0.05}{1.05} \\approx 0.047619$, so $d^{2} \\approx 0.00226757$.\nNumerator: ${}^2A_x - A_x^{2} = 0.09 - 0.25^{2} = 0.09 - 0.0625 = 0.0275$.\n$\\operatorname{Var}(Y) = \\dfrac{0.0275}{0.00226757} \\approx 12.1276$.",
      "tag": "Variance"
    },
    {
      "front": "Why does the variance of the annuity PV use ${}^2A_x$ (a double-rate insurance) rather than a 'double-rate annuity'?",
      "back": "Because $Y$ is a **linear** function of the single random variable $Z = v^{K+1}$: $Y=\\dfrac{1-Z}{d}$. Variance ignores additive constants and scales by the square of the multiplier, so $\\operatorname{Var}(Y)=\\dfrac{\\operatorname{Var}(Z)}{d^{2}}$.\nAll the randomness lives in $Z$, whose variance is ${}^2A_x - A_x^{2}$. There is no separate 'second-moment annuity' — the annuity's spread is entirely inherited from the insurance.",
      "tag": "Variance"
    },
    {
      "front": "State the variance of the present value of a continuous whole life annuity $\\bar a_x$.",
      "back": "With $\\bar Y = \\dfrac{1 - \\bar Z}{\\delta}$ and $\\bar Z = v^{T} = e^{-\\delta T}$,\n$\\operatorname{Var}(\\bar Y) = \\dfrac{{}^2\\bar A_x - (\\bar A_x)^{2}}{\\delta^{2}}$,\nwhere ${}^2\\bar A_x$ is the continuous insurance second moment, evaluated at force of interest $2\\delta$.",
      "tag": "Variance"
    },
    {
      "front": "Given $\\bar A_x = 0.35$, ${}^2\\bar A_x = 0.16$, and $\\delta = 0.06$, find $\\operatorname{Var}(\\bar Y)$ and the standard deviation.",
      "back": "Numerator: ${}^2\\bar A_x - \\bar A_x^{2} = 0.16 - 0.35^{2} = 0.16 - 0.1225 = 0.0375$.\n$\\delta^{2} = 0.06^{2} = 0.0036$.\n$\\operatorname{Var}(\\bar Y) = \\dfrac{0.0375}{0.0036} \\approx 10.4167$.\nStandard deviation $= \\sqrt{10.4167} \\approx 3.2275$.",
      "tag": "Variance"
    },
    {
      "front": "For an $n$-year temporary annuity-due, what insurance moments give $\\operatorname{Var}(Y)$?",
      "back": "$Y = \\dfrac{1 - Z}{d}$ where $Z$ is now the $n$-year **endowment** insurance PV with first moment $A_{x:\\overline{n|}}$ and second moment ${}^2A_{x:\\overline{n|}}$.\nThus $\\operatorname{Var}(Y) = \\dfrac{{}^2A_{x:\\overline{n|}} - \\left(A_{x:\\overline{n|}}\\right)^{2}}{d^{2}}$.\nThe endowment (not the term insurance) appears because the survival-to-$n$ payment is part of $Z$.",
      "tag": "Variance"
    },
    {
      "front": "Given $A_{x:\\overline{20|}} = 0.50$, ${}^2A_{x:\\overline{20|}} = 0.27$, and $i = 0.05$, find $\\operatorname{Var}(Y)$ for the 20-year temporary annuity-due.",
      "back": "$d = \\dfrac{0.05}{1.05}\\approx 0.047619$, $d^{2}\\approx 0.00226757$.\nNumerator: $0.27 - 0.50^{2} = 0.27 - 0.25 = 0.02$.\n$\\operatorname{Var}(Y) = \\dfrac{0.02}{0.00226757} \\approx 8.8200$.",
      "tag": "Variance"
    },
    {
      "front": "Build $\\ddot a_x$ from scratch over 3 years of survival then add a known tail. Use $i=0.05$, $q_x=0.04$, $q_{x+1}=0.05$, $q_{x+2}=0.06$, and $\\ddot a_{x+3}=12.00$.",
      "back": "Survival probabilities: $p_x=0.96$, $p_{x+1}=0.95$, $p_{x+2}=0.94$.\n${}_1p_x=0.96$, ${}_2p_x=0.96(0.95)=0.912$, ${}_3p_x=0.912(0.94)=0.85728$.\nTemporary piece: $\\ddot a_{x:\\overline{3|}} = 1 + v(0.96) + v^{2}(0.912)$ with $v=0.952381$, $v^{2}=0.907029$.\n$= 1 + 0.914286 + 0.827210 = 2.741496$.\nDeferred tail: ${}_3E_x\\,\\ddot a_{x+3} = v^{3}(0.85728)(12.00)$; $v^{3}=0.863838$, so ${}_3E_x=0.740551$, giving $0.740551(12.00)=8.886608$.\n$\\ddot a_x = 2.741497 + 8.886608 \\approx 11.6281$.",
      "tag": "Whole life annuity"
    },
    {
      "front": "Define the $m$-thly whole life annuity-due $\\ddot a_x^{(m)}$ and state what each payment is.",
      "back": "$\\ddot a_x^{(m)}$ pays $\\dfrac{1}{m}$ at the start of each $\\dfrac{1}{m}$-year period (total \\$1 per year), while $(x)$ survives:\n$\\ddot a_x^{(m)} = \\dfrac{1}{m}\\sum_{k=0}^{\\infty} v^{k/m}\\,{}_{k/m}p_x$.\nMore frequent payments **lower** the value slightly versus the annual annuity-due: the annual-due pays the full \\$1 at the start of each year, whereas the $m$-thly version delays most of the year's payment. By Woolhouse, $\\ddot a_x^{(m)}\\approx\\ddot a_x-\\frac{m-1}{2m}$, decreasing toward the continuous $\\bar a_x$ as $m\\to\\infty$.",
      "tag": "Continuous/mthly & Woolhouse"
    },
    {
      "front": "State the **Woolhouse** (3-term) approximation for the $m$-thly annuity-due $\\ddot a_x^{(m)}$.",
      "back": "$\\ddot a_x^{(m)} \\approx \\ddot a_x - \\dfrac{m-1}{2m} - \\dfrac{m^{2}-1}{12m^{2}}\\bigl(\\delta + \\mu_x\\bigr)$,\nwhere $\\delta=\\ln(1+i)$ is the force of interest and $\\mu_x$ is the force of mortality at age $x$.\nDropping the last term gives the 2-term Woolhouse; the full 3-term version is the standard FAM approximation.",
      "tag": "Continuous/mthly & Woolhouse"
    },
    {
      "front": "Apply 3-term Woolhouse to find $\\ddot a_x^{(12)}$ given $\\ddot a_x = 14.00$, $i = 0.05$, $\\mu_x = 0.02$.",
      "back": "$\\delta = \\ln 1.05 \\approx 0.048790$, $m=12$.\nTerm 2: $\\dfrac{m-1}{2m} = \\dfrac{11}{24} \\approx 0.458333$.\nTerm 3: $\\dfrac{m^{2}-1}{12m^{2}} = \\dfrac{143}{1728} \\approx 0.082755$; times $(\\delta+\\mu_x)=0.068790$ gives $0.082755(0.068790)\\approx 0.005693$.\n$\\ddot a_x^{(12)} \\approx 14.00 - 0.458333 - 0.005693 \\approx 13.5360$.",
      "tag": "Continuous/mthly & Woolhouse"
    },
    {
      "front": "State the Woolhouse approximation for the **continuous** annuity $\\bar a_x$ in terms of $\\ddot a_x$.",
      "back": "Let $m\\to\\infty$ in Woolhouse: $\\dfrac{m-1}{2m}\\to\\dfrac12$ and $\\dfrac{m^{2}-1}{12m^{2}}\\to\\dfrac{1}{12}$, giving\n$\\bar a_x \\approx \\ddot a_x - \\dfrac12 - \\dfrac{1}{12}\\bigl(\\delta + \\mu_x\\bigr)$.\nThe $-\\tfrac12$ is the continuous-vs-due timing correction; the $\\tfrac{1}{12}(\\delta+\\mu_x)$ term refines it.",
      "tag": "Continuous/mthly & Woolhouse"
    },
    {
      "front": "Apply Woolhouse to find $\\bar a_x$ given $\\ddot a_x = 14.00$, $i = 0.05$, $\\mu_x = 0.02$.",
      "back": "$\\delta = \\ln 1.05 \\approx 0.048790$, so $\\delta + \\mu_x = 0.068790$.\n$\\bar a_x \\approx \\ddot a_x - \\dfrac12 - \\dfrac{1}{12}(\\delta+\\mu_x)$\n$= 14.00 - 0.5 - \\dfrac{0.068790}{12} = 14.00 - 0.5 - 0.005733 \\approx 13.4943$.",
      "tag": "Continuous/mthly & Woolhouse"
    },
    {
      "front": "Use the 2-term Woolhouse approximation to find $\\ddot a_x^{(4)}$ given $\\ddot a_x = 13.50$.",
      "back": "2-term Woolhouse drops the $(\\delta+\\mu_x)$ term:\n$\\ddot a_x^{(m)} \\approx \\ddot a_x - \\dfrac{m-1}{2m}$.\nFor $m=4$: $\\dfrac{m-1}{2m} = \\dfrac{3}{8} = 0.375$.\n$\\ddot a_x^{(4)} \\approx 13.50 - 0.375 = 13.1250$.",
      "tag": "Continuous/mthly & Woolhouse"
    },
    {
      "front": "Relate the $m$-thly annuity-due $\\ddot a_x^{(m)}$ to the $m$-thly annuity-immediate $a_x^{(m)}$.",
      "back": "$a_x^{(m)} = \\ddot a_x^{(m)} - \\dfrac{1}{m}$.\nThe immediate version pays at the **end** of each $\\tfrac1m$-year period, so it omits the single time-$0$ payment of $\\tfrac1m$ that the due version makes. (Compare the annual case $a_x = \\ddot a_x - 1$.)",
      "tag": "Due vs immediate"
    },
    {
      "front": "Given $\\ddot a_x^{(12)} = 13.536$, find the monthly annuity-immediate $a_x^{(12)}$.",
      "back": "$a_x^{(12)} = \\ddot a_x^{(12)} - \\dfrac{1}{m} = 13.536 - \\dfrac{1}{12} = 13.536 - 0.083333 \\approx 13.4527$.",
      "tag": "Due vs immediate"
    },
    {
      "front": "Define the increasing whole life annuity-due $(I\\ddot a)_x$ and give its EPV as a sum.",
      "back": "$(I\\ddot a)_x$ pays $1$ at time $0$, $2$ at time $1$, $3$ at time $2$, and generally $k+1$ at time $k$, while $(x)$ survives:\n$(I\\ddot a)_x = \\sum_{k=0}^{\\infty} (k+1)\\,v^{k}\\,{}_kp_x$.\nIt is the EPV of a life annuity-due whose annual payment increases by \\$1 each year.",
      "tag": "Whole life annuity"
    },
    {
      "front": "Compute $(I\\ddot a)_{x:\\overline{3|}}$, the 3-year increasing temporary annuity-due, given $i=0.05$, $p_x=0.97$, $p_{x+1}=0.96$.",
      "back": "Payments $1,2,3$ at times $0,1,2$ while alive. ${}_0p_x=1$, ${}_1p_x=0.97$, ${}_2p_x=0.97(0.96)=0.9312$; $v=0.952381$, $v^{2}=0.907029$.\n$(I\\ddot a)_{x:\\overline{3|}} = 1(1)(1) + 2(0.952381)(0.97) + 3(0.907029)(0.9312)$\n$= 1 + 1.847619 + 2.533876 \\approx 5.38150$.",
      "tag": "Whole life annuity"
    },
    {
      "front": "Given $\\ddot a_x = 16.00$ and $i = 0.04$, what level annual benefit does a \\$100{,}000 single premium buy as a whole life annuity-due?",
      "back": "An annuity-due paying $B$ per year has EPV $B\\,\\ddot a_x$. Set equal to the premium:\n$B\\,\\ddot a_x = 100{,}000 \\Rightarrow B = \\dfrac{100{,}000}{\\ddot a_x} = \\dfrac{100{,}000}{16.00} = \\$6{,}250.00$ per year.",
      "tag": "Whole life annuity"
    },
    {
      "front": "A whole life insurance on $(x)$ pays \\$1 at end of year of death. Given $\\ddot a_x = 13.80$ and $i = 0.05$, find the level annual net premium for a fully discrete policy.",
      "back": "By equivalence principle, premium $P$ solves $P\\,\\ddot a_x = A_x$, so $P = \\dfrac{A_x}{\\ddot a_x}$.\nGet $A_x$ from the relation: $d=\\dfrac{0.05}{1.05}\\approx0.047619$, $A_x = 1 - d\\,\\ddot a_x = 1 - 0.047619(13.80) = 1 - 0.657143 = 0.342857$.\n$P = \\dfrac{0.342857}{13.80} \\approx 0.024845$ per \\$1, i.e. \\$24.85 per \\$1{,}000 of insurance.",
      "tag": "Insurance-annuity relations"
    },
    {
      "front": "Compute the EPV of a 2-year temporary annuity-immediate $a_{x:\\overline{2|}}$ paying \\$1 at the end of years 1 and 2, given $i = 0.06$, $p_x = 0.98$, $p_{x+1} = 0.97$.",
      "back": "$a_{x:\\overline{2|}} = v\\,{}_1p_x + v^{2}\\,{}_2p_x$ with ${}_1p_x=0.98$, ${}_2p_x=0.98(0.97)=0.9506$.\n$v=\\dfrac{1}{1.06}\\approx0.943396$, $v^{2}\\approx0.889996$.\n$= 0.943396(0.98) + 0.889996(0.9506)$\n$= 0.924528 + 0.846030 \\approx 1.77056$.",
      "tag": "Due vs immediate"
    },
    {
      "front": "Given $A_x = 0.20$, ${}^2A_x = 0.07$, and $i = 0.06$, find the standard deviation of the loss-free whole life annuity-due PV $Y$.",
      "back": "$d = \\dfrac{0.06}{1.06}\\approx 0.056604$, $d^{2}\\approx 0.00320399$.\nNumerator: ${}^2A_x - A_x^{2} = 0.07 - 0.20^{2} = 0.07 - 0.04 = 0.03$.\n$\\operatorname{Var}(Y) = \\dfrac{0.03}{0.00320399}\\approx 9.3633$.\nStandard deviation $= \\sqrt{9.3633}\\approx 3.0600$.",
      "tag": "Variance"
    },
    {
      "front": "A 65-year-old buys a 10-year certain-and-life annuity-due of \\$2{,}000/yr. Given $\\ddot a_{\\overline{10|}} = 8.1078$ and ${}_{10|}\\ddot a_{65} = 5.9488$, find the single premium.",
      "back": "The EPV per \\$1 of annual benefit is $\\ddot a_{\\overline{10|}} + {}_{10|}\\ddot a_{65} = 8.1078 + 5.9488 = 14.0566$.\nSingle premium $= 2{,}000 \\times 14.0566 = \\$28{,}113.20$.\nThe first 10 payments are guaranteed; payments from age 75 on continue only while the annuitant survives.",
      "tag": "Temporary & deferred"
    },
    {
      "front": "Why is a whole life annuity-due always worth more than the corresponding annuity-immediate, and by how much for the temporary case?",
      "back": "The due annuity pays each installment one period earlier, so each payment is discounted less — strictly more valuable.\nFor the $n$-year temporary case the gap is the time-$0$ payment minus the (missing) time-$n$ payment:\n$\\ddot a_{x:\\overline{n|}} - a_{x:\\overline{n|}} = 1 - {}_nE_x = 1 - v^{n}\\,{}_np_x$.\n(For whole life, ${}_nE_x\\to0$ and the gap is exactly $1$.)",
      "tag": "Due vs immediate"
    }
  ]
}