{
  "deckName": "Exam FAM — Premiums & Reserves",
  "examCode": "Exam FAM",
  "cards": [
    {
      "front": "State the **equivalence principle** and what it produces.",
      "back": "The equivalence principle sets the expected loss at issue to zero: $E[L_0]=0$, i.e. $\\text{APV future benefits}=\\text{APV future premiums}$.\nSolving this equation for the level premium gives the **net (benefit) premium**, the premium that exactly funds the benefits in expectation with no margin for expenses or profit. All the $P$ formulas below are just $E[L_0]=0$ rearranged.",
      "tag": "Equivalence principle"
    },
    {
      "front": "Define the **loss-at-issue random variable** $L_0$ for a fully discrete whole life insurance of $1$ with level annual premium $P$.",
      "back": "$L_0 = v^{K+1} - P\\,\\ddot a_{\\overline{K+1|}}$, where $K=K_x$ is the curtate future lifetime, so the benefit $1$ is paid at the end of the year of death ($K+1$) and a premium is paid at the start of each of the $K+1$ years lived.\n$L_0$ is the present value, at issue, of benefits paid out minus premiums received. The equivalence principle chooses $P$ so that $E[L_0]=0$.",
      "tag": "Loss at issue & variance"
    },
    {
      "front": "Give the **net annual premium** for a fully discrete whole life insurance of $1$ on $(x)$.",
      "back": "$P_x=\\dfrac{A_x}{\\ddot a_x}$.\nThis is $E[L_0]=0$ rearranged: APV of benefits $A_x$ equals APV of premiums $P_x\\,\\ddot a_x$.\nUsing the identity $A_x = 1 - d\\,\\ddot a_x$, an equivalent form is $P_x=\\dfrac{1}{\\ddot a_x}-d$.",
      "tag": "Net premiums"
    },
    {
      "front": "A fully discrete whole life policy of $1$ on $(x)$ has $A_x=0.24441$ and $\\ddot a_x=15.8166$. Find the net annual premium $P_x$.",
      "back": "By the equivalence principle, $P_x=\\dfrac{A_x}{\\ddot a_x}=\\dfrac{0.24441}{15.8166}$.\n$P_x\\approx 0.0154528$.\nFor a face amount of $\\$100{,}000$ the annual net premium is $100{,}000\\times 0.0154528\\approx \\$1{,}545.28$.",
      "tag": "Net premiums"
    },
    {
      "front": "Give the net annual premium formulas for fully discrete **$n$-year term** and **$n$-year endowment** insurances of $1$.",
      "back": "Both pay premiums for $n$ years (an $n$-year temporary annuity-due):\nTerm: $P^{1}_{x:\\overline{n|}}=\\dfrac{A^{1}_{x:\\overline{n|}}}{\\ddot a_{x:\\overline{n|}}}$.\nEndowment: $P_{x:\\overline{n|}}=\\dfrac{A_{x:\\overline{n|}}}{\\ddot a_{x:\\overline{n|}}}$, where $A_{x:\\overline{n|}}=A^{1}_{x:\\overline{n|}}+{}_nE_x$ includes the pure endowment.\nIn each case the numerator is the APV of that contract's benefits and the denominator is the APV of $\\$1$ of premium.",
      "tag": "Net premiums"
    },
    {
      "front": "A fully discrete $20$-year endowment insurance of $\\$1{,}000$ has $A_{x:\\overline{20|}}=0.72270$ and $\\ddot a_{x:\\overline{20|}}=5.8100$. Find the net annual premium.",
      "back": "$P_{x:\\overline{20|}}=\\dfrac{A_{x:\\overline{20|}}}{\\ddot a_{x:\\overline{20|}}}=\\dfrac{0.72270}{5.8100}\\approx 0.1243890$ per $\\$1$ of benefit.\nFor a $\\$1{,}000$ benefit: $1{,}000\\times 0.1243890\\approx \\$124.39$ per year, payable for at most $20$ years.",
      "tag": "Net premiums"
    },
    {
      "front": "What changes for a **limited-payment (\\,$n$-pay\\,) whole life** net premium, and give its formula.",
      "back": "Benefits are unchanged (whole life, $A_x$), but premiums are paid for only $n$ years, so the premium annuity is the $n$-year temporary annuity-due:\n${}_nP_x=\\dfrac{A_x}{\\ddot a_{x:\\overline{n|}}}$.\nSince $\\ddot a_{x:\\overline{n|}}<\\ddot a_x$, the limited-pay premium exceeds the ordinary $P_x$ — you fund the same benefit with fewer payments.",
      "tag": "Net premiums"
    },
    {
      "front": "A $\\$1{,}000$ whole life policy is paid up in $10$ years. Given $A_x=0.30$ and $\\ddot a_{x:\\overline{10|}}=7.5$, find the net annual premium.",
      "back": "${}_{10}P_x=\\dfrac{A_x}{\\ddot a_{x:\\overline{10|}}}=\\dfrac{0.30}{7.5}=0.04$ per $\\$1$.\nFor $\\$1{,}000$: $1{,}000\\times 0.04=\\$40.00$ per year for $10$ years.\nThis is larger than the lifetime-pay premium $P_x=1000(0.30)/\\ddot a_x$ would be, since the same $A_x$ is funded over fewer payments.",
      "tag": "Net premiums"
    },
    {
      "front": "Give the **fully continuous** net premium for a whole life insurance of $1$, and its two equivalent forms.",
      "back": "Premiums are paid continuously at annual rate $\\bar P(\\bar A_x)$, the benefit $1$ is paid at the moment of death:\n$\\bar P(\\bar A_x)=\\dfrac{\\bar A_x}{\\bar a_x}=\\dfrac{\\delta\\,\\bar A_x}{1-\\bar A_x}$.\nThe second form uses $\\bar a_x=\\dfrac{1-\\bar A_x}{\\delta}$. Equivalently $\\bar P(\\bar A_x)=\\dfrac{1}{\\bar a_x}-\\delta$.",
      "tag": "Net premiums"
    },
    {
      "front": "For a fully continuous whole life insurance of $\\$1{,}000$, $\\bar A_x=0.40$ and the force of interest is $\\delta=0.05$. Find the continuous net premium rate.",
      "back": "First $\\bar a_x=\\dfrac{1-\\bar A_x}{\\delta}=\\dfrac{1-0.40}{0.05}=12.0$.\n$\\bar P(\\bar A_x)=\\dfrac{\\bar A_x}{\\bar a_x}=\\dfrac{0.40}{12.0}\\approx 0.0333333$ per $\\$1$.\nCheck via $\\dfrac{\\delta\\bar A_x}{1-\\bar A_x}=\\dfrac{0.05(0.40)}{0.60}=0.0333333$. For $\\$1{,}000$: about $\\$33.33$ per year, paid continuously.",
      "tag": "Net premiums"
    },
    {
      "front": "What is a **semicontinuous** premium, and how is its net premium computed?",
      "back": "**Semicontinuous** means the death benefit is paid at the *moment* of death (continuous insurance, $\\bar A_x$) but premiums are paid *annually* in advance (discrete annuity-due, $\\ddot a_x$).\nNet premium: $P(\\bar A_x)=\\dfrac{\\bar A_x}{\\ddot a_x}$.\nIt mixes a continuous numerator with a discrete denominator — useful when claims settle on death but billing is annual.",
      "tag": "Net premiums"
    },
    {
      "front": "A semicontinuous whole life of $\\$1{,}000$ has $\\bar A_x=0.35$ and $\\ddot a_x=14.0$. Find the net annual premium.",
      "back": "$P(\\bar A_x)=\\dfrac{\\bar A_x}{\\ddot a_x}=\\dfrac{0.35}{14.0}=0.025$ per $\\$1$.\nFor $\\$1{,}000$: $1{,}000\\times 0.025=\\$25.00$ per year.\nThe benefit is paid at the moment of death (hence $\\bar A_x$), but the premium is billed annually (hence the discrete $\\ddot a_x$).",
      "tag": "Net premiums"
    },
    {
      "front": "Under a constant force of interest, how do you approximate $\\bar A_x$ from a discrete $A_x$, and why?",
      "back": "Under UDD (or the standard claims-acceleration adjustment), $\\bar A_x\\approx \\dfrac{i}{\\delta}\\,A_x$.\nMoving the benefit from the end of the year of death to the moment of death advances it on average about half a year, so it is discounted less — multiplying by $\\dfrac{i}{\\delta}>1$ inflates the EPV.\nExample: $A_x=0.30$, $i=0.06$ so $\\delta=\\ln 1.06\\approx 0.0582689$, giving $\\bar A_x\\approx \\dfrac{0.06}{0.0582689}(0.30)\\approx 0.30891$.",
      "tag": "Net premiums"
    },
    {
      "front": "A fully discrete whole life of $1$ is funded by net premium $P$ with $i=0.05$. A policyholder dies in policy year $5$ (so $K=4$, benefit paid at time $5$), and $P=0.02$. Compute the realized loss $L_0$.",
      "back": "$v=\\dfrac{1}{1.05}$, $d=\\dfrac{0.05}{1.05}\\approx 0.0476190$. Benefit PV: $v^{5}=1.05^{-5}\\approx 0.783526$.\nPremium annuity over $5$ years: $\\ddot a_{\\overline{5|}}=\\dfrac{1-v^{5}}{d}=\\dfrac{1-0.783526}{0.0476190}\\approx 4.545951$.\n$L_0=v^{5}-P\\,\\ddot a_{\\overline{5|}}=0.783526-0.02(4.545951)\\approx 0.783526-0.090919=0.692607$.\nA positive $L_0$: an early death is a loss to the insurer.",
      "tag": "Loss at issue & variance"
    },
    {
      "front": "Why is $L_0>0$ for early deaths and $L_0<0$ for late deaths on a whole life policy?",
      "back": "$L_0=v^{K+1}-P\\,\\ddot a_{\\overline{K+1|}}$ is a **decreasing** function of $K$: an early death pays the benefit almost immediately (large $v^{K+1}$) after collecting few premiums (small annuity) → loss.\nA late death discounts the benefit heavily and collects many premiums → gain.\nThe equivalence principle balances these so the probability-weighted average $E[L_0]=0$; individual outcomes are rarely zero.",
      "tag": "Loss at issue & variance"
    },
    {
      "front": "State the **variance of the loss at issue** $Var(L_0)$ for a fully discrete whole life insurance of $1$ funded by the net premium $P$.",
      "back": "$Var(L_0)=\\left({}^2A_x - A_x^2\\right)\\left(1+\\dfrac{P}{d}\\right)^2$.\nEquivalently, since $1+\\dfrac{P}{d}=\\dfrac{1}{d\\,\\ddot a_x}$, $Var(L_0)=\\dfrac{{}^2A_x-A_x^2}{\\left(d\\,\\ddot a_x\\right)^2}$.\nHere ${}^2A_x$ is the EPV of the benefit computed at *double* the force of interest. The factor $\\left(1+\\tfrac{P}{d}\\right)^2$ rescales the variance of $v^{K+1}$.",
      "tag": "Loss at issue & variance"
    },
    {
      "front": "A fully discrete whole life of $1$ has $A_x=0.30$, ${}^2A_x=0.12$, and $i=0.05$. Find the net premium $P_x$ and $Var(L_0)$.",
      "back": "$d=\\dfrac{0.05}{1.05}\\approx 0.0476190$; $\\ddot a_x=\\dfrac{1-A_x}{d}=\\dfrac{0.70}{0.0476190}=14.7$; so $P_x=\\dfrac{A_x}{\\ddot a_x}=\\dfrac{0.30}{14.7}\\approx 0.0204082$.\n$\\left(1+\\dfrac{P}{d}\\right)=1+\\dfrac{0.0204082}{0.0476190}\\approx 1.428571$.\n$Var(L_0)=(0.12-0.30^2)(1.428571)^2=(0.03)(2.040816)\\approx 0.0612245$. The standard deviation is $\\sqrt{0.0612245}\\approx 0.247436$.",
      "tag": "Loss at issue & variance"
    },
    {
      "front": "For a benefit of $b$ (not $1$) funded at the net premium, how does $Var(L_0)$ scale, and what is the standard deviation for $b=\\$1{,}000$ in the previous example ($A_x=0.30$, ${}^2A_x=0.12$, $i=0.05$)?",
      "back": "Loss scales linearly with $b$, so variance scales by $b^2$: $Var(L_0)=b^2\\left({}^2A_x-A_x^2\\right)\\left(1+\\tfrac{P}{d}\\right)^2$.\nFrom before the per-unit variance is $0.0612245$.\n$Var=1000^2(0.0612245)=61{,}224.49$, so the standard deviation is $1000\\times 0.247436\\approx \\$247.44$.\nThe per-policy loss has a large spread relative to its zero mean — the rationale for pooling many lives.",
      "tag": "Loss at issue & variance"
    },
    {
      "front": "State the **variance of $L_0$** for a fully continuous whole life insurance of $1$.",
      "back": "$Var(L_0)=\\left({}^2\\bar A_x-\\bar A_x^2\\right)\\left(1+\\dfrac{\\bar P}{\\delta}\\right)^2=\\dfrac{{}^2\\bar A_x-\\bar A_x^2}{\\left(\\delta\\,\\bar a_x\\right)^2}$,\nwhere ${}^2\\bar A_x$ uses force of interest $2\\delta$ and $\\bar P=\\bar P(\\bar A_x)$.\nIt is the continuous analogue of the discrete formula, with $\\delta$ replacing $d$ and bars on the insurance EPVs.",
      "tag": "Loss at issue & variance"
    },
    {
      "front": "A fully continuous whole life of $1$ has $\\bar A_x=0.40$, ${}^2\\bar A_x=0.20$, and $\\delta=0.05$. Find $\\bar P(\\bar A_x)$ and $Var(L_0)$.",
      "back": "$\\bar a_x=\\dfrac{1-0.40}{0.05}=12.0$, so $\\bar P=\\dfrac{0.40}{12.0}\\approx 0.0333333$ and $1+\\dfrac{\\bar P}{\\delta}=1+\\dfrac{0.0333333}{0.05}=1.666667$.\n$Var(L_0)=(0.20-0.40^2)(1.666667)^2=(0.04)(2.777778)\\approx 0.111111$.\nStandard deviation $=\\sqrt{0.111111}\\approx 0.333333$.",
      "tag": "Loss at issue & variance"
    },
    {
      "front": "Set up the net annual premium for a **$3$-year term** insurance of $\\$1{,}000$ (death benefit at end of year) from explicit mortality. Given $q_x=0.02$, $q_{x+1}=0.025$, $q_{x+2}=0.03$, $i=0.06$.",
      "back": "Survival: ${}_0p_x=1$, ${}_1p_x=0.98$, ${}_2p_x=0.98(0.975)=0.9555$. With $v=1/1.06$:\nAPV benefits $=1000\\sum_{k=0}^{2}v^{k+1}\\,{}_kp_x\\,q_{x+k}$\n$=1000[v(0.02)+v^2(0.98)(0.025)+v^3(0.9555)(0.03)]\\approx 1000[0.018868+0.021805+0.024068]=64.7405$.\nAPV premiums $=\\sum_{k=0}^{2}v^k\\,{}_kp_x=1+v(0.98)+v^2(0.9555)\\approx 2.774920$.\n$P=\\dfrac{64.7405}{2.774920}\\approx \\$23.33$.",
      "tag": "Equivalence principle"
    },
    {
      "front": "Find the net annual premium for a **$3$-year endowment** insurance of $\\$10{,}000$ (death benefit end of year, or $\\$10{,}000$ on survival to $3$). Given $q_x=0.04$, $q_{x+1}=0.05$, $q_{x+2}=0.06$, $i=0.06$.",
      "back": "Survival: ${}_1p_x=0.96$, ${}_2p_x=0.96(0.95)=0.912$, ${}_3p_x=0.912(0.94)=0.85728$. $v=1/1.06$.\nAPV death $=10000[v(0.04)+v^2(0.96)(0.05)+v^3(0.912)(0.06)]\\approx 1263.996$.\nAPV pure endowment $=10000\\,v^3\\,(0.85728)\\approx 7197.888$. Total benefit APV $\\approx 8461.885$.\nAPV premiums $=1+v(0.96)+v^2(0.912)\\approx 2.717337$.\n$P=\\dfrac{8461.885}{2.717337}\\approx \\$3{,}114.04$.",
      "tag": "Equivalence principle"
    },
    {
      "front": "Define the **net premium reserve** ${}_tV$ and give its **prospective** formula.",
      "back": "${}_tV$ is the expected value at time $t$ (for an in-force policy) of future losses: ${}_tV=E[L_t]$ under the equivalence-principle net premium.\nProspective: ${}_tV=\\text{APV future benefits at }x+t-\\text{APV future net premiums at }x+t$.\nIt is the fund the insurer must hold so that, together with future premiums, future benefits are covered in expectation.",
      "tag": "Net premium reserves"
    },
    {
      "front": "Give the prospective net premium reserve for a fully discrete **whole life** insurance of $1$, and the annuity-ratio shortcut.",
      "back": "${}_tV=A_{x+t}-P_x\\,\\ddot a_{x+t}$.\nUsing $A_{x+t}=1-d\\,\\ddot a_{x+t}$ and $P_x=\\dfrac{1}{\\ddot a_x}-d$, this collapses to\n${}_tV=1-\\dfrac{\\ddot a_{x+t}}{\\ddot a_x}$.\nReserves grow from ${}_0V=0$ toward $1$ as $\\ddot a_{x+t}$ shrinks with age.",
      "tag": "Net premium reserves"
    },
    {
      "front": "A fully discrete whole life of $1$ has $\\ddot a_x=15$. At duration $t$, $\\ddot a_{x+t}=12$. Find ${}_tV$ using the annuity-ratio identity.",
      "back": "${}_tV=1-\\dfrac{\\ddot a_{x+t}}{\\ddot a_x}=1-\\dfrac{12}{15}=1-0.8=0.20$.\nSo the reserve per $\\$1$ of benefit is $0.20$; for a $\\$100{,}000$ policy the held reserve is $\\$20{,}000$.\nThis identity needs only the two annuity-due values — no separate insurance EPV is required.",
      "tag": "Net premium reserves"
    },
    {
      "front": "Compute the prospective whole life reserve directly. Net premium uses $A_x=0.25$, $\\ddot a_x=15$; at time $t$, $A_{x+t}=0.35$ and $\\ddot a_{x+t}=13$.",
      "back": "First the net premium: $P_x=\\dfrac{A_x}{\\ddot a_x}=\\dfrac{0.25}{15}\\approx 0.0166667$.\nProspective reserve: ${}_tV=A_{x+t}-P_x\\,\\ddot a_{x+t}=0.35-0.0166667(13)$\n$=0.35-0.216667=0.133333$.\nPer $\\$1$ of benefit the reserve is about $0.133333$ (e.g. $\\$13{,}333$ on a $\\$100{,}000$ policy).",
      "tag": "Net premium reserves"
    },
    {
      "front": "State the **retrospective** reserve formula and when it equals the prospective reserve.",
      "back": "Retrospective: ${}_tV=\\dfrac{1}{{}_tp_x}\\Big[P\\,\\ddot a_{x:\\overline{t|}}-A^{1}_{x:\\overline{t|}}\\Big](1+i)^{t}$ — premiums collected minus benefits paid over $[0,t]$, accumulated with interest and survivorship.\nIt **equals the prospective reserve whenever the premium was set by the equivalence principle on the same basis** (same mortality and interest). The prospective view looks forward; the retrospective view looks back; the equivalence principle guarantees they agree.",
      "tag": "Net premium reserves"
    },
    {
      "front": "Give the fully **continuous** whole life reserve and its annuity-ratio form.",
      "back": "${}_t\\bar V(\\bar A_x)=\\bar A_{x+t}-\\bar P(\\bar A_x)\\,\\bar a_{x+t}=1-\\dfrac{\\bar a_{x+t}}{\\bar a_x}$.\nIt is the continuous analogue of ${}_tV=1-\\dfrac{\\ddot a_{x+t}}{\\ddot a_x}$, using continuous annuities $\\bar a$.\nExample: $\\bar a_x=12$, $\\bar a_{x+t}=9$ gives ${}_t\\bar V=1-\\dfrac{9}{12}=0.25$.",
      "tag": "Net premium reserves"
    },
    {
      "front": "Compute the reserve for the $3$-year endowment of $\\$10{,}000$ (premium $\\$3{,}114.04$; $q_{x+1}=0.05$, $q_{x+2}=0.06$, $i=0.06$) at the end of year $1$, **prospectively**.",
      "back": "At time $1$ a $2$-year endowment remains. Survival from $x+1$: ${}_1p_{x+1}=0.95$. $v=1/1.06$.\nAPV future benefits $=10000\\big[v(0.05)+v^2(0.95)(0.06)+v^2(0.95)(0.94)\\big]$\n$\\approx 10000[0.047170+0.050730+0.794767]=8926.66$.\nAPV future premiums $=3114.04\\big[1+v(0.95)\\big]\\approx 3114.04(1.896226)=5904.92$.\n${}_1V=8926.66-5904.92\\approx \\$3{,}021.74$.",
      "tag": "Net premium reserves"
    },
    {
      "front": "Verify the previous $\\$3{,}021.74$ reserve by the **recursion** from ${}_0V=0$. (Premium $\\$3{,}114.04$, $q_x=0.04$, benefit $\\$10{,}000$, $i=0.06$.)",
      "back": "Recursion: $({}_0V+P)(1+i)=q_x\\,b+p_x\\,{}_1V$.\n$(0+3114.04)(1.06)=0.04(10000)+0.96\\,{}_1V$\n$3300.88=400+0.96\\,{}_1V$\n${}_1V=\\dfrac{3300.88-400}{0.96}=\\dfrac{2900.88}{0.96}\\approx \\$3{,}021.75$.\nMatches the prospective $\\$3{,}021.74$ to the penny — the recursion and prospective methods agree (the cent gap is just rounding of the $\\$3{,}114.04$ premium).",
      "tag": "Reserve recursions"
    },
    {
      "front": "State the general **reserve recursion** for a fully discrete policy and name each piece.",
      "back": "$({}_tV+P)(1+i)=q_{x+t}\\,b_{t+1}+p_{x+t}\\,{}_{t+1}V$.\nLeft side: the reserve in hand plus the premium just received, accumulated one year with interest.\nRight side: it must fund the death benefit $b_{t+1}$ (with probability $q_{x+t}$) plus the reserve needed for survivors $\\,{}_{t+1}V$ (with probability $p_{x+t}$).\nSolve forward for ${}_{t+1}V$ given ${}_tV$, or backward.",
      "tag": "Reserve recursions"
    },
    {
      "front": "Given ${}_tV=0.20$ (per $\\$1$), net premium $P=0.02$, $i=0.06$, $q_{x+t}=0.03$, benefit $b_{t+1}=1$, find ${}_{t+1}V$.",
      "back": "$({}_tV+P)(1+i)=(0.20+0.02)(1.06)=0.22(1.06)=0.2332$.\nThen $0.2332=q_{x+t}b+p_{x+t}\\,{}_{t+1}V=0.03(1)+0.97\\,{}_{t+1}V$.\n${}_{t+1}V=\\dfrac{0.2332-0.03}{0.97}=\\dfrac{0.2032}{0.97}\\approx 0.209485$.\nThe reserve grew from $0.20$ to about $0.209485$ over the year.",
      "tag": "Reserve recursions"
    },
    {
      "front": "Build the first two reserves of a fully discrete whole life of $\\$1{,}000$ with level net premium $P=\\$20$, $i=0.06$, $q_x=0.010$, $q_{x+1}=0.012$. (${}_0V=0$.)",
      "back": "Year 1: $(0+20)(1.06)=21.20=q_x(1000)+p_x\\,{}_1V=10+0.990\\,{}_1V$, so ${}_1V=\\dfrac{21.20-10}{0.990}\\approx \\$11.31$.\nYear 2: $(11.31+20)(1.06)=31.31(1.06)=33.1886=q_{x+1}(1000)+p_{x+1}\\,{}_2V=12+0.988\\,{}_2V$,\nso ${}_2V=\\dfrac{33.1886-12}{0.988}\\approx \\$21.45$.\nThe reserve accumulates each year as premiums plus interest exceed the cost of insurance.",
      "tag": "Reserve recursions"
    },
    {
      "front": "Rewrite the reserve recursion to split the premium into a **savings premium** and a **cost-of-insurance premium**.",
      "back": "Rearranging $({}_tV+P)(1+i)=q_{x+t}b+p_{x+t}\\,{}_{t+1}V$ gives\n$P=\\underbrace{v\\,{}_{t+1}V-{}_tV}_{\\text{savings premium }P^s}+\\underbrace{v\\,q_{x+t}\\,(b-{}_{t+1}V)}_{\\text{cost of insurance }P^i}$.\nThe **savings** part funds the increase in the reserve; the **cost-of-insurance** part pays for the *net amount at risk* $b-{}_{t+1}V$. So $P=P^s+P^i$.",
      "tag": "Reserve recursions"
    },
    {
      "front": "For a whole life of $\\$1{,}000$ with ${}_tV=\\$150$, $P=\\$20$, $i=0.06$, $q_{x+t}=0.02$, find ${}_{t+1}V$, the net amount at risk, the cost of insurance, and the savings premium.",
      "back": "$({}_tV+P)(1+i)=(150+20)(1.06)=180.20=q_{x+t}(1000)+p_{x+t}\\,{}_{t+1}V=20+0.98\\,{}_{t+1}V$, so ${}_{t+1}V=\\dfrac{180.20-20}{0.98}\\approx \\$163.47$.\nNet amount at risk $=b-{}_{t+1}V=1000-163.47=\\$836.53$.\nCost of insurance $P^i=v\\,q_{x+t}(b-{}_{t+1}V)=\\dfrac{1}{1.06}(0.02)(836.53)\\approx \\$15.78$.\nSavings premium $P^s=P-P^i=20-15.78\\approx \\$4.22$ (and indeed $v\\,{}_{t+1}V-{}_tV=\\dfrac{163.47}{1.06}-150\\approx 4.22$).",
      "tag": "Reserve recursions"
    },
    {
      "front": "Verify a reserve **retrospectively** from the recursion: whole life of $\\$1{,}000$, $P=\\$20$, $i=0.06$, $q_x=0.010$. Find ${}_1V$ from premiums-less-benefits accumulated.",
      "back": "Over year 1 the insurer collects $P=20$ at issue and pays $\\$1{,}000$ to the fraction $q_x$ who die. Per survivor, accumulate to year-end:\n${}_1V=\\dfrac{(P-v\\,q_x\\,b)(1+i)}{p_x}=\\dfrac{\\big(20-\\frac{1}{1.06}(0.010)(1000)\\big)(1.06)}{0.990}$\n$=\\dfrac{(20-9.4340)(1.06)}{0.990}=\\dfrac{11.2000}{0.990}\\approx \\$11.31$.\nSame as the forward recursion — retrospective and prospective agree under the equivalence principle.",
      "tag": "Reserve recursions"
    },
    {
      "front": "Distinguish a **net (benefit) premium** from a **gross (expense-loaded) premium**.",
      "back": "The **net premium** is set by $E[L_0]=0$ on benefits *only* — it funds claims with no allowance for expenses, commissions, taxes, or profit.\nThe **gross (expense-loaded) premium** $G$ is set so that APV of premiums covers APV of benefits *plus* APV of expenses (and any profit loading). $G$ is what the policyholder actually pays; the difference $G-P$ is the **expense loading**.",
      "tag": "Gross premiums & expenses"
    },
    {
      "front": "List the standard **expense types** in a gross-premium calculation and how each enters the equivalence equation.",
      "back": "Typical categories:\n- **Per-premium** (e.g. commissions, premium tax): a percent of $G$, charged each premium → enters as $cG\\,\\ddot a$.\n- **Per-policy** (e.g. maintenance/admin, fixed \\$ each year): $\\to e\\,\\ddot a$.\n- **Acquisition / first-year** (\\$ at issue only): a one-time term $\\to$ constant.\n- **Settlement / claim** expense (\\$ paid at death): $\\to s\\,A_x$.\nThe gross-premium equation is $G\\,\\ddot a = \\text{APV benefits}+\\text{APV all expenses}$.",
      "tag": "Gross premiums & expenses"
    },
    {
      "front": "State the **equivalence-principle gross premium** equation for a whole life with level annual expenses and a settlement expense.",
      "back": "$G\\,\\ddot a_x = b\\,A_x + (c\\,G + e)\\,\\ddot a_x + s\\,A_x$,\nwhere $c$ = per-premium fraction, $e$ = per-policy annual expense, $s$ = settlement expense paid at death.\nSolving for $G$: $G=\\dfrac{b\\,A_x+e\\,\\ddot a_x+s\\,A_x}{(1-c)\\,\\ddot a_x}$.\nThe $cG$ term moves to the left to give the $(1-c)$ factor.",
      "tag": "Gross premiums & expenses"
    },
    {
      "front": "Find the **gross premium** for a $\\$100{,}000$ whole life. Net basis $A_x=0.30$, $\\ddot a_x=14.7$. Expenses: per-policy $\\$50$/yr, per-premium $10\\%$ of $G$ each year, settlement $\\$200$ at death.",
      "back": "Equation: $G(14.7)=100000(0.30)+50(14.7)+0.10\\,G(14.7)+200(0.30)$.\n$G(14.7)(1-0.10)=30000+735+60=30795$.\n$G=\\dfrac{30795}{0.90(14.7)}=\\dfrac{30795}{13.23}\\approx \\$2{,}327.66$.\nFor comparison the net premium is $\\dfrac{100000(0.30)}{14.7}\\approx \\$2{,}040.82$, so the expense loading is about $\\$286.84$.",
      "tag": "Gross premiums & expenses"
    },
    {
      "front": "Find the gross premium for a $\\$50{,}000$ whole life with a separate **first-year acquisition** cost. $A_x=0.25$, $\\ddot a_x=15$. Expenses: $\\$500$ acquisition at issue (year 1 only), $8\\%$ of $G$ per premium (all years), $\\$30$ per policy (all years).",
      "back": "Equation: $G(15)=50000(0.25)+500+0.08\\,G(15)+30(15)$.\n$G(15)(1-0.08)=12500+500+450=13450$.\n$G=\\dfrac{13450}{0.92(15)}=\\dfrac{13450}{13.8}\\approx \\$974.64$.\nThe one-time $\\$500$ acquisition cost enters as a flat constant (not multiplied by $\\ddot a$); the net premium here is $\\dfrac{50000(0.25)}{15}\\approx \\$833.33$.",
      "tag": "Gross premiums & expenses"
    },
    {
      "front": "What is the **expense reserve**, and how does it relate to the gross and net premium reserves?",
      "back": "The **gross premium reserve** is APV(future benefits + future expenses) − APV(future gross premiums). The **expense reserve** is its excess over the net premium reserve:\n$\\text{expense reserve}={}_tV^{\\,\\text{gross}}-{}_tV^{\\,\\text{net}}=\\text{APV future expenses}-\\text{APV future expense loadings}$.\nBecause heavy first-year (acquisition) costs are recovered by a level loading spread over all years, the expense reserve is typically **negative** in early durations — the insurer has front-loaded costs not yet recouped.",
      "tag": "Gross premiums & expenses"
    }
  ]
}