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Exam FAM — Premiums & Reserves Flashcards

Net (benefit) and gross premiums for life-contingent contracts on SOA Exam FAM: the equivalence principle $E[L_0]=0$, fully discrete / fully continuous / semicontinuous net premium formulas, the loss-at-issue random variable $L_0$ and its variance, net premium reserves by the prospective, retrospective, and $1-\ddot a_{x+t}/\ddot a_x$ methods, the year-by-year reserve recursion and its split into savings premium and cost of insurance, and expense-loaded gross premiums via the equivalence principle — with fully worked numeric examples built from explicit $A$, $\ddot a$, $q_x$, and $i$ inputs.

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  1. Equivalence principle
    State the **equivalence principle** and what it produces.
    The equivalence principle sets the expected loss at issue to zero: $E[L_0]=0$, i.e. $\text{APV future benefits}=\text{APV future premiums}$. Solving 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.
  2. Loss at issue & variance
    Define the **loss-at-issue random variable** $L_0$ for a fully discrete whole life insurance of $1$ with level annual premium $P$.
    $L_0 = v^{K+1} - P\,\ddot a_{\overline{K+1|}}$, 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. $L_0$ is the present value, at issue, of benefits paid out minus premiums received. The equivalence principle chooses $P$ so that $E[L_0]=0$.
  3. Net premiums
    Give the **net annual premium** for a fully discrete whole life insurance of $1$ on $(x)$.
    $P_x=\dfrac{A_x}{\ddot a_x}$. This is $E[L_0]=0$ rearranged: APV of benefits $A_x$ equals APV of premiums $P_x\,\ddot a_x$. Using the identity $A_x = 1 - d\,\ddot a_x$, an equivalent form is $P_x=\dfrac{1}{\ddot a_x}-d$.
  4. Net premiums
    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 $P_x$.
    By the equivalence principle, $P_x=\dfrac{A_x}{\ddot a_x}=\dfrac{0.24441}{15.8166}$. $P_x\approx 0.0154528$. For a face amount of $\$100{,}000$ the annual net premium is $100{,}000\times 0.0154528\approx \$1{,}545.28$.
  5. Net premiums
    Give the net annual premium formulas for fully discrete **$n$-year term** and **$n$-year endowment** insurances of $1$.
    Both pay premiums for $n$ years (an $n$-year temporary annuity-due): Term: $P^{1}_{x:\overline{n|}}=\dfrac{A^{1}_{x:\overline{n|}}}{\ddot a_{x:\overline{n|}}}$. Endowment: $P_{x:\overline{n|}}=\dfrac{A_{x:\overline{n|}}}{\ddot a_{x:\overline{n|}}}$, where $A_{x:\overline{n|}}=A^{1}_{x:\overline{n|}}+{}_nE_x$ includes the pure endowment. In each case the numerator is the APV of that contract's benefits and the denominator is the APV of $\$1$ of premium.
  6. Net premiums
    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.
    $P_{x:\overline{20|}}=\dfrac{A_{x:\overline{20|}}}{\ddot a_{x:\overline{20|}}}=\dfrac{0.72270}{5.8100}\approx 0.1243890$ per $\$1$ of benefit. For a $\$1{,}000$ benefit: $1{,}000\times 0.1243890\approx \$124.39$ per year, payable for at most $20$ years.
  7. Net premiums
    What changes for a **limited-payment (\,$n$-pay\,) whole life** net premium, and give its formula.
    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: ${}_nP_x=\dfrac{A_x}{\ddot a_{x:\overline{n|}}}$. Since $\ddot a_{x:\overline{n|}}<\ddot a_x$, the limited-pay premium exceeds the ordinary $P_x$ — you fund the same benefit with fewer payments.
  8. Net premiums
    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.
    ${}_{10}P_x=\dfrac{A_x}{\ddot a_{x:\overline{10|}}}=\dfrac{0.30}{7.5}=0.04$ per $\$1$. For $\$1{,}000$: $1{,}000\times 0.04=\$40.00$ per year for $10$ years. This 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.
  9. Net premiums
    Give the **fully continuous** net premium for a whole life insurance of $1$, and its two equivalent forms.
    Premiums are paid continuously at annual rate $\bar P(\bar A_x)$, the benefit $1$ is paid at the moment of death: $\bar P(\bar A_x)=\dfrac{\bar A_x}{\bar a_x}=\dfrac{\delta\,\bar A_x}{1-\bar A_x}$. The second form uses $\bar a_x=\dfrac{1-\bar A_x}{\delta}$. Equivalently $\bar P(\bar A_x)=\dfrac{1}{\bar a_x}-\delta$.
  10. Net premiums
    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.
    First $\bar a_x=\dfrac{1-\bar A_x}{\delta}=\dfrac{1-0.40}{0.05}=12.0$. $\bar P(\bar A_x)=\dfrac{\bar A_x}{\bar a_x}=\dfrac{0.40}{12.0}\approx 0.0333333$ per $\$1$. Check 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.
  11. Net premiums
    What is a **semicontinuous** premium, and how is its net premium computed?
    **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, $\ddot a_x$). Net premium: $P(\bar A_x)=\dfrac{\bar A_x}{\ddot a_x}$. It mixes a continuous numerator with a discrete denominator — useful when claims settle on death but billing is annual.
  12. Net premiums
    A semicontinuous whole life of $\$1{,}000$ has $\bar A_x=0.35$ and $\ddot a_x=14.0$. Find the net annual premium.
    $P(\bar A_x)=\dfrac{\bar A_x}{\ddot a_x}=\dfrac{0.35}{14.0}=0.025$ per $\$1$. For $\$1{,}000$: $1{,}000\times 0.025=\$25.00$ per year. The benefit is paid at the moment of death (hence $\bar A_x$), but the premium is billed annually (hence the discrete $\ddot a_x$).
  13. Net premiums
    Under a constant force of interest, how do you approximate $\bar A_x$ from a discrete $A_x$, and why?
    Under UDD (or the standard claims-acceleration adjustment), $\bar A_x\approx \dfrac{i}{\delta}\,A_x$. Moving 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. Example: $A_x=0.30$, $i=0.06$ so $\delta=\ln 1.06\approx 0.0582689$, giving $\bar A_x\approx \dfrac{0.06}{0.0582689}(0.30)\approx 0.30891$.
  14. Loss at issue & variance
    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 $L_0$.
    $v=\dfrac{1}{1.05}$, $d=\dfrac{0.05}{1.05}\approx 0.0476190$. Benefit PV: $v^{5}=1.05^{-5}\approx 0.783526$. Premium annuity over $5$ years: $\ddot a_{\overline{5|}}=\dfrac{1-v^{5}}{d}=\dfrac{1-0.783526}{0.0476190}\approx 4.545951$. $L_0=v^{5}-P\,\ddot a_{\overline{5|}}=0.783526-0.02(4.545951)\approx 0.783526-0.090919=0.692607$. A positive $L_0$: an early death is a loss to the insurer.
  15. Loss at issue & variance
    Why is $L_0>0$ for early deaths and $L_0<0$ for late deaths on a whole life policy?
    $L_0=v^{K+1}-P\,\ddot a_{\overline{K+1|}}$ 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. A late death discounts the benefit heavily and collects many premiums → gain. The equivalence principle balances these so the probability-weighted average $E[L_0]=0$; individual outcomes are rarely zero.
  16. Loss at issue & variance
    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 $P$.
    $Var(L_0)=\left({}^2A_x - A_x^2\right)\left(1+\dfrac{P}{d}\right)^2$. Equivalently, since $1+\dfrac{P}{d}=\dfrac{1}{d\,\ddot a_x}$, $Var(L_0)=\dfrac{{}^2A_x-A_x^2}{\left(d\,\ddot a_x\right)^2}$. Here ${}^2A_x$ 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 $v^{K+1}$.
  17. Loss at issue & variance
    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 $Var(L_0)$.
    $d=\dfrac{0.05}{1.05}\approx 0.0476190$; $\ddot a_x=\dfrac{1-A_x}{d}=\dfrac{0.70}{0.0476190}=14.7$; so $P_x=\dfrac{A_x}{\ddot a_x}=\dfrac{0.30}{14.7}\approx 0.0204082$. $\left(1+\dfrac{P}{d}\right)=1+\dfrac{0.0204082}{0.0476190}\approx 1.428571$. $Var(L_0)=(0.12-0.30^2)(1.428571)^2=(0.03)(2.040816)\approx 0.0612245$. The standard deviation is $\sqrt{0.0612245}\approx 0.247436$.
  18. Loss at issue & variance
    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 ($A_x=0.30$, ${}^2A_x=0.12$, $i=0.05$)?
    Loss scales linearly with $b$, so variance scales by $b^2$: $Var(L_0)=b^2\left({}^2A_x-A_x^2\right)\left(1+\tfrac{P}{d}\right)^2$. From before the per-unit variance is $0.0612245$. $Var=1000^2(0.0612245)=61{,}224.49$, so the standard deviation is $1000\times 0.247436\approx \$247.44$. The per-policy loss has a large spread relative to its zero mean — the rationale for pooling many lives.
  19. Loss at issue & variance
    State the **variance of $L_0$** for a fully continuous whole life insurance of $1$.
    $Var(L_0)=\left({}^2\bar A_x-\bar A_x^2\right)\left(1+\dfrac{\bar P}{\delta}\right)^2=\dfrac{{}^2\bar A_x-\bar A_x^2}{\left(\delta\,\bar a_x\right)^2}$, where ${}^2\bar A_x$ uses force of interest $2\delta$ and $\bar P=\bar P(\bar A_x)$. It is the continuous analogue of the discrete formula, with $\delta$ replacing $d$ and bars on the insurance EPVs.
  20. Loss at issue & variance
    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 $Var(L_0)$.
    $\bar a_x=\dfrac{1-0.40}{0.05}=12.0$, so $\bar P=\dfrac{0.40}{12.0}\approx 0.0333333$ and $1+\dfrac{\bar P}{\delta}=1+\dfrac{0.0333333}{0.05}=1.666667$. $Var(L_0)=(0.20-0.40^2)(1.666667)^2=(0.04)(2.777778)\approx 0.111111$. Standard deviation $=\sqrt{0.111111}\approx 0.333333$.
  21. Equivalence principle
    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 $q_x=0.02$, $q_{x+1}=0.025$, $q_{x+2}=0.03$, $i=0.06$.
    Survival: ${}_0p_x=1$, ${}_1p_x=0.98$, ${}_2p_x=0.98(0.975)=0.9555$. With $v=1/1.06$: APV benefits $=1000\sum_{k=0}^{2}v^{k+1}\,{}_kp_x\,q_{x+k}$ $=1000[v(0.02)+v^2(0.98)(0.025)+v^3(0.9555)(0.03)]\approx 1000[0.018868+0.021805+0.024068]=64.7405$. APV premiums $=\sum_{k=0}^{2}v^k\,{}_kp_x=1+v(0.98)+v^2(0.9555)\approx 2.774920$. $P=\dfrac{64.7405}{2.774920}\approx \$23.33$.
  22. Equivalence principle
    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 $q_x=0.04$, $q_{x+1}=0.05$, $q_{x+2}=0.06$, $i=0.06$.
    Survival: ${}_1p_x=0.96$, ${}_2p_x=0.96(0.95)=0.912$, ${}_3p_x=0.912(0.94)=0.85728$. $v=1/1.06$. APV death $=10000[v(0.04)+v^2(0.96)(0.05)+v^3(0.912)(0.06)]\approx 1263.996$. APV pure endowment $=10000\,v^3\,(0.85728)\approx 7197.888$. Total benefit APV $\approx 8461.885$. APV premiums $=1+v(0.96)+v^2(0.912)\approx 2.717337$. $P=\dfrac{8461.885}{2.717337}\approx \$3{,}114.04$.
  23. Net premium reserves
    Define the **net premium reserve** ${}_tV$ and give its **prospective** formula.
    ${}_tV$ 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. Prospective: ${}_tV=\text{APV future benefits at }x+t-\text{APV future net premiums at }x+t$. It is the fund the insurer must hold so that, together with future premiums, future benefits are covered in expectation.
  24. Net premium reserves
    Give the prospective net premium reserve for a fully discrete **whole life** insurance of $1$, and the annuity-ratio shortcut.
    ${}_tV=A_{x+t}-P_x\,\ddot a_{x+t}$. Using $A_{x+t}=1-d\,\ddot a_{x+t}$ and $P_x=\dfrac{1}{\ddot a_x}-d$, this collapses to ${}_tV=1-\dfrac{\ddot a_{x+t}}{\ddot a_x}$. Reserves grow from ${}_0V=0$ toward $1$ as $\ddot a_{x+t}$ shrinks with age.
  25. Net premium reserves
    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.
    ${}_tV=1-\dfrac{\ddot a_{x+t}}{\ddot a_x}=1-\dfrac{12}{15}=1-0.8=0.20$. So the reserve per $\$1$ of benefit is $0.20$; for a $\$100{,}000$ policy the held reserve is $\$20{,}000$. This identity needs only the two annuity-due values — no separate insurance EPV is required.
  26. Net premium reserves
    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 $\ddot a_{x+t}=13$.
    First the net premium: $P_x=\dfrac{A_x}{\ddot a_x}=\dfrac{0.25}{15}\approx 0.0166667$. Prospective reserve: ${}_tV=A_{x+t}-P_x\,\ddot a_{x+t}=0.35-0.0166667(13)$ $=0.35-0.216667=0.133333$. Per $\$1$ of benefit the reserve is about $0.133333$ (e.g. $\$13{,}333$ on a $\$100{,}000$ policy).
  27. Net premium reserves
    State the **retrospective** reserve formula and when it equals the prospective reserve.
    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. It **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.
  28. Net premium reserves
    Give the fully **continuous** whole life reserve and its annuity-ratio form.
    ${}_t\bar V(\bar A_x)=\bar A_{x+t}-\bar P(\bar A_x)\,\bar a_{x+t}=1-\dfrac{\bar a_{x+t}}{\bar a_x}$. It is the continuous analogue of ${}_tV=1-\dfrac{\ddot a_{x+t}}{\ddot a_x}$, using continuous annuities $\bar a$. Example: $\bar a_x=12$, $\bar a_{x+t}=9$ gives ${}_t\bar V=1-\dfrac{9}{12}=0.25$.
  29. Net premium reserves
    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**.
    At time $1$ a $2$-year endowment remains. Survival from $x+1$: ${}_1p_{x+1}=0.95$. $v=1/1.06$. APV future benefits $=10000\big[v(0.05)+v^2(0.95)(0.06)+v^2(0.95)(0.94)\big]$ $\approx 10000[0.047170+0.050730+0.794767]=8926.66$. APV future premiums $=3114.04\big[1+v(0.95)\big]\approx 3114.04(1.896226)=5904.92$. ${}_1V=8926.66-5904.92\approx \$3{,}021.74$.
  30. Reserve recursions
    Verify the previous $\$3{,}021.74$ reserve by the **recursion** from ${}_0V=0$. (Premium $\$3{,}114.04$, $q_x=0.04$, benefit $\$10{,}000$, $i=0.06$.)
    Recursion: $({}_0V+P)(1+i)=q_x\,b+p_x\,{}_1V$. $(0+3114.04)(1.06)=0.04(10000)+0.96\,{}_1V$ $3300.88=400+0.96\,{}_1V$ ${}_1V=\dfrac{3300.88-400}{0.96}=\dfrac{2900.88}{0.96}\approx \$3{,}021.75$. Matches 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).
  31. Reserve recursions
    State the general **reserve recursion** for a fully discrete policy and name each piece.
    $({}_tV+P)(1+i)=q_{x+t}\,b_{t+1}+p_{x+t}\,{}_{t+1}V$. Left side: the reserve in hand plus the premium just received, accumulated one year with interest. Right 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 $p_{x+t}$). Solve forward for ${}_{t+1}V$ given ${}_tV$, or backward.
  32. Reserve recursions
    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 ${}_{t+1}V$.
    $({}_tV+P)(1+i)=(0.20+0.02)(1.06)=0.22(1.06)=0.2332$. Then $0.2332=q_{x+t}b+p_{x+t}\,{}_{t+1}V=0.03(1)+0.97\,{}_{t+1}V$. ${}_{t+1}V=\dfrac{0.2332-0.03}{0.97}=\dfrac{0.2032}{0.97}\approx 0.209485$. The reserve grew from $0.20$ to about $0.209485$ over the year.
  33. Reserve recursions
    Build the first two reserves of a fully discrete whole life of $\$1{,}000$ with level net premium $P=\$20$, $i=0.06$, $q_x=0.010$, $q_{x+1}=0.012$. (${}_0V=0$.)
    Year 1: $(0+20)(1.06)=21.20=q_x(1000)+p_x\,{}_1V=10+0.990\,{}_1V$, so ${}_1V=\dfrac{21.20-10}{0.990}\approx \$11.31$. Year 2: $(11.31+20)(1.06)=31.31(1.06)=33.1886=q_{x+1}(1000)+p_{x+1}\,{}_2V=12+0.988\,{}_2V$, so ${}_2V=\dfrac{33.1886-12}{0.988}\approx \$21.45$. The reserve accumulates each year as premiums plus interest exceed the cost of insurance.
  34. Reserve recursions
    Rewrite the reserve recursion to split the premium into a **savings premium** and a **cost-of-insurance premium**.
    Rearranging $({}_tV+P)(1+i)=q_{x+t}b+p_{x+t}\,{}_{t+1}V$ gives $P=\underbrace{v\,{}_{t+1}V-{}_tV}_{\text{savings premium }P^s}+\underbrace{v\,q_{x+t}\,(b-{}_{t+1}V)}_{\text{cost of insurance }P^i}$. The **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 $P=P^s+P^i$.
  35. Reserve recursions
    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.
    $({}_tV+P)(1+i)=(150+20)(1.06)=180.20=q_{x+t}(1000)+p_{x+t}\,{}_{t+1}V=20+0.98\,{}_{t+1}V$, so ${}_{t+1}V=\dfrac{180.20-20}{0.98}\approx \$163.47$. Net amount at risk $=b-{}_{t+1}V=1000-163.47=\$836.53$. Cost of insurance $P^i=v\,q_{x+t}(b-{}_{t+1}V)=\dfrac{1}{1.06}(0.02)(836.53)\approx \$15.78$. Savings premium $P^s=P-P^i=20-15.78\approx \$4.22$ (and indeed $v\,{}_{t+1}V-{}_tV=\dfrac{163.47}{1.06}-150\approx 4.22$).
  36. Reserve recursions
    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.
    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: ${}_1V=\dfrac{(P-v\,q_x\,b)(1+i)}{p_x}=\dfrac{\big(20-\frac{1}{1.06}(0.010)(1000)\big)(1.06)}{0.990}$ $=\dfrac{(20-9.4340)(1.06)}{0.990}=\dfrac{11.2000}{0.990}\approx \$11.31$. Same as the forward recursion — retrospective and prospective agree under the equivalence principle.
  37. Gross premiums & expenses
    Distinguish a **net (benefit) premium** from a **gross (expense-loaded) premium**.
    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. The **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**.
  38. Gross premiums & expenses
    List the standard **expense types** in a gross-premium calculation and how each enters the equivalence equation.
    Typical categories: - **Per-premium** (e.g. commissions, premium tax): a percent of $G$, charged each premium → enters as $cG\,\ddot a$. - **Per-policy** (e.g. maintenance/admin, fixed \$ each year): $\to e\,\ddot a$. - **Acquisition / first-year** (\$ at issue only): a one-time term $\to$ constant. - **Settlement / claim** expense (\$ paid at death): $\to s\,A_x$. The gross-premium equation is $G\,\ddot a = \text{APV benefits}+\text{APV all expenses}$.
  39. Gross premiums & expenses
    State the **equivalence-principle gross premium** equation for a whole life with level annual expenses and a settlement expense.
    $G\,\ddot a_x = b\,A_x + (c\,G + e)\,\ddot a_x + s\,A_x$, where $c$ = per-premium fraction, $e$ = per-policy annual expense, $s$ = settlement expense paid at death. Solving for $G$: $G=\dfrac{b\,A_x+e\,\ddot a_x+s\,A_x}{(1-c)\,\ddot a_x}$. The $cG$ term moves to the left to give the $(1-c)$ factor.
  40. Gross premiums & expenses
    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.
    Equation: $G(14.7)=100000(0.30)+50(14.7)+0.10\,G(14.7)+200(0.30)$. $G(14.7)(1-0.10)=30000+735+60=30795$. $G=\dfrac{30795}{0.90(14.7)}=\dfrac{30795}{13.23}\approx \$2{,}327.66$. For comparison the net premium is $\dfrac{100000(0.30)}{14.7}\approx \$2{,}040.82$, so the expense loading is about $\$286.84$.
  41. Gross premiums & expenses
    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).
    Equation: $G(15)=50000(0.25)+500+0.08\,G(15)+30(15)$. $G(15)(1-0.08)=12500+500+450=13450$. $G=\dfrac{13450}{0.92(15)}=\dfrac{13450}{13.8}\approx \$974.64$. The one-time $\$500$ acquisition cost enters as a flat constant (not multiplied by $\ddot a$); the net premium here is $\dfrac{50000(0.25)}{15}\approx \$833.33$.
  42. Gross premiums & expenses
    What is the **expense reserve**, and how does it relate to the gross and net premium reserves?
    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: $\text{expense reserve}={}_tV^{\,\text{gross}}-{}_tV^{\,\text{net}}=\text{APV future expenses}-\text{APV future expense loadings}$. Because 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.