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Exam FAM — Life Insurance Flashcards

Actuarial present values of life-insurance benefits for SOA Exam FAM: discrete (end-of-year-of-death) whole life, term, pure endowment, endowment, and deferred insurances; their continuous and $m$-thly analogues; second moments and variances via the rule of moments (evaluate at double the force of interest); backward recursions; UDD and constant-force relationships such as $\bar A_x=\tfrac{i}{\delta}A_x$ and $A_x=\tfrac{q}{q+i}$; and increasing/decreasing insurances — all illustrated with self-contained worked APV and variance computations from explicit $q_x$ and interest values.

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  1. Whole life
    Define the present-value random variable $Z$ for a discrete whole life insurance of $1$, and write its actuarial present value (EPV) $A_x$.
    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 $Z=v^{K_x+1}$. The EPV is $A_x=E[Z]=\sum_{k=0}^{\infty} v^{k+1}\,{}_kp_x\,q_{x+k}$, where $v=\frac{1}{1+i}$, ${}_kp_x$ is the probability of surviving $k$ years, and $q_{x+k}$ is the probability of dying in the following year.
  2. Term & endowment
    Write the EPV of a discrete $n$-year **term** insurance $A^{1}_{x:\overline{n|}}$ and say what the PV variable looks like.
    The unit benefit is paid at the end of the year of death **only if death occurs within $n$ years**: $Z=\begin{cases} v^{K_x+1}, & K_x \le n-1\\ 0, & K_x \ge n\end{cases}$ so $A^{1}_{x:\overline{n|}}=\sum_{k=0}^{n-1} v^{k+1}\,{}_kp_x\,q_{x+k}$. The little $1$ over the $x$ marks that the benefit is contingent on **death** (the life status) before the term expires.
  3. Pure endowment & deferred
    Write the EPV of an $n$-year **pure endowment** ${}_nE_x$ and describe the benefit.
    A pure endowment pays $1$ at time $n$ **only if the life survives** to age $x+n$; nothing is paid on death. ${}_nE_x = v^{n}\,{}_np_x$. It 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.
  4. Term & endowment
    Write the EPV of a discrete $n$-year **endowment** insurance $A_{x:\overline{n|}}$ and decompose it.
    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 $n$. $A_{x:\overline{n|}} = A^{1}_{x:\overline{n|}} + {}_nE_x$, the sum of the $n$-year term insurance and the $n$-year pure endowment. Its PV variable is $Z=v^{\min(K_x+1,\;n)}$.
  5. Pure endowment & deferred
    Write the EPV of an $n$-year **deferred whole life** insurance ${}_{n|}A_x$ and give its product form.
    Coverage starts after $n$ years: the unit is paid at the end of the year of death **only if death occurs after age $x+n$**. ${}_{n|}A_x = \sum_{k=n}^{\infty} v^{k+1}\,{}_kp_x\,q_{x+k} = {}_nE_x\,A_{x+n} = v^{n}\,{}_np_x\,A_{x+n}$. You survive $n$ years (factor ${}_nE_x$) and then hold an ordinary whole life on $(x+n)$.
  6. Pure endowment & deferred
    Give the basic identity linking term, deferred, and whole life insurance.
    Whole life splits into the term piece plus the deferred piece: $A_x = A^{1}_{x:\overline{n|}} + {}_{n|}A_x = A^{1}_{x:\overline{n|}} + {}_nE_x\,A_{x+n}$. Death 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.
  7. Continuous/mthly & relations
    Write the EPV of a **continuous** whole life insurance $\bar A_x$, where the benefit is paid at the exact moment of death.
    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. $\bar A_x = E[Z]=\int_0^{\infty} v^{t}\,{}_tp_x\,\mu_{x+t}\,dt = \int_0^{\infty} e^{-\delta t}\,{}_tp_x\,\mu_{x+t}\,dt$, where ${}_tp_x\,\mu_{x+t}$ is the density of the time of death.
  8. Continuous/mthly & relations
    Write the EPV of an $m$-thly whole life insurance $A_x^{(m)}$ and explain the benefit timing.
    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, $A_x^{(m)} = \sum_{j=0}^{\infty} v^{(j+1)/m}\,{}_{j/m}p_x\,{}_{1/m}q_{x+j/m}$. As $m\to\infty$, $A_x^{(m)}\to\bar A_x$. Faster payment means a larger EPV: $A_x \le A_x^{(m)} \le \bar A_x$.
  9. Moments & variance
    State the **rule of moments**: how do you get the second moment ${}^2A_x$ of a life-insurance PV?
    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 $(1+i)^2-1$. Same holds for term, endowment, and deferred forms: ${}^2A^{1}_{x:\overline{n|}}$, ${}^2A_{x:\overline{n|}}$, etc., each evaluated at the doubled force.
  10. Moments & variance
    Give the variance of the PV of a unit life insurance, and of a benefit of $b$.
    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. For 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 $b$. The same template works with term/endowment $A$'s.
  11. Recursions
    State the backward **recursion** for discrete whole life insurance $A_x$.
    $A_x = v\,q_x + v\,p_x\,A_{x+1}$. Intuition: 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. To move forward instead, rearrange: $A_{x+1}=\dfrac{A_x - v q_x}{v\,p_x}$.
  12. Recursions
    State the recursions for $n$-year **term** and **endowment** insurance.
    Term: $A^{1}_{x:\overline{n|}} = v\,q_x + v\,p_x\,A^{1}_{x+1:\overline{n-1|}}$, with $A^{1}_{x:\overline{0|}}=0$. Endowment: $A_{x:\overline{n|}} = v\,q_x + v\,p_x\,A_{x+1:\overline{n-1|}}$, with $A_{x:\overline{0|}}=1$ (the survival payment is due immediately at term end). Both peel off the first year: pay on death now, otherwise discount one year and hold a contract one term shorter on $(x+1)$.
  13. Continuous/mthly & relations
    State the relationship between whole life insurance $A_x$ and the annuity-due $\ddot a_x$.
    $A_x = 1 - d\,\ddot a_x$, where $d=\dfrac{i}{1+i}=iv$ is the effective discount rate. Equivalently $\ddot a_x = \dfrac{1-A_x}{d}$. The endowment version is $A_{x:\overline{n|}} = 1 - d\,\ddot a_{x:\overline{n|}}$. The continuous analogue is $\bar A_x = 1 - \delta\,\bar a_x$.
  14. Continuous/mthly & relations
    Under the **UDD** assumption, relate the continuous insurance $\bar A_x$ to the discrete $A_x$.
    Under a uniform distribution of deaths within each year of age, $\bar A_x = \dfrac{i}{\delta}\,A_x$, where $\delta=\ln(1+i)$. 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).
  15. Continuous/mthly & relations
    Under **UDD**, relate the $m$-thly insurance $A_x^{(m)}$ to the annual $A_x$.
    $A_x^{(m)} = \dfrac{i}{i^{(m)}}\,A_x$, where $i^{(m)} = m\big[(1+i)^{1/m}-1\big]$ 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 $A_x^{(m)}\to\bar A_x$.
  16. Continuous/mthly & relations
    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 $Var(Z)$.
    With ${}_tp_x=e^{-\mu t}$ and density $\mu e^{-\mu t}$, $\bar A_x = \int_0^\infty e^{-\delta t}\,e^{-\mu t}\mu\,dt = \dfrac{\mu}{\mu+\delta}$. Doubling the force of interest gives ${}^2\bar A_x = \dfrac{\mu}{\mu+2\delta}$. So $Var(Z) = \dfrac{\mu}{\mu+2\delta} - \left(\dfrac{\mu}{\mu+\delta}\right)^2$.
  17. Whole life
    Under a **constant** annual mortality rate $q$ (so $q_{x+k}=q$ for all $k$), give the closed form for discrete whole life $A_x$.
    Then ${}_kp_x\,q_{x+k}=(1-q)^k q$, a geometric distribution, so $A_x = \sum_{k=0}^\infty v^{k+1}(1-q)^k q = \dfrac{vq}{1-v(1-q)} = \dfrac{q}{q+i}$. The second moment uses interest $(1+i)^2-1$ in place of $i$: ${}^2A_x = \dfrac{q}{q+\big[(1+i)^2-1\big]}$, and $Var(Z)={}^2A_x - A_x^2$.
  18. Whole life
    Define the **increasing** whole life insurance $(IA)_x$ and the increasing $n$-year term $(IA)^{1}_{x:\overline{n|}}$.
    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 $Z=(K_x+1)v^{K_x+1}$. $(IA)_x = \sum_{k=0}^{\infty}(k+1)\,v^{k+1}\,{}_kp_x\,q_{x+k}$, and the $n$-year term version truncates the sum at $k=n-1$.
  19. Whole life
    Define the **decreasing** $n$-year term insurance $(DA)^{1}_{x:\overline{n|}}$ and give the identity with the increasing term.
    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 $n$): $(DA)^{1}_{x:\overline{n|}} = \sum_{k=0}^{n-1}(n-k)\,v^{k+1}\,{}_kp_x\,q_{x+k}$. Useful 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 $n+1$.
  20. Whole life
    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 $A_x$.
    $v=\frac{1}{1.06}\approx 0.943396$. Survival: ${}_0p_x=1$, ${}_1p_x=0.90$, ${}_2p_x=0.90(0.85)=0.765$. Death-year terms $v^{k+1}\,{}_kp_x\,q_{x+k}$: $k=0:\;0.943396(1)(0.10)=0.094340$ $k=1:\;0.943396^{2}(0.90)(0.15)=0.889996(0.135)=0.120150$ $k=2:\;0.943396^{3}(0.765)(1.00)=0.839619(0.765)=0.642309$ $A_x = 0.094340+0.120150+0.642309 = \boxed{0.856798}$.
  21. Moments & variance
    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.
    Double the force of interest: use rate $j=(1.06)^2-1=0.1236$, so $v_j=\frac{1}{1.1236}\approx 0.889996$. ${}^2A_x = \sum v_j^{k+1}\,{}_kp_x\,q_{x+k}$: $k=0:\;0.889996(0.10)=0.089000$ $k=1:\;0.889996^{2}(0.90)(0.15)=0.792092(0.135)=0.106932$ $k=2:\;0.889996^{3}(0.765)=0.704958(0.765)=0.539293$ ${}^2A_x = 0.089000+0.106932+0.539293 = 0.735225$. $Var(Z) = 0.735225 - 0.856798^2 = 0.735225 - 0.734103 = \boxed{0.001122}$.
  22. Whole life
    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.
    $v=0.943396$. Survival: ${}_1p_x=0.92$, ${}_2p_x=0.92(0.88)=0.8096$, ${}_3p_x=0.8096(0.80)=0.64768$. Death terms: $k=0:\;0.943396(0.08)=0.075472$ $k=1:\;0.889996(0.92)(0.12)=0.098255$ $k=2:\;0.839619(0.8096)(0.20)=0.135950$ $k=3:\;0.792094(0.64768)(1.00)=0.513025$ $A_x=0.075472+0.098255+0.135950+0.513025=0.822702$. APV $=100{,}000(0.822702)=\boxed{\$82{,}270.20}$.
  23. Term & endowment
    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 $A^{1}_{x:\overline{3|}}$.
    $v=\frac{1}{1.05}\approx 0.952381$. Survival: ${}_1p_x=0.96$, ${}_2p_x=0.96(0.94)=0.9024$. Death terms $v^{k+1}\,{}_kp_x\,q_{x+k}$: $k=0:\;0.952381(0.04)=0.038095$ $k=1:\;0.907029(0.96)(0.06)=0.052245$ $k=2:\;0.863838(0.9024)(0.09)=0.070157$ $A^{1}_{x:\overline{3|}} = 0.038095+0.052245+0.070157 = \boxed{0.160498}$.
  24. Term & endowment
    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 $A_{x:\overline{3|}}$.
    ${}_3p_x = 0.96(0.94)(0.91)=0.821184$ and $v^3=0.952381^3\approx 0.863838$. ${}_3E_x = v^3\,{}_3p_x = 0.863838(0.821184) = \boxed{0.709370}$. Endowment $=$ term $+$ pure endowment: $A_{x:\overline{3|}} = A^{1}_{x:\overline{3|}} + {}_3E_x = 0.160498 + 0.709370 = \boxed{0.869868}$.
  25. Term & endowment
    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.
    $v=0.952381$. ${}_1p_x=0.97$. Death terms: $k=0:\;v\,q_x = 0.952381(0.03)=0.028571$ $k=1:\;v^2\,{}_1p_x\,q_{x+1}=0.907029(0.97)(0.05)=0.043991$ $A^{1}_{x:\overline{2|}} = 0.028571+0.043991 = 0.072562$. APV $= 50{,}000(0.072562) = \boxed{\$3{,}628.12}$.
  26. Pure endowment & deferred
    Compute a $5$-year pure endowment ${}_5E_x$ at $i=0.04$ when the yearly survival probabilities are $p_x=0.97,\,p_{x+1}=0.96,\,p_{x+2}=0.95,\,p_{x+3}=0.93,\,p_{x+4}=0.90$.
    Survival to $5$ years: ${}_5p_x = 0.97(0.96)(0.95)(0.93)(0.90)$. $0.97(0.96)=0.9312$; $\times 0.95 = 0.88464$; $\times 0.93 = 0.822715$; $\times 0.90 = 0.740444$. Discount: $v^5 = 1.04^{-5} \approx 0.821927$. ${}_5E_x = v^5\,{}_5p_x = 0.821927(0.740444) = \boxed{0.608592}$.
  27. Pure endowment & deferred
    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 $A_{x+2}=0.454545$.
    Pure-endowment factor: ${}_2p_x = 0.95(0.93)=0.8835$ and $v^2 = 1.06^{-2}\approx 0.889996$, so ${}_2E_x = v^2\,{}_2p_x = 0.889996(0.8835)=0.786312$. Deferred insurance $=$ survive $2$ years, then hold a whole life on $(x+2)$: ${}_{2|}A_x = {}_2E_x\,A_{x+2} = 0.786312(0.454545) = \boxed{0.357415}$.
  28. Recursions
    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 $A_{x+1}=0.40$.
    $v = \frac{1}{1.05}\approx 0.952381$, $p_x = 0.97$. $A_x = v q_x + v p_x A_{x+1}$ $= 0.952381(0.03) + 0.952381(0.97)(0.40)$ $= 0.028571 + 0.952381(0.388)$ $= 0.028571 + 0.369524 = \boxed{0.398095}$.
  29. Recursions
    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.
    $v=0.952381$, $p_x=0.96$. A $1$-year endowment always pays $1$ at year-end (death or survival), so $A_{x+1:\overline{1|}} = v = 0.952381$. Recursion: $A_{x:\overline{2|}} = v q_x + v p_x A_{x+1:\overline{1|}} = 0.952381(0.04) + 0.952381(0.96)(0.952381)$ $= 0.038095 + 0.870748 = \boxed{0.908844}$. Direct 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
  30. Moments & variance
    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.
    Constant-$q$ closed form: $A_x = \dfrac{q}{q+i} = \dfrac{0.04}{0.04+0.06} = \boxed{0.40}$. Second moment doubles the force of interest; use $j=(1.06)^2-1 = 0.1236$: ${}^2A_x = \dfrac{q}{q+j} = \dfrac{0.04}{0.04+0.1236} = \dfrac{0.04}{0.1636} = 0.244499$. $Var(Z) = {}^2A_x - A_x^2 = 0.244499 - 0.40^2 = 0.244499 - 0.16 = \boxed{0.084499}$.
  31. Moments & variance
    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.
    $\bar A_x = \dfrac{\mu}{\mu+\delta} = \dfrac{0.02}{0.02+0.05} = \dfrac{0.02}{0.07} = \boxed{0.285714}$. Second moment doubles $\delta$: ${}^2\bar A_x = \dfrac{\mu}{\mu+2\delta} = \dfrac{0.02}{0.02+0.10} = \dfrac{0.02}{0.12} = 0.166667$. $Var(Z) = 0.166667 - 0.285714^2 = 0.166667 - 0.081633 = \boxed{0.085034}$.
  32. Moments & variance
    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.
    $v=\frac{1}{1.1}\approx 0.909091$; ${}_1p_x=0.95$, ${}_2p_x=0.95(0.90)=0.855$. First moment: $k=0:\;0.909091(0.05)=0.045455$ $k=1:\;0.826446(0.95)(0.10)=0.078512$ $k=2:\;0.751315(0.855)(0.20)=0.128475$ $A^{1}_{x:\overline{3|}} = 0.252442$. Second moment at $j=(1.1)^2-1=0.21$, $v_j=\frac{1}{1.21}\approx 0.826446$: $0.826446(0.05)=0.041322$; $0.683013(0.95)(0.10)=0.064886$; $0.564474(0.855)(0.20)=0.096525$; sum $={}^2A^{1}_{x:\overline{3|}}=0.202733$. $Var(Z)=0.202733-0.252442^2 = 0.202733-0.063727 = \boxed{0.139006}$.
  33. Moments & variance
    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.
    Add the pure endowment to the term piece ($A^{1}_{x:\overline{3|}}=0.252442$). ${}_3p_x=0.95(0.90)(0.80)=0.684$, $v^3=0.751315$. ${}_3E_x = 0.751315(0.684)=0.513900$, so $A_{x:\overline{3|}}=0.252442+0.513900=0.766342$. Second moment: term part $0.202733$ (from before) plus endowment part at doubled interest $v_j^3\,{}_3p_x=0.564474(0.684)=0.386100$. ${}^2A_{x:\overline{3|}}=0.202733+0.386100=0.588833$. $Var(Z)=0.588833-0.766342^2 = 0.588833-0.587280 = \boxed{0.001553}$. Note the variance is far smaller than the pure term case — the guaranteed survival payment removes most of the dispersion.
  34. Continuous/mthly & relations
    Given $A_x = 0.35$ and $i=0.06$, find the continuous insurance $\bar A_x$ under UDD.
    Force of interest $\delta = \ln(1.06) \approx 0.058269$. Under UDD, $\bar A_x = \dfrac{i}{\delta}\,A_x = \dfrac{0.06}{0.058269}\,(0.35)$. $\dfrac{0.06}{0.058269} \approx 1.029709$, so $\bar A_x \approx 1.029709(0.35) = \boxed{0.360398}$. The continuous EPV exceeds the discrete one because the benefit is paid on average half a year sooner.
  35. Continuous/mthly & relations
    Given $A_x = 0.35$ and $i=0.06$, find the **monthly** ($m=12$) insurance $A_x^{(12)}$ under UDD.
    Nominal rate: $i^{(12)} = 12\big[(1.06)^{1/12}-1\big]$. Since $(1.06)^{1/12}\approx 1.004868$, $i^{(12)}\approx 12(0.004868)=0.058411$. Under UDD, $A_x^{(12)} = \dfrac{i}{i^{(12)}}\,A_x = \dfrac{0.06}{0.058411}(0.35)$. $\dfrac{0.06}{0.058411}\approx 1.027211$, so $A_x^{(12)} \approx 1.027211(0.35) = \boxed{0.359524}$. This sits between $A_x=0.35$ and $\bar A_x\approx 0.360398$, as expected.
  36. Continuous/mthly & relations
    Given $\ddot a_x = 14.20$ and $i=0.06$, find $A_x$ using the annuity-insurance relationship.
    Effective discount rate $d = \dfrac{i}{1+i} = \dfrac{0.06}{1.06} \approx 0.056604$. $A_x = 1 - d\,\ddot a_x = 1 - 0.056604(14.20) = 1 - 0.803774 = \boxed{0.196226}$. Check: a large annuity value (long expected lifetime) goes with a small insurance value — death is far off and heavily discounted.
  37. Whole life
    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 $(IA)^{1}_{x:\overline{3|}}$.
    $v=0.943396$; ${}_1p_x=0.90$, ${}_2p_x=0.90(0.85)=0.765$. Benefit in year $k+1$ is $k+1$: $k=0:\;1\cdot 0.943396(0.10)=0.094340$ $k=1:\;2\cdot 0.889996(0.90)(0.15)=2(0.120149)=0.240299$ $k=2:\;3\cdot 0.839619(0.765)(0.20)=3(0.128462)=0.385385$ $(IA)^{1}_{x:\overline{3|}} = 0.094340+0.240299+0.385385 = \boxed{0.720024}$.
  38. Whole life
    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.
    Benefit in year $k+1$ is $n-k = 3,2,1$: $k=0:\;3\cdot 0.943396(0.10)=3(0.094340)=0.283019$ $k=1:\;2\cdot 0.889996(0.90)(0.15)=0.240299$ $k=2:\;1\cdot 0.839619(0.765)(0.20)=0.128462$ $(DA)^{1}_{x:\overline{3|}} = 0.283019+0.240299+0.128462 = \boxed{0.651780}$. Check: 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
  39. Moments & variance
    Why does an $n$-year **endowment** insurance always have a far smaller variance than the corresponding $n$-year **term** insurance (same $i$, same mortality)?
    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. Numerically (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.
  40. Term & endowment
    Distinguish $A^{1}_{x:\overline{n|}}$, $A_{x:\overline{n|}}^{\;\;1}$, and $A_{x:\overline{n|}}$ by where the "$1$" sits.
    $A^{1}_{x:\overline{n|}}$ — the $1$ is over **$x$**: pays on **death** within $n$ years ($n$-year **term** insurance). $A_{x:\overline{n|}}^{\;\;1}$ — the $1$ is over the **$n$**: pays on **survival** to time $n$ ($n$-year **pure endowment**, $={}_nE_x$). $A_{x:\overline{n|}}$ — **no** $1$: pays on whichever comes first, death or survival ($n$-year **endowment** insurance $=$ sum of the other two).
  41. Continuous/mthly & relations
    Order $A_x$, $A_x^{(m)}$, and $\bar A_x$ for a fixed life and interest rate, and explain the ordering.
    $A_x \;\le\; A_x^{(m)} \;\le\; \bar A_x$. All 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 $m\to\infty$, $A_x^{(m)}\to\bar A_x$.
  42. Continuous/mthly & relations
    A life has **constant force** $\mu=0.03$ and $\delta=0.06$. Find the EPV of a **continuous** $10$-year term insurance $\bar A^{1}_{x:\overline{10|}}$.
    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 $n$). $\dfrac{\mu}{\mu+\delta} = \dfrac{0.03}{0.09} = 0.333333$. With $n=10$: $(\mu+\delta)n = 0.9$, $e^{-0.9}\approx 0.406570$. $\bar A^{1}_{x:\overline{10|}} = 0.333333(1 - 0.406570) = 0.333333(0.593430) = \boxed{0.197810}$.
  43. Moments & variance
    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.
    Unit variance via rule of moments: $Var(Z) = {}^2A_x - A_x^2 = 0.11 - 0.30^2 = 0.11 - 0.09 = 0.02$. The payout PV is $250{,}000\,Z$, so its variance is $250{,}000^2(0.02)$ and its standard deviation is $250{,}000\sqrt{0.02} = 250{,}000(0.141421) = \boxed{\$35{,}355.34}$. Standard deviation scales linearly with the benefit; variance scales with its square.