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Exam FAM — Life Annuities Flashcards

Expected present values of life annuities for SOA Exam FAM: whole life, temporary, and deferred annuities in both annuity-due and annuity-immediate forms, guaranteed certain-and-life annuities, the fundamental insurance–annuity relations ($\ddot a_x=\frac{1-A_x}{d}$ and continuous analogue), the backward recursion, the variance of the annuity present value from the insurance second moment, $m$-thly annuities with the Woolhouse approximation, and increasing annuities — with fully worked numeric examples built from explicit survival probabilities and interest rates.

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  1. Whole life annuity
    Define the random variable $Y$ for a whole life annuity-due and give its EPV $\ddot a_x$ as a sum.
    $Y=\ddot a_{\overline{K+1|}}$ pays $1$ at the start of each year while $(x)$ survives, where $K=K_x$ is the curtate future lifetime. Its EPV is $\ddot a_x = E[Y] = \sum_{k=0}^{\infty} v^{k}\,{}_kp_x$, the 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 $1$.)
  2. Due vs immediate
    State the relation between the whole life annuity-immediate $a_x$ and the annuity-due $\ddot a_x$.
    $a_x = \ddot a_x - 1$. The 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 $1,2,\dots$
  3. Due vs immediate
    Given $\ddot a_x = 14.20$, find the annuity-immediate $a_x$.
    $a_x = \ddot a_x - 1 = 14.20 - 1 = 13.20$. The difference of exactly $1$ is the time-$0$ payment that the due annuity makes but the immediate annuity does not.
  4. Insurance-annuity relations
    State the fundamental insurance–annuity relation linking $\ddot a_x$ and $A_x$, and solve it both ways.
    $\ddot a_x = \dfrac{1 - A_x}{d}$, equivalently $A_x = 1 - d\,\ddot a_x$, where $d=\dfrac{i}{1+i}=1-v$ is the effective annual discount rate. The identity comes from $1 = d\,\ddot a_{\overline{K+1|}} + v^{K+1}$ taken in expectation: $1 = d\,\ddot a_x + A_x$.
  5. Insurance-annuity relations
    Given $A_x = 0.30$ and $i = 0.05$, compute $\ddot a_x$.
    First $d = \dfrac{i}{1+i} = \dfrac{0.05}{1.05} \approx 0.047619$. $\ddot a_x = \dfrac{1 - A_x}{d} = \dfrac{1 - 0.30}{0.047619} = \dfrac{0.70}{0.047619} \approx 14.7000$. Then $a_x = \ddot a_x - 1 \approx 13.7000$.
  6. Insurance-annuity relations
    Given $\ddot a_x = 12.50$ and $i = 0.06$, find $A_x$.
    $d = \dfrac{0.06}{1.06} \approx 0.056604$. $A_x = 1 - d\,\ddot a_x = 1 - 0.056604(12.50) = 1 - 0.707547 \approx 0.29245$.
  7. Insurance-annuity relations
    State the continuous insurance–annuity relation for $\bar a_x$ and $\bar A_x$.
    $\bar a_x = \dfrac{1 - \bar A_x}{\delta}$, equivalently $\bar A_x = 1 - \delta\,\bar a_x$, where $\delta = \ln(1+i)$ 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. It mirrors the discrete case with $\delta$ in place of $d$ and $\bar A_x$ in place of $A_x$.
  8. Insurance-annuity relations
    Given $\bar A_x = 0.40$ and $\delta = 0.05$, find $\bar a_x$.
    $\bar a_x = \dfrac{1 - \bar A_x}{\delta} = \dfrac{1 - 0.40}{0.05} = \dfrac{0.60}{0.05} = 12.0000$.
  9. Whole life annuity
    State the backward recursion for the whole life annuity-due $\ddot a_x$.
    $\ddot a_x = 1 + v\,p_x\,\ddot a_{x+1}$. The \$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 $v$. This is the workhorse for building $\ddot a_x$ down a life table from an endpoint value.
  10. Whole life annuity
    A life table gives $\ddot a_{x+1} = 13.00$, $p_x = 0.99$, and $i = 0.05$. Find $\ddot a_x$.
    $v = \dfrac{1}{1.05} \approx 0.952381$. $\ddot a_x = 1 + v\,p_x\,\ddot a_{x+1} = 1 + 0.952381(0.99)(13.00)$ $= 1 + 0.952381(12.87) = 1 + 12.25714 \approx 13.2571$.
  11. Temporary & deferred
    Define the $n$-year temporary life annuity-due $\ddot a_{x:\overline{n|}}$ as a sum.
    $\ddot a_{x:\overline{n|}} = \sum_{k=0}^{n-1} v^{k}\,{}_kp_x$, paying \$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 $n-1$.
  12. Temporary & deferred
    Compute the 3-year temporary annuity-due $\ddot a_{x:\overline{3|}}$ given $i = 0.05$, $p_x = 0.97$, $p_{x+1} = 0.96$.
    Need ${}_0p_x = 1$, ${}_1p_x = 0.97$, ${}_2p_x = 0.97(0.96) = 0.9312$, and $v=0.952381$. $\ddot a_{x:\overline{3|}} = v^{0}(1) + v^{1}(0.97) + v^{2}(0.9312)$ $= 1 + 0.952381(0.97) + 0.907029(0.9312)$ $= 1 + 0.923810 + 0.844626 \approx 2.76844$.
  13. Insurance-annuity relations
    Relate the $n$-year temporary annuity-due to the $n$-year endowment insurance: $\ddot a_{x:\overline{n|}} = ?$
    $\ddot a_{x:\overline{n|}} = \dfrac{1 - A_{x:\overline{n|}}}{d}$, equivalently $A_{x:\overline{n|}} = 1 - d\,\ddot a_{x:\overline{n|}}$, where $A_{x:\overline{n|}}$ 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.
  14. Insurance-annuity relations
    Given $A_{x:\overline{20|}} = 0.45$ and $i = 0.06$, find $\ddot a_{x:\overline{20|}}$.
    $d = \dfrac{0.06}{1.06} \approx 0.056604$. $\ddot a_{x:\overline{20|}} = \dfrac{1 - A_{x:\overline{20|}}}{d} = \dfrac{1 - 0.45}{0.056604} = \dfrac{0.55}{0.056604} \approx 9.71667$.
  15. Temporary & deferred
    Define the pure endowment ${}_nE_x$ and state how it splits a whole life annuity-due into temporary + deferred pieces.
    ${}_nE_x = v^{n}\,{}_np_x$ is the EPV of \$1 paid at time $n$ only if $(x)$ survives to $x+n$ — the $n$-year discount factor. It links the pieces by $\ddot a_x = \ddot a_{x:\overline{n|}} + {}_{n|}\ddot a_x$, where the deferred annuity is ${}_{n|}\ddot a_x = {}_nE_x\,\ddot a_{x+n}$.
  16. Temporary & deferred
    Define the $n$-year deferred whole life annuity-due ${}_{n|}\ddot a_x$ and give its EPV.
    ${}_{n|}\ddot a_x$ pays \$1 at the start of each year **beginning at age $x+n$**, while $(x)$ survives. Its EPV is ${}_{n|}\ddot a_x = {}_nE_x\,\ddot a_{x+n} = v^{n}\,{}_np_x\,\ddot a_{x+n} = \sum_{k=n}^{\infty} v^{k}\,{}_kp_x$. Equivalently ${}_{n|}\ddot a_x = \ddot a_x - \ddot a_{x:\overline{n|}}$.
  17. Temporary & deferred
    Given ${}_{10}p_x = 0.85$, $i = 0.05$, and $\ddot a_{x+10} = 11.40$, find the 10-year deferred annuity ${}_{10|}\ddot a_x$.
    First the pure endowment: ${}_{10}E_x = v^{10}\,{}_{10}p_x = (1.05)^{-10}(0.85)$. $(1.05)^{-10} \approx 0.613913$, so ${}_{10}E_x \approx 0.613913(0.85) = 0.521826$. ${}_{10|}\ddot a_x = {}_{10}E_x\,\ddot a_{x+10} = 0.521826(11.40) \approx 5.94882$.
  18. Temporary & deferred
    Given $\ddot a_x = 15.00$ and $\ddot a_{x:\overline{10|}} = 7.80$, find the 10-year deferred annuity ${}_{10|}\ddot a_x$.
    $ {}_{10|}\ddot a_x = \ddot a_x - \ddot a_{x:\overline{10|}} = 15.00 - 7.80 = 7.20$. The whole life annuity is exactly the temporary piece (years $0$–$9$) plus the deferred piece (years $10,11,\dots$).
  19. Temporary & deferred
    Define an $n$-year **guaranteed (certain-and-life)** annuity-due $\ddot a_{\overline{x:\overline{n|}}}$ and give its EPV.
    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. EPV: $\ddot a_{\overline{x:\overline{n|}}} = \ddot a_{\overline{n|}} + {}_{n|}\ddot a_x = \ddot a_{\overline{n|}} + {}_nE_x\,\ddot a_{x+n}$, where $\ddot a_{\overline{n|}} = \dfrac{1-v^{n}}{d}$ is the ordinary annuity-certain-due for the first $n$ guaranteed payments.
  20. Temporary & deferred
    Compute the 10-year certain-and-life annuity-due given $i = 0.05$, ${}_{10}p_x = 0.85$, and $\ddot a_{x+10} = 11.40$.
    Annuity-certain-due: $\ddot a_{\overline{10|}} = \dfrac{1-v^{10}}{d}$ with $v^{10}\approx0.613913$, $d=\dfrac{0.05}{1.05}\approx0.047619$. $\ddot a_{\overline{10|}} = \dfrac{1-0.613913}{0.047619} = \dfrac{0.386087}{0.047619} \approx 8.10783$. Deferred life piece: ${}_{10}E_x\,\ddot a_{x+10} = 0.613913(0.85)(11.40) \approx 5.94882$. Total $= 8.10783 + 5.94882 \approx 14.0567$.
  21. Variance
    State the variance of the present value $Y$ of a whole life annuity-due in terms of insurance moments.
    Since $Y = \dfrac{1 - v^{K+1}}{d} = \dfrac{1 - Z}{d}$ where $Z=v^{K+1}$ is the whole life insurance PV, $\operatorname{Var}(Y) = \dfrac{\operatorname{Var}(Z)}{d^{2}} = \dfrac{{}^2A_x - (A_x)^{2}}{d^{2}}$, where ${}^2A_x$ is the second moment of $Z$ (the insurance evaluated at double the force of interest, i.e. at rate $i'=2i+i^{2}$).
  22. Variance
    Given $A_x = 0.25$, ${}^2A_x = 0.09$, and $i = 0.05$, find $\operatorname{Var}(Y)$ for the whole life annuity-due.
    $d = \dfrac{0.05}{1.05} \approx 0.047619$, so $d^{2} \approx 0.00226757$. Numerator: ${}^2A_x - A_x^{2} = 0.09 - 0.25^{2} = 0.09 - 0.0625 = 0.0275$. $\operatorname{Var}(Y) = \dfrac{0.0275}{0.00226757} \approx 12.1276$.
  23. Variance
    Why does the variance of the annuity PV use ${}^2A_x$ (a double-rate insurance) rather than a 'double-rate annuity'?
    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 $\operatorname{Var}(Y)=\dfrac{\operatorname{Var}(Z)}{d^{2}}$. All 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.
  24. Variance
    State the variance of the present value of a continuous whole life annuity $\bar a_x$.
    With $\bar Y = \dfrac{1 - \bar Z}{\delta}$ and $\bar Z = v^{T} = e^{-\delta T}$, $\operatorname{Var}(\bar Y) = \dfrac{{}^2\bar A_x - (\bar A_x)^{2}}{\delta^{2}}$, where ${}^2\bar A_x$ is the continuous insurance second moment, evaluated at force of interest $2\delta$.
  25. Variance
    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.
    Numerator: ${}^2\bar A_x - \bar A_x^{2} = 0.16 - 0.35^{2} = 0.16 - 0.1225 = 0.0375$. $\delta^{2} = 0.06^{2} = 0.0036$. $\operatorname{Var}(\bar Y) = \dfrac{0.0375}{0.0036} \approx 10.4167$. Standard deviation $= \sqrt{10.4167} \approx 3.2275$.
  26. Variance
    For an $n$-year temporary annuity-due, what insurance moments give $\operatorname{Var}(Y)$?
    $Y = \dfrac{1 - Z}{d}$ where $Z$ is now the $n$-year **endowment** insurance PV with first moment $A_{x:\overline{n|}}$ and second moment ${}^2A_{x:\overline{n|}}$. Thus $\operatorname{Var}(Y) = \dfrac{{}^2A_{x:\overline{n|}} - \left(A_{x:\overline{n|}}\right)^{2}}{d^{2}}$. The endowment (not the term insurance) appears because the survival-to-$n$ payment is part of $Z$.
  27. Variance
    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.
    $d = \dfrac{0.05}{1.05}\approx 0.047619$, $d^{2}\approx 0.00226757$. Numerator: $0.27 - 0.50^{2} = 0.27 - 0.25 = 0.02$. $\operatorname{Var}(Y) = \dfrac{0.02}{0.00226757} \approx 8.8200$.
  28. Whole life annuity
    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 $\ddot a_{x+3}=12.00$.
    Survival probabilities: $p_x=0.96$, $p_{x+1}=0.95$, $p_{x+2}=0.94$. ${}_1p_x=0.96$, ${}_2p_x=0.96(0.95)=0.912$, ${}_3p_x=0.912(0.94)=0.85728$. Temporary piece: $\ddot a_{x:\overline{3|}} = 1 + v(0.96) + v^{2}(0.912)$ with $v=0.952381$, $v^{2}=0.907029$. $= 1 + 0.914286 + 0.827210 = 2.741496$. Deferred tail: ${}_3E_x\,\ddot a_{x+3} = v^{3}(0.85728)(12.00)$; $v^{3}=0.863838$, so ${}_3E_x=0.740551$, giving $0.740551(12.00)=8.886608$. $\ddot a_x = 2.741497 + 8.886608 \approx 11.6281$.
  29. Continuous/mthly & Woolhouse
    Define the $m$-thly whole life annuity-due $\ddot a_x^{(m)}$ and state what each payment is.
    $\ddot a_x^{(m)}$ pays $\dfrac{1}{m}$ at the start of each $\dfrac{1}{m}$-year period (total \$1 per year), while $(x)$ survives: $\ddot a_x^{(m)} = \dfrac{1}{m}\sum_{k=0}^{\infty} v^{k/m}\,{}_{k/m}p_x$. More 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 $m\to\infty$.
  30. Continuous/mthly & Woolhouse
    State the **Woolhouse** (3-term) approximation for the $m$-thly annuity-due $\ddot a_x^{(m)}$.
    $\ddot a_x^{(m)} \approx \ddot a_x - \dfrac{m-1}{2m} - \dfrac{m^{2}-1}{12m^{2}}\bigl(\delta + \mu_x\bigr)$, where $\delta=\ln(1+i)$ is the force of interest and $\mu_x$ is the force of mortality at age $x$. Dropping the last term gives the 2-term Woolhouse; the full 3-term version is the standard FAM approximation.
  31. Continuous/mthly & Woolhouse
    Apply 3-term Woolhouse to find $\ddot a_x^{(12)}$ given $\ddot a_x = 14.00$, $i = 0.05$, $\mu_x = 0.02$.
    $\delta = \ln 1.05 \approx 0.048790$, $m=12$. Term 2: $\dfrac{m-1}{2m} = \dfrac{11}{24} \approx 0.458333$. Term 3: $\dfrac{m^{2}-1}{12m^{2}} = \dfrac{143}{1728} \approx 0.082755$; times $(\delta+\mu_x)=0.068790$ gives $0.082755(0.068790)\approx 0.005693$. $\ddot a_x^{(12)} \approx 14.00 - 0.458333 - 0.005693 \approx 13.5360$.
  32. Continuous/mthly & Woolhouse
    State the Woolhouse approximation for the **continuous** annuity $\bar a_x$ in terms of $\ddot a_x$.
    Let $m\to\infty$ in Woolhouse: $\dfrac{m-1}{2m}\to\dfrac12$ and $\dfrac{m^{2}-1}{12m^{2}}\to\dfrac{1}{12}$, giving $\bar a_x \approx \ddot a_x - \dfrac12 - \dfrac{1}{12}\bigl(\delta + \mu_x\bigr)$. The $-\tfrac12$ is the continuous-vs-due timing correction; the $\tfrac{1}{12}(\delta+\mu_x)$ term refines it.
  33. Continuous/mthly & Woolhouse
    Apply Woolhouse to find $\bar a_x$ given $\ddot a_x = 14.00$, $i = 0.05$, $\mu_x = 0.02$.
    $\delta = \ln 1.05 \approx 0.048790$, so $\delta + \mu_x = 0.068790$. $\bar a_x \approx \ddot a_x - \dfrac12 - \dfrac{1}{12}(\delta+\mu_x)$ $= 14.00 - 0.5 - \dfrac{0.068790}{12} = 14.00 - 0.5 - 0.005733 \approx 13.4943$.
  34. Continuous/mthly & Woolhouse
    Use the 2-term Woolhouse approximation to find $\ddot a_x^{(4)}$ given $\ddot a_x = 13.50$.
    2-term Woolhouse drops the $(\delta+\mu_x)$ term: $\ddot a_x^{(m)} \approx \ddot a_x - \dfrac{m-1}{2m}$. For $m=4$: $\dfrac{m-1}{2m} = \dfrac{3}{8} = 0.375$. $\ddot a_x^{(4)} \approx 13.50 - 0.375 = 13.1250$.
  35. Due vs immediate
    Relate the $m$-thly annuity-due $\ddot a_x^{(m)}$ to the $m$-thly annuity-immediate $a_x^{(m)}$.
    $a_x^{(m)} = \ddot a_x^{(m)} - \dfrac{1}{m}$. The 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 $a_x = \ddot a_x - 1$.)
  36. Due vs immediate
    Given $\ddot a_x^{(12)} = 13.536$, find the monthly annuity-immediate $a_x^{(12)}$.
    $a_x^{(12)} = \ddot a_x^{(12)} - \dfrac{1}{m} = 13.536 - \dfrac{1}{12} = 13.536 - 0.083333 \approx 13.4527$.
  37. Whole life annuity
    Define the increasing whole life annuity-due $(I\ddot a)_x$ and give its EPV as a sum.
    $(I\ddot a)_x$ pays $1$ at time $0$, $2$ at time $1$, $3$ at time $2$, and generally $k+1$ at time $k$, while $(x)$ survives: $(I\ddot a)_x = \sum_{k=0}^{\infty} (k+1)\,v^{k}\,{}_kp_x$. It is the EPV of a life annuity-due whose annual payment increases by \$1 each year.
  38. Whole life annuity
    Compute $(I\ddot a)_{x:\overline{3|}}$, the 3-year increasing temporary annuity-due, given $i=0.05$, $p_x=0.97$, $p_{x+1}=0.96$.
    Payments $1,2,3$ at times $0,1,2$ while alive. ${}_0p_x=1$, ${}_1p_x=0.97$, ${}_2p_x=0.97(0.96)=0.9312$; $v=0.952381$, $v^{2}=0.907029$. $(I\ddot a)_{x:\overline{3|}} = 1(1)(1) + 2(0.952381)(0.97) + 3(0.907029)(0.9312)$ $= 1 + 1.847619 + 2.533876 \approx 5.38150$.
  39. Whole life annuity
    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?
    An annuity-due paying $B$ per year has EPV $B\,\ddot a_x$. Set equal to the premium: $B\,\ddot a_x = 100{,}000 \Rightarrow B = \dfrac{100{,}000}{\ddot a_x} = \dfrac{100{,}000}{16.00} = \$6{,}250.00$ per year.
  40. Insurance-annuity relations
    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.
    By equivalence principle, premium $P$ solves $P\,\ddot a_x = A_x$, so $P = \dfrac{A_x}{\ddot a_x}$. Get $A_x$ from the relation: $d=\dfrac{0.05}{1.05}\approx0.047619$, $A_x = 1 - d\,\ddot a_x = 1 - 0.047619(13.80) = 1 - 0.657143 = 0.342857$. $P = \dfrac{0.342857}{13.80} \approx 0.024845$ per \$1, i.e. \$24.85 per \$1{,}000 of insurance.
  41. Due vs immediate
    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 $i = 0.06$, $p_x = 0.98$, $p_{x+1} = 0.97$.
    $a_{x:\overline{2|}} = v\,{}_1p_x + v^{2}\,{}_2p_x$ with ${}_1p_x=0.98$, ${}_2p_x=0.98(0.97)=0.9506$. $v=\dfrac{1}{1.06}\approx0.943396$, $v^{2}\approx0.889996$. $= 0.943396(0.98) + 0.889996(0.9506)$ $= 0.924528 + 0.846030 \approx 1.77056$.
  42. Variance
    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 $Y$.
    $d = \dfrac{0.06}{1.06}\approx 0.056604$, $d^{2}\approx 0.00320399$. Numerator: ${}^2A_x - A_x^{2} = 0.07 - 0.20^{2} = 0.07 - 0.04 = 0.03$. $\operatorname{Var}(Y) = \dfrac{0.03}{0.00320399}\approx 9.3633$. Standard deviation $= \sqrt{9.3633}\approx 3.0600$.
  43. Temporary & deferred
    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.
    The EPV per \$1 of annual benefit is $\ddot a_{\overline{10|}} + {}_{10|}\ddot a_{65} = 8.1078 + 5.9488 = 14.0566$. Single premium $= 2{,}000 \times 14.0566 = \$28{,}113.20$. The first 10 payments are guaranteed; payments from age 75 on continue only while the annuitant survives.
  44. Due vs immediate
    Why is a whole life annuity-due always worth more than the corresponding annuity-immediate, and by how much for the temporary case?
    The due annuity pays each installment one period earlier, so each payment is discounted less — strictly more valuable. For the $n$-year temporary case the gap is the time-$0$ payment minus the (missing) time-$n$ payment: $\ddot a_{x:\overline{n|}} - a_{x:\overline{n|}} = 1 - {}_nE_x = 1 - v^{n}\,{}_np_x$. (For whole life, ${}_nE_x\to0$ and the gap is exactly $1$.)