Guaranteed-Period Annuity Reserve Recursions

Why a Guaranteed Period Changes the Recursion

A life annuity normally stops paying when the annuitant dies. A life annuity with a guaranteed period has a different liability: payments during the guarantee continue even if the annuitant dies, while payments after the guarantee remain contingent on survival. A reserve recursion must preserve that distinction in its death branch. Applying the ordinary life-annuity recursion throughout the guarantee can omit payments that are still owed.

This tutorial uses annual payments in advance. The reserve at integer time $t$ is measured immediately before the payment due at time $t$. Thus the payment at time $t$ is included whether it is guaranteed or payable only if the annuitant is then alive. Death during year $t$ means death in the interval $(t,t+1]$. A payment at $t+1$ is guaranteed only if $t+1$ is still within the guarantee period.

Definition: Suppose an annuity pays $b$ at times $0,1,\ldots$, with the first $m$ payments guaranteed. The payments at times $0,\ldots,m-1$ are due whether or not the annuitant is alive. Payments at times $m,m+1,\ldots$ are due only if the annuitant is alive at the payment time. Let $V_t$ be the reserve immediately before the payment at time $t$, conditional on the annuitant being alive then, and let $H_t$ be the value at time $t$ of the remaining guaranteed payments alone. For $t<m$, $$ H_t=b\sum_{j=0}^{m-t-1}v^j, $$ and $H_t=0$ for $t\ge m$. Here $v=1/(1+i)$, with $i$ the effective annual interest rate.

The reserve $V_t$ includes both the guaranteed payments still due and any later life-contingent payments. The quantity $H_t$ includes only the guaranteed payments remaining from time $t$. In particular, $H_t$ is not generally equal to the undiscounted total of those payments: it is their value at time $t$.

Deriving the One-Year Reserve Recursion

Write $p_{x+t}={}_1p_{x+t}$ for the probability that a life aged $x+t$ survives one year, and $q_{x+t}={}_1q_{x+t}=1-p_{x+t}$ for the probability of dying in that year. Conditional on being alive at time $t$, the payment due at $t$ is included in $V_t$. The value at time $t$ of payments and liabilities from time $t+1$ onward depends on whether the annuitant survives the year.

Include the payment due now.
At time $t$, add $b$, since the reserve is measured immediately before the payment at $t$.

Value the survival branch.
With probability $p_{x+t}$, the annuitant is alive at $t+1$, so the continuation value is $V_{t+1}$.

Value the death branch.
With probability $q_{x+t}$, the annuitant dies during the year. The remaining guaranteed payments are still owed, with value $H_{t+1}$ at time $t+1$. If the guarantee has ended by then, this value is zero.

Discount the continuation value.
Both branch values are measured at $t+1$, so their probability-weighted value is discounted by $v$ to time $t$.

Theorem: Under annual effective interest $i$, annual payments in advance, and one-year survival probability $p_{x+t}$, the recursion is $$ V_t=b+v\left(p_{x+t}V_{t+1}+q_{x+t}H_{t+1}\right),\qquad t<m. $$ For $t\ge m$, the ordinary life-annuity-due recursion applies: $$ V_t=b+v p_{x+t}V_{t+1}. $$ This post-guarantee recursion presumes the annuity has a payment due at each integer time while the annuitant is alive. At a finite contract termination, the appropriate terminal value replaces $V_{t+1}$.

The death-branch term is the essential adjustment. During the guarantee, death does not erase the payments that are guaranteed after the death but before the guarantee ends. After the guarantee, there are no remaining guaranteed payments, so $H_{t+1}=0$ and death leaves no continuation liability.

Prospective check: For $t<m$, the reserve can also be written as the value of the guaranteed payments plus the value of the later annuity, conditional on survival to the guarantee boundary: $$ V_t=b\sum_{j=0}^{m-t-1}v^j+v^{m-t}{}_{m-t}p_{x+t}V_m. $$ This expression assumes that the annuity after time $m-1$ continues only while the annuitant is alive. It does not apply unchanged if the contract has additional benefits after death or a different payment condition.

The prospective expression follows by grouping payments into two sets. First, each guaranteed payment from $t$ through $m-1$ is due regardless of future survival, so its present value is $b v^j$ at time $t$. Second, the post-guarantee liability is $V_m$ at time $m$, but it is payable only if the annuitant survives from $t$ to $m$. Its expected value at time $t$ is therefore $v^{m-t}{}_{m-t}p_{x+t}V_m$. The recursion and this grouped present value describe the same future cash flows and should agree.

Worked Recursions

Worked Example: A Two-Payment Guarantee With a Life-Annuity Tail

A hypothetical annuity pays $100$ at times $0,1,2,\ldots$. The payments at times 0 and 1 are guaranteed; payments from time 2 onward require survival. The effective annual interest rate is $i=25\%$. Assume independent annual mortality with constant $q=0.20$, so $p=0.80$ in every year. Find the reserve at time 0. The post-guarantee annuity is whole life, with no terminal payment.

Method and timing. Use the death/survival recursion during the two-payment guarantee, then solve the constant-recursion equation for the post-guarantee whole-life annuity-due reserve. This avoids treating the payment at time 1 as life-contingent after a death during year 0.

Here $v=1/1.25=0.80$. Let $V_2$ be the reserve just before the time-2 payment. Because mortality and interest are constant, the post-guarantee recursion gives $V_2=100+vpV_3$, and $V_3=V_2$. Therefore, $$ V_2=100+(0.80)(0.80)V_2, \qquad (1-0.64)V_2=100, \qquad V_2=\frac{100}{0.36}=277.7778. $$

At time 1, the payment of $100$ is the last guaranteed payment. Death during year 1 does not create a time-2 payment, so the death-branch continuation value is zero: $$ V_1=100+(0.80)(0.80)V_2 =100+0.64(277.7778) =277.7778. $$

At time 0, if death occurs during year 0, the payment at time 1 remains guaranteed. Its value at time 1 is $H_1=100$. Thus, $$ \begin{aligned} V_0 &=100+0.80\left[(0.80)(277.7778)+(0.20)(100)\right]\\ &=100+0.80(222.2222+20)\\ &=293.7778. \end{aligned} $$ The reserve is $\boxed{\$293.78}$, rounded to the nearest cent.

Check. The prospective value is $100$ now, $100v=80$ at time 1, and a post-guarantee reserve at time 2 only if the life survives two years. The last component is $v^2p^2V_2=(0.64)(0.64)(277.7778)=113.7778$. The total is $100+80+113.7778=293.7778$, agreeing with the recursion.

Worked Example: A Finite Annuity With Two Payments After the Guarantee

A hypothetical annuity pays $100$ at times 0 through 4 and has its first three payments guaranteed. Payments at times 3 and 4 are payable only if the annuitant is alive. The effective annual interest rate is $25\%$, and the annual mortality model has $q=0.20$ and $p=0.80$ in each year. Find the reserve at time 0. There are no payments after time 4.

Method and timing. Work backward from the final payment, using the guaranteed-only value in the death branch while the guarantee remains in force. This avoids applying survival to the payments at times 1 and 2, which remain due even after death.

The discount factor is $v=0.80$. At time 4, the last payment is due and there is no later liability, so $V_4=100$. At time 3, the time-4 payment is payable only if the life survives: $$ V_3=100+vpV_4 =100+(0.80)(0.80)(100) =164. $$

At time 2, the payment at time 2 is guaranteed, but time 3 is not. Death during year 2 therefore leaves no payment at time 3, and $$ V_2=100+vpV_3 =100+0.64(164) =204.96. $$ At time 1, the payment at time 2 remains guaranteed even if death occurs during year 1. Its value at time 2 is $H_2=100$: $$ \begin{aligned} V_1 &=100+0.80\left[(0.80)(204.96)+(0.20)(100)\right]\\ &=100+0.80(163.968+20)\\ &=247.1744. \end{aligned} $$

At time 0, death during year 0 does not remove the guaranteed payments at times 1 and 2. Their value at time 1 is $$ H_1=100+100(0.80)=180. $$ Consequently, $$ \begin{aligned} V_0 &=100+0.80\left[(0.80)(247.1744)+(0.20)(180)\right]\\ &=100+0.80(197.73952+36)\\ &=286.991616. \end{aligned} $$ The reserve is $\boxed{\$286.99}$, rounded to the nearest cent.

Check. The guaranteed payments have present value $100+100(0.80)+100(0.80)^2=244$. The payment at time 3 has expected present value $100(0.80)^3(0.80)^3=26.2144$, since survival to time 3 and discounting each contribute a factor of $0.80^3$. The time-4 payment has expected present value $100(0.80)^4(0.80)^4=16.777216$. Their sum is $244+26.2144+16.777216=286.991616$, matching the recursion.

Worked Example: Unequal Mortality Across the Guarantee Boundary

A hypothetical annuity pays $120$ at times 0, 1, 2, and 3. The payments at times 0 and 1 are guaranteed; those at times 2 and 3 require survival. The effective annual interest rate is $5\%$. The one-year survival probabilities are $p_x=0.90$, $p_{x+1}=0.80$, and $p_{x+2}=0.75$. There are no payments after time 3. Find the reserve at time 0.

Method and timing. Use the appropriate survival probability for each policy year, and keep the death-branch value only while guaranteed payments remain. This avoids using one average mortality rate across years or incorrectly guaranteeing the payment at time 2.

The discount factor is $v=1/1.05=20/21$. At time 3, the final payment is due, so $V_3=120$. At time 2, the time-3 payment requires survival with probability $0.75$: $$ V_2=120+\frac{20}{21}(0.75)(120) =120+\frac{600}{7} =\frac{1440}{7} =205.7143. $$

At time 1, the payment at time 1 is guaranteed, but the time-2 payment is not. A death during year 1 leaves no later guaranteed payment, so $H_2=0$: $$ \begin{aligned} V_1 &=120+\frac{20}{21}(0.80)V_2\\ &=120+\frac{16}{21}\left(\frac{1440}{7}\right)\\ &=120+\frac{7680}{49}\\ &=\frac{13560}{49}=276.7347. \end{aligned} $$ At time 0, death during year 0 leaves the guaranteed time-1 payment. Its value at time 1 is $H_1=120$. Therefore, $$ \begin{aligned} V_0 &=120+\frac{20}{21}\left[(0.90)\left(\frac{13560}{49}\right)+(0.10)(120)\right]\\ &=120+\frac{20}{21}\left(\frac{12792}{49}\right)\\ &=368.6297. \end{aligned} $$ The reserve is $\boxed{\$368.63}$, rounded to the nearest cent.

Check. The guaranteed payments have value $120+120v=234.2857$. The time-2 payment has expected present value $120v^2(0.90)(0.80)=78.3673$. The time-3 payment has expected present value $120v^3(0.90)(0.80)(0.75)=55.9767$. These sum to $368.6297$, as required.

Common Mistakes and Practical Checks

The most important exam trap is using the wrong death-branch value. During a guarantee, the death branch is not automatically zero. It is the present value, at the next valuation time, of the guaranteed payments that remain. Conversely, after the guarantee ends, a death branch should not include a payment merely because the annuity has paid in earlier years.

For example, in the two-payment guarantee example above, a common incorrect calculation is $$ V_0^{\text{wrong}}=100+v p V_1 =100+(0.80)(0.80)(277.7778) =277.7778. $$ This incorrectly sets the death-branch value to zero, omitting the guaranteed payment at time 1 when death occurs during year 0. The corrected calculation includes $qH_1$: $$ V_0=100+v(pV_1+qH_1) =100+0.80\left[(0.80)(277.7778)+(0.20)(100)\right] =293.7778. $$

A second frequent error is counting a number of years rather than a number of payments. Under the timing convention here, an $m$-payment guarantee means payments at times $0$ through $m-1$. It does not mean that the payment at time $m$ is guaranteed. Before using a recursion, write down the payment times and mark which ones are guaranteed; this makes the boundary unambiguous.

Finally, distinguish an annuity-due reserve from an annuity-immediate reserve. Here the reserve is immediately before a payment, so $b$ is included at the current time. If a problem instead defines the reserve immediately after a payment, the cash-flow timing and recursion must be adjusted consistently. Do not change the timing convention midway through a calculation.

Key takeaway: While guaranteed payments remain, the recursion is $V_t=b+v(p_{x+t}V_{t+1}+q_{x+t}H_{t+1})$. The death branch carries only the remaining guaranteed payments; the survival branch carries the full reserve for a life that is still alive. After the guarantee boundary, the ordinary life-annuity recursion applies.

Check Your Understanding

For each problem, identify which payments survive a death before applying the recursion.

A hypothetical annuity pays $200$ at times 0, 1, and 2, all guaranteed. The effective annual interest rate is $10\%$. What is the reserve at time 0? (Answer: $\$547.11$)
A hypothetical annuity pays $50$ annually in advance, with the payments at times 0 and 1 guaranteed and a whole-life annuity thereafter. Assume $i=25\%$ and constant annual $q=0.20$. Find the reserve at time 0. (Answer: $\$146.89$)
A hypothetical annuity pays $80$ at times 0 and 1 regardless of survival, then pays $80$ at times 2 and 3 only if alive. There are no later payments. Assume $i=0$ and annual survival probability $p=0.50$ in each year. Find the reserve at time 0. (Answer: $\$190.00$)
A hypothetical annuity pays $100$ at times 0, 1, and 2, with the first three payments guaranteed. After time 2, there is only one possible further payment: $100$ at time 3 if alive. Find the reserve immediately before the time-1 payment, given $i=0$, $p_{x+1}=0.80$, and $p_{x+2}=0.50$. (Answer: $\$240.00$)

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