Boundary Cases for Guaranteed Annuity Periods

Why the Boundary Cases Matter

A guarantee changes what happens to annuity payments if the annuitant dies. During the guaranteed period, scheduled payments remain payable even if the annuitant is no longer alive; after that period, a payment is made only if the annuitant survives to its payment date. Valuing a guaranteed annuity therefore requires two steps: identify the payment dates covered by the guarantee, then value the remaining payments as life-contingent.

The endpoints deserve particular care. A zero-year guarantee does not make any future payment certain. If the guarantee period equals the deferment, it reaches the first payment date under the convention used here, but not the later dates. If the guarantee extends through the last date of a temporary annuity, every payment in that temporary term is certain. These statements depend on how payment dates and the guarantee period are defined, so those conventions must be made explicit.

Definition: Consider an annual deferred temporary annuity-due valued at time 0. It pays one unit at times \(m,m+1,\ldots,m+n-1\), where \(m\geq 1\) is an integer deferment and \(n\geq 1\) is the number of payments. A guarantee period of \(g\) years, measured from time 0, makes each scheduled payment at time \(k\leq g\) payable regardless of whether the annuitant survives to time \(k\). A scheduled payment at time \(k>g\) is made only if the annuitant is alive at time \(k\). We take \(g\) to be a nonnegative integer.

The guarantee concerns whether a scheduled payment is made, not its amount or timing. For this tutorial, there are no expenses or other benefits, and all payments are level and annual. The present value is per unit payment. Let \(i\) be the effective annual interest rate and \(v=1/(1+i)\). The survival probability to time \(k\) is \({}_kp_x\). For guaranteed payments, the value includes no survival factor: they are payable even after death. For a payment outside the guarantee period, the expected present value does include \({}_kp_x\).

Counting the Guaranteed Payments

The first payment date is \(m\), and the last is \(m+n-1\). A date is guaranteed if it is no later than \(g\). Thus the number \(q\) of guaranteed payments is

q=\min\!\bigl(n,\max(0,g-m+1)\bigr).

The \(+1\) matters because the endpoint is included: if \(g=m\), the date \(m\) is guaranteed. If \(g<m\), no payment date falls within the guarantee period. If \(g\geq m+n-1\), all \(n\) payment dates are guaranteed.

Present-value formula: Under the annual payment and guarantee conventions above, the value per unit payment is

V=\sum_{j=0}^{q-1}v^{m+j} +\sum_{j=q}^{n-1}v^{m+j}\,{}_{m+j}p_x.

An empty sum is defined to be zero. The first sum is the present value of guaranteed payments. The second is the present value of payments made only if the annuitant survives to the relevant payment date. When \(q=0\), the formula reduces to the ordinary deferred temporary life-annuity-due value. When \(q=n\), the second sum is empty, and mortality does not affect the value.

Why the Formula Holds

For a payment of one unit at time \(k\), the discount factor is \(v^k\). If \(k\leq g\), the payment is made with probability 1, so its expected present value is \(v^k\). If \(k>g\), it is made with probability \({}_kp_x\), so its expected present value is \(v^k{}_kp_x\). Adding the expected present values of all scheduled payments gives the annuity value.

List the payment dates.
Write the dates as \(m,m+1,\ldots,m+n-1\), including both endpoints.

Mark the guaranteed dates.
A scheduled date \(k\) is guaranteed exactly when \(k\leq g\). The number of such dates is \(q\).

Value each part.
Discount guaranteed payments without a survival factor; discount all other payments with their survival probabilities.

This derivation also shows why a guarantee period must not be confused with a deferment. The deferment determines the first payment date; the guarantee period determines which payment dates are certain. Neither quantity by itself tells us how many payments are guaranteed unless both are known.

Worked Boundary Cases

Worked Example: Zero Guarantee Period

A hypothetical annuity pays one unit annually at times 2, 3, and 4. The annual effective interest rate is \(i=5\%\). Assume the annuitant has an independent 10% probability of dying in each year, so \({}_kp_x=0.9^k\). The guarantee period is \(g=0\). Find the value.

Method. There are \(m=2\) years of deferment and \(n=3\) payments. Since \(g=0<m\), none of the future payment dates is guaranteed, and \(q=0\). The value is the sum of the survival-contingent payment values.

The discount factor is \(v=1/1.05=20/21\). For each date \(k\), \(v^k{}_kp_x=(20/21)^k(0.9)^k=(6/7)^k\). Therefore,

\begin{aligned} V &=\left(\frac{6}{7}\right)^2 +\left(\frac{6}{7}\right)^3 +\left(\frac{6}{7}\right)^4\\ &=\frac{36}{49}+\frac{216}{343}+\frac{1296}{2401}\\ &=\frac{1764+1512+1296}{2401} =\frac{4572}{2401} \approx 1.9042066. \end{aligned}

Rounded to four decimal places, the value is 1.9042 units. As a check, each term includes survival to its payment date; omitting survival would incorrectly value this as a certain annuity.

Worked Example: Guarantee Period Equals the Deferment

A hypothetical annuity pays one unit at times 2, 3, 4, and 5. The effective annual interest rate is \(5\%\), and \({}_kp_x=0.9^k\). The guarantee period is \(g=2\) years from time 0. Find the value and compare it with the value when \(g=0\).

Method. Here \(m=2\) and \(n=4\). Since \(g=m=2\), only the payment at time 2 is guaranteed: \(q=\min(4,\max(0,2-2+1))=1\). The payments at times 3, 4, and 5 remain survival-contingent.

As in the previous example, \(v=20/21\), and the contingent payment terms are \((6/7)^k\). The guaranteed payment at time 2 is discounted but is not multiplied by \({}_2p_x\):

\begin{aligned} V &=v^2+\left(\frac{6}{7}\right)^3 +\left(\frac{6}{7}\right)^4 +\left(\frac{6}{7}\right)^5\\ &=0.9070295+0.6297376+0.5397751+0.4626644\\ &\approx 2.5392065. \end{aligned}

Rounded to four decimal places, the value is 2.5392 units. With no guarantee, the time-2 term would instead be \((6/7)^2=0.7346939\). Replacing it with \(v^2=0.9070295\) increases the value by \(0.9070295-0.7346939=0.1723356\). This agrees with \(v^2(1-{}_2p_x)=0.9070295(1-0.81)=0.1723356\).

Common mistake avoided: Setting \(g=m\) does not make all payments guaranteed. Under the stated convention, it includes the payment exactly at time \(m\); later dates exceed \(g\).

Worked Example: Guarantee Extends Beyond the Temporary Term

A hypothetical annuity pays one unit at times 2, 3, and 4. The effective annual interest rate is \(5\%\), the annual survival probabilities are \({}_kp_x=0.9^k\), and the guarantee period is \(g=6\) years from time 0. Find the value.

Method. Here \(m=2\), \(n=3\), and the final scheduled payment is at time \(m+n-1=4\). Since \(g=6\geq4\), all three payments are guaranteed, so \(q=3\). Mortality does not affect this temporary annuity's value.

\begin{aligned} V &=v^2+v^3+v^4\\ &=\left(\frac{20}{21}\right)^2 +\left(\frac{20}{21}\right)^3 +\left(\frac{20}{21}\right)^4\\ &=0.9070295+0.8638376+0.8227025\\ &\approx 2.5935696. \end{aligned}

Rounded to four decimal places, the value is 2.5936 units. The result is a discounted certain annuity value: multiplying any of these terms by a survival probability would understate the value because every payment date falls within the guarantee period.

Boundary Cases at a Glance

Condition	Guaranteed payments	Valuation consequence
\(g=0\), with \(m\geq1\)	None	Every payment is survival-contingent.
\(g=m\)	Payment at time \(m\) only	The first payment is certain; later payments depend on survival.
\(g\geq m+n-1\)	All \(n\) payments	The temporary annuity is certain; survival probabilities are unnecessary.

A guarantee can never reduce the value when the payment schedule and interest and mortality assumptions are fixed: it changes some payments from survival-contingent to certain, and \({}_kp_x\leq1\). Specifically, making a single payment at time \(k\) guaranteed instead of contingent increases its expected present value by \(v^k(1-{}_kp_x)\), which is nonnegative. This provides a useful check when comparing two guarantee periods.

Common Exam Traps

Counting the endpoint incorrectly. At \(g=m\), the payment at time \(m\) is included because the rule is \(k\leq g\). It is one guaranteed payment, not zero.
Using survival for a guaranteed payment. A guaranteed payment is payable regardless of survival, so its present value is \(v^k\), not \(v^k{}_kp_x\).
Confusing the guarantee's reference date. Here \(g\) is measured from time 0, not from the first payment date. If a contract instead states a guarantee period from commencement, convert that duration into actual payment dates before applying the formula.
Assuming every payment is guaranteed because the guarantee exceeds the deferment. Compare \(g\) with the last payment date \(m+n-1\), not just with \(m\).

Key takeaway: List the scheduled payment dates first. A payment at date \(k\) contributes \(v^k\) when \(k\leq g\), and \(v^k{}_kp_x\) when \(k>g\). The boundary \(g=m\) guarantees exactly the first payment; full certainty requires \(g\geq m+n-1\).

Check Your Understanding

For each problem, use the convention that the guarantee period is measured from time 0 and includes payments at dates \(k\leq g\). Round final values to four decimal places unless the answer is exact.

A hypothetical annuity pays one unit at times 1, 2, and 3, with \(i=4\%\), \({}_kp_x=0.8^k\), and \(g=0\). Find its value. (Answer: 1.8161 units)
A hypothetical annuity pays one unit at times 2, 3, and 4, with \(i=5\%\), \({}_kp_x=0.9^k\), and \(g=2\). Find its value. (Answer: 2.0765 units)
A four-payment annuity pays at times 1, 2, 3, and 4. At \(i=4\%\), its guarantee period is \(g=4\). Find the value, regardless of the mortality assumption. (Answer: 3.6299 units)
A five-payment annuity pays at times 2 through 6. Let \(i=0\%\), \({}_kp_x=0.8^k\), and \(g=5\). Identify how many payments are guaranteed and calculate the value. (Answer: 4 guaranteed payments; value \(4.2621\) units)
For payments at times 2, 3, 4, and 5, let \(i=5\%\) and \({}_kp_x=0.9^k\). By how much does the value increase when the guarantee changes from \(g=0\) to \(g=2\)? (Answer: 0.1723 units)

What You'll Learn