Life Annuity Values from Life-Table Columns

Life-Table Columns and Annual Annuity Payments

An annual life annuity pays only while the annuitant is alive. To value it from a life table, combine two ingredients: the probability the annuitant is alive on each payment date, and the discount factor for that date. Getting the dates right matters: an annuity-due pays at the beginning of each year, while an annuity-immediate pays at the end.

This tutorial uses integer-age life-table columns. It does not require an assumption about deaths within a year: the table supplies the probabilities of surviving from one integer age to the next. All examples assume level annual payments of 1, an effective annual interest rate \(i\), and mortality represented by the stated life table. Multiply the resulting value by the actual annual payment amount when it is not 1.

Definition: In a life table, \(l_x\) is the number of lives surviving to exact age \(x\), and \(d_x=l_x-l_{x+1}\) is the number dying between ages \(x\) and \(x+1\). For an individual aged \(x\), the probability of surviving \(k\) full years is \({}_kp_x=l_{x+k}/l_x\), provided \(l_x>0\). Thus the probability of being alive at the annual payment date \(k\) is found from the survivor count at age \(x+k\).

The death column can also be used to recover a future survivor count. For \(k\geq 1\),

l_{x+k}=l_x-\sum_{j=0}^{k-1}d_{x+j}.

This subtraction includes deaths over each of the \(k\) years from age \(x\) to age \(x+k\). If a life table gives both columns, check that \(d_x=l_x-l_{x+1}\) before using it.

Temporary Annuities: Match Each Payment to Its Survival Probability

A temporary annuity-immediate with term \(n\) makes possible payments at times \(1,2,\ldots,n\). The payment at time \(k\) is made only if the annuitant is alive then, so it is weighted by \({}_kp_x\) and discounted by \(v^k\), where \(v=1/(1+i)\). A temporary annuity-due with \(n\) payments pays at times \(0,1,\ldots,n-1\). Its time-zero payment is certain for a person alive at issue; each later payment is conditional on survival to its payment date.

Formula: For unit annual payments, the temporary annuity-immediate value is \(a_{x:\overline{n}|}=\sum_{k=1}^{n}v^k\,{}_kp_x\). The temporary annuity-due value is \(\ddot{a}_{x:\overline{n}|}=\sum_{k=0}^{n-1}v^k\,{}_kp_x\), where \({}_0p_x=1\). These formulas use annual payments at the stated integer times; they do not value payments at fractional ages.

Worked Example: A Three-Year Temporary Annuity-Immediate

A hypothetical life table has \(l_{44}=1{,}000\), \(l_{45}=900\), \(l_{46}=700\), and \(l_{47}=400\). Find the value at age 44 of a three-year temporary annuity-immediate with annual payment 1 and effective annual interest rate \(i=25\%\). Use the table as the mortality basis.

Method. The payments are at times 1, 2, and 3. First find each survival probability from the corresponding survivor count, then discount that payment. Here \(v=1/1.25=0.8\).

At time 1, \({}_1p_{44}=900/1{,}000=0.9\), so the present-value contribution is \(0.8(0.9)=0.72\).

At time 2, \({}_2p_{44}=700/1{,}000=0.7\), so the contribution is \(0.8^2(0.7)=0.64(0.7)=0.448\).

At time 3, \({}_3p_{44}=400/1{,}000=0.4\), so the contribution is \(0.8^3(0.4)=0.512(0.4)=0.2048\).

Adding the contributions gives \(0.72+0.448+0.2048=1.3728\). The value is 1.3728 per unit of annual payment. As a check, each survival probability is at most 1, and later payments have both lower survival probabilities and stronger discounting.

Common mistake avoided: A time-1 payment in an annuity-immediate is conditional on survival to age 45. It is not paid with probability \(d_{44}/l_{44}=0.1\), which is the probability of dying during the year.

Worked Example: A Three-Payment Temporary Annuity-Due

A hypothetical table gives \(l_{55}=400\), \(l_{56}=360\), and \(l_{57}=300\). Calculate the value at age 55 of a three-payment temporary annuity-due with annual payment 1 and \(i=5\%\). Assume the annuitant is alive at age 55.

Method. The three payment dates are 0, 1, and 2, not 1, 2, and 3. The time-zero payment is certain. The later survival probabilities are \(360/400=0.9\) and \(300/400=0.75\). The discount factor is \(v=1/1.05\).

The time-zero contribution is \(1\). The time-1 contribution is \(0.9/1.05=0.8571428571\). The time-2 contribution is \(0.75/(1.05)^2=0.75/1.1025=0.6802721088\).

Therefore the value is \(1+0.8571428571+0.6802721088=2.5374149659\), or 2.5374 rounded to four decimal places. This is the value per unit annual payment.

Common mistake avoided: Using ages 56, 57, and 58 would value three payments at times 1, 2, and 3. That is not this annuity-due: its first payment occurs immediately, and its last occurs two years after issue.

Whole-Life Values from Survivor and Death Columns

A whole-life annuity continues for as long as the annuitant is alive. If the table has a finite ultimate age at which the survivor count reaches zero, the sum has only finitely many nonzero terms. Otherwise, the value is an infinite sum, provided that the expected discounted payments converge. A table that ends while \(l_x\) is still positive does not, by itself, provide enough information to calculate the complete whole-life value.

Formula: For unit annual payments, the whole-life annuity-immediate value is \(a_x=\sum_{k\geq 1}v^k(l_{x+k}/l_x)\), and the whole-life annuity-due value is \(\ddot{a}_x=\sum_{k\geq 0}v^k(l_{x+k}/l_x)\), with \(l_x/l_x=1\) for the time-zero term. For a finite table, stop when subsequent survivor counts are zero. The time-zero term gives \(\ddot{a}_x=1+a_x\) under these annual-payment definitions.

The formulas follow from the expected present value of the payments. Let the payment at time \(k\) be 1 if the life is then alive and 0 otherwise. Its present value is \(v^k\) times that indicator. The expectation of the indicator is the probability of survival to time \(k\), namely \({}_kp_x\). Therefore the expected present value at that date is \(v^k{}_kp_x\). Adding those expected contributions over the payment dates gives the annuity value by linearity of expectation. No independence assumption between survival at different dates is needed.

Set the payment dates.
For an immediate annuity, start at time 1. For a due annuity, include the certain time-zero payment.

Find survival to each date.
Use \(l_{x+k}/l_x\), or reconstruct \(l_{x+k}\) by subtracting the intervening \(d\) values from \(l_x\).

Discount and add.
Multiply each payment-date survival probability by \(v^k\), then sum the contributions.

Worked Example: Whole-Life Values Using Death Columns

A hypothetical life table has \(l_{60}=1{,}000\), \(d_{60}=200\), \(d_{61}=300\), and \(d_{62}=250\), followed by \(d_{63}=250\). Find the whole-life annuity-immediate and annuity-due values at age 60 when \(i=25\%\). Payments are 1 annually and mortality follows this table.

Method. Reconstruct survivor counts using the death columns. We have \(l_{61}=1{,}000-200=800\), \(l_{62}=800-300=500\), \(l_{63}=500-250=250\), and \(l_{64}=250-250=0\). Thus the probabilities of survival to times 1, 2, and 3 are \(0.8\), \(0.5\), and \(0.25\); survival to time 4 is zero. The discount factor is \(v=1/1.25=0.8\).

For the annuity-immediate, the time-1 contribution is \(0.8(0.8)=0.64\). The time-2 contribution is \(0.8^2(0.5)=0.64(0.5)=0.32\). The time-3 contribution is \(0.8^3(0.25)=0.512(0.25)=0.128\). The time-4 contribution is zero because \(l_{64}=0\).

Therefore \(a_{60}=0.64+0.32+0.128=1.088\). The annuity-due includes one additional certain payment at time zero, so \(\ddot{a}_{60}=1+1.088=2.088\). The results are 1.088 and 2.088, respectively. As a check, their difference is exactly 1, the value of the due annuity’s extra immediate payment.

Common mistake avoided: The \(d_x\) entries count deaths during a year; they are not the probabilities of receiving a life-annuity payment. Payments depend on being alive at the payment date, so use the survivor counts \(l_{x+k}\).

Exam Checks and Practical Use

A reliable calculation begins by writing the payment dates before entering numbers. Then label each probability as survival to a particular date, calculate each discount factor using the same time unit as the interest rate, and add the terms. This makes the annuity type visible in the calculation rather than leaving it implicit.

Check the term. An \(n\)-payment annuity-due has payment dates 0 through \(n-1\); an \(n\)-payment annuity-immediate has dates 1 through \(n\).
Check the denominator. For a life aged \(x\), divide future survivor counts by \(l_x\), not by a later-age survivor count.
Check survival versus death. \(l_{x+k}/l_x\) is survival to time \(k\); a \(d\) entry concerns deaths over an interval.
Check the table boundary. Do not treat a truncated table as an ultimate-age table unless the remaining survival information is supplied.

These calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. The same discipline—identify payment timing, obtain the matching survival probability, and discount by the correct number of years—supports more involved life-annuity and premium calculations.

Check Your Understanding

For each problem, assume annual payments of 1 and use the stated life table as the mortality basis. Round values to four decimal places where needed.

A hypothetical table has \(l_{40}=800\), \(l_{41}=760\), and \(l_{42}=680\). Find the value at age 40 of a two-year temporary annuity-immediate at \(i=4\%\). (Answer: 1.6993)
A hypothetical table has \(l_{50}=500\), \(l_{51}=450\), and \(l_{52}=400\). Find the value at age 50 of a three-payment temporary annuity-due at \(i=5\%\). (Answer: 2.5828)
A hypothetical table has \(l_{60}=1{,}000\), \(l_{61}=900\), \(l_{62}=600\), \(l_{63}=300\), and \(l_{64}=0\). Find the whole-life annuity-immediate value at age 60 when \(i=10\%\). (Answer: 1.5394)
At age 70, a hypothetical table gives \(l_{70}=600\), \(d_{70}=60\), \(d_{71}=90\), and \(d_{72}=150\). Find the value of a three-year temporary annuity-immediate at \(i=5\%\). (Answer: 1.9693)
A hypothetical table has \(l_{30}=200\), \(l_{31}=160\), \(l_{32}=100\), and \(l_{33}=0\). At \(i=25\%\), find both whole-life annuity values at age 30 and verify their difference. (Answer: \(a_{30}=0.9600\), \(\ddot{a}_{30}=1.9600\); difference \(=1.0000\))

What You'll Learn