Why a Finite Table Needs a Terminal-Age Assumption
A whole-life insurance benefit is payable whenever death occurs, so its value depends on mortality throughout the insured’s future lifetime. A finite mortality table, however, lists lives only up to a final age. To calculate a whole-life value from that table, you need to know what the table assumes about survivors at its endpoint. In particular, the probability of reaching that age is not the same as the probability of dying at that age.
This tutorial treats a discrete whole-life insurance with a unit benefit paid at the end of the year of death. The key task is to include the final probability of survival to the table’s terminal age, then apply the stated assumption about what happens to those survivors. Without that assumption, the table alone may not determine a whole-life value.
Some tables use a different endpoint convention: they call \(\omega\) the age by which all lives have died, so that \(\ell_\omega=0\). Under that convention, the final age with survivors is \(\omega-1\). Always identify which ages have lives and which final death interval the table represents before setting up the present value.
Whole-Life Value From a Finite Table
For an insured aged \(x\), let \(A_x\) denote the actuarial present value of a unit benefit paid at the end of the year of death. If the effective annual interest rate is \(i\), then \(v=1/(1+i)\). Under the terminal-age convention above, the value is a finite sum:
For life-table counts, the probability of dying between ages \(x+k\) and \(x+k+1\), conditional on being alive at age \(x\), is \((\ell_{x+k}-\ell_{x+k+1})/\ell_x\). This is also \({}_kp_xq_{x+k}\). Therefore the same value can be calculated directly from the table:
The final term is especially important. The probability of reaching age \(\omega\) is \(\ell_\omega/\ell_x\). Under the assumption that everyone alive at age \(\omega\) dies in the next year, that is also the unconditional probability, from age \(x\), of receiving the benefit in that final interval. Its present value is \(v^{\omega-x+1}\ell_\omega/\ell_x\).
Why the Formula Holds
The formula follows by splitting the possible death years into mutually exclusive events. Death during the year of age \(x+k\) requires surviving \(k\) full years from age \(x\), with probability \({}_kp_x\), and then dying during the next year, with conditional probability \(q_{x+k}\). The event therefore has probability \({}_kp_xq_{x+k}\). Since its benefit is paid \(k+1\) years after issue, its present value when it occurs is \(v^{k+1}\). Multiplying the payment’s present value by the event probability gives its contribution to the expected present value. Adding the contributions over all possible death years gives \(A_x\).
At the terminal age, \(q_\omega=1\), so the final event probability becomes \({}_{\omega-x}p_xq_\omega={}_{\omega-x}p_x\). The terminal survivors are not omitted; the assumption assigns all of their deaths to the final listed interval. If instead some survive beyond age \(\omega\), their later death probabilities and payment times must be specified before a whole-life value can be calculated.
Read the terminal age and confirm whether the table sets \(q_\omega=1\), or uses another endpoint convention.
Use \((\ell_y-\ell_{y+1})/\ell_x\) for deaths in the year beginning at age \(y\), or use \({}_kp_xq_{x+k}\).
Deaths in the year beginning at age \(x+k\) produce a benefit at the end of year \(k+1\), so multiply by \(v^{k+1}\).
Under \(q_\omega=1\), add the probability of reaching age \(\omega\), discounted to the end of the final death year.
Worked Values From Life-Table Counts
Worked Example: Valuing a Table With a Certain Final Death Interval
A hypothetical table has \(\ell_{60}=1{,}000\), \(\ell_{61}=900\), \(\ell_{62}=720\), \(\ell_{63}=360\), and \(\ell_{64}=0\). Age 63 is the terminal age, so all lives reaching it die before age 64. Calculate \(A_{60}\) at an effective annual interest rate of 25%. Give the value to four decimal places.
Method. Calculate each death probability from the age-60 cohort, discount the end-of-year benefit, and add the four contributions. This avoids treating the 360 lives reaching age 63 as if they had already died before that age.
The number dying in each interval is \(1{,}000-900=100\), \(900-720=180\), \(720-360=360\), and \(360-0=360\). Dividing each by \(\ell_{60}=1{,}000\) gives the probabilities \(0.10\), \(0.18\), \(0.36\), and \(0.36\). They sum to \(1.00\), as required under the terminal-age assumption.
Here \(v=1/1.25=0.8\). The benefits are paid at the ends of years 1, 2, 3, and 4, respectively:
Thus \(A_{60}=0.5270\) to four decimal places. As a check, the final term uses the \(0.36\) probability of reaching age 63 and then dying in the final interval; its payment is discounted four years, not three.
Worked Example: Recovering the Terminal Death Probability
A hypothetical table gives \(\ell_{40}=800\), \(\ell_{41}=600\), \(\ell_{42}=300\), and \(\ell_{43}=80\). Age 43 is terminal, with \(q_{43}=1\) and \(\ell_{44}=0\). Calculate \(A_{40}\) at an effective annual interest rate of 25%, to five decimal places. Also identify the probability of surviving from age 40 to the terminal age.
Method. Use the cohort counts to find the unconditional death probabilities, including the terminal group. This directly accounts for the remaining survival probability at age 43.
The numbers dying from one age to the next are \(800-600=200\), \(600-300=300\), \(300-80=220\), and \(80-0=80\). Dividing by 800 gives \(0.25\), \(0.375\), \(0.275\), and \(0.10\). The probability of reaching age 43 is \(80/800=0.10\); because \(q_{43}=1\), that is also the probability of dying in the final interval.
With \(v=0.8\), the present value is:
Therefore \(A_{40}=0.62176\), and the probability of surviving to age 43 is \(0.10\). The probability check is \(0.25+0.375+0.275+0.10=1\). A common mistake is to stop after the age-42 interval; that would omit the \(0.10\) of the cohort reaching age 43.
The Value Depends on What Happens After the Table Ends
A terminal age is an assumption about the timing of deaths, not merely a formatting detail. If a table stops reporting mortality at age \(\omega\) but does not say that \(q_\omega=1\), then the remaining probability of survival cannot simply be discarded. Any continuation of mortality beyond the last reported age changes the probability distribution of payment times and may change the whole-life value.
The following comparison uses the same known table entries and interest rate but changes the assumption for those reaching age 62.
Worked Example: Comparing Two Terminal-Age Assumptions
A hypothetical table has \(\ell_{60}=1{,}000\), \(\ell_{61}=800\), and \(\ell_{62}=400\), and provides no later counts. Use an effective annual interest rate of 25%. Calculate \(A_{60}\) under (a) the assumption that age 62 is terminal and \(q_{62}=1\), and (b) the assumption that \(q_{62}=0.5\) and all survivors reaching age 63 die before age 64. Give both values to five decimal places.
Method. First use the known counts to calculate the probabilities of death at ages 60 and 61. Then allocate the \(0.40\) probability of reaching age 62 according to each terminal assumption. This avoids silently treating an incomplete table as if its endpoint convention were known.
For both assumptions, the probabilities of death at ages 60 and 61 are \((1{,}000-800)/1{,}000=0.20\) and \((800-400)/1{,}000=0.40\). The probability of reaching age 62 is \(400/1{,}000=0.40\).
(a) Age 62 is terminal. All \(0.40\) who reach age 62 die in that year, with payment at the end of year 3. With \(v=0.8\):
(b) Half of those at age 62 survive to age 63. Of the original cohort, \(0.40(0.5)=0.20\) die during the year of age 62, and \(0.20\) survive to age 63. Under the stated final assumption, that remaining \(0.20\) die in the year of age 63, with payment at the end of year 4:
The values are \(0.62080\) under (a) and \(0.60032\) under (b), both to five decimal places. The probabilities in each case sum to 1. Moving part of the death probability to a later year lowers its present-value contribution at a positive interest rate. The two values differ because the terminal-age assumptions differ, not because the known table entries changed.
Common Mistakes and Practical Checks
- Omitting terminal survivors. If \(\ell_\omega>0\), those lives represent positive probability of reaching the final age. Under \(q_\omega=1\), include their deaths in the last interval.
- Using the wrong payment time. A death during the year beginning at age \(x+k\) is paid at the end of year \(k+1\), so the discount factor is \(v^{k+1}\).
- Assuming the table specifies more than it does. A last listed count does not, by itself, establish that everyone at that age dies within the next year. State the endpoint assumption, or specify the mortality continuation.
- Mixing endpoint conventions. If the table sets \(\ell_\omega=0\), then there are no survivors at age \(\omega\); do not also add a terminal group at that age.
Handling finite tables correctly is useful whenever a valuation must be completed from a bounded set of life-table counts. The essential checks are that all death probabilities, including the terminal group, sum to 1; that the terminal convention is explicit; and that each benefit is discounted to its correct year-end. These methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, identify the terminal assumption and include the probability of reaching the terminal age in the calculation.
- A hypothetical table has \(\ell_{35}=1{,}000\), \(\ell_{36}=900\), and \(\ell_{37}=600\). Age 37 is terminal, with \(q_{37}=1\). At an effective annual interest rate of 25%, find \(A_{35}\) to four decimal places. (Answer: \(0.5792\))
- A hypothetical table has \(\ell_{50}=400\), \(\ell_{51}=300\), and \(\ell_{52}=120\). Age 52 is terminal, with \(q_{52}=1\). At an effective annual interest rate of 25%, calculate \(A_{50}\) to four decimal places. (Answer: \(0.6416\))
- For an insured aged 40, a hypothetical table gives \(\ell_{40}=1{,}000\), \(\ell_{41}=700\), \(\ell_{42}=250\), and \(\ell_{43}=150\). Age 43 is terminal. At an effective annual interest rate of 25%, calculate \(A_{40}\) to five decimal places. (Answer: \(0.64064\))
- For an insured aged 55, suppose \(q_{55}=0.20\), \(q_{56}=0.25\), and all lives reaching age 57 die in the following year. At an effective annual interest rate of 25%, calculate \(A_{55}\) to four decimal places. (Answer: \(0.5952\))
- A hypothetical table has \(\ell_{60}=1{,}000\), \(\ell_{61}=700\), and \(\ell_{62}=350\). At an effective annual interest rate of 25%, compare \(A_{60}\) if (a) \(q_{62}=1\), and (b) \(q_{62}=0.4\) with \(q_{63}=1\). Give each value to four decimal places. (Answer: (a) \(0.6432\); (b) \(0.6217\))