Using Life Tables to Calculate Probabilities
A life table may report the number of lives remaining at each integer age, annual mortality rates, or only selected entries of either kind. The calculation task is to translate those entries into the probability asked for: death within one year, survival for several years, or death over an interval conditional on reaching its starting age. A reliable solution starts by identifying the exact ages and years involved, then uses the table entries for those intervals.
The calculations below use integer ages and one-year mortality rates. They do not require an assumption about deaths at fractional ages. Unless stated otherwise, a life table is treated as describing the mortality probabilities for the hypothetical life under consideration.
For a life aged \(x\), the probability of surviving \(t\) complete years is the ratio of the lives at the endpoint to the lives at the starting age. Equivalently, it is the product of the one-year survival probabilities for all intervening ages. The probability of death during those \(t\) years is the complement.
Why the Life Table Ratios Give Probabilities
The life table counts can be read as proportions of a starting group. Of the \(\ell_x\) lives at age \(x\), exactly \(\ell_{x+t}\) remain at age \(x+t\). Thus the proportion of the starting group surviving is \(\ell_{x+t}/\ell_x\). The same result follows by multiplying one-year survival probabilities: survival through each year requires survival through the preceding years as well.
For a conditional probability, the starting group changes. Given survival to age \(x+m\), the relevant group has size \(\ell_{x+m}\), not \(\ell_x\). Therefore the probability of death before age \(x+m+n\), conditional on reaching age \(x+m\), is
This is the same calculation as an unconditional multi-year probability, but with the conditional starting age used as the denominator and the number of years measured from that age.
Worked Example: One-Year, Multi-Year, and Conditional Probabilities
A hypothetical life table gives \(\ell_{40}=10{,}000\), \(\ell_{41}=9{,}900\), \(\ell_{42}=9{,}750\), and \(\ell_{43}=9{,}600\). Calculate (a) \(q_{40}\), (b) the probability that a life aged 40 dies before age 43, and (c) the probability of death before age 43 given survival to age 41.
Method. Use the ratio of endpoint to starting counts for each interval. In part (c), the condition changes the starting age to 41.
(a) One-year death probability. The number dying from age 40 to 41 is \(d_{40}=10{,}000-9{,}900=100\). Hence \(q_{40}=100/10{,}000=0.01\).
(b) Death before age 43 from age 40. The probability of survival to age 43 is \(\ell_{43}/\ell_{40}=9{,}600/10{,}000=0.96\). Therefore the probability of death before age 43 is \(1-0.96=0.04\).
(c) Conditional death before age 43 given survival to age 41. Starting at age 41, survival to age 43 has probability \(9{,}600/9{,}900=32/33\). Thus the required probability is \(1-32/33=1/33=0.030303\ldots\), or 0.0303 to four decimal places.
Check. The conditional calculation uses 9,900 as its denominator because only lives reaching age 41 are in the conditioned group. The result is smaller than the unconditional two-year death probability from age 40, \(1-9{,}600/10{,}000=0.04\), as expected after excluding deaths in the first year.
Calculating from One-Year Rates
Sometimes a table supplies \(q_x\) values rather than \(\ell_x\) counts. For a life aged \(x\), calculate survival through each year as \(p_x=1-q_x\), multiply the relevant survival probabilities, and take a complement if the question asks for death. For a conditional probability, begin with the rate at the age when the condition says the life has survived.
Worked Example: Multi-Year Probabilities from Annual Rates
A hypothetical table gives \(q_{60}=0.02\), \(q_{61}=0.03\), and \(q_{62}=0.05\). For a life aged 60, calculate (a) the probability of surviving to age 62, (b) the probability of death before age 63, and (c) the probability of death between ages 61 and 63 given survival to age 61.
Method. Convert each annual death probability to survival, multiply over the required years, and use a complement for death.
(a) Survival to age 62. The one-year survival probabilities are \(p_{60}=1-0.02=0.98\) and \(p_{61}=1-0.03=0.97\). Therefore \({}_2p_{60}=0.98(0.97)=0.9506\).
(b) Death before age 63. Include the third year, for which \(p_{62}=1-0.05=0.95\). Survival to age 63 is \(0.98(0.97)(0.95)=0.9506(0.95)=0.90307\). The probability of death before age 63 is \(1-0.90307=0.09693\), or 0.0969 to four decimal places.
(c) Conditional death between ages 61 and 63. Given survival to age 61, the relevant two-year survival probability is \(p_{61}p_{62}=0.97(0.95)=0.9215\). The required conditional death probability is \(1-0.9215=0.0785\).
Check. The conditional interval starts at age 61, so \(q_{60}\) is not part of part (c). The product in part (b) includes all three years because that probability starts at age 60 and ends at age 63.
Working with Incomplete Life Table Entries
A table need not list every count or rate. The identities \(d_x=\ell_x-\ell_{x+1}\), \(q_x=d_x/\ell_x\), and \(\ell_{x+1}=\ell_x(1-q_x)\) allow a missing entry to be recovered when the other relevant entries are known. Before calculating a requested probability, check whether an intermediate count must be found.
Worked Example: Recovering a Missing Count and Rate
A hypothetical table gives \(\ell_{70}=5{,}000\), \(\ell_{71}=4{,}900\), \(q_{71}=0.04\), and \(\ell_{73}=4{,}550\), but does not list \(\ell_{72}\) or \(q_{72}\). Calculate (a) the missing entries and (b) the probability of death before age 73 given survival to age 71.
Method. First use the supplied one-year rate to recover the count at age 72. Then use the age-72 to age-73 count change to find \(q_{72}\). Finally calculate the conditional probability over the full two-year interval.
(a) Recover \(\ell_{72}\). Since \(p_{71}=1-q_{71}=1-0.04=0.96\), \(\ell_{72}=\ell_{71}p_{71}=4{,}900(0.96)=4{,}704\).
The number dying between ages 72 and 73 is \(d_{72}=4{,}704-4{,}550=154\). Therefore \(q_{72}=d_{72}/\ell_{72}=154/4{,}704=0.032738\ldots\), or 0.0327 to four decimal places.
(b) Conditional death before age 73 given survival to age 71. The conditional survival probability over the two years is \(\ell_{73}/\ell_{71}=4{,}550/4{,}900=0.928571\ldots\). The required probability is \(1-4{,}550/4{,}900=350/4{,}900=0.071428\ldots\), or 0.0714 to four decimal places.
Check. The recovered count gives \(q_{71}=(4{,}900-4{,}704)/4{,}900=196/4{,}900=0.04\), matching the supplied rate. The two-year conditional death probability also equals \(1-(1-q_{71})(1-q_{72})\).
Using Select and Ultimate Rates
A select-and-ultimate table can give different rates for lives recently selected at the same age and for lives after the select period has ended. The duration index matters: use each select rate for its stated year, then use ultimate rates at the attained ages that follow. Do not treat a select rate as applying indefinitely.
Worked Example: Crossing from Select to Ultimate Mortality
A hypothetical table has a two-year select period. For a life selected at age 50, the rates are \(q_{[50]}=0.01\) and \(q_{[50]+1}=0.02\). The ultimate rates are \(q_{52}=0.04\) and \(q_{53}=0.05\). Calculate (a) the probability of death before age 54 for this life and (b) the probability of death during the two years from age 52 to 54, conditional on survival to age 52.
Method. Apply the two select rates first, followed by the ultimate rates for attained ages 52 and 53. For the conditional probability in part (b), start at age 52 and use only the two ultimate years.
(a) Death before age 54. The four annual survival probabilities are \(1-0.01=0.99\), \(1-0.02=0.98\), \(1-0.04=0.96\), and \(1-0.05=0.95\). Survival from selection to age 54 is \(0.99(0.98)(0.96)(0.95)=0.9702(0.96)(0.95)=0.931392(0.95)=0.8848224\). Thus the probability of death before age 54 is \(1-0.8848224=0.1151776\), or 0.1152 to four decimal places.
(b) Conditional death from age 52 to 54. Given survival to age 52, the relevant survival probability is \(0.96(0.95)=0.912\). The conditional probability of death is \(1-0.912=0.0880\).
Check. The first two rates apply at select durations zero and one. The last two rates are ultimate rates at attained ages 52 and 53. Part (b) correctly excludes both select rates because the condition places the life at age 52.
Common Errors and a Reliable Procedure
A frequent error is to use the wrong starting group. For example, when asked for death in the next two years given survival to age 61, using \(1-\ell_{63}/\ell_{60}\) calculates an unconditional three-year probability from age 60. The correction is to use \(1-\ell_{63}/\ell_{61}\). Similarly, multiplying death probabilities across years is not the method for a multi-year death probability: the product of survival probabilities gives survival, and death is its complement.
For select-and-ultimate calculations, another trap is to use the ultimate rate at attained age \(x\) in place of the first select rate \(q_{[x]}\). Follow the table's select duration before switching to ultimate rates. A compact workflow helps prevent these mistakes:
Write down the starting age, ending age, and any condition such as survival to an intermediate age.
Use counts, annual rates, or select-duration rates that correspond to each year in the interval.
Take an endpoint-to-starting count ratio or multiply the one-year survival probabilities.
Subtract the survival probability from one, using the conditioned starting age when applicable.
Check Your Understanding
For each hypothetical table, identify the relevant interval before calculating. Probabilities are rounded to four decimal places where needed.
- A table gives \(\ell_{35}=8{,}000\), \(\ell_{36}=7{,}920\), and \(\ell_{37}=7{,}800\). Find the probability of death before age 37 for a life aged 35. (Answer: 0.0250)
- A table gives \(q_{45}=0.01\), \(q_{46}=0.02\), and \(q_{47}=0.03\). Find the probability that a life aged 45 survives to age 48. (Answer: 0.9411)
- A table gives \(\ell_{60}=10{,}000\), \(\ell_{61}=9{,}800\), and \(\ell_{62}=9{,}600\). Find the probability of death before age 62 given survival to age 61. (Answer: 0.0204)
- A two-year select table gives \(q_{[40]}=0.02\) and \(q_{[40]+1}=0.03\), followed by ultimate \(q_{42}=0.04\). Find the probability that a life selected at age 40 survives the first two years and dies during the third year. (Answer: 0.0380)
- A table gives \(q_{70}=0.02\) and a two-year survival probability from age 70 to 72 of \(0.95\). Find the missing rate \(q_{71}\). (Answer: 0.0306)
These calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.