Finite-Term Probabilities
A finite-term probability asks about one specified period measured from the current time origin. For a life aged \(x\), the question might be, “What is the probability that this life survives beyond 8 years?” or “What is the probability of death within 8 years?” The two probabilities describe complementary outcomes at the same horizon. Getting the horizon and the event right is essential: a survival probability is not the probability of death, and neither is an annual mortality rate.
Let \(T_x\) be the future lifetime of a life aged \(x\), measured in years from the present. For a nonnegative future lifetime, the distribution function \(F_x(t)\) gives the probability that \(T_x\) is at most \(t\). The survival function \(S_x(t)\) gives the probability that \(T_x\) is greater than \(t\). In the usual continuous-lifetime setting, using “at least \(t\)” instead of “greater than \(t\)” gives the same probability because there is no probability mass at exactly \(t\).
Converting Between the Two Functions
At any fixed \(t\), either the life has died by that time or it has survived beyond it. These events do not overlap and together cover all possible outcomes. Therefore their probabilities add to one. This gives the conversion rule:
The rule is a complement calculation, not a new mortality model. It works whether the supplied information is a formula, a table, or a stated probability, provided the value refers to the same life, time origin, and duration. If a model permits a point mass at \(t\), the definitions above still make the events \(T_x\leq t\) and \(T_x>t\) exact complements. A convention using \(T_x\geq t\) would then need to be handled with care.
Read the duration and its time unit from the question. Use the function value at that duration.
Death by \(t\) corresponds to \(F_x(t)\); survival beyond \(t\) corresponds to \(S_x(t)\).
Subtract the stated probability from one to obtain the other event’s probability.
Confirm both probabilities lie between zero and one and sum to one.
The starting age \(x\) identifies whose future lifetime is being modelled, while \(t\) is the length of the period being considered. For example, a 55-year-old’s 8-year survival probability concerns the years from age 55 to age 63; it is not a probability for a life starting at age 63. No interest rate is needed for these probability conversions.
Worked Example: Converting a Lifetime Distribution Function
A hypothetical model for a life aged 40 assumes a future lifetime uniformly distributed from 0 to 20 years, with no probability mass at the endpoints. Thus, for \(0\leq t\leq20\), \(F_{40}(t)=t/20\). Calculate the probabilities of death by 6 years and survival beyond 12 years. State the complementary probabilities as a check.
Method: evaluate the distribution function at each requested duration, then use the complement rule. The mortality assumption is the stated uniform distribution on \([0,20]\); no interest assumption is involved.
At \(t=6\), the probability of death by 6 years is:
Therefore, the probability of survival beyond 6 years is:
At \(t=12\), first evaluate the distribution function at that horizon:
The requested probability of survival beyond 12 years is its complement:
Check: At 6 years, \(0.30+0.70=1.00\). At 12 years, \(0.60+0.40=1.00\). Both pairs are valid probabilities. A common mistake is to report \(F_{40}(12)=0.60\) as the probability of surviving 12 years; it is the probability of death by 12 years.
Reading Values From a Survival Table
A table may give survival probabilities directly instead of giving a formula for \(S_x(t)\). If the exact duration requested appears in the table, use that entry and take its complement only when the question asks for death by the term. Do not interpolate between entries unless a stated model or instruction justifies it. The complement calculation does not supply missing table values.
| Question wording | Probability to use | If the other function is supplied |
|---|---|---|
| Death by \(t\) years | \(F_x(t)\) | \(1-S_x(t)\) |
| Survival beyond \(t\) years | \(S_x(t)\) | \(1-F_x(t)\) |
Worked Example: Auditing a Report From a Survival Table
A hypothetical table for a life aged 55 gives \(S_{55}(0)=1.00\), \(S_{55}(5)=0.91\), \(S_{55}(10)=0.78\), and \(S_{55}(15)=0.60\). A draft report states that the probability of surviving beyond 10 years is 0.22 and the probability of death by 10 years is 0.78. Determine which values are correct, using only the table entry at the requested duration.
Method: read the survival probability at exactly 10 years, then calculate its complement. The model assumption is the hypothetical table as given; the calculation requires no interpolation.
The table directly gives the probability of surviving beyond 10 years:
The probability of death by 10 years is:
Thus the draft report has swapped the two values. The corrected report gives 0.78 for survival and 0.22 for death by 10 years.
Check: \(0.78+0.22=1.00\). The 5- and 15-year entries are not needed; using them to alter the 10-year value would be unjustified. The common mistake avoided here is treating a table entry for survival as if it were the complementary probability of death.
Why the Complement Rule Holds
The conversion follows directly from the definition of a distribution function and the fact that a lifetime either has or has not reached the specified threshold. For a fixed \(t\), the events \(\{T_x\leq t\}\) and \(\{T_x>t\}\) are disjoint: the same lifetime cannot satisfy both inequalities. They are also exhaustive: every value of \(T_x\) is either at most \(t\) or greater than \(t\). The probability of their union is therefore one.
Substituting \(F_x(t)=P(T_x\leq t)\) and \(S_x(t)=P(T_x>t)\) gives \(F_x(t)+S_x(t)=1\). Rearranging produces both conversion formulas. This justification also explains why the calculation needs the same \(t\) in both terms: probabilities at different horizons do not represent complementary events.
Worked Example: Using a Parametric Survival Function
For a hypothetical life aged 50, suppose the future lifetime has survival function \(S_{50}(t)=(1-t/24)^3\) for \(0\leq t\leq24\), and \(S_{50}(t)=0\) for \(t\geq24\). Calculate the probabilities of survival beyond 8 years and death by 8 years. Then calculate both probabilities at 12 years. Give probabilities to four decimal places.
Method: evaluate the stated survival model at each duration and use \(F_{50}(t)=1-S_{50}(t)\). The mortality assumption is the supplied survival function, with its stated 24-year endpoint; there is no interest assumption.
At 8 years:
The probability of death by 8 years is:
At 12 years:
The probability of death by 12 years is:
Check: At 8 years, \(0.2963+0.7037=1.0000\). At 12 years, \(0.1250+0.8750=1.0000\). The common mistake avoided is subtracting the duration from a probability or using the 8-year value when the requested term is 12 years; evaluate the model at the stated horizon first.
Common Interpretation Traps
The arithmetic is simple, but the wording and time scale can make a finite-term question easy to misread. Before calculating, identify exactly what is being measured and which event is requested.
- Reversing the functions: \(F_x(t)\) is death by \(t\), while \(S_x(t)\) is survival beyond \(t\). If one is given and the other requested, use one minus the given value.
- Using a different duration: A 5-year value does not answer a 10-year question. Substitute the requested \(t\) into the formula or use the matching table entry.
- Changing time units: A duration stated in months must be converted to years if the model’s time variable is in years. For example, 18 months is \(18/12=1.5\) years.
- Calling a probability a rate: A probability of death by 10 years covers the whole 10-year period; it is not an annual death rate.
- Interpolating without a model: Two table entries do not, by themselves, determine the survival probability at a duration between them.
Finite-term probabilities are useful because many actuarial questions begin by specifying a fixed coverage period or a fixed future horizon. The central discipline is to keep the time origin, duration, and event aligned. Once those are clear, the relationship between death by the term and survival beyond it is a direct complement.
Check Your Understanding
For each problem, identify the requested event and use the probability at the stated duration. Answers are rounded to four decimals where needed.
- A hypothetical lifetime model gives \(F_{60}(4)=0.13\). Find the probability of surviving beyond 4 years. (Answer: \(0.8700\))
- A survival table gives \(S_{45}(7)=0.68\). Find the probability of death by 7 years and verify the two probabilities sum to one. (Answer: death by 7 years \(=0.3200\); \(0.68+0.32=1.00\))
- A hypothetical table gives \(S_{52}(3)=0.94\) and \(S_{52}(8)=0.79\). Find the probability of death by 8 years. Do not interpolate. (Answer: \(0.2100\))
- For a hypothetical life aged 35, \(F_{35}(t)=t/16\) for \(0\leq t\leq16\). Find the probabilities of death by 6.4 years and survival beyond 6.4 years. (Answer: death \(=0.4000\); survival \(=0.6000\))
- A hypothetical survival model is \(S_{70}(t)=(1-t/20)^2\) for \(0\leq t\leq20\). Find the probabilities of survival beyond 5 years and death by 5 years. (Answer: survival \(=0.5625\); death \(=0.4375\))