Introduction
A lifetime statement needs a clear starting point. “Survives for 10 years” is incomplete unless we know from what age those 10 years are counted. In a survival model, \(T_x\) records the future lifetime of a person who is alive at age \(x\): it measures elapsed time from that starting age, not the person’s age.
Keeping the starting age and the elapsed duration separate is essential when reading actuarial notation. In particular, survival for \(t\) years from age \(x\) is the event \(T_x>t\). The strict inequality matters: it says the future lifetime exceeds the stated duration.
Starting Age and Elapsed Time
There are two separate quantities in the notation. The \(x\) in \(T_x\) is the person’s age at the start of the measurement. The \(t\) in \(T_x>t\) is the number of years measured from that point. If a person is age \(x\) now, then after \(t\) years the person’s age, if alive, is \(x+t\). Thus \(x+t\) is an age, while \(t\) is a duration.
For example, for a person alive at age 40, \(T_{40}>6\) means that the person’s future lifetime exceeds six years. It does not mean that the person is older than age 6, nor does it refer to six years measured from birth. The six-year duration starts at age 40.
The standard actuarial notation for the probability of surviving more than \(t\) years from age \(x\) is \({}_tp_x\). This notation describes a probability, while \(T_x\) is the random variable whose event is being considered.
Why the Survival Event Uses \(T_x>t\)
The relationship follows directly from how future lifetime is measured. Set the time origin at the instant the person is age \(x\). At elapsed time \(t\), the person has reached age \(x+t\) if still alive. The person is alive beyond that elapsed time precisely when the amount of time remaining at the origin is greater than \(t\). That amount of time remaining is \(T_x\).
The model begins when the person is alive at age \(x\). Future durations are counted from this point.
Choose a nonnegative duration \(t\). After \(t\) years, the corresponding age is \(x+t\).
Being alive beyond the duration means the remaining lifetime exceeds it: \(T_x>t\). Its probability is \({}_tp_x\).
The strict inequality also determines what happens at the boundary. If \(T_x=t\), death occurs exactly at duration \(t\), so the event \(T_x>t\) does not occur. Its complement is \(T_x\leq t\), and therefore $$ P(T_x\leq t)=1-{}_tp_x. $$ This complement includes the boundary value \(T_x=t\). In a model with no probability of death at one exact duration, the distinction may not change a numerical probability, but the event definitions remain different.
Worked Example: Reading a Future-Lifetime Distribution
A hypothetical simplified model for people alive at age 36 assigns \(T_{36}\) the following values and probabilities: 2 years with probability 0.20, 7 years with probability 0.30, and 12 years with probability 0.50. Find the probability of surviving more than 5 years from age 36, and identify the age reached after those 5 years if the person is alive. This is a discrete illustrative lifetime distribution; there is no interest rate or financial discounting in the question.
Step 1: Translate the duration into an event. “More than 5 years from age 36” is \(T_{36}>5\). The starting age is 36; the threshold is the elapsed duration 5.
Step 2: Select the outcomes in the event. Of the three possible future lifetimes, 2 does not exceed 5, while 7 and 12 do. Therefore, $$ P(T_{36}>5)=P(T_{36}=7)+P(T_{36}=12)=0.30+0.50=0.80. $$
Step 3: Convert duration to age. Five years after age 36 is age \(36+5=41\). The answer is \({}_5p_{36}=0.80\), and the checkpoint age is 41.
Check and mistake avoided. The probabilities in the supplied model sum to \(0.20+0.30+0.50=1.00\). A common error is to include the 2-year outcome because it is a possible lifetime in the model; it does not belong to the event \(T_{36}>5\). The method is to compare each lifetime value with the stated duration.
Using \({}_tp_x\) Without Changing the Time Origin
When a survival probability is supplied directly, first identify the age attached to its subscript and the duration attached to its leading \(t\). The probability \({}_9p_{52}\), for instance, concerns nine years measured from age 52. It is not a probability measured from birth, and it is not automatically the survival probability for nine years from some other starting age.
If the question gives a calendar date as well, use it to locate the starting point in the scenario. The probability notation still records age and duration: a person age \(x\) at the stated starting date who survives \(t\) years reaches age \(x+t\) at the corresponding later date. Unless a model explicitly includes calendar-time effects, the date itself does not replace either \(x\) or \(t\).
Worked Example: Converting a Duration to an Attained Age
A hypothetical mortality basis gives \({}_9p_{52}=0.81\). A person is alive at exact age 52 on a specified policy date. Find the probability of being alive beyond the nine-year duration, give the age corresponding to the end of that duration, and find the probability that the future lifetime is at most nine years. No interest rate is relevant because this is a probability question, not a present-value calculation.
Step 1: Identify the origin and duration. The subscript 52 gives the starting age. The leading 9 gives the elapsed duration. By definition, $$ P(T_{52}>9)={}_9p_{52}=0.81. $$
Step 2: Find the age after nine years. Add the duration to the starting age: \(52+9=61\). Thus the survival probability is described as survival beyond nine years from age 52, corresponding to the age-61 checkpoint.
Step 3: Use the complement for “at most.” The event \(T_{52}\leq 9\) is the complement of \(T_{52}>9\). Hence $$ P(T_{52}\leq 9)=1-0.81=0.19. $$
Check and mistake avoided. The two probabilities add to \(0.81+0.19=1.00\). A common error is to report 52 as the age at the end of nine years; 52 is the starting age, while 61 is the age corresponding to that duration.
Comparing Different Starting Ages
The subscript cannot be dropped when comparing future-lifetime probabilities. \({}_5p_{40}\) and \({}_5p_{45}\) both refer to a five-year duration, but their time origins differ. Each probability concerns a person alive at the specified starting age. A numerical comparison between the two values is valid if the values are supplied, but it does not make the underlying starting ages interchangeable.
Similarly, a five-year duration from age 40 ends at age 45, whereas a five-year duration from age 45 ends at age 50. The durations match; the age intervals do not. A useful habit is to annotate every probability in words before calculating: “survive \(t\) years, starting at age \(x\).”
Worked Example: Comparing Two Five-Year Probabilities
A hypothetical table gives \({}_5p_{40}=0.92\) and \({}_5p_{45}=0.87\). For two separate groups, each selected while alive at its stated starting age, identify the probability and end age for each group, then calculate the difference between the supplied survival probabilities. Treat the table values as given; no interest rate or additional mortality model is required.
Step 1: Interpret the first entry. \({}_5p_{40}=P(T_{40}>5)=0.92\). The starting age is 40, and the five-year checkpoint corresponds to age \(40+5=45\).
Step 2: Interpret the second entry. \({}_5p_{45}=P(T_{45}>5)=0.87\). The starting age is 45, and the five-year checkpoint corresponds to age \(45+5=50\).
Step 3: Compare the values. Subtract the second supplied probability from the first: $$ 0.92-0.87=0.05. $$ The first five-year survival probability is 0.05, or 5 percentage points, higher.
Check and mistake avoided. The values are probabilities between 0 and 1, and their difference is 0.05. A common error is to describe both entries as survival from age 40 to age 45. Only \({}_5p_{40}\) has that origin and endpoint; \({}_5p_{45}\) starts at age 45 and corresponds to age 50 after five years.
Common Time-Origin Errors
Most errors with \(T_x\) come from confusing one of three things: the starting age \(x\), the elapsed duration \(t\), or the event’s endpoint. Before using a probability, write down all three in context.
- Using age where duration is required. In \({}_6p_{43}\), the duration is 6 years, not 43 years.
- Using duration as the attained age. Six years from age 43 corresponds to age 49, not age 6.
- Changing the starting age without changing the notation. A probability from age 43 is not automatically a probability from age 49.
- Ignoring the endpoint. \(T_x>t\) excludes a lifetime equal to \(t\); its complement is \(T_x\leq t\).
- Adding a financial interpretation that is not present. \(T_x\) describes a lifetime duration. Interest and discounting are needed only when the problem also asks for a financial value.
Check Your Understanding
For each question, identify the starting age and duration before calculating.
- A person is alive at age 29, and a hypothetical model gives \({}_6p_{29}=0.93\). State the event represented, the age corresponding to the six-year checkpoint, and the probability of \(T_{29}\leq 6\). (Answer: \(P(T_{29}>6)=0.93\); checkpoint age 35; \(P(T_{29}\leq6)=0.07\).)
- In a hypothetical discrete model at age 40, \(T_{40}\) is 1 year with probability 0.20, 5 years with probability 0.30, and 10 years with probability 0.50. Find \(P(T_{40}>4)\) and the age corresponding to a four-year duration. (Answer: \(0.30+0.50=0.80\); age 44.)
- A table gives \({}_7p_{54}=0.84\). Find the probability that future lifetime from age 54 is at most seven years, and state the age corresponding to the seven-year checkpoint. (Answer: \(1-0.84=0.16\); age 61.)
- Two hypothetical entries are \({}_3p_{60}=0.96\) and \({}_8p_{60}=0.83\). What is the probability represented by the second entry, and what age corresponds to its duration? (Answer: \(P(T_{60}>8)=0.83\); age 68.)
- A table gives \({}_5p_{38}=0.91\) and \({}_5p_{43}=0.88\). Interpret both entries, identify both checkpoint ages, and find the difference in the probabilities. (Answer: survival beyond five years from age 38, corresponding to age 43, is 0.91; survival beyond five years from age 43, corresponding to age 48, is 0.88; difference \(0.91-0.88=0.03\), or 3 percentage points.)