The Distribution Function F_x(t)

Introduction

A lifetime probability is meaningful only after its time origin is clear. If a person is currently age 60, “death within 5 years” refers to the next five years, not to the person’s first five years of life. The future-lifetime distribution function records the probability of death by a specified duration measured from the person’s current age.

The function is useful because it turns a duration into a probability statement. For example, a value at 5 years answers the question “What is the probability this person dies within five years?” The starting age matters: two people can have different future-lifetime distributions even when the duration is the same. This tutorial develops the definition and its interpretation, aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: For a life currently aged $x$, let $T_x$ denote that life’s future lifetime, measured in years from age $x$. The future-lifetime distribution function is $$ F_x(t)=P(T_x\le t). $$ For $t\ge 0$, $F_x(t)$ is the probability that the life dies within $t$ years of age $x$, including death at duration $t$. Under the usual assumption that the life is alive at age $x$ and has strictly positive future lifetime, $F_x(0)=0$. Since future lifetime cannot be negative, $F_x(t)=0$ for $t<0$.

Reading the Time Scale Correctly

The argument $t$ is a duration, not an attained age. If the starting age is $x$ and the question concerns death by attained age $y$, the duration to use is $y-x$. Thus the probability of death by age $y$, for $y\ge x$, is $F_x(y-x)$. In contrast, $F_x(y)$ would refer to death within $y$ years after age $x$, which is a different time horizon.

The inequality in the definition is also important. $F_x(t)$ concerns $T_x\le t$: death at exactly duration $t$, if the model assigns positive probability to that event, is included. The function is a cumulative probability. As $t$ increases, the event “death by duration $t$” can only stay the same or become larger. Consequently, $F_x(t)$ cannot decrease as $t$ increases.

t<0:\quad F_x(t)=0, \qquad t\ge 0:\quad F_x(t)=P(T_x\le t).

The definition does not require a particular mortality model. A model may specify a continuous distribution of future lifetimes or a discrete set of possible durations. In either case, the same question is answered: what is the probability that future lifetime is no greater than the stated duration? The age subscript $x$ identifies the life’s age at the time the future-lifetime clock starts.

Why the Function Is Cumulative

The basic ordering property follows directly from the events in the definition. If $0\le s\le t$, then every outcome in which death occurs by duration $s$ is also an outcome in which death occurs by duration $t$. Therefore the first event is contained in the second, and its probability cannot be larger.

Identify the two events.
For $s\le t$, the events are $\{T_x\le s\}$ and $\{T_x\le t\}$.

Compare the events.
If $T_x\le s$ and $s\le t$, then $T_x\le t$. Hence $\{T_x\le s\}\subseteq\{T_x\le t\}$.

Compare their probabilities.
Probability preserves event inclusion, so $F_x(s)=P(T_x\le s)\le P(T_x\le t)=F_x(t)$.

This argument also provides a practical check on supplied values: a proposed value at a later duration must not be smaller than a value at an earlier duration. Each value must be a probability between 0 and 1. These checks do not determine an entire lifetime model, but they can reveal an inconsistent interpretation or calculation.

Evaluating the Distribution Function

A problem may give a formula for $F_x(t)$, specify a distribution for $T_x$, or provide selected values of the function. In every case, first identify the starting age and the requested duration. Then apply the definition or the given model. Keep the phrase “by duration $t$” tied to $T_x\le t$, rather than switching to an attained-age argument without converting it.

Worked Example: A Uniform Future-Lifetime Model

Model and assumptions. For a hypothetical life aged 58, assume $T_{58}$ has a uniform distribution on $[0,20]$ years. Thus the stipulated distribution function is $F_{58}(t)=t/20$ for $0\le t\le20$, with $F_{58}(t)=0$ for $t<0$ and $F_{58}(t)=1$ for $t>20$. Find the probability of death by age 64, by age 70, and within a negative duration of 2 years. Give probabilities to four decimal places.

Method. Convert each attained age to a duration from age 58, then evaluate the stated distribution function. The negative-duration case is handled from the nonnegativity of future lifetime.

By age 64, the duration is $64-58=6$ years. Since $0\le6\le20$, $$ F_{58}(6)=\frac{6}{20}=0.3000. $$

By age 70, the duration is $70-58=12$ years. Therefore $$ F_{58}(12)=\frac{12}{20}=0.6000. $$

A duration of $-2$ years is negative, so $$ F_{58}(-2)=0. $$

Check. The durations satisfy $6<12$, and the corresponding values satisfy $0.3000\le0.6000$, as a cumulative probability must. The calculation avoids treating attained age 70 as the argument of $F_{58}$; the correct argument is 12 years.

Worked Example: A Discrete Future-Lifetime Distribution

Model and assumptions. A hypothetical life aged 45 has future lifetime $T_{45}$ taking the values 2, 5, 9, or 14 years, with probabilities 0.10, 0.25, 0.40, and 0.25, respectively. These probabilities sum to $0.10+0.25+0.40+0.25=1.00$. Find the probability of death by age 54, by age 53, and by age 59. Report probabilities to four decimal places.

Method. Convert each target age to a duration from age 45. For each duration, add the probabilities at all support points no greater than that duration, because the definition uses $T_{45}\le t$.

Age 54 is $54-45=9$ years after age 45. The support points no greater than 9 are 2, 5, and 9, so $$ F_{45}(9)=0.10+0.25+0.40=0.7500. $$

Age 53 is $53-45=8$ years after age 45. The support points no greater than 8 are 2 and 5, so $$ F_{45}(8)=0.10+0.25=0.3500. $$

Age 59 is $59-45=14$ years after age 45. All four support points are no greater than 14, so $$ F_{45}(14)=0.10+0.25+0.40+0.25=1.0000. $$

Check. The target durations are 8, 9, and 14 years, and the resulting values $0.3500$, $0.7500$, and $1.0000$ do not decrease. The calculation includes the probability at duration 9 when evaluating $F_{45}(9)$, because death at exactly 9 years satisfies $T_{45}\le9$.

Using Age-Specific Values in a Group

For a group made up of lives at different ages, each person’s probability must be read from the function appropriate to that person’s starting age. If a group contains $n_j$ lives of a particular age and each has the same probability $p_j$ of death within the stated duration, the expected number of deaths in that group is $n_jp_j$. Adding these expected counts gives the expected total. This calculation uses linearity of expectation; independence among the lives is not required.

Worked Example: Expected Deaths Across Three Age Groups

Model and assumptions. Consider a hypothetical group of 120 lives. There are 40 lives aged 60, 30 aged 65, and 50 aged 70. The stipulated age-specific future-lifetime distributions give $F_{60}(5)=0.08$, $F_{65}(5)=0.12$, and $F_{70}(5)=0.20$. These are the probabilities of death within five years for the respective ages. Find the expected number of deaths within five years.

Method. Multiply each age-group count by its own five-year death probability, then add the expected counts. No assumption of independence is needed.

For the 40 lives aged 60, the expected number is $$ 40(0.08)=3.20. $$ For the 30 lives aged 65, it is $$ 30(0.12)=3.60. $$ For the 50 lives aged 70, it is $$ 50(0.20)=10.00. $$ Therefore the expected total is $$ 3.20+3.60+10.00=16.80. $$

Check. The group counts sum to $40+30+50=120$, and the expected total is between 0 and 120. The answer is an expected count, not a claim that exactly 16.80 people will die. The calculation avoids applying one age group’s $F_x(5)$ value to every life.

Common Exam Trap: Confusing Age With Duration

Suppose a life is age 67 and the problem gives $F_{67}(3)=0.28$, $F_{67}(7)=0.63$, and $F_{67}(10)=0.82$. To find the probability of death by age 77, an incorrect calculation might use $F_{67}(7)=0.63$, perhaps because 77 is mistakenly treated as a duration of 7 years. But the correct duration is $77-67=10$ years. Therefore the requested probability is $$ P(\text{death by age 77})=P(T_{67}\le10)=F_{67}(10)=0.82. $$ The value 0.63 corresponds to death by duration 7, or by attained age 74, not by age 77. The reliable check is to write the conversion explicitly before selecting a function value: target duration equals target attained age minus starting age.

Another error is to interpret “by $t$” as “before $t$, but not at $t$.” The definition here is $T_x\le t$, so an outcome at exactly $t$ is included. In a discrete model, this distinction can change the answer, as the probability at a support point may be positive.

Key takeaway: $F_x(t)$ is a probability indexed by a duration from age $x$. For negative durations it is zero; for nonnegative durations it is the probability of death by that time. When a problem names an attained age, subtract the starting age before evaluating the function.

Check Your Understanding

For each question, identify the duration from the stated starting age before using the given distribution information.

A life is currently age 54. What is $F_{54}(-1.5)$ under the standard future-lifetime assumption? (Answer: $0.0000$)
A hypothetical life aged 52 has a uniform future lifetime on $[0,16]$ years. Find the probability of death by age 60. (Answer: $0.5000$)
A hypothetical life aged 60 has $T_{60}$ equal to 1, 4, 6, or 10 years with probabilities 0.15, 0.20, 0.35, and 0.30, respectively. Find the probability of death by age 66. (Answer: $0.7000$)
For a hypothetical life aged 70, the given value is $F_{70}(9)=0.40$. What is the probability of death by age 79? (Answer: $0.4000$)
A hypothetical group contains 25 lives aged 61 with five-year death probability 0.08, 40 lives aged 66 with five-year death probability 0.15, and 35 lives aged 71 with five-year death probability 0.24. Find the expected number of deaths within five years. (Answer: $16.40$ deaths)

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