The Survival Function S_x(t)

Why Survival Probabilities Matter

Many actuarial questions are naturally phrased in terms of how long a person survives: the probability of surviving another five years, or the proportion of a hypothetical group expected to be alive after a specified duration. The survival function records these probabilities directly. It is also closely tied to the distribution function, so a value given in one form can be converted to the other.

The key detail is the boundary: survival beyond duration $t$ means that future lifetime is strictly greater than $t$. This matters whenever the lifetime distribution assigns positive probability to exactly $t$. The rule connecting survival and the distribution function remains exact even in that case.

Definition: Let $T_x$ be the future lifetime of a life currently aged $x$, measured from the present in the same time units as $t$. The survival function is $$ S_x(t)=P(T_x>t). $$ The distribution function is $F_x(t)=P(T_x\le t)$. For every real $t$, $$ S_x(t)=1-F_x(t). $$ For a nonnegative future lifetime, $S_x(t)=1$ and $F_x(t)=0$ when $t<0$.

Interpreting the Definition and Its Complement

The subscript $x$ identifies the current age of the life; the argument $t$ is a duration measured from that current age. Thus $S_{40}(10)$, for example, describes survival for more than ten years from age 40. It is not the probability of surviving to age 10.

The events $\{T_x>t\}$ and $\{T_x\le t\}$ are complements: every possible value of $T_x$ either exceeds $t$ or is at most $t$, and the two events cannot occur together. Their probabilities therefore sum to one. This gives the complement relationship without requiring any assumption that the lifetime distribution is continuous.

P(T_x>t)+P(T_x\le t)=1, \qquad S_x(t)=1-F_x(t), \qquad F_x(t)=1-S_x(t).

The strict inequality in $S_x(t)$ is part of its definition. If $P(T_x=t)>0$, then that probability belongs to $F_x(t)$, not to $S_x(t)$. Consequently, $P(T_x\ge t)$ is not always the same as $S_x(t)$: the event $T_x\ge t$ includes lives whose future lifetime is exactly $t$. If there is no probability at the boundary, the two probabilities agree, but that conclusion requires an additional fact about the distribution.

As $t$ increases, the event $\{T_x>t\}$ can only become more restrictive. If $t_2>t_1$, then $\{T_x>t_2\}$ is contained in $\{T_x>t_1\}$, so $S_x(t_2)\le S_x(t_1)$. Thus a survival function is nonincreasing and its values lie between zero and one. This is a useful consistency check on calculations.

Reading Survival from Distribution Information

The complement rule is often a one-step calculation, but care is needed to identify what the supplied information represents. A cumulative probability through duration $t$ is $F_x(t)$; the probability remaining beyond that duration is $S_x(t)$. The following examples use hypothetical lifetime distributions and make the boundary convention explicit.

Worked Example: A Discrete Future-Lifetime Distribution

A hypothetical model for a life aged 52 assigns future lifetime $T_{52}$ the following probabilities: $P(T_{52}=1)=0.20$, $P(T_{52}=3)=0.50$, and $P(T_{52}=5)=0.30$, with durations measured in years. Find $F_{52}(3)$, $S_{52}(3)$, and $S_{52}(2)$. Check that the complement relationship holds at three years.

Method. First add the probabilities at or below the requested duration to obtain $F_{52}(t)$. Then subtract that value from one. The common mistake is to include the probability at exactly three years in $S_{52}(3)$, even though survival at that boundary is defined by $T_{52}>3$.

At three years, the values at or below the threshold are 1 and 3. Therefore,

F_{52}(3)=P(T_{52}\le 3)=0.20+0.50=0.70.

The complement gives

S_{52}(3)=1-F_{52}(3)=1-0.70=0.30.

This also follows directly: the only outcome greater than three is five years, which has probability $0.30$. At two years, only the one-year outcome is at or below the threshold, so

F_{52}(2)=0.20, \qquad S_{52}(2)=1-0.20=0.80.

Check. At three years, $F_{52}(3)+S_{52}(3)=0.70+0.30=1.00$. The probability $0.50$ at exactly three years is included in $F_{52}(3)$, which is why it is excluded from $S_{52}(3)$.

Worked Example: Converting a Piecewise Distribution Function

For a hypothetical life aged 61, the future-lifetime distribution function is specified by $$ F_{61}(t)= \begin{cases} 0, & t<0,\\ 0.10+0.10t, & 0\le t<4,\\ 0.50+0.075(t-4), & 4\le t<8,\\ 1, & t\ge 8. \end{cases} $$ Calculate $S_{61}(0)$, $S_{61}(3)$, $S_{61}(4)$, and $S_{61}(8)$. Also verify that the specified distribution function does not decrease at the boundaries.

Method. Select the piece of $F_{61}$ that includes each requested value, evaluate it, and use $S_{61}(t)=1-F_{61}(t)$. The common mistake is to use the piece for $t<4$ at $t=4$, although the formula assigns $t=4$ to the next interval.

At $t=0$, the first nonnegative-duration piece applies: $$ F_{61}(0)=0.10+0.10(0)=0.10, \qquad S_{61}(0)=1-0.10=0.90. $$ At $t=3$, the same piece gives $$ F_{61}(3)=0.10+0.10(3)=0.40, \qquad S_{61}(3)=1-0.40=0.60. $$ At $t=4$, use the piece for $4\le t<8$: $$ F_{61}(4)=0.50+0.075(4-4)=0.50, \qquad S_{61}(4)=1-0.50=0.50. $$ At $t=8$, the final piece applies: $$ F_{61}(8)=1, \qquad S_{61}(8)=1-1=0. $$ For the boundary check, the left-hand formula approaches $0.10+0.10(4)=0.50$ at four years, matching $F_{61}(4)=0.50$. The formula for $4\le t<8$ approaches $0.50+0.075(8-4)=0.80$ as $t$ approaches eight from below, and $F_{61}(8)=1$. Neither boundary produces a decrease. The resulting survival values $0.90, 0.60, 0.50, 0$ are also nonincreasing, as required.

Worked Example: Recovering Missing Probabilities from Survival Values

A hypothetical model for $T_{70}$ allows exactly three future lifetimes: 2, 5, and 9 years. The probability of a two-year future lifetime is $0.25$, and the model states that $S_{70}(5)=0.45$. Find the probabilities of future lifetimes of 5 and 9 years, then calculate $F_{70}(5)$ and $S_{70}(2)$. Check that all three outcome probabilities sum to one.

Method. Translate the survival statement using its strict boundary: $S_{70}(5)$ includes only outcomes greater than five. The common mistake is to include the five-year outcome in $S_{70}(5)$.

The only allowed lifetime greater than five is nine years. Therefore, $$ P(T_{70}=9)=S_{70}(5)=0.45. $$ The probabilities of all three outcomes must sum to one. Hence the probability of a five-year lifetime is $$ P(T_{70}=5)=1-P(T_{70}=2)-P(T_{70}=9) =1-0.25-0.45=0.30. $$ At five years, the outcomes at or below five are 2 and 5, so $$ F_{70}(5)=P(T_{70}\le5)=0.25+0.30=0.55. $$ The complement check is $1-F_{70}(5)=1-0.55=0.45=S_{70}(5)$. At two years, only the two-year outcome is at or below the boundary: $$ F_{70}(2)=0.25, \qquad S_{70}(2)=1-0.25=0.75. $$ Finally, the outcome probabilities check as $0.25+0.30+0.45=1.00$.

Common Boundary Errors

Most errors with the survival function are not difficult probability calculations; they are errors in translating the event. Before calculating, write the event in words or symbols and identify whether the threshold value itself is included.

Using the wrong complement. Since $F_x(t)=P(T_x\le t)$, its complement is $P(T_x>t)$, not $P(T_x\ge t)$.
Confusing age and duration. The subscript $x$ is the current age, while $t$ is elapsed time. A statement about surviving to age $x+t$ corresponds to future lifetime exceeding $t$, subject to the chosen boundary convention.
Ignoring units. If $t$ is measured in years, a duration stated in months must be converted before substituting it into a formula expressed in years.
Assuming continuity without evidence. The identity $S_x(t)=1-F_x(t)$ holds whether or not the distribution has probability at a single point. But replacing $P(T_x>t)$ by $P(T_x\ge t)$ needs justification.

These distinctions are useful whenever a problem supplies cumulative probabilities but asks for the share surviving beyond a duration. State the event first, apply the complement rule, and then check that the result lies in $[0,1]$ and that survival does not increase as the duration grows.

Check Your Understanding

For each problem, use the strict definition $S_x(t)=P(T_x>t)$, and pay particular attention to outcomes exactly at the stated duration.

A hypothetical distribution has $P(T_{45}=1)=0.15$, $P(T_{45}=4)=0.35$, and $P(T_{45}=7)=0.50$. Find $F_{45}(4)$ and $S_{45}(4)$. (Answer: $F_{45}(4)=0.50,\ S_{45}(4)=0.50$)
For a life aged 58, $F_{58}(2)=0.12$ and $F_{58}(6)=0.47$. Find $S_{58}(2)$ and $S_{58}(6)$, and state which survival probability is larger. (Answer: $S_{58}(2)=0.88,\ S_{58}(6)=0.53$; $S_{58}(2)$ is larger.)
A hypothetical future lifetime takes values 3, 6, and 10 years with probabilities $0.20$, $0.30$, and $0.50$, respectively. Find $S_{63}(6)$ and $F_{63}(6)$, then check their sum. (Answer: $S_{63}(6)=0.50,\ F_{63}(6)=0.50$; their sum is $1.00$.)
A distribution function is $F_{66}(t)=0.08+0.12t$ for $0\le t<5$, and $F_{66}(t)=0.68+0.04(t-5)$ for $5\le t<10$. Find $S_{66}(3)$, $S_{66}(5)$, and $S_{66}(9)$. (Answer: $S_{66}(3)=0.56,\ S_{66}(5)=0.32,\ S_{66}(9)=0.16$)
A hypothetical life aged 72 can have future lifetime 2, 5, or 8 years. The probabilities at 2 and 8 years are $0.18$ and $0.42$. Find the probability at 5 years, then find $S_{72}(5)$ and $F_{72}(5)$. (Answer: $P(T_{72}=5)=0.40,\ S_{72}(5)=0.42,\ F_{72}(5)=0.58$)

What You'll Learn