Conditional Lifetime Probabilities After Survival

Why Condition on Survival?

A survival probability answers a question such as, “What is the probability that a person survives at least five years from now?” A different question arises when the person is already known to have survived for some time: “Given that the person has survived two years, what is the probability of surviving three more?” The second probability is conditional. Its denominator must account for the fact that only people who survived the first two years remain in the group under consideration.

This distinction matters whenever a survival estimate is used after a period of observed survival. The conditioning event is not an additional assumption about future mortality; it is information about the person’s lifetime up to the present point. The calculation uses the survival function at two times: the end of the full period and the time at which survival is already known.

Definition: Let $T_x$ be the future lifetime of a life currently aged $x$, measured from age $x$, and let $S_x(t)=P(T_x>t)$ be its survival function. For $s,t\geq 0$, if $S_x(s)>0$, the conditional probability of surviving at least $t$ further years given survival for $s$ years is $$ P(T_x>s+t\mid T_x>s)=\frac{S_x(s+t)}{S_x(s)}. $$ In actuarial notation, ${}_tp_x=P(T_x>t)=S_x(t)$, so the same probability is ${}_{s+t}p_x/{}_sp_x$.

Reading the Conditioning Event

The event $T_x>s$ means that the life has survived beyond $s$ years after age $x$, or, equivalently, has reached age $x+s$. The event $T_x>s+t$ requires survival beyond the later time $s+t$. Because survival beyond $s+t$ necessarily includes survival beyond $s$, the numerator event is contained in the conditioning event.

The ratio follows directly from the definition of conditional probability. For events $A$ and $B$ with $P(B)>0$, $P(A\mid B)=P(A\cap B)/P(B)$. Here, take $A=\{T_x>s+t\}$ and $B=\{T_x>s\}$. Since $A\subseteq B$, their intersection is just $A$, giving the stated ratio. The denominator must be positive: if survival to $s$ has probability zero, conditioning on that event is not defined by this formula.

P(T_x>s+t\mid T_x>s) =\frac{P(T_x>s+t)}{P(T_x>s)} =\frac{S_x(s+t)}{S_x(s)}.

The conditional probability of dying during the following $t$ years is the complement. Under the strict-survival convention above, the event is $s<T_x\leq s+t$: the life has survived beyond $s$, but not beyond $s+t$. Subtracting the later survival probability from the earlier one gives

P(s<T_x\leq s+t\mid T_x>s) =1-\frac{S_x(s+t)}{S_x(s)} =\frac{S_x(s)-S_x(s+t)}{S_x(s)}.

These formulas apply to a valid survival function whether or not the lifetime distribution is continuous. When a distribution has point masses, the strict and non-strict inequalities in event definitions matter; the formulas above consistently use $S_x(t)=P(T_x>t)$. In a continuous lifetime model, changing a single endpoint does not change a probability, but the event definitions are still useful for keeping the conditioning clear.

Worked Example: Uniform Lifetime and Two-Year Prior Survival

Hypothetical model. Suppose $T_x$, measured in years from the life’s current age $x$, has a uniform distribution on $[0,10]$. Thus $S_x(u)=1-u/10$ for $0\leq u\leq10$, and $S_x(u)=0$ for $u\geq10$. Given that the life has survived for $s=2$ years, calculate the probability it survives three more years and the probability it dies during those three years.

Method. Use the ratio of survival at the end of the full five-year period to survival at the conditioning time. The relevant times are $s=2$ and $s+t=2+3=5$.

First evaluate the survival probabilities:

S_x(2)=1-\frac{2}{10}=0.8,\qquad S_x(5)=1-\frac{5}{10}=0.5.

The conditional probability of surviving the next three years is therefore

P(T_x>5\mid T_x>2) =\frac{S_x(5)}{S_x(2)} =\frac{0.5}{0.8} =0.625.

The conditional probability of dying between years 2 and 5 is the complement:

$$ 1-0.625=0.375. $$

To four decimal places, the answers are $0.6250$ for further survival and $0.3750$ for death in the interval. They sum to $1.0000$, as required. The common mistake is to report $S_x(5)=0.5$ as the conditional survival probability; that is the probability of surviving five years from the original starting age, not the probability given survival through year 2.

Using Survival Values from a Table

The survival function need not be given by an algebraic formula. A life table or a problem statement may instead provide survival probabilities at selected durations. The calculation is unchanged: identify the conditioning duration $s$, identify the total duration $s+t$, and divide the later survival probability by the earlier one. Do not divide by the probability of dying before $s$, and do not treat $t$ as the total time since age $x$.

Worked Example: Conditional Death Probability from Survival Values

Hypothetical model. A survival table for a specified life currently aged $x$ gives $S_x(0)=1.00$, $S_x(5)=0.90$, $S_x(10)=0.72$, and $S_x(15)=0.54$. Treat these as the model’s survival probabilities at the listed durations. Given survival for 5 years, calculate (a) the probability of surviving another 10 years and (b) the probability of dying during the next 5 years.

Method. For (a), the total duration from age $x$ is $5+10=15$ years. For (b), the next 5 years end at total duration $5+5=10$.

(a) The conditional probability of surviving another 10 years is

P(T_x>15\mid T_x>5) =\frac{S_x(15)}{S_x(5)} =\frac{0.54}{0.90} =0.60.

(b) The conditional probability of dying in the interval after year 5 through year 10 is

\frac{S_x(5)-S_x(10)}{S_x(5)} =\frac{0.90-0.72}{0.90} =\frac{0.18}{0.90} =0.20.

To four decimal places, the answers are $0.6000$ and $0.2000$. As a check, the probability of surviving the next 5 years, $S_x(10)/S_x(5)=0.72/0.90=0.8000$, plus the probability of dying in that interval, $0.2000$, equals $1.0000$. Note that part (a) asks about the next 10 years, while part (b) asks about the next 5 years; the corresponding total durations are different.

A Conditional Probability Is Not an Unconditional One

The ratio adjusts for the information that survival to $s$ is already known. In general, $S_x(s+t)$ and $S_x(s+t)/S_x(s)$ answer different questions. The first is measured among all lives at the original starting point; the ratio is measured among lives that have survived to $s$. There is no general rule that the conditional probability equals the unconditional survival probability over $t$ years.

This can be seen directly from the definition: $S_x(s+t)$ includes both surviving to $s$ and then surviving a further $t$, whereas the conditional probability removes the initial survival requirement by dividing by $S_x(s)$. The ratio is only valid when that denominator is positive. It also cannot be replaced by $S_x(t)$ merely because $t$ is the length of the future interval; $S_x(t)$ describes survival for $t$ years from the original age $x$, not survival for $t$ more years after $s$.

Worked Example: A Nonlinear Survival Function

Hypothetical model. Let the future lifetime at age $x$ have survival function $S_x(u)=(1+u/4)^{-2}$ for $u\geq0$. Given that the life has survived for 2 years, find the probability of surviving 2 additional years and the probability of dying in that next 2-year interval.

Method. Evaluate the survival function at total durations 2 and 4, then take the ratio. This model is specified by its survival function; no constant-rate assumption is being made.

At duration 2,

S_x(2)=\left(1+\frac{2}{4}\right)^{-2} =(1.5)^{-2} =\frac{1}{2.25} =\frac{4}{9}.

At duration 4,

S_x(4)=\left(1+\frac{4}{4}\right)^{-2} =2^{-2} =\frac{1}{4}.

Thus, the probability of surviving the additional 2 years is

P(T_x>4\mid T_x>2) =\frac{S_x(4)}{S_x(2)} =\frac{1/4}{4/9} =\frac{9}{16} =0.5625.

The probability of dying during the interval is $1-0.5625=0.4375$. To four decimal places, the answers are $0.5625$ and $0.4375$. As a check, the unconditional probability $S_x(4)=0.2500$ is not the conditional answer: it applies to the original group, before learning that the life survived the first two years.

A Reliable Calculation Process

A short sequence of checks helps prevent most errors in conditional lifetime calculations:

Translate the condition.
“Survived for $s$ years” means condition on $T_x>s$, so the denominator is $S_x(s)$.

Find the total duration.
If the question asks about $t$ more years, the endpoint is $s+t$, not $t$.

Form the ratio or complement.
Further survival is $S_x(s+t)/S_x(s)$; death in the interval is one minus that ratio.

Check the result.
For a valid survival function, $S_x(s+t)\leq S_x(s)$. If $S_x(s)>0$, the ratio must be between 0 and 1, and interval death plus further survival must equal 1.

The key exam trap is to ignore the conditioning event. A plausible-looking value such as $S_x(t)$ may be a valid survival probability but still answer the wrong question. Write the event in words first, convert “$t$ more years” into the total duration $s+t$, and only then substitute survival values. Conditional survival calculations are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical model, identify the conditioning time and total endpoint before calculating. Probabilities in the answers are rounded to four decimal places.

A lifetime has survival function $S_x(u)=1-u/12$ for $0\leq u\leq12$, and zero thereafter. Given survival for 3 years, find the probability of surviving 3 more years. (Answer: $0.6667$)
A hypothetical survival table gives $S_x(4)=0.88$ and $S_x(9)=0.66$. Given survival for 4 years, find the probability of dying during the next 5 years. (Answer: $0.2500$)
A lifetime has survival function $S_x(u)=(1+u/5)^{-1}$ for $u\geq0$. Given survival for 5 years, calculate the probability of surviving 5 additional years. (Answer: $0.6667$)
A hypothetical model has $S_x(u)=1-u/20$ for $0\leq u\leq20$. Given survival for 8 years, find both the probability of surviving the next 6 years and the probability of dying in that interval. (Answer: $0.5000$ and $0.5000$)
A survival table for a life currently aged $x$ gives $S_x(10)=0.80$ and $S_x(20)=0.52$. Given survival for 10 years, find the probability of dying during the following 10 years. (Answer: $0.3500$)

What You'll Learn