Interval Death Probabilities from CDF and Survival

Finding the Probability of Death Within an Interval

A lifetime question often asks whether death occurs between two specified times, rather than by one particular time. For a future lifetime $T_x$, the event $s<T_x\le t$ means that the individual survives beyond time $s$ but does not survive beyond time $t$. The interval probability can be calculated by subtracting cumulative probabilities or by subtracting survival probabilities. The distinction between strict and inclusive endpoints is important in general, but it disappears when the lifetime distribution is continuous.

Definition: For a future lifetime $T_x$, its distribution function is $F_x(t)=P(T_x\le t)$, and its survival function is $S_x(t)=P(T_x>t)$. For $s<t$, the probability of a lifetime in the interval $s<T_x\le t$ is $$ P(s<T_x\le t)=F_x(t)-F_x(s)=S_x(s)-S_x(t). $$

The probability $F_x(t)$ includes lifetimes equal to $t$. In contrast, $S_x(t)$ includes only lifetimes strictly greater than $t$. Accordingly, the formula above describes an interval that excludes its lower endpoint and includes its upper endpoint. That convention is exact for any distribution; continuity is not required for this particular event.

Why the Difference Formula Works

The event $T_x\le t$ consists of two disjoint parts: $T_x\le s$ and $s<T_x\le t$. Subtracting the probability of the first part from the probability of the whole event leaves the probability of the interval. The survival-function version follows by partitioning the event $T_x>s$.

\begin{aligned} \{T_x\le t\}&=\{T_x\le s\}\,\cup\,\{s<T_x\le t\},\\ F_x(t)&=F_x(s)+P(s<T_x\le t),\\ P(s<T_x\le t)&=F_x(t)-F_x(s). \end{aligned}

For the survival version, every outcome with $T_x>s$ either has $s<T_x\le t$ or has $T_x>t$. These two events are disjoint, so $S_x(s)=P(s<T_x\le t)+S_x(t)$. Rearranging gives $P(s<T_x\le t)=S_x(s)-S_x(t)$. Both derivations rely on the same event partition, not on a particular lifetime model.

Identify the interval.
Write down its lower and upper times and whether each endpoint is included.

Choose the available function.
Use $F_x(t)-F_x(s)$ if the CDF values are given, or $S_x(s)-S_x(t)$ if the survival values are given.

Check the result.
The probability must be between 0 and 1. It cannot be negative when $s<t$, because a valid CDF is nondecreasing and a valid survival function is nonincreasing.

Endpoint Conventions

A CDF uses a less-than-or-equal sign: $F_x(u)=P(T_x\le u)$. Therefore, subtracting $F_x(s)$ removes the outcomes at $s$ as well as all outcomes before $s$. This is why $F_x(t)-F_x(s)$ gives $P(s<T_x\le t)$, not $P(s\le T_x\le t)$.

For a distribution that may assign positive probability to a particular time, changing whether an endpoint is included can change the answer. The left limit $F_x(u-)$ means the probability $P(T_x<u)$. The corresponding formulas show which endpoint convention is needed.

Interval event	Probability from the CDF
$s<T_x\le t$	$F_x(t)-F_x(s)$
$s\le T_x\le t$	$F_x(t)-F_x(s-)$
$s<T_x<t$	$F_x(t-)-F_x(s)$
$s\le T_x<t$	$F_x(t-)-F_x(s-)$

A continuous lifetime has no point mass at any individual time: $P(T_x=u)=0$. Consequently, $F_x(u-)=F_x(u)$, and including or excluding either endpoint does not change an interval probability. This is the endpoint property used throughout the continuous-lifetime examples below. It does not mean that the interval has probability zero; it means only that the endpoints themselves contribute zero probability.

Key result: If $T_x$ has a continuous distribution and $s<t$, then all four endpoint choices for the interval from $s$ to $t$ have the same probability: $$ P(s<T_x\le t)=P(s\le T_x\le t)=P(s<T_x<t)=P(s\le T_x<t)=F_x(t)-F_x(s)=S_x(s)-S_x(t). $$

Worked Examples

Worked Example: Exponential Lifetime Over a Specified Interval

Hypothetical model. Suppose a future lifetime $T_x$, measured in years, has an exponential distribution with constant parameter $\lambda=0.2$ per year. Thus $S_x(u)=e^{-0.2u}$ for $u\ge0$, and $F_x(u)=1-e^{-0.2u}$. Find the probability of death after 2 years but by 5 years. Give the answer to four decimal places.

Method. The event is $2<T_x\le5$, so use the CDF difference or the survival difference. From the CDF,

\begin{aligned} F_x(2)&=1-e^{-0.2(2)}=1-e^{-0.4}\approx1-0.670320=0.329680,\\ F_x(5)&=1-e^{-0.2(5)}=1-e^{-1}\approx1-0.367879=0.632121,\\ P(2<T_x\le5)&=F_x(5)-F_x(2)\\ &=0.632121-0.329680=0.302441. \end{aligned}

As a check, the survival calculation gives $S_x(2)-S_x(5)=0.670320-0.367879=0.302441$, the same result. Rounded only at the final step, the probability is 0.3024. The common mistake is to subtract the two survival values in the opposite order; that would give a negative number, contradicting the fact that an interval probability cannot be negative.

Worked Example: Endpoint Inclusion Under a Continuous Model

Hypothetical model. A future lifetime $T_x$, in years, is uniformly distributed on the interval from 0 to 12. Its CDF is $F_x(u)=u/12$ for $0\le u\le12$, and the lifetime is continuous. Find the probability of death from year 3 through year 9, including both endpoints. Then find the probability of being outside that interval, $T_x\le3$ or $T_x>9$.

Method. Since the lifetime is continuous, the endpoint probabilities at 3 and 9 are zero. The inclusive interval therefore has the same probability as $3<T_x\le9$. Apply the CDF difference:

F_x(3)=\frac{3}{12}=0.25,\qquad F_x(9)=\frac{9}{12}=0.75,\qquad P(3<T_x\le9)=0.75-0.25=0.50.

For the complement, the two disjoint parts have probabilities $P(T_x\le3)=F_x(3)=0.25$ and $P(T_x>9)=S_x(9)=1-F_x(9)=0.25$. Thus the probability of being outside the interval is $0.25+0.25=0.50$. The check is $0.50+0.50=1.00$, as required for an event and its complement. The common mistake is to treat “including both endpoints” as a different calculation even though a continuous lifetime has zero probability at either endpoint.

Worked Example: Combining Adjacent Intervals From a Piecewise CDF

Hypothetical model. Suppose $T_x$ is a continuous lifetime whose CDF is piecewise linear through the points $(0,0)$, $(2,0.08)$, $(6,0.31)$, $(11,0.72)$, and $(16,1)$, with $F_x(u)=0$ before 0 and $F_x(u)=1$ after 16. The increasing knot values define a valid continuous CDF. Find the probability of death in $2<T_x\le11$, and verify it by splitting at year 6.

Method. Read the endpoint CDF values from the specified model. The direct calculation is

P(2<T_x\le11)=F_x(11)-F_x(2)=0.72-0.08=0.64.

For the split calculation, first find the probability in $2<T_x\le6$:

P(2<T_x\le6)=F_x(6)-F_x(2)=0.31-0.08=0.23.

Next find the probability in $6<T_x\le11$:

P(6<T_x\le11)=F_x(11)-F_x(6)=0.72-0.31=0.41.

The split intervals are disjoint and together form $2<T_x\le11$, so their probabilities add: $0.23+0.41=0.64$. This agrees with the direct calculation. Continuity means the year-6 endpoint has probability zero, so placing that single endpoint in either adjacent interval would not change the total. The common mistake is to add $F_x(6)$ directly to $F_x(11)$; CDF values are cumulative, so interval probabilities require differences.

Common Checks and Exam Traps

Interval calculations are short, but a small notation error can reverse the event or count the wrong endpoint. Before calculating, translate the verbal statement into an inequality for $T_x$. For example, “after $s$ but no later than $t$” is $s<T_x\le t$, while “at least $s$ and before $t$” is $s\le T_x<t$. For a continuous lifetime these have the same probability, but writing the event first helps prevent mistakes in settings where continuity has not been established.

Reversing the order. For $s<t$, use $F_x(t)-F_x(s)$, or $S_x(s)-S_x(t)$. Reversing either subtraction produces a negative value.
Subtracting the wrong cumulative quantities. $F_x(t)$ alone is the probability of death by $t$, not the probability of death between $s$ and $t$.
Assuming endpoint conventions never matter. They do not affect probabilities under a continuous lifetime model, but they can matter if a distribution has a point mass at an endpoint.
Mixing the CDF and survival interpretations. $F_x(u)$ counts $T_x\le u$, whereas $S_x(u)$ counts $T_x>u$. Keep each definition attached to its inequality.
Ignoring a basic probability check. A valid interval probability is between 0 and 1. If $s<t$, the CDF difference and survival difference should be nonnegative.

These formulas provide a direct way to turn a lifetime distribution into the probability of death during any specified interval. The essential steps are to identify the event, respect the CDF or survival-function convention, and use continuity only when making claims about endpoint equivalence. This distinction is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each hypothetical model, write the interval event carefully and use the stated distribution information. Answers are given to four decimal places where needed.

A future lifetime $T_x$ is exponential with parameter $\lambda=0.1$ per year. Find $P(1<T_x\le4)$, using the survival function. (Answer: 0.2345)
A future lifetime $T_x$ is uniform on $[0,15]$ years. Find the probability of death between years 3 and 9, including both endpoints, and state whether the endpoint convention changes the result. (Answer: 0.4000; no, the continuous model assigns zero probability to either endpoint)
A continuous lifetime model has $F_x(5)=0.18$, $F_x(9)=0.42$, and $F_x(12)=0.63$. Find $P(5<T_x\le12)$ directly and by splitting at year 9. (Answer: 0.4500 directly; $0.2400+0.2100=0.4500$ by splitting)
A future lifetime $T_x$ is exponential with parameter $\lambda=0.25$ per year. Find $P(2<T_x\le6)$ and the probability that death occurs outside this interval. (Answer: 0.3834 in the interval; 0.6166 outside)
A continuous hypothetical lifetime CDF is specified at three times by $F_x(3)=0.12$, $F_x(8)=0.46$, and $F_x(14)=0.81$. Find the probability of death in $3<T_x\le14$, then verify it using the two adjacent intervals split at year 8. (Answer: 0.6900 directly; $0.3400+0.3500=0.6900$ by splitting)

Interval event	Probability from the CDF
\(s<T_x\le t\)	\(F_x(t)-F_x(s)\)
\(s\le T_x\le t\)	\(F_x(t)-F_x(s-)\)
\(s<T_x<t\)	\(F_x(t-)-F_x(s)\)
\(s\le T_x<t\)	\(F_x(t-)-F_x(s-)\)

What You'll Learn