Tail Integrals for Survival Probabilities

Why Integrate the Tail?

A survival probability asks whether a future lifetime exceeds a specified duration. If a hypothetical policyholder aged $x$ has future lifetime $T_x$, then surviving at least $t$ time units means the lifetime lies to the right of $t$ on the time axis. When $T_x$ has a density, the probability of that event is the area under the density over that right-hand tail. This turns a picture of a lifetime distribution into a direct calculation.

The direction of integration matters: integrating from zero up to $t$ gives the probability of a lifetime at or below $t$, not the probability of surviving beyond $t$. The support of the density matters too. A density may extend indefinitely, or it may be zero beyond a finite endpoint. The integration limits must reflect the model in either case.

Definition: Suppose the future lifetime $T_x$ is continuous, has density $f_x(u)$ for $u\geq 0$, and that density integrates to 1 over its support. Its survival function is $S_x(t)=P(T_x>t)$. For $t\geq 0$, $$ S_x(t)=\int_t^\infty f_x(u)\,du. $$ If the density is zero beyond a finite endpoint, integrating to infinity is still valid: the part of the integral beyond that endpoint contributes zero.

Why the Tail Integral Gives Survival

A density assigns probability to intervals through the area under its curve. In particular, for a continuous lifetime with density $f_x$, the probability that the lifetime lies between $t$ and a larger value $b$ is the integral of $f_x(u)$ from $t$ to $b$. Taking the right endpoint through the entire remaining support gives the probability that the lifetime exceeds $t$. That is precisely the survival probability.

Equivalently, split the total probability into the area to the left of $t$ and the area to its right. Since the total area is 1, the area to the right is 1 minus the area to the left. Using the distribution function $F_x(t)$, defined as $P(T_x\leq t)$, this gives the same result:

1=\int_0^t f_x(u)\,du+\int_t^\infty f_x(u)\,du, \qquad S_x(t)=\int_t^\infty f_x(u)\,du=1-F_x(t).

This justification uses the assumptions that the lifetime has a density on its support and that its total probability is 1. The integral is an improper integral when the support is unbounded; it must converge. If a model instead assigns probability to isolated lifetime values, integrating only its density would not account for those point probabilities. The formulas here are for the continuous lifetime models specified in the examples.

A Reliable Calculation Method

Before integrating, identify the event, the density's support, and the units of time. Then use the right-hand tail, splitting the integral wherever the density's formula changes. A compact workflow is:

Translate the event.
“Survives more than $t$” means $T_x>t$, so the desired probability is the area to the right of $t$.

Check the support.
Write the interval on which the density is nonzero. If it has a finite endpoint, the tail ends there in practice.

Integrate each applicable piece.
For a piecewise density, split at every formula change between $t$ and the end of the support.

Check the result.
A survival probability must lie between 0 and 1. It cannot exceed the total area under the density.

Worked Example: Exponential Lifetime Tail

A hypothetical lifetime model has density $f_x(u)=0.2e^{-0.2u}$ for $u\geq 0$, with $u$ measured in years and zero density for $u<0$. Calculate the probability that the future lifetime exceeds 3 years. Give the probability to four decimal places.

Method. Integrate the specified density from the threshold, 3 years, to the end of its support. The parameter $0.2$ is in inverse years, so the product $0.2u$ is dimensionless.

\begin{aligned} S_x(3) &=\int_3^\infty 0.2e^{-0.2u}\,du\\ &=\left[-e^{-0.2u}\right]_3^\infty\\ &=0-(-e^{-0.6})\\ &=e^{-0.6}\\ &\approx 0.5488. \end{aligned}

The probability is $0.5488$ to four decimal places. As a check, the integral from 0 to 3 is $1-e^{-0.6}\approx 0.4512$; the left-tail and right-tail probabilities sum to 1. A common mistake is to report $0.4512$ as the probability of surviving beyond 3 years. That is the area to the left of 3, so the correction is to integrate from 3 toward infinity.

Worked Example: A Density with a Finite Endpoint

For a hypothetical coverage model, suppose future lifetime has density $f_x(u)=u/8$ for $0\leq u\leq 4$ years and zero otherwise. Calculate the probability of a lifetime longer than 2 years. First check that the stated function is a valid density.

Method. Verify the total area, then integrate the density from 2 to the finite endpoint 4. Integrating to infinity would give the same value because the density is zero after year 4.

\int_0^4 \frac{u}{8}\,du =\left[\frac{u^2}{16}\right]_0^4 =\frac{16}{16}-0 =1.

The density is normalized. Now compute the right-hand tail:

\begin{aligned} S_x(2) &=\int_2^4 \frac{u}{8}\,du\\ &=\left[\frac{u^2}{16}\right]_2^4\\ &=\frac{16}{16}-\frac{4}{16}\\ &=\frac{12}{16}\\ &=0.75. \end{aligned}

Thus, the probability is $0.75$, or $0.7500$ to four decimal places. The omitted tail above 4 contributes zero because the density has no support there. A common mistake is to integrate $u/8$ from 2 to infinity as though that formula continued indefinitely; it would not be a valid density on that unbounded interval. The stated support must be respected.

Worked Example: Integrating Across a Change in Density

A hypothetical model describes future lifetime $T_x$, in years, by the density $f_x(u)=1/6$ for $0\leq u\leq 3$, $f_x(u)=1/12$ for $3<u\leq 9$, and zero otherwise. Verify the normalization and calculate the probability that $T_x>2$. Give the final probability to four decimal places.

Method. Check the area on each interval. For the survival calculation, split the tail at year 3, where the density changes.

\int_0^3 \frac{1}{6}\,du+\int_3^9 \frac{1}{12}\,du =\frac{3}{6}+\frac{6}{12} =\frac{1}{2}+\frac{1}{2} =1.

The density is normalized. From year 2 to year 3, the density is $1/6$, so that part of the tail has area $(3-2)/6=1/6$. From year 3 to year 9, the density is $1/12$, giving area $(9-3)/12=6/12=1/2$. Therefore,

S_x(2)=\int_2^3 \frac{1}{6}\,du+\int_3^9 \frac{1}{12}\,du =\frac{1}{6}+\frac{1}{2} =\frac{2}{3} \approx 0.6667.

The result is $0.6667$ to four decimal places. The remaining area to the left of year 2 is $\int_0^2 (1/6)\,du=1/3$, and $1/3+2/3=1$, as required. A common mistake is to use the first density over the entire tail from year 2 through year 9. The formula changes at year 3, so the tail must be split there.

Common Traps and Practical Checks

Tail integration is simple once the event and support are clear, but errors in either can reverse or distort the answer. In actuarial calculations, a survival probability may be an input to a later calculation involving benefits or premiums. An incorrect tail can therefore affect more than one subsequent step.

Using the left tail. $\int_0^t f_x(u)\,du$ is the probability of a lifetime at or below $t$. For survival beyond $t$, use the remaining area $\int_t^\infty f_x(u)\,du$.
Ignoring a change in formula. If $f_x$ is piecewise, split the tail at each change that lies above $t$. Do not extend one piece beyond its stated interval.
Ignoring finite support. If the density becomes zero after a finite time, the tail ends at that time. The notation with infinity is still correct, but it does not mean the density's formula continues forever.
Mixing time units. A density expressed per year must be integrated with a time variable in years. If the threshold is given in months, convert it to years before using a per-year density.
Skipping the total-area check. Integrating the density over its full support should give 1. If it does not, recheck the density or its support before interpreting a tail probability.

Two quick checks are especially useful: a computed tail must be between 0 and 1, and the tail plus the area to the left of its threshold must equal 1. These checks do not replace careful integration, but they often reveal a reversed bound, a missing interval, or an arithmetic error.

Key takeaway: For a continuous future lifetime with density $f_x$, survival beyond $t$ is the right-tail area $S_x(t)=\int_t^\infty f_x(u)\,du$. Integrate only over the density's support, split at any formula changes, and check that the result is a probability.

Check Your Understanding

For each model, identify the right-tail integral before evaluating it. Answers are rounded to four decimal places where appropriate.

A hypothetical lifetime has density $f_x(u)=1/5$ for $0\leq u\leq5$ years and zero otherwise. Find the probability the lifetime exceeds 3.2 years. (Answer: 0.3600)
A hypothetical model has density $f_x(u)=u/18$ for $0\leq u\leq6$ years and zero otherwise. Verify normalization and find $S_x(3)$. (Answer: normalized; $S_x(3)=0.7500$)
A lifetime density is $1/10$ for $0\leq u\leq4$, $1/20$ for $4<u\leq16$, and zero otherwise, with time in years. Verify the total area and calculate $S_x(10)$. (Answer: total area $1$; $S_x(10)=0.3000$)
A hypothetical exponential lifetime has density $f_x(u)=0.25e^{-0.25u}$ for $u\geq0$, where $u$ is in years. Calculate the probability of exceeding 4 years. (Answer: $e^{-1}\approx0.3679$)
A density is $1/12$ for $0\leq u\leq6$ years, $1/24$ for $6<u\leq18$ years, and zero otherwise. Verify normalization and calculate the probability that the lifetime exceeds 4 years, splitting the integral where needed. (Answer: normalized; $S_x(4)=2/3\approx0.6667$)

What You'll Learn