Differentiating Survival to Obtain Density

Introduction

A survival function records the probability that a future lifetime exceeds each possible time. When that function is differentiable, its slope describes how quickly survival probability is decreasing. The corresponding density is the negative of that slope: the sign matters because a valid survival curve decreases rather than increases.

This tutorial focuses on recovering the density from a given survival curve and using the curve to calculate probabilities. The examples use hypothetical future lifetimes measured in years. This treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

From Survival Function to Density

Definition: For a future lifetime $T_x$ measured from age $x$, the survival function is $S_x(t)=P(T_x>t)$, for $t\geq 0$. If the distribution has a density $f_x$ and $S_x$ is differentiable at $t$, then $$ f_x(t)=-\frac{d}{dt}S_x(t). $$ The density is measured in inverse units of time; if $t$ is measured in years, $f_x(t)$ is measured per year.

The minus sign follows directly from the definitions. The distribution function is $F_x(t)=P(T_x\leq t)$. The events $T_x\leq t$ and $T_x>t$ are complements, so $F_x(t)=1-S_x(t)$. Differentiating this equality at a point where the functions are differentiable gives the density.

f_x(t)=\frac{d}{dt}F_x(t) =\frac{d}{dt}\bigl(1-S_x(t)\bigr) =-S_x'(t).

This derivation also gives a useful sign check: where the derivative exists, a nonincreasing survival function has $S_x'(t)\leq 0$, so its density cannot be negative. A positive derivative of survival, or a negative result for the density, is a warning to check the sign or the assumed curve.

The density is not itself a probability at a single time. Probabilities over intervals are obtained by integrating the density. Equivalently, for $0\leq a<b$, the survival probability lost between $a$ and $b$ is

P(a<T_x\leq b)=S_x(a)-S_x(b) =\int_a^b f_x(t)\,dt.

The survival difference is often the quickest calculation when the survival curve is given explicitly. The integral provides a separate check when the density is straightforward to integrate. The equality uses the definitions of survival and interval probability; endpoint distinctions do not affect these calculations for the continuous models used below.

Worked Calculations

Worked Example: Differentiate a Survival Curve and Find an Interval Probability

A hypothetical future lifetime $T_x$, measured in years, has survival function $S_x(t)=(1-t/8)^2$ for $0\leq t\leq 8$, and $S_x(t)=0$ for $t>8$. Find $f_x(2)$ and $P(1<T_x\leq4)$. Check the density over its full support.

Method. Differentiate the given survival function and reverse the sign. Then calculate the interval probability as the decrease in survival. The model is the stated continuous lifetime distribution; no interest assumption is involved.

For $0<t<8$, apply the chain rule:

S_x'(t)=2(1-t/8)(-1/8) =-\frac{1}{4}(1-t/8), \qquad f_x(t)=\frac{1}{4}(1-t/8).

At $t=2$, $1-2/8=3/4$, so

f_x(2)=\frac{1}{4}\cdot\frac{3}{4} =\frac{3}{16}=0.1875\text{ per year}.

For the requested probability, evaluate survival at both endpoints:

S_x(1)=(7/8)^2=\frac{49}{64}, \qquad S_x(4)=(1/2)^2=\frac{16}{64}, $$ $$ P(1<T_x\leq4)=S_x(1)-S_x(4) =\frac{49-16}{64} =\frac{33}{64}=0.515625.

As a check on the density, its integral over the support is $\int_0^8 \frac14(1-t/8)\,dt =\frac14(8-4)=1$. The density is nonnegative on the support and has total integral one. The common mistake avoided here is reporting $S_x(4)-S_x(1)$, which reverses the order and gives a negative interval probability.

Worked Example: Use the Derivative and Survival Drop as Independent Checks

A hypothetical future lifetime, in years, has $S_x(t)=(1-t/6)^3$ for $0\leq t\leq6$, and zero after year 6. Find the density at year 3 and the probability of surviving beyond year 2 but not beyond year 5. Verify the interval probability by integrating the density.

Method. Differentiate the survival curve to obtain the density. Calculate the interval probability first from the survival difference and then from the density integral. The distribution assumption is exactly the stated continuous survival model, with support from zero through six years.

By the chain rule,

S_x'(t)=3(1-t/6)^2(-1/6) =-\frac{1}{2}(1-t/6)^2, \qquad f_x(t)=\frac{1}{2}(1-t/6)^2.

At $t=3$, $1-3/6=1/2$, and therefore

f_x(3)=\frac{1}{2}(1/2)^2 =\frac{1}{8}=0.125\text{ per year}.

For the interval probability, the endpoint values are $S_x(2)=(2/3)^3=8/27$ and $S_x(5)=(1/6)^3=1/216$. Thus

P(2<T_x\leq5) =\frac{8}{27}-\frac{1}{216} =\frac{64-1}{216} =\frac{63}{216} =\frac{7}{24} \approx0.2917.

To check by integration, an antiderivative of $\frac12(1-t/6)^2$ is $- (1-t/6)^3$. Hence

\int_2^5 f_x(t)\,dt =\left[-(1-t/6)^3\right]_2^5 =\frac{8}{27}-\frac{1}{216} =\frac{7}{24}.

The two methods agree. A common error is to substitute the endpoints into the survival curve but subtract in the wrong direction; the earlier survival value must be larger, so the probability is $S_x(2)-S_x(5)$.

Worked Example: Recover a Density from a Quadratic Survival Curve

For a hypothetical future lifetime in years, suppose $S_x(t)=1-(t/5)^2$ for $0\leq t\leq5$, and $S_x(t)=0$ thereafter. Find $f_x(2)$ and $P(1<T_x\leq4)$. Check the probability using the density.

Method. Differentiate the quadratic curve, apply the negative sign, and use both the survival difference and the density integral. The specified curve is the full model; there are no additional mortality or interest assumptions.

Rewrite the survival function as $S_x(t)=1-t^2/25$. For $0<t<5$,

S_x'(t)=-\frac{2t}{25}, \qquad f_x(t)=\frac{2t}{25}.

At $t=2$, the density is

f_x(2)=\frac{2(2)}{25} =\frac{4}{25}=0.16\text{ per year}.

The survival probabilities at the interval endpoints are $S_x(1)=1-1/25=24/25$ and $S_x(4)=1-16/25=9/25$. Therefore

P(1<T_x\leq4) =\frac{24}{25}-\frac{9}{25} =\frac{15}{25} =\frac{3}{5}=0.6000.

The density check gives $\int_1^4 (2t/25)\,dt=[t^2/25]_1^4=(16-1)/25=3/5$, the same probability. Also, integrating over the full support gives $[t^2/25]_0^5=1$. The common mistake avoided is differentiating $1-t^2/25$ and treating its negative derivative as the density; the density is the negative of the survival derivative, so it is positive.

A Reliable Calculation Process

When a question supplies a survival curve and asks for a density or probability, keeping the operations in order helps prevent sign, endpoint, and unit errors. The following process applies at points where the curve is differentiable.

Identify the applicable formula.
Write $f_x(t)=-S_x'(t)$. Check the specified time range before differentiating or substituting values.

Differentiate with respect to time.
Apply the chain rule where needed, and retain the negative sign that converts the survival slope to a density.

Calculate the requested quantity.
For a density value, substitute the time into $f_x$. For an interval probability, use $S_x(a)-S_x(b)$, or integrate $f_x$ from $a$ to $b$.

Check the result.
A density should be nonnegative where the model is differentiable; an interval probability should be between zero and one. Confirm that its units and rounding match the question.

Common Exam Traps

Most errors in this calculation are not difficult calculus errors; they come from confusing what the survival function measures with what the density measures. In particular, the density is a rate of probability change with respect to time, not the remaining survival probability.

Dropping the negative sign. If $S_x'(t)=-0.04$ per year, then $f_x(t)=0.04$ per year, not $-0.04$.
Using a survival value as a density. $S_x(2)$ is a probability of surviving beyond two years; $f_x(2)$ is a density value per unit time. They are different quantities.
Reversing an interval difference. For $a<b$, survival at $a$ is at least survival at $b$, and the interval probability is $S_x(a)-S_x(b)$, not the reverse.
Ignoring the curve's stated range. A formula may apply only on a finite interval. Differentiate it on that interval and use any separately stated survival values outside it.
Omitting units. If time is given in years, report a density in inverse years, such as $0.125$ per year. Probabilities, by contrast, have no units.

Key takeaway: Where the survival function is differentiable, its density is $f_x(t)=-S_x'(t)$. For an interval, subtract the later survival probability from the earlier one: $P(a<T_x\leq b)=S_x(a)-S_x(b)$.

Check Your Understanding

For each hypothetical lifetime, use the stated survival model and show the derivative or survival values needed before checking the answer.

A lifetime in years has $S_x(t)=(1-t/12)^2$ for $0\leq t\leq12$, and zero thereafter. Find $f_x(3)$ and $P(2<T_x\leq6)$. (Answer: $f_x(3)=1/8=0.1250$ per year; probability $=4/9\approx0.4444$.)
A lifetime in years has $S_x(t)=(1-t/9)^3$ for $0\leq t\leq9$, and zero thereafter. Find $f_x(3)$ and $P(1<T_x\leq6)$. (Answer: $f_x(3)=4/27\approx0.1481$ per year; probability $=485/729\approx0.6653$.)
A lifetime in years has $S_x(t)=1-(t/8)^2$ for $0\leq t\leq8$, and zero thereafter. Find $f_x(4)$ and $P(2<T_x\leq6)$. (Answer: $f_x(4)=1/8=0.1250$ per year; probability $=1/2=0.5000$.)
A lifetime in years has $S_x(t)=(1-t/a)^2$ for $0\leq t\leq a$, and zero thereafter. Given $S_x(3)=25/49$, find $a$, $f_x(7)$, and $P(2<T_x\leq7)$. (Answer: $a=10.5$ years; $f_x(7)=4/63\approx0.0635$ per year; probability $=80/147\approx0.5442$.)
A lifetime in years has $S_x(t)=(1-t/4)^2$ for $0\leq t\leq4$, and zero thereafter. Find $f_x(1)$ and $P(1<T_x\leq3)$. (Answer: $f_x(1)=3/8=0.3750$ per year; probability $=1/2=0.5000$.)

What You'll Learn