Failure Densities and Interval Failure Probabilities

From Lifetime to Failure Probability

For a nonnegative lifetime $T$, the density $f$ describes how probability is distributed over time, while the survival function $S$ describes the probability that failure has not yet occurred. The hazard compares failure density at a time with the probability of having survived up to that time. Together, these quantities let us answer questions such as: what is the probability of failure between times $a$ and $b$, and how does that probability change if the unit is known to have survived to $a$?

Throughout this tutorial, assume $T$ has an absolutely continuous distribution on $[0,\infty)$, with no atom at zero. Let $f$ be a density version, $F(t)=P(T\leq t)$, and $S(t)=P(T>t)=1-F(t)$ for $t\geq 0$. The continuous-distribution assumption means that including or excluding a single time endpoint does not change a probability. Density identities below are understood almost everywhere unless a particular version and additional regularity are specified.

Definition: The lifetime density is a nonnegative function $f$ satisfying $F(t)=\int_0^t f(u)\,du$ for $t\geq 0$. The survival function is $S(t)=P(T>t)=1-F(t)$. At times where $S(t)>0$, the hazard function is represented almost everywhere by $$ h(t)=\frac{f(t)}{S(t)}. $$ The cumulative hazard is $H(t)=\int_0^t h(u)\,du$, wherever this integral is defined.

The hazard is a rate, not itself a probability. Informally, it measures failure density at time $t$ relative to the fraction still at risk just before that time. A precise small-interval interpretation is available when $f$ and $S$ are continuous near $t$ and $S(t)>0$: $$ P(t<T\leq t+\Delta\mid T>t) =\frac{\int_t^{t+\Delta}f(u)\,du}{S(t)} \sim h(t)\Delta \quad\text{as }\Delta\downarrow 0. $$ The ratio becomes a probability only after it is integrated over a time interval in the appropriate conditional calculation.

Density, Survival, and Hazard

Because $F(t)=\int_0^t f(u)\,du$, the fundamental theorem for Lebesgue integrals gives $F'(t)=f(t)$ for almost every $t$. Consequently $S'(t)=-f(t)$ almost everywhere. This is an almost-everywhere statement: a density is defined only up to changes on sets of measure zero, so its value at an individual point need not equal $-S'(t)$ there.

Theorem: Suppose $T$ is nonnegative and absolutely continuous, $S(0)=1$, and $S(t)>0$ on $[0,\tau]$. For a density version $f$, define $h=f/S$ almost everywhere on this interval. Then $$ S'(t)=-f(t)\quad\text{almost everywhere},\qquad H(t)=\int_0^t h(u)\,du=-\log S(t),\qquad S(t)=e^{-H(t)} $$ for $0\leq t\leq\tau$.

Proof. Absolute continuity of $F$ implies absolute continuity of $S=1-F$, and $S'=-f$ almost everywhere. Since $S$ is continuous and strictly positive on the compact interval $[0,\tau]$, it has a positive minimum there. The function $\log S$ is therefore absolutely continuous on that interval, and the chain rule for absolutely continuous functions gives $$ \frac{d}{dt}\log S(t)=\frac{S'(t)}{S(t)} =-\frac{f(t)}{S(t)}=-h(t) \quad\text{almost everywhere}. $$ Integrating from $0$ to $t$, and using $S(0)=1$, yields $$ \log S(t)-\log S(0)=-\int_0^t h(u)\,du, \qquad \log S(0)=0. $$ Thus $\log S(t)=-H(t)$, so $S(t)=e^{-H(t)}$. $\square$

This result explains how the three functions fit together: the density gives the rate of probability accumulation, survival gives the probability still at risk, and hazard is their ratio. The identity is valid on intervals where survival remains positive. If survival reaches zero at a finite endpoint, the logarithm and hazard ratio need not remain finite at that endpoint; the result applies on intervals strictly before it.

Worked Example: A Uniform Lifetime

Suppose $T$ is uniformly distributed on $(0,10)$. Find its density, survival function, hazard, and probability of failure between times 4 and 7. Then calculate that interval probability conditional on surviving to time 4.

The density is $f(t)=1/10$ for $0<t<10$, and zero outside that interval. For $0\leq t\leq10$, $$ S(t)=P(T>t)=\int_t^{10}\frac{1}{10}\,du=\frac{10-t}{10}. $$ Therefore, for $0\leq t<10$, the hazard is $$ h(t)=\frac{f(t)}{S(t)} =\frac{1/10}{(10-t)/10} =\frac{1}{10-t}. $$ It increases as $t$ approaches 10 because fewer lifetimes remain at risk near the endpoint.

The unconditional probability of failure in $(4,7]$ is the density integral: $$ P(4<T\leq7)=\int_4^7\frac{1}{10}\,du=\frac{3}{10}. $$ Equivalently, $S(4)-S(7)=6/10-3/10=3/10$. Given survival beyond 4, the conditional probability is $$ P(4<T\leq7\mid T>4) =\frac{P(4<T\leq7)}{S(4)} =\frac{3/10}{6/10} =\frac12. $$

Failure Probabilities on an Interval

The event of failure during $(a,b]$, for $0\leq a<b$, is exactly the event that the lifetime exceeds $a$ but does not exceed $b$. Its probability can be computed by integrating the density or by subtracting survival probabilities. If the cumulative hazard is known, the survival representation gives another route to the answer.

Theorem: For $0\leq a<b$, an absolutely continuous lifetime satisfies $$ P(a<T\leq b)=\int_a^b f(t)\,dt=F(b)-F(a)=S(a)-S(b). $$ If $S(a)>0$, then $$ P(a<T\leq b\mid T>a) =\frac{S(a)-S(b)}{S(a)} =1-\frac{S(b)}{S(a)}. $$ On an interval where $S(t)=e^{-H(t)}$, this conditional probability is $$ 1-\exp\bigl(-(H(b)-H(a))\bigr). $$

Proof. By absolute continuity, the probability of the interval is the integral of the density: $$ P(a<T\leq b)=\int_a^b f(t)\,dt=F(b)-F(a). $$ Since $S=1-F$, this difference equals $S(a)-S(b)$. The conditioning event $\{T>a\}$ has probability $S(a)$, which is positive by assumption, and $\{a<T\leq b\}$ is a subset of it. The definition of conditional probability therefore gives $$ P(a<T\leq b\mid T>a) =\frac{S(a)-S(b)}{S(a)} =1-\frac{S(b)}{S(a)}. $$ When $S=e^{-H}$ at both endpoints, the survival ratio is $$ \frac{S(b)}{S(a)} =\frac{e^{-H(b)}}{e^{-H(a)}} =e^{-(H(b)-H(a))}. $$ Substitution proves the stated cumulative-hazard expression. $\square$

The unconditional and conditional probabilities answer different questions. The unconditional probability counts failures in the original population. The conditional probability uses only units that survived past $a$ as its reference group. In particular, if $S(a)=0$, conditioning on survival past $a$ is not defined by the usual conditional-probability ratio.

Worked Example: From a Density to Conditional Failure

Consider a lifetime with density $$ f(t)=\frac{t}{(1+t^2)^{3/2}},\qquad t>0. $$ Verify its survival function and hazard, then find the probability of failure between times 1 and 3 both unconditionally and given survival past time 1.

Integrating the density from zero to $t$ gives $$ F(t)=\int_0^t\frac{u}{(1+u^2)^{3/2}}\,du =1-\frac{1}{\sqrt{1+t^2}}. $$ Here the antiderivative follows because the derivative of $-(1+u^2)^{-1/2}$ is $u(1+u^2)^{-3/2}$. Thus $$ S(t)=1-F(t)=\frac{1}{\sqrt{1+t^2}}. $$ This is a valid lifetime distribution: $S(0)=1$, $S(t)$ is nonincreasing, and $S(t)\to0$ as $t\to\infty$. Its hazard is $$ h(t)=\frac{f(t)}{S(t)} =\frac{t}{1+t^2}. $$

At the endpoints, $S(1)=1/\sqrt{2}$ and $S(3)=1/\sqrt{10}$. Therefore $$ P(1<T\leq3)=S(1)-S(3) =\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{10}} \approx 0.3909. $$ Since $P(T>1)=1/\sqrt{2}$, the conditional probability is $$ P(1<T\leq3\mid T>1) =1-\frac{S(3)}{S(1)} =1-\frac{\sqrt{2}}{\sqrt{10}} =1-\frac{1}{\sqrt{5}} \approx 0.5528. $$ The conditional probability is larger than the unconditional one because the latter includes the entire original population in its denominator.

Using Cumulative Hazard for Conditional Risk

If cumulative hazard is available, it is often simpler to use its increment over the interval than to integrate a density from scratch. The conditional failure probability depends on $H(b)-H(a)$, not on $H(b)$ alone. This subtraction reflects that the unit has already survived through time $a$.

Worked Example: An Interval Specified by Cumulative Hazard

Suppose a lifetime has cumulative hazard $H(t)=t^2/20$ for $t\geq0$. Find the survival probability at time 2, the unconditional probability of failure between times 2 and 4, and the conditional probability of failure in that interval given survival beyond 2.

The survival function is $$ S(t)=e^{-H(t)}=e^{-t^2/20}. $$ In particular, $H(2)=4/20=0.2$ and $H(4)=16/20=0.8$, so $S(2)=e^{-0.2}$ and $S(4)=e^{-0.8}$. The unconditional interval probability is $$ P(2<T\leq4)=S(2)-S(4) =e^{-0.2}-e^{-0.8} =e^{-0.2}(1-e^{-0.6}) \approx 0.3694. $$

For units that have survived beyond 2, the conditional probability is $$ P(2<T\leq4\mid T>2) =1-\frac{S(4)}{S(2)} =1-e^{-(0.8-0.2)} =1-e^{-0.6} \approx 0.4512. $$ Multiplying this conditional probability by $P(T>2)=e^{-0.2}\approx0.8187$ recovers the unconditional value: $0.8187\times0.4512\approx0.3694$. This checks that the two calculations use consistent denominators.

Common Mistakes and Practical Checks

Claiming a pointwise density identity. It is generally correct that $S'(t)=-f(t)$ almost everywhere, not that every chosen density version equals $-S'(t)$ at every point. Changing a density at one point leaves its distribution unchanged.
Treating hazard as an interval probability. The value $h(t)$ is a rate. Over a short interval of length $\Delta$, the conditional failure probability is approximately $h(t)\Delta$ under suitable continuity; an exact interval probability requires integration or a survival difference.
Using the wrong denominator after conditioning. Given survival beyond $a$, divide the interval probability by $S(a)$. Do not divide by $S(b)$, and do not report the unconditional probability as the conditional one.
Forgetting the survival requirement. A conditional probability given $T>a$ requires $S(a)>0$. If no lifetimes can survive to $a$, the conditioning event has probability zero.
Using the cumulative hazard from zero instead of its increment. Conditional failure on $(a,b]$ depends on $H(b)-H(a)$. The expression $1-e^{-H(b)}$ applies to failure by $b$ from time zero, not generally to failure after already surviving to $a$.

Key takeaway: For an absolutely continuous lifetime, $S'=-f$ almost everywhere and $h=f/S$ wherever survival is positive. Interval failure probability is $S(a)-S(b)$; conditional on surviving past $a$, it is $1-S(b)/S(a)$, or $1-e^{-(H(b)-H(a))}$ when expressed through cumulative hazard.

Check Your Understanding

For each question, distinguish the unconditional probability from the probability conditional on survival.

Why is $S'(t)=-f(t)$ stated almost everywhere rather than as a pointwise identity for every density version?
If $S(2)=0.8$ and $S(5)=0.3$, calculate $P(2<T\leq5)$ and $P(2<T\leq5\mid T>2)$.
Suppose $H(1)=0.15$ and $H(4)=0.55$. Find the conditional probability of failure in $(1,4]$, given survival beyond 1.
What condition is needed for $P(a<T\leq b\mid T>a)$ to be defined by a conditional-probability ratio?
Explain why a hazard value at one time is not itself the probability of failure over an interval.