Reliability Survival and Hazard Functions

Introduction

In reliability, a lifetime is the time until a component fails. In survival analysis, it may be the time until an event such as recovery or recurrence. In either setting, two related questions matter: what is the probability that the event has not happened by time \(t\), and how quickly is it happening among units that have lasted until \(t\)? The first question leads to the survival function; the second leads to the hazard rate.

These quantities describe different aspects of the same lifetime distribution. Survival is an ordinary probability of lasting beyond a specified time. Hazard is a conditional rate at a particular time, and so it depends on having survived up to that time. Keeping that distinction clear is essential when interpreting reliability calculations.

Survival and Hazard: Definitions

Let \(T\) be a nonnegative random variable representing a lifetime. Write \(F(t)=P(T\leq t)\) for its distribution function. The survival function gives the probability of lasting longer than \(t\).

Definition: The survival function of \(T\) is \(S(t)=P(T>t)=1-F(t)\). For a nonnegative lifetime, \(S(t)=1\) when \(t<0\). At time zero, \(S(0)=P(T>0)\), which equals one if the lifetime is strictly positive with probability one.

Survival is nonincreasing: if \(s<t\), then the event \(\{T>t\}\) is contained in \(\{T>s\}\), so \(S(t)\leq S(s)\). Its values lie between zero and one. For a lifetime that is finite with probability one, \(S(t)\) tends to zero as \(t\) tends to infinity. A value such as \(S(10)=0.8\) means that 80% of the population is expected to last longer than 10 time units; it is not a rate.

To define a hazard rate using a density, suppose \(T\) has a density \(f\), and consider a time \(t\) for which \(S(t)>0\). The hazard compares the probability of failure in a short interval after \(t\) with the interval’s length, among those that have lasted past \(t\).

Definition: At points where the density-based limit exists, the hazard rate is \(h(t)=\lim_{\Delta\downarrow 0}P(t<T\leq t+\Delta\mid T>t)/\Delta\). If \(f\) is continuous at \(t\) and \(S(t)>0\), then \(h(t)=f(t)/S(t)\). The hazard has units of inverse time.

Why the Density Quotient Is the Conditional Rate

The quotient \(f(t)/S(t)\) is not just an algebraic convention. It follows from conditioning on survival to \(t\). The following result states conditions under which that interpretation holds pointwise.

Theorem: Suppose \(T\) has a density \(f\) that is continuous at \(t\), and \(S(t)>0\). Then \(h(t)=f(t)/S(t)\), where \(h(t)\) is the limiting conditional failure probability per unit time defined above.

Proof. For \(\Delta>0\), conditioning on \(T>t\) gives \(P(t<T\leq t+\Delta\mid T>t)=P(t<T\leq t+\Delta)/S(t)\). Because \(T\) has density \(f\), the numerator is \(\int_t^{t+\Delta} f(u)\,du\). Therefore,

\frac{P(t<T\leq t+\Delta\mid T>t)}{\Delta} = \frac{1}{S(t)}\frac{1}{\Delta}\int_t^{t+\Delta}f(u)\,du.

By continuity of \(f\) at \(t\), its average over \([t,t+\Delta]\) tends to \(f(t)\) as \(\Delta\downarrow0\). Since \(S(t)>0\) is fixed, the displayed expression tends to \(f(t)/S(t)\). This proves the result. \(\square\)

For a sufficiently small interval of length \(\Delta\), the conditional probability of failure is thus approximately \(h(t)\Delta\). This is a local approximation, not an exact probability for a finite interval. A hazard of \(0.02\) per hour, for example, corresponds to about a 0.002 conditional failure probability over the next 0.1 hour when that interval is sufficiently short and the hazard does not change much within it.

If a lifetime distribution has atoms, the density formula may not describe all its failure probability. A point mass creates a jump in the distribution, while a density-based rate describes probability spread continuously over time. The formula \(f(t)/S(t)\) therefore applies to continuous lifetime models at times with positive survival and suitable density regularity. It should not be used to erase or approximate an atom without separately accounting for it.

Worked Lifetime Models

The following examples show how to obtain survival from a distribution and then calculate its hazard. The exponential and Weibull models are parameterized here by a rate and by shape and scale, respectively.

Worked Example: Exponential Lifetime

Suppose \(T\) has an exponential distribution with rate \(\lambda=0.2\) per hour, so its density is \(f(t)=0.2e^{-0.2t}\) for \(t\geq0\). Find its survival at 3 hours and its hazard rate.

The survival probability is the probability of exceeding 3:

S(3)=P(T>3)=\int_3^\infty 0.2e^{-0.2u}\,du =\left[-e^{-0.2u}\right]_3^\infty=e^{-0.6}\approx 0.5488.

For any \(t\geq0\), integrating the density from \(t\) to infinity gives \(S(t)=e^{-0.2t}\). Since this is positive at every finite \(t\), the density quotient is defined:

h(t)=\frac{f(t)}{S(t)} =\frac{0.2e^{-0.2t}}{e^{-0.2t}} =0.2 \text{ per hour}.

For example, conditional on lasting beyond 2 hours, the probability of failure by 2.5 hours is \(1-e^{-0.2(0.5)}=1-e^{-0.1}\approx0.0952\). Dividing by the interval length gives about \(0.1903\) per hour, close to the limiting rate \(0.2\) per hour.

Worked Example: Weibull Lifetime

Let \(T\) have a Weibull distribution with shape \(k=2\) and scale \(\eta=10\) hours. Its distribution function for \(t\geq0\) is \(F(t)=1-\exp[-(t/\eta)^k]\). Find the survival probability and hazard at 5 hours.

Subtracting the distribution function from one gives \(S(t)=\exp[-(t/10)^2]\). Hence \(S(5)=\exp[-(5/10)^2]=e^{-0.25}\approx0.7788\). Differentiating the distribution function gives the density:

f(t)=\frac{2}{10}\left(\frac{t}{10}\right) \exp\!\left[-\left(\frac{t}{10}\right)^2\right], \qquad t>0.

At \(t=5\), the density is \(f(5)=0.1e^{-0.25}\). Dividing by survival cancels the exponential factor:

h(5)=\frac{f(5)}{S(5)} =\frac{0.1e^{-0.25}}{e^{-0.25}} =0.1 \text{ per hour}.

More generally, the same quotient gives \(h(t)=\frac{k}{\eta}(t/\eta)^{k-1}\) for \(t>0\). Here that is \(h(t)=0.2(t/10)\) per hour. The calculation describes a conditional rate at each time; it is not a probability of failure over an interval of arbitrary length.

Worked Example: Uniform Lifetime

Suppose a component’s lifetime is uniformly distributed between 0 and 12 hours. Find its survival and hazard functions for times within its possible lifetime, and evaluate both at 6 hours.

The density is \(f(t)=1/12\) for \(0<t<12\). For \(0\leq t<12\), integrate the density from \(t\) to the endpoint:

S(t)=P(T>t)=\int_t^{12}\frac{1}{12}\,du =\frac{12-t}{12}.

Thus \(S(6)=(12-6)/12=1/2\). For \(0\leq t<12\), the hazard is

h(t)=\frac{1/12}{(12-t)/12}=\frac{1}{12-t}, \qquad h(6)=\frac{1}{6}\text{ per hour}.

At \(t=12\), survival is zero, so the conditional hazard definition given here does not apply: there are no units with positive probability of surviving beyond that time. The formula is used only where \(S(t)>0\).

Interpreting the Two Functions

Survival and hazard answer different questions, even though both come from the same distribution. Survival is an unconditional probability: it describes the fraction of the original population expected to last past \(t\). Hazard is conditional: it describes the instantaneous failure rate among units that have already lasted past \(t\). The survival value does not itself determine a short-term conditional rate without information about how failure probability is distributed in time.

This distinction is useful in reliability work, where an operating component may be known to have survived to the time it is inspected. It also matters with censored observations: a unit recorded as still operating at the end of follow-up supplies information that it survived to that time, but not an observed failure time. The distributional definitions describe the underlying lifetime model, not a claim that every observed unit’s failure time is known.

Hazard rates also carry units. If time is measured in hours, \(f(t)\) has units of inverse hours, \(S(t)\) is dimensionless, and \(h(t)\) has units of inverse hours. Changing the time unit changes the numerical rate: a rate expressed per hour must be converted before comparison with a rate expressed per day.

Common Mistake

Common mistake: Treating \(h(t)\) as the probability that a unit fails exactly at time \(t\), or as the probability it fails over the next hour. For a continuous lifetime, \(P(T=t)=0\); the hazard is a conditional probability rate, not a point probability. Over a short interval of length \(\Delta\), the conditional failure probability is approximately \(h(t)\Delta\), subject to the hazard being suitably stable over that interval.

A related error is to compute \(f(t)\) and call it the hazard. The density measures probability per unit time in the original population near \(t\). The hazard divides that density by \(S(t)\), accounting for the fact that its reference group consists only of units that have survived to \(t\). For example, in the Weibull calculation, the exponential factor appears in both density and survival and cancels in their ratio.

Key takeaway: For a continuous lifetime with density \(f\) and positive survival \(S(t)\), survival is \(S(t)=P(T>t)\), while the hazard is \(h(t)=f(t)/S(t)\). Survival is a probability of lasting beyond a time; hazard is a conditional instantaneous failure rate.

Check Your Understanding

Try these questions using the definitions and calculations above.

In words, what population is conditioned on when interpreting \(h(t)\) as an instantaneous failure rate?
A continuous lifetime has density \(f(t)=0.03\) per day and survival \(S(t)=0.6\) at a particular time. What is its hazard there, and what are its units?
For a uniform lifetime on \([0,8]\), calculate \(S(2)\) and \(h(2)\).
Why is a hazard rate not the probability of failure exactly at time \(t\) for a continuous lifetime?
What condition makes the density quotient \(f(t)/S(t)\) undefined as a conditional rate at a given time?