Exponential Lifetimes and Constant Hazard

From Constant Hazard to an Exponential Lifetime

For a lifetime \(T\), the hazard at time \(t\) describes the instantaneous failure rate among units that have survived until \(t\). A model with constant hazard says that this rate does not change with age. That simple assumption determines an entire lifetime distribution: the exponential distribution. It also has a distinctive implication: conditional on surviving to the present, the distribution of the additional waiting time is unchanged.

We derive that implication from the survival and hazard functions, then use it to calculate conditional probabilities. The model is useful when a constant rate is a reasonable approximation; it should not be mistaken for a universal description of how failure risk behaves over time.

For the derivation, assume \(T\) is absolutely continuous, has no atom at time zero, and has a proper distribution on \([0,\infty)\). Thus \(S(0)=1\), and \(S(t)>0\) at every finite \(t\) in the constant-hazard model. Since \(f(t)=-S'(t)\) almost everywhere, the hazard equation gives a differential equation for survival.

Proof. The definition \(h(t)=f(t)/S(t)\), together with \(f=-S'\) almost everywhere, gives \(S'(t)=-\lambda S(t)\) almost everywhere. Because \(S\) is absolutely continuous and positive on finite intervals, \(\log S\) is absolutely continuous there, with derivative \(S'(t)/S(t)=-\lambda\) almost everywhere. Integrating from \(0\) to \(t\) and using \(S(0)=1\),

Exponentiating gives \(S(t)=e^{-\lambda t}\). Therefore \(F(t)=1-S(t)=1-e^{-\lambda t}\); differentiating yields \(f(t)=\lambda e^{-\lambda t}\). Since \(\lambda>0\), \(S(t)\) tends to zero as \(t\) tends to infinity, so the resulting distribution is proper. \(\square\)

The rate \(\lambda\) has units of inverse time. If time is measured in hours, \(\lambda\) is failures per hour; changing the time unit changes the numerical rate. The density has units of inverse time, while survival and failure probabilities are dimensionless.

Memorylessness

For any lifetime with \(P(T>s)>0\), the probability of surviving a further \(t\) units given survival to time \(s\) is

For exponential survival, the ratio simplifies:

The conditional survival probability over the next \(t\) time units is the same as the survival probability for the first \(t\) time units. Equivalently, the conditional distribution of \(T-s\) given \(T>s\) is the same as the distribution of \(T\). This is the memoryless property.

Proof of the characterization. The exponential case follows from the survival-ratio calculation above. Conversely, suppose the stated class of lifetimes satisfies the memoryless property. Since \(S(s)>0\), multiplying the conditional probability identity by \(S(s)\) gives

Define \(g(t)=-\log S(t)\). Positivity of \(S\) makes this well-defined; continuity of the survival function makes \(g\) continuous. The product identity implies \(g(s+t)=g(s)+g(t)\). Set \(\lambda=g(1)\), using the chosen time unit. Additivity gives \(g(m/n)=(m/n)\lambda\) for nonnegative integers \(m\) and positive integers \(n\): first \(ng(1/n)=g(1)\), and then \(g(m/n)=mg(1/n)\). By continuity, approximating any \(t\geq0\) by nonnegative rationals gives \(g(t)=\lambda t\). Since \(S\) is nonincreasing, \(g\) is nondecreasing, so \(\lambda\geq0\). Properness requires \(S(t)\to0\) as \(t\to\infty\), excluding \(\lambda=0\). Thus \(S(t)=e^{-\lambda t}\) with \(\lambda>0\), which is the exponential survival function. \(\square\)

Conditional Failure Over an Interval

The density also gives the probability of failure during a particular interval. For \(0\leq a<b\), the unconditional probability is \(P(a<T\leq b)=F(b)-F(a)\). Conditional on surviving beyond \(a\), divide by \(S(a)\). Under the exponential model, this yields

Only the interval length \(b-a\) appears. The same formula applies whether the unit is new or has already survived for a long time, provided the constant-hazard model is appropriate.

Interpreting the Constant Rate Carefully

A constant hazard does not mean a fixed probability of failure in every interval, regardless of the interval’s length. Over an interval of length \(\Delta>0\), conditional on survival to its start, the failure probability is

This depends on \(\Delta\). For a short interval, the approximation \(1-e^{-\lambda\Delta}\approx\lambda\Delta\) follows from the first-order expansion of the exponential, but the exact probability is not \(\lambda\) unless the interval happens to give that value. The rate is an instantaneous quantity, not a probability by itself.

The model’s practical implication is also strong: surviving a long time does not, by itself, reduce or increase the modeled conditional failure risk over a future interval of fixed length. This may suit some simplified settings, but a changing hazard requires another model. The constant-hazard assumption should therefore be treated as a substantive modeling choice, not a consequence of lifetimes being positive.

• Using the wrong conditioning event. To find failure between \(a\) and \(b\) given survival to \(a\), divide the interval probability by \(S(a)\), or use the conditional formula directly.
• Calling \(\lambda\) a failure probability. It is a rate with units of inverse time. The probability of failure over length \(\Delta\) is \(1-e^{-\lambda\Delta}\).
• Assuming every lifetime is memoryless. The property follows from the exponential survival function; it does not hold for general lifetime distributions.