Hazard as an Instantaneous Rate
For a waiting time, the hazard rate describes the event’s instantaneous rate of occurrence, conditional on the event not having occurred yet. Unlike a probability, a hazard rate is measured per unit time and can vary with elapsed time. This makes it useful for deciding whether a proposed waiting-time model has the constant-rate structure of exponential interarrivals.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \(T\) be a nonnegative waiting time with a continuous density \(f\). Write \(S(t)=\mathbb{P}(T>t)\) for its survival function. Assume that \(S(t)>0\) at each time where the hazard is considered. This condition makes conditioning on survival through time \(t\) meaningful. Since \(T\) has a continuous distribution, \(\mathbb{P}(T=t)=0\), so using \(T\ge t\) in that conditioning gives the same probability.
The quotient formula follows from the definition. The conditional probability in the numerator is \((S(t)-S(t+\Delta))/S(t)\). Dividing by \(\Delta\) and taking the limit gives \(-S'(t)/S(t)\). Because \(S'(t)=-f(t)\), this is \(f(t)/S(t)\). Thus the hazard is not generally the density: it is the density divided by the probability of still waiting.
From Hazard to Survival
The relation \(h(t)=f(t)/S(t)\), together with \(S'(t)=-f(t)\), gives a differential equation: $$ S'(t)=-h(t)S(t). $$ For a proper nonnegative waiting time with a continuous distribution, \(S(0)=1\). If \(h\) is continuous and \(S\) remains positive, this differential equation determines the survival function.
Proof. Since \(S(t)>0\), the logarithm \(\log S(t)\) is defined. The differential equation \(S'(t)=-h(t)S(t)\) implies $$ \frac{d}{dt}\log S(t)=\frac{S'(t)}{S(t)}=-h(t). $$ Integrating from \(0\) to \(t\) gives \(\log S(t)-\log S(0)=-\int_0^t h(u)\,du\). As \(S(0)=1\), \(\log S(0)=0\); exponentiating proves the formula. \(\square\)
In particular, if \(h(t)=\lambda\) for all \(t\ge0\), then \(S(t)=e^{-\lambda t}\). For a proper waiting time with a finite constant hazard, \(\lambda\) must be strictly positive: if \(\lambda=0\), the formula gives \(S(t)=1\) for every finite \(t\), so \(T\) is not finite with probability one. Conversely, an exponential waiting time with rate \(\lambda>0\) has density \(f(t)=\lambda e^{-\lambda t}\), survival \(S(t)=e^{-\lambda t}\), and hazard \(f(t)/S(t)=\lambda\). This proves both directions of the characterization.
Proof. The survival-from-hazard result gives \(S(t)=e^{-\lambda t}\). A hazard rate is nonnegative, so \(\lambda\ge0\). If \(\lambda=0\), then \(S(t)=1\) for every finite \(t\), contradicting that \(T\) is proper, which requires \(\lim_{t\to\infty}S(t)=0\). Hence \(\lambda>0\). The distribution function is therefore \(\mathbb{P}(T\le t)=1-e^{-\lambda t}\) for \(t\ge0\), the exponential distribution with rate \(\lambda\). For the converse, an exponential waiting time has \(S(t)=e^{-\lambda t}\) and density \(f(t)=\lambda e^{-\lambda t}\); dividing gives \(h(t)=\lambda\) for every \(t\ge0\). \(\square\)
For a Poisson process with positive rate \(\lambda\), the interarrival times are exponential with rate \(\lambda\), as established in the earlier material on Poisson processes. The characterization explains the rate interpretation: while waiting, the instantaneous arrival rate remains \(\lambda\), independent of how long the current wait has lasted. The theorem concerns a single waiting-time distribution; it does not by itself establish independence among a sequence of interarrival times.
Worked Hazard Calculations
Worked Example: A Constant Hazard of Three Halves
Suppose a proper waiting time has constant hazard \(h(t)=\tfrac32\) per time unit. Find its survival probability at time \(2\), its probability of occurring by time \(1\), and its density at time \(2\).
The hazard is constant and positive, so the theorem gives \(S(t)=e^{-3t/2}\). Substituting \(t=2\) gives \(S(2)=e^{-(3/2)(2)}=e^{-3}\). Thus the probability of waiting longer than \(2\) time units is \(e^{-3}\). At time \(1\), \(S(1)=e^{-(3/2)(1)}=e^{-3/2}\), so \(\mathbb{P}(T\le1)=1-S(1)=1-e^{-3/2}\).
The density is \(f(t)=h(t)S(t)\). At \(t=2\), \(f(2)=\tfrac32e^{-3}\). Check: \(f(2)/S(2)=(\tfrac32e^{-3})/e^{-3}=\tfrac32\), the specified hazard.
Worked Example: Verifying an Exponential Hazard
Let \(T\) have an exponential distribution with rate \(2\). Compute its hazard and the probability that \(T\) falls in the interval \([1/2,1)\).
The survival function is \(S(t)=e^{-2t}\), and differentiating \(1-S(t)\) gives the density \(f(t)=2e^{-2t}\). Therefore, for every \(t\ge0\), $$ h(t)=\frac{f(t)}{S(t)}=\frac{2e^{-2t}}{e^{-2t}}=2. $$
For the interval probability, subtract the survival probability at the right endpoint from that at the left endpoint: \(\mathbb{P}(1/2\le T<1)=S(1/2)-S(1)\). Substituting the endpoints gives \(e^{-2(1/2)}-e^{-2(1)}=e^{-1}-e^{-2}\). Check: the corresponding distribution-function difference is \((1-e^{-2})-(1-e^{-1})=e^{-1}-e^{-2}\), as required.
Worked Example: A Time-Varying Hazard
Consider a waiting time with survival function \(S(t)=e^{-t^2}\) for \(t\ge0\). Find its density and hazard, and use them to show that this waiting time is not exponential.
Since \(S(0)=e^0=1\) and \(S(t)\to0\) as \(t\to\infty\), this is the survival function of a proper nonnegative continuous waiting time. Differentiating gives \(S'(t)=-2t e^{-t^2}\), so \(f(t)=-S'(t)=2t e^{-t^2}\). The hazard is $$ h(t)=\frac{2t e^{-t^2}}{e^{-t^2}}=2t. $$
For example, \(h(1)=2(1)=2\), while \(h(2)=2(2)=4\); the hazard is not constant, so this distribution is not exponential. Its survival probabilities also show the time dependence: \(S(1)=e^{-1}\) and \(S(2)=e^{-4}\). Hence \(\mathbb{P}(1\le T<2)=S(1)-S(2)=e^{-1}-e^{-4}\). Check: integrating the density over this interval gives \(\int_1^2 2t e^{-t^2}\,dt=[-e^{-t^2}]_1^2=e^{-1}-e^{-4}\).
Why the Hazard Distinction Matters
A constant hazard means that the conditional chance of an event in a very short interval, given survival to its start, is approximately the same rate times the interval’s length at every elapsed time. It does not mean that the unconditional density is constant. For an exponential waiting time, \(f(t)=\lambda e^{-\lambda t}\) decreases as \(t\) increases, even though \(h(t)=\lambda\) does not. The density accounts for how many waiting times remain at risk at time \(t\); the hazard is the event rate among those that have not yet ended.
Common mistake: treating the density as the hazard. In the preceding example, at \(t=2\) the density is \(f(2)=2(2)e^{-2^2}=4e^{-4}\). Calling this the hazard would give \(4e^{-4}\), which is not the correct rate. The survival probability at that time is \(S(2)=e^{-4}\), so the hazard is \(f(2)/S(2)=4e^{-4}/e^{-4}=4\). The correction is to divide the density by the probability of still waiting.
Another useful check is to recover survival from a proposed hazard. If \(h(t)=2t\), then \(\int_0^t h(u)\,du=\int_0^t 2u\,du=t^2\), so the survival-from-hazard formula gives \(S(t)=e^{-t^2}\). If \(h(t)=\lambda\), the same formula gives \(e^{-\lambda t}\). Thus the cumulative hazard \(\int_0^t h(u)\,du\) is the quantity that determines survival; a nonconstant hazard is allowed, but it generally produces a non-exponential waiting time.
Check Your Understanding
Use the definition \(h(t)=f(t)/S(t)\) where applicable, and show the calculation that supports each answer.
- A proper waiting time has constant hazard \(h(t)=3\). What is \(S(2)\), and what is \(\mathbb{P}(T\le2)\)?
- If \(S(t)=e^{-4t}\) for \(t\ge0\), find the density and verify the hazard.
- For \(S(t)=e^{-t^2}\), compute \(h(3)\). Is the distribution exponential?
- Explain why a proper waiting time cannot have constant hazard zero on all \(t\ge0\).
- For a waiting time with \(S(t)=e^{-t^2}\), compute \(\mathbb{P}(2\le T<3)\) by subtracting survival probabilities.