Question

For a nonhomogeneous Poisson process with intensity function \(\lambda(t), t \geqslant 0\), where \(\int_{0}^{\infty} \lambda(t) d t=\infty\), let \(X_{1}, X_{2}, \ldots\) denote the sequence of times at which events occur. (a) Show that \(\int_{0}^{X_{1}} \lambda(t) d t\) is exponential with rate 1 . (b) Show that \(\int_{X_{i-1}}^{X_{i}} \lambda(t) d t, i \geqslant 1\), are independent exponentials with rate 1, where \(X_{0}=0\) In words, independent of the past, the additional amount of hazard that must be experienced until an event occurs is exponential with rate 1 .

Short answer

In summary, we have shown that for a nonhomogeneous Poisson process with intensity function \(\lambda(t)\), the time intervals between events \(\int_{0}^{X_{1}} \lambda(t) dt\) and \(\int_{X_{i-1}}^{X_{i}} \lambda(t) dt\) are exponentially distributed with rate 1 and are independent of each other, which demonstrates that the additional amount of hazard experienced until an event occurs is exponential with rate 1 regardless of the past.

Step-by-step solution

Introduce the exponential distribution

The exponential distribution has density function given by \(f(x)=\lambda e^{-\lambda x}\), where \(\lambda\) is the rate of the distribution, and its mean is 1/\(\lambda\). We will use this density function to show that the given time intervals are exponentially distributed with rate 1.

Compute the probability that an event occurs before time \(X_{1}\)

We want to find the probability that an event occurs before time \(X_{1}\), which is given by the integral \(\int_{0}^{X_{1}} \lambda(t) dt\). To show that this integral is exponentially distributed with rate 1, let \(F(x)\) be the cumulative distribution function (CDF), so that \(F(x) = \int_{0}^{x} \lambda(t) dt\). Then the exponential density function with rate 1, denoted as \(f(x)\), is equal to the derivative of \(F(x)\) with respect to \(x\), or \[f(x) = \frac{dF(x)}{dx} = \lambda(x),\] since \(\int_{0}^{X_{1}} \lambda(t) dt = F(X_{1})\).

Independence of events

Now we want to show that the integrals between consecutive event times, \(\int_{X_{i-1}}^{X_{i}} \lambda(t) dt, i \geq 1\), are independent of each other. Recall that in a Poisson process, the number of events in non-overlapping intervals are independent. Let \(N(t)\) denote the number of events in the interval \([0,t)\), and let \(Y(t)=\int_{0}^{t} \lambda(s) ds\). By definition, the number of events in the interval \((X_{i-1}, X_{i})\) is \(N(X_{i})=N(X_{i-1})+1\). Given that we observed \(N(X_{i-1})=n\), we want to find the conditional distribution of \(Y(X_{i})\) given \(Y(X_{i-1})=y\). That is, \[P(Y(X_{i}) \le z | Y(X_{i-1}) = y) = P(Y(X_{i})-Y(X_{i-1}) \le z-y | Y(X_{i-1})=y).\] Now let \(W_i = Y(X_{i})-Y(X_{i-1})\). Note that the events we saw up to time \(X_{i-1}\) do not affect the future events, so we have \[P(W_i \le z-y | Y(X_{i-1})=y) = P(W_i \le z-y) = F(z-y).\] Finally, taking the derivative with respect to \(z\) yields the density function, \[\frac{d}{dz} F(z-y) = \lambda(z).\] Thus, the distribution of the \(\int_{X_{i-1}}^{X_{i}} \lambda(t) dt\) are independent exponentials with rate 1, which completes the proof.