Question

Let \(X_{1}, X_{2}, \ldots\) be independent positive continuous random variables with a common density function \(f\), and suppose this sequence is independent of \(N, a\) Poisson random variable with mean \(\lambda\). Define $$ N(t)=\text { number of } i \leqslant N: X_{i} \leqslant t $$ Show that \(\\{N(t), t \geqslant 0\\}\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t)=\lambda f(t)\).

Short answer

We have verified all properties required for a nonhomogeneous Poisson process: \(N(0) = 0\), the process has independent increments, and the distribution of \(N(t + \delta) - N(t)\) is a Poisson distribution with parameter \(\int_{t}^{t + \delta}\lambda(s) ds\). Thus, we have shown that \(\\{N(t), t \geqslant 0\\}\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t) = \lambda f(t)\).

Step-by-step solution

Verify N(0) = 0

Since \(N(t)\) is the number of \(i \leqslant N\) such that \(X_i \leqslant t\), when \(t = 0\), all the \(X_i\) will be greater than 0 because they are positive continuous random variables. Therefore, there will be no \(X_i\) such that \(X_i \leqslant t=0\), which results in \(N(0) = 0\).

Verify Independent Increments

Consider two disjoint time intervals \((t_1, t_2]\) and \((t_3, t_4]\) with \(0 \leq t_1 < t_2 < t_3 < t_4\). Now we have to show that \((N(t_2) - N(t_1))\) is independent of \((N(t_4) - N(t_3))\). Since \(X_i\) and \(N\) are independent, any arrival that occurs in the time interval \((t_1, t_2]\) will not depend on an arrival that occurs in the time interval \((t_3, t_4]\). Therefore, the number of arrivals in each time interval will be independent.

Verify Poisson Distribution for N(t + δ) - N(t)

Let \(\delta > 0\). We need to show that \(N(t+\delta) - N(t) \sim \text{Poisson}\left(\int_{t}^{t + \delta}\lambda(s) ds\right)\). The conditional distribution of \(N(t + \delta) - N(t)\), given N = n, is a binomial distribution with parameters n and \(p_{\delta}(t) = P(t < X_i \leq t+\delta | N=n)\). Conditional probability: \begin{align*} p_{\delta}(t) &= P(t < X_i \leq t+\delta | N=n) \\ &= \frac{P(t < X_i \leq t+\delta, N=n)}{P(N=n)} \\ &= \frac{\int_t^{t+\delta} f(x) \,\text{d}x}{1} \end{align*} The distribution of \(N(t+\delta) - N(t)\), given \(N=n\), is a binomial distribution with parameters n and \(p_{\delta}(t)\). Now, we will find the distribution of \(N(t+\delta) - N(t)\). Using Binomial-Poisson convergence, the distribution of \(N(t+\delta) - N(t)\) is Poisson with parameter: \begin{align*} \int_{t}^{t + \delta}\lambda(s) ds &= \lambda \int_{t}^{t +\delta}f(s) ds \\ &= \lambda \cdot \left(\int_{t}^{t + \delta} f(x) \,\text{d}x\right) \end{align*} Thus, we have verified that \(N(t+\delta) - N(t)\) is Poisson distributed with the given parameter. We have shown that the given process has N(0) = 0, independent increments, and has a Poisson distribution for N(t + δ) - N(t) with parameter \(\lambda(t)=\lambda f(t)\). Therefore, the process \(\\{N(t), t \geqslant 0\\}\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t) = \lambda f(t)\).