Question

Suppose that \(\left[N_{0}(t), t \geqslant 0\right\\}\) is a Poisson process with rate \(\lambda=1\). Let \(\lambda(t)\) denote a nonnegative function of \(t\), and let $$ m(t)=\int_{0}^{t} \lambda(s) d s $$ Define \(N(t)\) by $$ N(t)=N_{0}(m(t)) $$ Argue that \(\\{N(t), t \geqslant 0\\}\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t), t \geqslant 0\) Hint: Make use of the identity $$ m(t+h)-m(t)=m^{\prime}(t) h+o(h) $$

Short answer

In order to show that \(\{N(t), t \geq 0\}\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t)\), we note the following: 1. The initial condition is satisfied: \(N(0) = N_0(m(0)) = N_0(0) = 0\). 2. The increments of \(N(t)\) are independent as \(N_0(t)\) is a homogeneous Poisson process with rate 1, and the mapping \(m(t)\) is deterministic. 3. Using the given hint, we have \(m(t+s)-m(t) = \lambda(t)s + o(s)\), which shows that the number of events in the interval \((t,t+s]\) depends only on the length of the interval \(s\). Thus, we can conclude that \(\{N(t), t \geq 0\}\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t), t \geq 0\).

Step-by-step solution

Define a nonhomogeneous Poisson process

A nonhomogeneous Poisson process \(N(t)\) with intensity function \(\lambda(t)\) for \(t \geq 0\) is a counting process that satisfies the following conditions: 1. \(N(0) = 0\) 2. The increments are independent. Meaning, for any non-overlapping time intervals \((t_1, t_2], \dots, (t_{n-1}, t_n]\), the increments of \(N\) in these intervals are independent. 3. The number of events \(N(t+s)-N(t)\) in the interval \((t, t+s]\) depends only on the length of the interval \(s\). Now, we will use the definition of the Poisson process to prove that \(N(t) = N_0(m(t))\) is a nonhomogeneous Poisson process.

Confirm initial condition

As \(N_0(t)\) is a Poisson process with rate 1, we have \(N_0(0) = 0\). Since \(m(0) = \int_{0}^{0} \lambda(s)ds = 0\), we can confirm that the initial condition for \(N(t)\) is satisfied: $$ N(0) = N_0(m(0)) = N_0(0) = 0 $$

Check independence of increments

We need to show that for any \(0 \le t_1 < t_2 < \dots < t_n\), the increments \(N(t_i + s_i) - N(t_i)\) over non-overlapping intervals \((t_i,t_i + s_i], i = 1,\dots, n\), are independent: $$ N(t_1 + s_1) - N(t_1), \dots, N(t_n + s_n) - N(t_n) $$ By definition, we have that \(N(t) = N_0(m(t))\). Therefore: $$ N_0(m(t_1 + s_1)) - N_0(m(t_1)), \dots, N_0(m(t_n + s_n)) - N_0(m(t_n)) $$ As \(N_0(t)\) is a Poisson process with rate 1, its increments are independent. Since the \(m(t)\) mapping is deterministic, the increments of \(N(t)\) are also independent.

Use the given hint to show that the number of events depends only on the length of the interval

We need to show that the number of events in the interval \((t, t+s]\) depends only on the length of the interval \(s\): $$ N(t+s) - N(t) = N_0(m(t+s)) - N_0(m(t)) $$ To prove this, we can use the given hint: $$ m(t+s) - m(t) = m'(t)s + o(s) = \lambda(t)s + o(s) $$ As \(m'(t) = \lambda(t)\), we can deduce that the number of events in the interval \((t, t+s]\) depends only on the length of the interval \(s\). Therefore, we have shown that \(N(t) = N_0(m(t))\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t), t \geqslant 0\).