Let $X=(X_n)$ be a sequence of i.i.d. Exp($\lambda$) random variables and $Z=(Z_n)$ a sequence of i.i.d. Ber($p$) random variables independent of $X$ and let $T_n=\sum_{i=1}^n X_i$.
I want to prove that the random variables defined by
$$N'_t = \sum_{n=1}^\infty Z_n \mathbb 1_{(0,t]}(T_n) \tag{1}$$
form a Poisson process with intensity $p \lambda$.
I know how to prove this by applying that any process starting at $0$ almost surely that has independent, Poisson distributed increments is a Poisson process.
However, I have some trouble finding the same result while relying on this definition: $(N_t: t\geq0)$ is called a Poisson process if
\begin{equation}N_t = \max\{n\in \mathbb N_0: T_n \leq t \} \tag{2}\end{equation}
where $T_n$ is defined as above.
My reasoning so far goes like this:
Since the $T_n$ are increasing by one after each arrival, we may just count them and thus see that $(2)$ is equivalent to
$N_t = \sum_{n=1}^\infty \mathbb 1_{(0,t]}(T_n).$
We also have
$Z_n \mathbb 1_{(0,t]}(T_n) = \mathbb 1_{(0,t]}(Z_n T_n).$ So what I need to show is that $Z_n T_n = \sum_{i=1}^n Y_n$ for some i.i.d. $Y_i \sim \mathrm{Exp}(p \lambda)$. This means that I need to show that $Z_n T_n \sim \mathrm{Gamma}(p\lambda,n)$, and here is where I'm stuck. Can someone help me to see why this is true? Or did I make a mistake before?
Best Answer
If you know the infinitesimal description of the Poisson process then you can proceed in the following manner.
Without thinning, we have
$ \begin{align*} P(N_{t+h}-N_t=0)&=1-\lambda h+o(h), \\ P(N_{t+h}-N_t=1)&=\lambda h+o(h)\quad\text{and} \\ P(N_{t+h}-N_t\geq2)&=o(h) \end{align*} $
where $\lambda$ is the intensity. If you have thinning with probability $p$ of retaining the points then the intensity should become $p\lambda$. To show this we should show that
$ \begin{align*} P(N'_{t+h}-N'_t=0)&=1-p\lambda h+o(h), \\ P(N'_{t+h}-N'_t=1)&=p\lambda h+o(h)\quad\text{and} \\ P(N'_{t+h}-N'_t\geq2)&=o(h). \end{align*} $
Since these are similar to show, I will demonstrate the first one.
$ \begin{align*} P(N'_{t+h}-N'_t=0)&=P(N'_{t+h}-N'_{t}=0\mid N_{t+h}-N_t=0)(1-\lambda h+o(h)) \\ &\quad+P(N'_{t+h}-N'_t=0\mid N_{t+h}-N_t=1)(\lambda h+o(h)) \\ &\quad+P(N'_{t+h}-N'_t=0\mid N_{t+h}-N_t\geq2)(o(h)) \\ &=(1-\lambda h+o(h))+(1-p)(\lambda h+o(h))+o(h) \\ &=1-p\lambda h+o(h). \end{align*} $