How to prove that the thinned Poisson process (by a specific definition) has intensity $p \lambda$?

Question

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.

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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?

Answer 1

Let $X_n,T_n,Z_n$ be the sequences of RVs and $N_t, N_t^{\prime}$ be the processes as you gave. I would try to prove the $N_t, N_t^{\prime}$ are Poisson processes. The key point is following equalities: $$ \{T_n\le t\}=\{N_t\ge n\},\qquad \{T_n> t\}=\{N_t< n\}. \tag{1}$$ Firstly, $\{N_t, t\ge 0\}$ is a Poisson process: Let $\{\widetilde{N}_t, t\ge 0\}$ be a Poisson Process with $\mathsf{E}[\widetilde{N}_t]=\lambda t$ and $$ \widetilde{T}_n=\inf\{t: \widetilde{N}_t\ge n\},\qquad n\in \mathbb{N}. $$ Then $\widetilde{T}_n$ is the n-th jump of Poisson process $\{\widetilde{N}_t, t\ge 0\}$ and $\{\widetilde{T}_n,n\ge1\}\stackrel{d}{=}\{T_n,n\ge 1\}$. Therefore $$\begin{aligned} \mathsf{P}(0\le N_{t_i} \le k_i,1\le i\le n) &=\mathsf{P}(T_{k_i+1}>t_i,1\le i\le n)\qquad (\text{by (1)})\\ (\{\widetilde{T}_n,n\ge1\}\stackrel{d}{=}\{T_n,n\ge 1\})\qquad &=\mathsf{P}(\widetilde{T}_{k_i+1}>t_i,1\le i\le n)\\ (\{\widetilde{T}_n> t\}=\{\widetilde{N}_t< n\})\qquad &= \mathsf{P}(0\le \widetilde{N}_{t_i} \le k_i,1\le i\le n) \end{aligned}$$ This means that $\{N_t,t\ge 0\}\stackrel{d}{=}\{\widetilde{N}_t,t\ge 0\}$ and $\{N_t,t\ge 0\}$ is a Poisson Process with $\mathsf{E}[N_t]=\mathsf{E}[\widetilde{N}_t]=\lambda t$.

Secondly, $\{N_t^{\prime},t\ge 0\}$ is a Poisson process. According the definition of $N_t^{\prime}$, \begin{align}N_t^{\prime}&=\sum_{n=1}^\infty Z_n1_{(0,t]}(T_n)=\sum_{n=1}^\infty Z_n1_{\{T_n\le t\}}\\ &=\sum_{n=1}^\infty Z_n1_{\{n\le N_t\}}=\sum_{n=1}^{N_t}Z_n. \end{align} Using this express for $N_t^{\prime}$, we can deduce that the $\{N_t^{\prime},t\ge 0\}$ is a process with independent homogeneous increments(This could calculate similar to the following) and the distribution of increments may be calculate as following: \begin{align} \mathsf{P}(N_t^{\prime}=k)&=\sum_{j=0}^\infty\mathsf{P}\Bigl(\sum_{n=1}^jZ_n=k\Bigm|N(t)=j\Bigr)\mathsf{P}(N(t)=j)\\ &=\sum_{j=0}^\infty\mathsf{P}\Bigl(\sum_{n=1}^jZ_n=k\Bigr)\mathsf{P}(N(t)=j)\\ &=\sum_{j=k}^\infty \frac{j!}{k!(j-k)!}p^k(1-p)^{j-k}\frac{(\lambda t)^j}{j!}e^{-\lambda t}\\ &=\frac{(p\lambda t)^k}{k!}e^{-p\lambda t} \end{align} Hence $\{N_t^{\prime},t\ge 0\}$ is a Poisson process with intensity $p\lambda$.

Answer 2

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*} $