Showing a conditional Poisson Process does not have independent increments

Question

Suppose $N\left(\cdot\right)$ is a Poisson Process with rate $1$ and $Z$ is a positive non-constant random variable, define $N_{Z}\left(t\right)=N\left(Zt\right)$. I know that conditional on $Z$ this is a Poisson Process with rate $Z=z$ but without conditioning while it has stationary increments they are not independent. However I don't know how to show that they are not independent which is what I would like to do.

Following the comment, does it suffice to show that given $t_{0}<t_{1}<t_{2}$ $$\mathbb{P}\left(\left\{ N_{Z}\left(t_{1}\right)-N_{Z}\left(t_{0}\right)=0\,\wedge\, N_{Z}\left(t_{2}\right)-N_{Z}\left(t_{1}\right)=0\,|Z=z\right\} \right)\neq\mathbb{P}\left(N_{Z}\left(t_{1}\right)-N_{Z}\left(t_{0}\right)=0|Z=z\right)\mathbb{P}\left(N_{Z}\left(t_{2}\right)-N_{Z}\left(t_{1}\right)=0|Z=z\right)$$

Generally to show that there is no independence I would want to show that $$\mathbb{P}\left(\left\{ N_{Z}\left(t_{1}\right)-N_{Z}\left(t_{0}\right)=0\,\wedge\, N_{Z}\left(t_{2}\right)-N_{Z}\left(t_{1}\right)=0\right\} \right)\neq\mathbb{P}\left(N_{Z}\left(t_{1}\right)-N_{Z}\left(t_{0}\right)=0\right)\mathbb{P}\left(N_{Z}\left(t_{2}\right)-N_{Z}\left(t_{1}\right)=0\right)$$ I'm not sure I understand why it suffices to show it after conditioning on $Z$?

Following Ben's response I still have some questions:

1. I don't immediately see why $\mathbb{P}\left[N\left(Z\right)=0\right]=\mathbb{E}\left[e^{-Z}\right]$ and the same for the other equalities.

2. I'm not sure in what part exactly are we conditioning on $Z$, like you said conditioned on Z I won't get the desired result since $N|Z$ is a proper Poisson Process.

3. I'm also not sure why showing there's no equality in the last equation yoou wrote would give me what I want. Generally as far as I understood what I want to do is take two non-overlapping intervals and show the increments of $N_{Z}\left(\cdot\right)$ in those intervals are not independent.

I would extremely appreciate if you could clear this out for me, it's been a while since I took a course in Probability and I'm a bit lost now in this Random Processes course as a result.. :)

Answer 1

Let's prove that $N_Z$ is not a Poisson process by contradiction. If it were, the increments $ N_Z(1)-N_Z(0)=N_Z(1)$ and $N_Z(2)-N_Z(1)$ would be independent. So $$ \begin{align} \mathbb{P}[N_Z(1) = 0]\mathbb{P}[N_Z(2)-N_Z(1)=0] &= \mathbb{P}[N_Z(1) = 0 \cap N_Z(2)-N_Z(1)=0]\\ &= \mathbb{P}[N_Z(2)=0]. \end{align} $$

That is, we'd have $$ \mathbb{P}[N(Z) = 0]\mathbb{P}[N(2Z)-N(Z)=0] = \mathbb{P}[N(2Z)=0]\tag{A}, $$ by the definition of $N_Z.$

Note on conditioning: If $Z$ is independent of $N$, then you won't be able to show your result by conditioning on $Z$, since after conditioning, you in effect have a Poisson process with a deterministic rate.

So let's show $$ \mathbb{P}[N(Z)=0]\mathbb{P}[N(2Z)-N(Z)=0] \neq \mathbb{P}[N(2Z)=0], $$

instead of a conditional version of this.

Hint: By conditioning on $Z$, as you suggest, we see that

$$ \begin{align} \mathbb{P}[N(Z)=0] &= \mathbb{E}[\mathbb{P}[N(z)=0|Z=z]]\\ &= \mathbb{E}[e^{-Z}], \end{align} $$

since, N(z) has a Poisson distribution of parameter $z$ and the probability that a Poisson random variable of parameter $z$ is zero is $e^{-z}$. For fixed $z$, $N(2z)-N(z)$ has Poisson distribution of parameter $z$, by the definition of $N$. Thus

$$ \mathbb{P}[N(2Z)-N(Z)=0] = \mathbb{E}[e^{-Z}]\\ $$

and similarly

$$ \mathbb{P}[N(2Z) = 0] = \mathbb{E}[e^{-2Z}].\\ $$

So

$$ \begin{align} \mathbb{P}[N(Z)=0]\mathbb{P}[N(2Z)-N(Z)=0] &= (\mathbb{E}[e^{-Z}])^2\\ &\leq \mathbb{E}[e^{-2Z}]\\ &= \mathbb{P}[N(2Z)=0], \end{align} $$ by Jensen's inequality or Cauchy-Schwarz. If this inequality is strict, $(A)$ cannot hold. When is this inequality strict?