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Let $\lambda>0$ and $(N_t)_{t\ge0}$ be an almost surely right-continuous Poisson process on a probability space $(\Omega,\mathcal A,\operatorname P)$ with intensity $\lambda$.

How can we show that $$Z_n:=\inf\left\{t\ge0:N_t=n\right\}-\sum_{i=1}^{n-1}Z_i$$ is exponentially distributed with parameter $\lambda$ for all $n\in\mathbb N$?

Clearly, $$\operatorname P\left[Z_1>t\right]=\operatorname P\left[N_t=0\right]=e^{-\lambda t}=\lambda\int_t^\infty e^{-\lambda s}\:{\rm d}s\tag1$$ for all $t\in\mathbb R$ and hence, since $\{(t,\infty):t\in\mathbb R\}$ is a $\cap$-stable generator of the Borel $\sigma$-algebra $\mathcal B(\mathbb R)$, we can conclude that $Z_1\sim\operatorname{Exp}(\lambda)$.

Now, for $n\ge2$, we somehow need to use that $(N_t)_{t\ge0}$ has independent increments. I think it's crucial that we have assumed that $(N_t)_{t\ge0}$ is almost surely right-continuous, since this ensures that $(N_t)_{t\ge0}$ is almost surely nondecreasing. This allows us to conclude that \begin{equation}\begin{split}\operatorname P\left[Z_1\le s,Z_2>t\right]&=\operatorname P\left[N_s=1,N_t-N_s=0\right]\\&=\operatorname P\left[N_s=1\right]\operatorname P\left[N_t-N_s=0\right]\\&=\lambda e^{-\lambda t}\end{split}\tag2\end{equation} for all $t\ge s\ge0$. On the other hand, if $s>t\ge0$, then $\operatorname P\left[Z_1\le s,Z_2>t\right]=0$.

But is the distribution of $(Z_1,Z_2)$ uniquely determined by $\{(-\infty,s]\times(t,\infty):s,t\in\mathbb R\}$? And how do we need to argue for general $n$?

0xbadf00d
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  • Do you know that the restarted process $N_{t+\tau}-N_{\tau}$ is also a Poisson process? If so, then you can reduce everything to the case $n=1$. E.g. for $n=2$ define $\tilde{N}t := N{t+Z_1}-N_{Z_1}$ (which is a Poisson process) and note that $Z_2 = \inf{t \geq 0; \tilde{N}_t=1}$, which implies directly that it is expnentially distributed with parameter $\lambda$. – saz Oct 22 '20 at 15:04
  • @saz Do you have a reference for the claim about the restarted process? – 0xbadf00d Oct 22 '20 at 15:34
  • Depends very much on your definition of "Poisson process". if you know that the process has independent and stationary increments, then the proof is not so hard (it is then essentially the same reasoning as for Brownian motion) – saz Oct 22 '20 at 17:06
  • @saz I'm not sure, but I guess we can obtain a more general result and I would like to show that: If $L$ is a $\mathcal F$-Lévy process and $\tau$ a $\mathcal F$-stopping time, then $(L_{\tau+t}-L_\tau){t\ge0}$ is a $(\mathcal F{\tau+t})_{t\ge0}$-Lévy process and $L$ is strongy $\mathcal F$-Markov at $\tau$. Can we show this? At least if $\mathcal F$ is the filtration generated by $L$? – 0xbadf00d Oct 22 '20 at 17:45
  • @saz Please take a look at the question I've asked for this: https://math.stackexchange.com/q/3876350/47771. – 0xbadf00d Oct 22 '20 at 18:30
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    Yes, the strong Markov property holds indeed for any Lévy process. See e.g. Theorem 4.12 here – saz Oct 22 '20 at 18:50
  • @saz Please take a look at my other question: https://math.stackexchange.com/q/3876350/47771. There is only a single piece left, which I need to conclude. The main problem being that the conditional expectation is taken with respect to a $\sigma$-algebra which depends on a sequence index and I'm not sure whether we are allowed to take the limit and conclude. – 0xbadf00d Oct 28 '20 at 16:00

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