# A property of Poisson process

Let $Y_t$ be a centered Poisson process, why \begin{equation} \lim_{n \to \infty} \sup_{s<t} |n^{-1}Y(ns)| = 0 \qquad a.s. \qquad \forall t\ge 0 \end{equation} This is a fundamental step in the proof of the law of large number for continuous time Markov Chain. I'm following the proof on the book by Ethier and Kurtz but they don't explain, or show, the previous statement.

• Could you give a page number from Ethier and Kurtz? I'd like to take a look. – user940 Dec 16 '15 at 12:57
• @ByronSchmuland Sure. The page is 456, Chap. 11, theorem 2.1 and my question is relative to the equation (2.5). Thanks. – andreasvr Dec 16 '15 at 13:22
• I have sketched a solution without checking the details. Let me know if you have any questions. – user940 Dec 16 '15 at 15:33

## 1 Answer

Define $M_{n}(s)={Y(ns)\over n}={1\over n}(N(ns)-\lambda ns),$ where $N$ is a Poisson process with parameter $\lambda$. Then $M_n$ is a martingale, so by Doob's maximal inequality with $p=4$, we have $$\mathbb{E}\left(\sup_{0\leq s\leq t} M^4_n(s)\right)\leq c\,\mathbb{E}(M^4_n(t))\leq {c\,(\lambda n t)^2\over n^4}\leq {c\over n^2}.\tag1$$ We use the fact that the normalized fourth moment of a Poisson distribution with mean $\mu$ is bounded above by a constant times $\mu^2$.

Since the right hand side of (1) is summable, we get $\sum_{n=1}^\infty \mathbb{E}\left(\sup_{0\leq s\leq t} M^4_n(s)\right)<\infty,$ so

$$\sum_{n=1}^\infty \,\sup_{0\leq s\leq t} M^4_n(s) <\infty\quad \mbox{ almost surely},$$ which implies $$\sup_{0\leq s\leq t} M_n(s)\to 0 \quad \mbox{ almost surely}.$$

• Thank you for your attention. And sorry but i'm not able to understand the step in equation (1)...How do you use the inequality that you linked? there is a lot of confusion in literature about what is the Doob or the maximal inequality...I usually call Doob inequality $\mathbb{E}(\sup X_n) \le (\frac{p}{p-1})^p \sup \mathbb{E}[X_n]$...and also, sorry for the question but the mean of this process isn't 0? it is centered! And I can't see how the sum implies the last equation. – andreasvr Dec 16 '15 at 16:20
• The mean is zero, but not the mean of the fourth power. – user940 Dec 16 '15 at 17:09
• @andreasvr See math.stackexchange.com/questions/1530776/… but with 4th power instead of 2nd – user940 Dec 16 '15 at 18:17
• one last question: why works the last inequality? It has to be $(\lambda t)^2 \le 1$...but I didn't how to prove it or how I can conclude. And how I can find the proof that the fourth moment is bounded by a constant times the mean to the second power? – andreasvr Dec 17 '15 at 10:05
• @andreasvr You shouldn't take the constant $c$ literally. It is different each time that $c$ is used. This is a standard mathematical shortcut. If you don't like this, just replace $c$ above with $c_1, c_2, c_3$. – user940 Dec 17 '15 at 13:20