Let $N = \{N(t) \}_{t\geq 0 }$ be a Poisson process. I already know that $N(t)- \lambda t$ is a martingale where $\mathbb{E} [ N(t) ] = \lambda t$.
I want to prove that
$$ \frac{N(t)}{t} \rightarrow \lambda, \quad t \rightarrow \infty \quad \text{ a.s.}. $$
To prove this I received a hint that I should use the martingale law of large numbers which states that:
Theorem: (Martingale LLN)
Let $S= (S_n = \sum_{k=1}^n X_k)_{n=1,2,\ldots}$ be a martingale with respect to the filtration $(\mathcal{F}_n)_{n=1,2,\ldots}$. Then $\sum_{k=1}^n X_k/k$ converges almost surely on the set $\left\{ \sum_{k=1}^\infty k^{-2} \mathbb{E}\left[ X_k^2 \mid \mathcal{F}_{k-1} \right] < \infty \right\}$. Hence $S_n/n \rightarrow 0$ a.s. on the same set.
Now to use this theorem I think I need to use Doob-Meyer to decompose $N(t)$ as the sum of a martingale and an increasing process. Which would be the martingale $M(t) = N(t) – \lambda t$ and increasing process $A(t) = \lambda t$. But how is $M(t)$ the sum of a martingale? And I do not know how this exactly relates to the thing I have to prove. Thanks for any help.
Best Answer
I may have a solution to my problem but I think it doesn't need the martingale law of large numbers.
As we can note that for all $ n \in \mathbb{N}$ $$ N(n) = \sum_{1 \leq i \leq n} \left(N(i) - N(i-1)\right)$$ is the sum of $n$ iid random variables all with a Poisson$(\lambda)$ distribution and hence are integrable. Then applying the strong law of large number we obtain that $$ \frac{N(n)}{n} \rightarrow \lambda, \text{ a.s. as } n \rightarrow \infty $$ But notice that if we use the notation $[t]=$ integer part of t then, $$\frac{N(t)}{t} = \frac{N([t])}{[t]} \times \frac{[t]}{t} + \frac{N(t)-N([t])}{t}.$$ I guess it then suffices to show that $$ \sup_{n<t<n+1} \frac{N(t) - N(n)}{n} \rightarrow 0 \text{ as } n \rightarrow \infty.$$ To show this let $\xi_n = \sup_{n<t<n+1} N(t) - N(n) = N(n+1) - N(n)$. Now the $\xi_n$'s are integrable iid random variables thus $ \frac{\xi_1+ \cdots + \xi_n}{n} \rightarrow \lambda$ a.s. consequently $$ \frac{\xi_n}{n} \rightarrow 0 \text{ a.s..}$$