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Let $N_t = N([0,t])$ denote a Poisson process with rate $\lambda = 1$ on the interval $[0,1]$.

I am wondering how I can use the Law of Large Numbers to formally argue that:

$$\frac{N_t}{t} \rightarrow \lambda \quad \text{ a.s.} $$

As it stands, I can almost prove the required result but I have to to assume that $t \in Z_+$. With this assumption, I can define Poisson random variables on intervals of size $1$ as follows

$$N_i = N([i-1,i])$$


$$\mathbb{E}[N([i-1,i])] = \text{Var}[N([i-1,i])] = 1$$


$$N_t = N([0,t]) = \sum_{i=1}^t N([i-1,i]) = \sum_{i=1}^t N_i$$

Accordingly, we can use the Law of Large Numbers to state the result above...

Given that $t \in \mathbb{R}_+$, this proof needs to be tweaked in some way... But I'm not exactly sure how to do it.

Intuitively speaking, I believe that the correct approach would be to decompose $N[0,t]$ into $N[0,\lfloor t\rfloor]$ and $N[\lfloor t\rfloor, t]$, and argue that the latter term $\rightarrow 0$ almost surely.

However, I'm not sure how to formally state this.

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up vote 2 down vote accepted

If $n\leqslant t\lt n+1$, then $N_n\leqslant N_t\leqslant N_{n+1}$ hence $$ \frac{n}t\cdot\frac{N_{n}}{n}\leqslant\frac{N_t}t\leqslant\frac{n+1}t\cdot\frac{N_{n+1}}{n+1}. $$ When $t\to\infty$, $\frac{n}t\to1$ and $\frac{n+1}t\to1$ because $n$ is the integer part of $n$ hence $t-1\lt n\leqslant t$. Furthermore, $\frac{N_n}n\to\lambda$ (the result you showed) because $n\to\infty$, and $\frac{N_{n+1}}{n+1}\to\lambda$ for the same reason. You are done.

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