Setting
Given $S_{\lambda} \overset{d}{\sim} \operatorname{Poisson}(\lambda)$. Let $G_{\lambda}(t)$ be the distribution function of $\frac{S_{\lambda}}{\lambda}$. I need to determine $\lim_{\lambda\rightarrow \infty} G_{\lambda}(t)$ when
- $t>1$
- $t<1$
- $t = 1$
Solution
Taking hint from the way the question is asked, I let $\lambda = n$, so that
$$S_{n} = S_{\lambda} = X_1 + \ldots X_n$$
Where $X_j \overset{\mathrm{iid}}{\sim} \operatorname{Poisson}(1)$ for all $j$. So I can see I'm in a CLT setting. However, I am not sure where to proceed from here. Ideas?
Edit I presume $\frac{S_{\lambda}}{\lambda} \rightarrow 1$, but what kind of convergence is this? Furthermore, I assume $G_{\lambda}(t) \rightarrow^d \Phi(t)$ where $\Phi(t)$ is the normal distribution centered at $1$, but does the variance change as $n \rightarrow \infty$?
The CLT does hold here:
$\frac{S_n-\lambda}{\sqrt{\lambda}}\to \mathcal{N}(0,1) \implies S_n\dot{\sim} \mathcal{N}(\lambda,\lambda)\implies \frac{S_n}{\lambda} \dot{\sim} \mathcal{N}(1,\lambda^{-1})\implies \frac{S_n}{\lambda}\xrightarrow{p} 1$
So, your ratio converges to $1$ in probability; but we can prove more:
Using your construction: $Var(S_n)=n \implies \sum_{i\in \mathbb{N}} \frac{Var(X_i)}{n^2}=\sum_{i\in \mathbb{N}} \frac{1}{n^2}<\infty \implies \frac{S_n-n}{n}\to 0\;a.s.\implies\frac{S_n}{n} \to 1\; a.s.$
So the ratio converges almost surely.
However, we only need convergence in probability to answer your questions:
$G_{\lambda}(t) \xrightarrow {d} \mathcal{N}(1,\lambda^{-1}) \to \mathbf{1}_{\geq1}(t)$
Thus: $t<1 \implies G_{\lambda}(t) \to 0,\;t=1 \implies G_{\lambda}(t) \to 0.5,\;t>1G_{\lambda}(t) \to 1$