I am following High-Dimensional Probability book by Roman Vershynin, available here, and in particular exercise 2.3.8 (page 21):
Let $X\sim \text{Pois}(\lambda)$. Show that, as $\lambda\to\infty$, we have:
$$\frac{X-\lambda}{\sqrt{\lambda}}\overset{(d)}{\longrightarrow}N(0,1)$$
I managed to prove it using characteristic functions, but it seems that the author is suggesting an alternative way to proceed:
Hint: derive this from the central limit theorem. Use the fact that the sum of independent Poisson distributions is a Poisson distribution.
However, I can't find this alternative proof. What I tried is the following:
Let's discretize: we consider some sequence $(\lambda_n)_{n\in\mathbb{N}}$ such that $\lambda_n\to\infty$, $\mu_n=\lambda_n-\lambda_{n-1}$ and $\lambda_{-1}=0$. Let $Y_n\sim \text{Pois}(\mu_n)$. Now it is enough to prove that:
$$\frac{Y_1+\ldots+Y_n-\lambda_n}{\sqrt{\lambda_n}}\overset{(d)}{\longrightarrow} N(0,1)$$
I would like to use CLT at this point, but the $(Y_n)$ are not identically distributed. I know there exist variants of CLT without this assumption, but they are not mentioned before that in the book. Furthermore, this is categorized as a relatively easy exercise, so I think I am just missing some elementary idea here.
Let $n = \lfloor \lambda\rfloor$, and define $S_n = Y_1 + \cdots + Y_n$, where $Y_1, \ldots, Y_n$ are i.i.d. Poisson($1$) variables and $r \sim \text{Poisson}(\lambda - n)$, where $r$ is taken to be independent of $S_n$. Clearly, $S_n + r$ has the same distribution of $X$, so it is equivalent to prove that $$\frac{S_n + r - \lambda}{\sqrt{\lambda}} \to_d N(0, 1). $$
By classical central limit theorem, we have $$\frac{S_n - n}{\sqrt{n}} \to_d N(0, 1)$$
In view of $$\frac{S_n + r - \lambda}{\sqrt{\lambda}} = \frac{S_n - n}{\sqrt{n}}\frac{\sqrt{n}}{\sqrt{\lambda}} + \frac{r + n - \lambda}{\sqrt{\lambda}}$$ and Slutsky's theorem, it suffices to show $(r + n - \lambda)/\sqrt{\lambda} \to_p 0$ as $\lambda \to \infty$. This is due to \begin{align} & E\left[\frac{r + n - \lambda}{\sqrt{\lambda}}\right] = 0 \\ & \mathrm{Var}\left(\frac{r + n - \lambda}{\sqrt{\lambda}}\right) = \frac{\lambda - n}{\lambda} \to 0 \end{align} as $\lambda \to \infty$.