$X_n$ Poisson of parameter $\lambda_n \to +\infty$ tends to $N(0,1)$

Question

Let $\lambda_n \in \mathbb R^+$ with $\lambda_n \to +\infty$. Let $X_n$ a sequence of random variables with distribution Poisson with parameter $\lambda_n$. Show that $$\frac{X_n-\lambda_n}{\sqrt{\lambda_n}}$$ tends to a Standard Normal distribution.

Attempt: $F_n(t) =\mathbb P(X_n \leq (t+\lambda_n{\sqrt{\lambda_n}})) = \sum_{k=0}^{\lfloor (t+\lambda_n{\sqrt{\lambda_n}}) \rfloor} e^{-\lambda_n}\frac{\lambda_n^k}{k!}$

and I would like to take $n \to \infty$.

Thanks!

Answer 1

Use the characteristic function instead. Start from the fact that $X_n$ has characteristic function $$\phi_n(t)=\exp(\lambda_n (e^{it}-1))$$ so the ratio has characteristic function

$$\psi_n(t)=\exp(\lambda_n (e^{it/\sqrt{\lambda_n}}-1))e^{-it\sqrt{\lambda_n}}$$ Taylor expansion gives:

$$\psi_n(t)=\exp\left[\lambda_n\left(\frac{it}{\sqrt{\lambda_n}}-\frac{t^2}{2\lambda_n}+O(\lambda_n^{-3/2})\right)-it\sqrt{\lambda_n}\right]\to\exp(-t^2/2)$$

which is precisely the characteristic function of a standard normal.

Answer 2

Recall that $X_n \sim Poiss(\lambda_n)$ can be expressed as $$ X_n = \sum_{i=1}^n X_i \sim Poiss(\sum_{i=1}^n\lambda_i = \lambda_n), $$ namely, $X_n$ is basically a sum of $n$ Poisson i.i.d. ($X_i$) where each $\lambda_i = \lambda_n/n$. Moreover, $\operatorname{Var}(X_i) = \lambda_n/n $, thus $S_n^2 = \lambda_n$ and $S_n = \sqrt{\lambda_n}$. Define, $$ Y_i = X_i - \lambda_n/n, $$ therefore, $\mathbb{E}Y_i = 0$, $\forall i$, and $T_n = \sum_{i=1}^n Y_i = \sum_{i=1}^n X_i - \lambda_n = X_n - \lambda_n$. Hence, as $\operatorname{Var}(Y_n) =\lambda_n$, then by the Lyapunov CLT (given that the sequence satisfies Lyapunov condition, namely, $\exists \delta >0$, $\sum\mathbb{E}|Y_i|^{2+\delta}/S_n^{2+\delta} \to 0$) we have that $$ Z_n = \frac{Y_n}{\sqrt{\lambda_n}} \xrightarrow{D} \mathcal{N}(0,1). $$