$Y_n = \frac{X_n - n}{\sqrt{n}} $, Levy's continuity theorem

Question

Let $X_n$ be $Po(n)$-distributed. How do I show that $Y_n = \frac{X_n - n}{\sqrt{n}} $ for $n \to \infty$ converges to a standard normal random variable? I think that I have to use Levy's continuity theorem but I don't really know how.

Answer 1

As Brian pointed out, you can use the fact that $X_n=\sum_{i=1}^{n}\tilde{X}_k$ where $\tilde{X}_k$ are i.i.d. $Poi(1)$ random variable. Now the central limit theorem tells you that $\frac{X_n-n}{\sqrt{n}}$ converges weakly to standard normal.

Alternatively, let $\phi_n(t):=\mathbb{E}(Y_n)$ be the characteristic function of $Y_n$. We can compute $$\phi_n(t)=e^{-it\sqrt{n}}\mathbb{E}\left(e^{i\frac{t}{\sqrt{n}}X_n}\right)=e^{-it\sqrt{n}}e^{n\left(e^{i\frac{t}{\sqrt{n}}}-1\right)}.$$ Use the fact that $e^{i\frac{t}{\sqrt{n}}}-1=i\frac{t}{\sqrt{n}}-i\frac{t^2}{2n}+o(\frac{1}{n})$. To rewrite the characteristic function as $$\phi_n(t)=e^{-it\sqrt{n}}e^{i\sqrt{n}t-t^2/2+o(1)}=e^{-\frac{t^2}{2}+o(1)}.$$ It follows that $\phi_n(t)\to e^{-t^2/2}=:\phi(t)$. We already know that $\phi(t)$ is the characteristic function of standard normal. Levy's theorem tells you that $Y_n\to Z$ weakly where $Z\sim N(0, 1)$.