# Showing convergence by using Levy's continuity theorem

Let $X_n \sim \mathrm{Po} (n)$. I want to show that $Y_n = \displaystyle \frac{X_n -n}{ \sqrt{n}}$ for $n \rightarrow \infty$ converges in distribution to a random variable that is standard normal distributed by usig Levy's continuity theorem.

So i have to calculate $\lim_{n \to \infty} \Phi_{Y_n} = e^{ \displaystyle \frac{-t^2}{2}}$, where $\Phi_{Y_n}$ is the characteristic function of $Y_n$. I know that the characteristic function of $X_n$ is given by $e^{eit - 1}$, but I'm struggeling with the one for $Y_n$. Can anybody help?

Hint: using the definition of characteristic function, show that if $\phi(t)$ is the characteristic function of a random variable $X$, then the characteristic function of $(X-a)/b$ is $e^{-ita/b} \phi(t/b)$.
Also, you have the wrong characteristic function for $X_n$; it's not $e^{eit-1}$ but $e^{n(e^{it}-1)}$.
• So I got $e^{-itn / \sqrt{n}} e^{n(e^{it / \sqrt{n}} -1)}$, but I don't know how to simplify to get the result – xxx Jan 19 '15 at 19:28
• @xxx: Now it's just a matter of computing that limit as $n \to \infty$ - simple calculus. You need to compute $$\lim_{n \to \infty} n(e^{it/\sqrt{n}}-1)-it\sqrt{n}$$ for which you may find it helpful to use L'Hospital's rule, or the Taylor expansion of $e^{it/\sqrt{n}}$. – Nate Eldredge Jan 19 '15 at 19:33