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

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.

• Alternatively, you can use the central limit theorem. Do you know what the distribution of $\sum_{k=1}^n Y_k$ is when $Y_i$ are i.i.d. $Po(1)$ random variables? Commented Jan 16, 2022 at 3:26
• Your edit makes this question nonsensical. Commented Jan 17, 2022 at 10:16

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 $$Z\sim N(0, 1)$$.