Let $(X_n)_{n\geq1}$ be a sequence of Poisson random variables of parameter $\lambda_n = n$
Prove $Z_n = \frac{X_n-n}{\sqrt{n}} \longrightarrow^{\omega} Z $~$ N(0,1) $ where $\omega$ denotes the convergence is weak.
I believe I should use Levy's Continuity Theorem to prove convergence in probability, and given that the latter implies weak convergence.
This is the theorem:
If $F_n(x)$ are the cdfs and $M_n(t)$ the mgfs. Let $X$ be a rv with pdf $F(x)$ and mgf $M(t)$.
If $M_n(t) \longrightarrow M(t) \ \forall t$ in an open interval containing zero, then $F_n(x) \longrightarrow F(x)$ at all continuity points of $F$. That is $X_n$ converges in distribution to $X$.
The issue for me is how to compute the mgf of $Z_n$ and how to prove the convergence of the mgf to the mgf of a standard rv.
Best Answer
$X_n$ has the same distribution as the sum of $n$ i.i.d. Poisson random variables of parameter $1$ (so mean and variance $1$), so you could apply the Central Limit Theorem to say that $Z_n$ converges in distribution to $N(0,1)$
Alternatively, the mgf of $X_n$ is $\exp(n(e^t-1))$ so the mgf of $X_n-n$ is $\exp(n(e^t-1-t))$ and the mgf of $Z_n = \frac{X_n-n}{\sqrt{n}}$ is $$\exp\left(n\left(e^{t\sqrt{n}}-1-\frac{t}{\sqrt{n}}\right)\right)$$ and with some effort you can show that for any $t$ this converges to $\exp(\frac12t^2)$ as $n$ increases, the mgf of a standard normal distribution