Converge in Distribution For Poisson Random Variables Let $(X_n)$ be a sequence of i.i.d. random variables with $X_1$~$Pois(1)$ and $S_n=X_1+X_2+...+X_n$. Prove that $\frac{S_n-n}{\sqrt{n}}$ converges in distribution to $N(0,1)$ (normal distribution). Use this result to prove $\frac{Y_\lambda-\lambda}{\sqrt{\lambda}}$ converges in distribution to $N(0,1)$ for $Y_\lambda$~$Pois(\lambda)$
I think this problem most likely will use the central limit theorem, but I am not sure how to apply it.
 A: Since $S_n = X_1+\cdots + X_n$ is the sum of IID random variables having finite variance, the standard CLT implies that
$$
\frac{S_n - E(S_n)}{SD(S_n)} = \frac{S_n - n}{\sqrt{n}} \stackrel{D}{\to}N(0,1), \mbox{ as } n\to \infty. 
$$
By using for example moment generating functions, you can show that $S_n \sim Poi(n)$, so that this result in your notation defining $Y_\lambda$ can be interpreted that
$$\frac{Y_n - n}{\sqrt{n}} \stackrel{D}{\to}N(0,1), \mbox{ as } n\to \infty. $$
In particular the final result holds when the limit as $\lambda \to \infty$ is taken for integer values of $\lambda$. To put everything together, notice that for any $\lambda >0$, there must exist a natural number $n$ so that $\lambda \in [n,n+1]$. You can check that for all $x \in \mathbb{R}$, $P(Y_{n+1} \le x ) \le P(Y_\lambda \le x) \le P(Y_n \le x)$. Moreover,
$$P\left(\frac{Y_{n+1}-n}{\sqrt{n}} \le x \right) \le P\left(\frac{Y_{\lambda}-\lambda}{\sqrt{\lambda}} \le x \right) \le P\left(\frac{Y_{n}-(n+1)}{\sqrt{n+1}} \le x \right).$$
Using the first result, we have that the left and right hand sides of the above both converge to $\Phi(x)$, the standard normal CDF, as $\lambda \to \infty$. Hence
$$
P\left(\frac{Y_{\lambda}-\lambda}{\sqrt{\lambda}} \le x \right) \to \Phi(x), \mbox{ as } \lambda\to \infty. 
$$
This gives the second result.
