examples on sequence of poisson random variables 
Let $X_n$ be distributed as a poisson random variable with parameter $n$. Then which of the following are true ?
1.$\underset{n\rightarrow \infty}{\lim} \mathbb{P} (X_n > n + \sqrt n)=0 $
2.$\underset{n\rightarrow \infty}{\lim} \mathbb{P} (X_n \le n)=1 $
3.$\underset{n\rightarrow \infty}{\lim}\mathbb{P} (X_n \le n)=\frac {1}{2} $

The answer is 3. I am not able to think how to start on this. Please any suggestions.
 A: *

*Let $\lambda$ be the lamda for your sum of independent Poisson random variables ($X_n$)

*Since $\lambda$ approaches to infinity, we should be able to approximate it with $N(\lambda,\lambda)$.

*The normal approximation centres on $\lambda$ and is symmetric. You get your answer.

A: First of all, by elimination the first two are solvable (you can look at the variance and mean of the distribution).
But I strongly agree 3 is tricky, but I came up with an explanation:
Poisson distribution has a nice property: $x_i - \operatorname{Poisson}(\lambda_i) \Rightarrow \sum_ix_i-\operatorname{Poisson}(\sum_i\lambda_i) $
Therefore:
$X_n = \sum_{i=1}^{n}X_i, X_i-\operatorname{Poisson}(1)\rightarrow_{n\rightarrow\infty}N\left(\mathbb{E}\left[\sum_iX_i\right],\sum_i\lambda_i\right)=N(n,n) \\\Rightarrow \mathbb{P}(X_n\leq n)=0.5$
The last part is true since Gaussian distribution is symmetric.
BTW - I encourage you to use the same way the calculate the actual probability in 2 and 3
A: The limiting distribution of $X_{n}$ as $n\rightarrow\infty$ is normal. We can show this using the moment generating function.
Consider a standardized Poisson random variable:
$$\frac{X_{n}-n}{\sqrt{n}}$$
With moment generating function:
$$M_{X_{n}}(t)=E\big[e^{tX_{n}}\big]=e^{n(e^{t}-1)}$$
Taking the limit:
$$\begin{align}
\lim_{n\rightarrow\infty}M_{(X_{n}-n)/\sqrt{n}}(t)&=\lim_{n\rightarrow\infty}E\bigg[\text{exp}\bigg(t\cdot\frac{X_{n}-n}{\sqrt{n}}\bigg)\bigg]\\
&=\lim_{n\rightarrow\infty}\text{exp}\big(-t\sqrt{n}\big)E\bigg[\text{exp}\bigg(\frac{tX_{n}}{\sqrt{n}}\bigg)\bigg]\\
&=\lim_{n\rightarrow\infty}\text{exp}\big(-t\sqrt{n}\big)\text{exp}\big(n(e^{t/\sqrt{n}}-1)\big)\\
&=\lim_{n\rightarrow\infty}\text{exp}\big(-t\sqrt{n}+n(tn^{-1/2}+t^{2}n^{-1}/2+t^{3}n^{-3/2}/6+\ldots)\big)\\
&=\lim_{n\rightarrow\infty}\text{exp}\big(t^{2}/2+t^{3}n^{-1/2}/6+\ldots\big)\\
&=\text{exp}\big(t^{2}/2\big)\\
&=M_{Z}(t)
\end{align}$$
Which is the moment generating function of the standard normal random variable, $Z\sim N(0,1)$. Thus, the limiting distribution of our Poisson random variable is simply:
$$\sqrt{n}Z+n\sim N(n,n)$$
So for your options above:


*

*$$\begin{align}\mathbb{P}(X_{n}>n+\sqrt{n})&=\mathbb{P}(N(n,n)>n+\sqrt{n})\\
&=\mathbb{P}(N(0,1)>1)\\
&=1-\Phi(1)\\
&\ne 0
\end{align}$$

*$$\begin{align}\mathbb{P}(X_{n}<n)&=\mathbb{P}(N(n,n)<n)\\
&=\mathbb{P}(N(0,1)<0)\\
&=\Phi(0)\\
&\ne 1
\end{align}$$

*$$\begin{align}\mathbb{P}(X_{n}<n)&=\mathbb{P}(N(n,n)<n)\\
&=\mathbb{P}(N(0,1)<0)\\
&=\Phi(0)\\
&=0.5
\end{align}$$
