# Show that $\lim\limits_{n\rightarrow\infty} e^{-n}\sum\limits_{k=0}^n \frac{n^k}{k!}=\frac{1}{2}$

Show that $\displaystyle\lim_{n\rightarrow\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}=\frac{1}{2}$ using the fact that if $X_j$ are independent and identically distributed as Poisson(1), and $S_n=\sum\limits_{j=1}^n X_j$ then $\displaystyle\frac{S_n-n}{\sqrt{n}}\rightarrow N(0,1)$ in distribution.

I know that $\displaystyle\frac{e^{-n} n^k}{k!}$ is the pdf of $K$ that is Poisson(n) and $k=\sum\limits_{j=1}^n X_j$ is distributed Poisson(n), but I don't know what I can do next.

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## 2 Answers

Hint: $e^{-n} \sum_{k=0}^n \dfrac{n^k}{k!}$ is the probability of what?

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Is it $P(k<n)$? – lightfish Nov 25 '13 at 22:26
I got it, thank you! – lightfish Nov 25 '13 at 22:38

If $X_1,\ldots, X_n$ are independent Poisson random variables having parameter $1$, then $Y=X_1+\cdots+X_n$ is also a Poisson r.v. with parameter $n$. We have also;

$\mathbf E(Y)=\operatorname{Var}(Y)=n$

The central limit theorem then implies that;

$$Z:=\frac{Y-\mathbf E(Y)}{\sqrt{\operatorname{Var}(Y)}}=\frac{Y-n}{\sqrt{n}}\tag I$$

is approximately normal and

$$\Pr(Z\le z)\approx\Phi(z) \tag{II}$$

($\Phi$ is cdf of a standard normal r.v.)

Substituting $(I)$ in $(II)$ gives

$$\Pr\left(\frac{Y-n}{\sqrt{n}}\le z\right)=\Pr(Y\le z\sqrt{n}+n))\approx\Phi(z)$$

$$\Rightarrow\Pr(Y\le y)\approx\Phi\bigg(\frac{y-n}{\sqrt{n}}\bigg)$$

but since $\displaystyle e^{-n}\sum\limits_{k=1}^{n}\frac{n^k}{k!}=\Pr(Y\le n)$

Hence

$$\Pr(Y\le n)\approx\Phi\bigg(\frac{n-n}{\sqrt{n}}\bigg)=\Phi(0)=\frac12$$

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