How Does the Limit of the Sum Approach One Half in POTW #43?

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem!

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Problem: Show that $\displaystyle\lim_{n\to\infty}e^{-n}\sum_{k=0}^n \frac{n^k}{k!}=\frac{1}{2}$.

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Hint:

Spoiler

Let $X_n$ be a Poisson random variable with mean $n$. Use the Central Limit Theorem to show that $\mathbb{P}\{X_n\leq n\}\rightarrow \frac{1}{2}$.

 

[Chris L T521]

This week's question was correctly answered by Sudharaka. You can find his solution below:

Spoiler

Let \(X_n\) be a Poisson random variable with mean \(n\). Then,

\[\mathbb{P}(X_n=k) = \frac{n^k e^{-n}}{k!}\]

\[\therefore \mathbb{P}(X_n\leq n)= \sum_{k\leq n}\mathbb{P}(X_n=k)=e^{-n}\sum_{k=0}^{n}\frac{n^k}{k!}~~~~~~~~~~~~(1)\]

By the central limit theorem we get,

\begin{eqnarray}

\lim_{n\rightarrow\infty}\mathbb{P}(X_n\leq n)&=&F_\mathrm{normal}(x;\mu=n,\sigma^2=n)\\

&=&\int_{0}^{\infty}\frac{1}{\sqrt{2\pi n}} e^{ -\frac{1}{2}\left(\frac{x-n}{\sqrt{n}}\right)^2 }\\

&=&\frac{1}{2}~~~~~~~~~~~~(2)

\end{eqnarray}

Therefore by (1) and (2) we get,

\[\lim_{n\rightarrow\infty}e^{-n}\sum_{k=0}^{n}\frac{n^k}{k!}=\frac{1}{2}\]