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Let$\ X_1,X_2,...,X_n$ be independent Poisson random variables with parameter$\ λ=1$, use the Central Limit Theorem to prove:

$\ \lim_{n→∞} \frac{1}{e^n} \sum_{k=0}^n \frac{n^k}{k!} =\frac{1}{2}$

My idea:

It holds: $\ E(S_n)=n$ and$\ V(S_n)=n$

$$ \lim_n P(\frac{S_n-n}{\sqrt n} \overset{} \leq z)\longrightarrow \Phi(z) $$ How do I get to $ \frac{1}{2}$ ?

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  • $\begingroup$ Look at your given sum --- what is the set of values of $S_n$ that it (before taking the limit) corresponds to ? $\endgroup$ – user10354138 Jun 16 '19 at 13:58
  • $\begingroup$ I don't exactly know what you mean? $\endgroup$ – Sarah Jun 16 '19 at 14:04
  • $\begingroup$ Our famous question: math.stackexchange.com/questions/160248/…. $\endgroup$ – StubbornAtom Jun 16 '19 at 14:37
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You are very close. Take the expression that you have, $\lim_n P(\frac{S_n-n}{\sqrt{n}}\leq z) \to \Phi(z)$ by CLT. Note that $\Phi(0)=\frac12.$ Therefore, $$ \frac12=\Phi(0)\overset{CLT}{=}\lim_n P(\frac{S_n-n}{\sqrt{n}}\leq 0)=\lim_n P(\frac{S_n}{\sqrt{n}}\leq \frac{n}{\sqrt{n}})= \lim_n P(S_n\leq n). $$ Since the sum of Poissons is Poisson, $S_n$ is Poisson with $\lambda=n.$ The PMF is $P(S_n=k)=\frac{e^{-n}n^k}{k!}$. Hence, $$ P(S_n\leq n)=\sum_{k=0}^n \frac{e^{-n}n^k}{k!}. $$

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  • $\begingroup$ Thank you for your help:) $\endgroup$ – Sarah Jun 16 '19 at 14:39

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