# Central limit theorem/ poisson distribution

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 $$\frac{1}{2}$$ ?

• Look at your given sum --- what is the set of values of $S_n$ that it (before taking the limit) corresponds to ? – user10354138 Jun 16 '19 at 13:58
• I don't exactly know what you mean? – Sarah Jun 16 '19 at 14:04
• Our famous question: math.stackexchange.com/questions/160248/…. – StubbornAtom Jun 16 '19 at 14:37

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, $$P(S_n\leq n)=\sum_{k=0}^n \frac{e^{-n}n^k}{k!}.$$