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}$ ?

  • $\begingroup$ 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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.