3

Exercise:

Compute the following limit by using the central limit theorem: $$ \lim_{n\to\infty}e^{-n}\sum_{k=0}^n \frac{n^k}{k!} $$

Solution (from textbook):

Let $X_1,X_2,\ldots$ i.i.d. with $X_i\sim Pois(1)$. Then $\mu=E[X_i]=1=Var[X_i]=\sigma^2$. Further let $S_n=X_1+\ldots+X_n$, thus $S\sim Pois(n)$. With the central limit theorem we get $$ e^{-n}\sum_{k=0}^n \frac{n^k}{k!}= \sum_{k=0}^n e^{-n}\frac{n^k}{k!}= \sum_{k=0}^n Pr[S_n=k] = Pr[S_n\leq n] = Pr\left[ \frac{S_n-n\mu}{\sqrt{n}\sigma} \leq \frac{n-n\mu}{\sqrt{n}\sigma} \right]=\ldots $$ Now this is the step I don't understand: How does term on the right side of "$\leq$" disappear? $$ \ldots=Pr\left[ \frac{S_n-n\mu}{\sqrt{n}\sigma} \leq 0 \right] \stackrel{n\to\infty}{\longrightarrow} \Phi(0)=\frac{1}{2} $$ Could you explain me the step in between?

ndrizza
  • 1,348
  • 5
    $\mu=1$ (first line of the solution) so $n-n\mu$ is zero. – GEdgar Jan 04 '16 at 14:28
  • 1
    See also http://math.stackexchange.com/a/493480/147873 One of the answers solves it using the CLT. – Winther Jan 04 '16 at 14:29
  • @GEdgar I can see that $\mu=1$ for each of the random variables $X_i$. However, I thought that the purpose of the fractions inside the probability is to standardize $S_n$ in order to be able to use $\Phi$. So we should take the $\mu$ of $S_n$ (the notation is confusing). In this case isn't $E[S_n]=\mu=n=Var[S_n]=\sigma^2$? – ndrizza Jan 04 '16 at 14:34
  • @GEdgar Now I got you. You subtract the rhs of the inequality, and still use the $\mu$ of the $X_i$ since $E[S_n]=n\mu=n\cdot 1 = n$. An then $S_n$ is perfectly standardized. Thank you! – ndrizza Jan 04 '16 at 14:38
  • @GEdgar Could you post your short sentence as an answer? I'll then mark the question as solved then and leave it for reference. – ndrizza Jan 04 '16 at 14:42
  • @Lucian Not a duplicate, this question asks about a specific step of a proof, not for a proof of the result. – Did Jan 04 '16 at 15:54

1 Answers1

3

$\mu=1$ (first line of the solution) so $n−n\mu = 0$.

GEdgar
  • 111,679