The problem I have is the following:
By applying the central limit theorem to a sequence of random variables with the Possion distribution, prove that
$$
e^{-n}\left(1+n+\frac{n^2}{2!} + \dots + \frac{n^n}{n!}\right)\to\frac{1}{2}, \mbox{ as } n\to\infty
$$
What I did so far:
If we chose $X \sim Poisson(n)$, then we have that
$$
P(X = k) = \frac{n^k}{k!}e^{-n}
$$
Hence,
$$
P(X\leq n) = e^{-n}\left(1+n+\frac{n^2}{2!} + \dots + \frac{n^n}{n!}\right)
$$
And, we know that $E[X] = n$ and that $Var(X) = n$, hence
$$
P(X\leq n) = P\left (\frac{X-E[X]}{\sqrt{var(X)}\sqrt{n}}\leq 0\right )
$$
Now, I know that we cannot take the limit in the probability as $X$ is not, yet a sum of independent random variables, but if we could write $X$ as a sum of independent random variables, then we would be able to have $P(X\leq n)\to \phi(0) = \frac{1}{2}$.
My understanding of the question is that it is required to find $X_1, X_2, \dots$ random variables, each a particular type of a Poisson distribution, so that added up, namely $X_1 + \dots + X_n = X$. Also, I know that a Poisson distribution can be written as a sum of independent Poisson distributions, however given that we have $X \sim Poisson(n)$ can we find exactly $n$ such distributions so that they add up to $X$ ? If so:
How do I find these Poisson distributions?
Best Answer
The idea is to take a sequence of independent random variables $X_1,X_2,\ldots$ each Poisson with mean $1$. Then $S_n=X_1+\cdots+X_n$ is Poisson with mean $n$ and one can then apply CLT.