# Expectation and confidence intervals of a Poisson process

A Poisson process has PDF

$$P(X=k)=\frac{e^{-\lambda t}(\lambda t)^k}{k!}$$

I'm trying to find an expression for:

1. $E[X | \lambda, t]$
2. Confidence intervals (i.e. find $\delta$ such that $P(\bar{x}-\delta<X<\bar{x}+\delta)=c$ for some $c$)

To find the expectation, I've noticed that

$$E[X]=\sum_{k=0}^{\infty} \frac{e^{-\lambda t}(\lambda t)^k k}{k!}=e^{-\lambda t}\sum_{k=0}^{\infty}\frac{(\lambda t)^k}{(k-1)!}$$

But I have no idea how to go further. I read on wikipedia that the mean of a Poisson distribution is $\lambda$ - does this mean that the mean of a Poisson process is just $\lambda t$? And similarly its variance I suppose?

EDIT: I made some more progress based on procrastinator's comment, but I think I made a mistake.

Let $x=\lambda t$ and define $(-n)!=(n!)^{-1}$. Then we have

$$e^{x}\sum_{k=0}^{\infty}\frac{x^k}{(k-1)!}=e^{x}x\sum_{k=0}^{\infty}\frac{x^{k-1}}{(k-1)!}=e^{x}x\left(x^{-1}+e^{x}\right)=e^x+xe^{2x}$$

Which does not appear to equal $x$ as it should. Where did I go wrong?

-
You are correct, the mean and variance are $\lambda t$. The proof follows analogously as in the case where the mean and variance are $\lambda$. The basic idea is to factorise $\lambda t$ outside the series and re-arranging the indices. –  user10525 Dec 22 '12 at 19:51
@Procrastinator: Thanks this helps! Can you see the error in my edit? –  Xodarap Dec 22 '12 at 22:14
There is a little mistake, for $k=0$, the first terms vanishes. Just takes this into account while rewriting the series. –  user10525 Dec 22 '12 at 22:28
Oh I see, it's sort of $0/0!$, which my rewriting as $(k-1)!$ masked. (and it's $e^{-x}$, not $e^x$). If you want to put that as an answer, I'll accept. –  Xodarap Dec 22 '12 at 22:59
I would prefer you to write the solution you have got. I would certainly upvote it. –  user10525 Dec 22 '12 at 23:03
Let $x=\lambda t$ for simplicity. Then we have:
$$\begin{eqnarray} \sum_{k=0}^{\infty}\frac{x^{k}e^{-x}k}{k!} &=& e^{-x}\sum_{k=1}^{\infty}\frac{x^{k}k}{k!} \\ &=& e^{-x}x\sum_{k=1}^{\infty}\frac{x^{k-1}}{(k-1)!} \\ &=& e^{-x}x\sum_{k=0}^{\infty}\frac{x^{k}}{k!} \\ &=& x \end{eqnarray}$$
Where the last step uses the Taylor Series expansion for $e^x$.