Here is the question, it is SOA number 280: The number of burglaries occurring on Burlington Street during a one-year period is Poisson distributed with mean $1$.
Calculate the expected number of burglaries on Burlington Street in a one-year period, given that there are at least two burglaries.
My work so far:
$$E(X|x=2)={\sum_{x=2}^{\infty}xp(x)\over 1-p(0)-p(1)} $$
$$={\sum_{x=0}^{\infty}xp(x)-0p(0)-p(1)\over 1-p(0)-p(1)}$$
$$={\sum_{x=0}^{\infty}xp(x)-p(1)}\over 1-p(0)-p(1)$$
My question is solving this summation. I've checked my work with the solution and it is right, but I don't understand how they solve the summation. Admittedly I've always had difficulty with these things. Any hints would be greatly appreciated!
Edit: also any hints on how to make my equations not so tiny so people can actually see them would be great.
From just looking at it, I decided to try this
$$p(x)={e^{-1}1^x\over x!}$$
so I get
$$\sum_{x=0}^{\infty}{xe^{-1}1^{x}\over x!}-e^{-1}=\sum_{x=0}^{\infty}e^{-1}({x1^{x}\over x!}-1)$$
But from here I am stuck.
In your numerator you have $\sum_{x=0}^\infty x p(x)$. But $p$ is the pdf of a Poisson random variable with mean 1, so $\sum_{x=0}^\infty x p(x)$ is the expectation of a Poisson random variable with mean $1$, which is 1.