The general calculation for the expected value of Poisson is clear to me.
$$\sum_{x=0}^\infty x*\frac{(\lambda^{x}*e^{-\lambda})}{x!} =...= \lambda$$
Generally proofs declare as their first step that $x=0$ results in a zero term, therefore the above can be written as:
$$\sum_{x=1}^\infty x*\frac{(\lambda^{x}*e^{-\lambda})}{x!} $$
Can't I simply leave x going from zero to $\infty$? My intuitive approach would have been:
$$\sum_{x=0}^\infty x*\frac{(\lambda^{x}*e^{-\lambda})}{x!} =e^{-\lambda}*\sum_{x=0}^\infty x*\frac{(\lambda^{x})}{x!} $$
Using the power series I can then proceed to:
$$e^{-\lambda}*\sum_{x=0}^\infty x*e^\lambda = \sum_{x=0}^\infty x $$
I definitely made a mistake somewhere along the way because the above is not $\lambda$. However, I'm not sure at what stages my thinking went wrong. Any pointers would be greatly appreciated.
Thanks!
Comment posted as answer, as requested:
It went wrong in going from $\sum_{x=0}^\infty x \frac{\lambda^x}{x!}$ to $\sum_{x=0}^\infty x e^\lambda$. I don't know why you would think this is the case. $e^\lambda$ is not $\lambda x/x!$, it's the sum of that from $x=0$ to $\infty$.