$X$ ~ $P_o(\lambda)$. Find $E(X)$

Question

$X$ ~ $Poisson(\lambda)$. Find $E(X)$

$P_X(x) = \frac{\lambda^{x}e^{-\lambda}}{x!}$

$e^{-\lambda}\sum_{x=1}^{\infty} \frac{\lambda^x}{(x-1)!}$

$= e^{-\lambda} \lambda \sum_{y=0}^{\infty} \frac{\lambda^{y}}{y!} $

$= e^{-\lambda} \lambda e^{\lambda} = \lambda$

They let $y = x-1$.

I don't understand how $\sum_{y=0}^{\infty} \frac{\lambda^y}{y!} = e^{\lambda}$.

Answer 1

The sum $$e^x = \sum_{n = 0}^{\infty} \frac{x^n}{n!}$$ is just a Taylor series for the exponential function.