I have found the following task and I need a hint to solve it:
Be $X\text{~}Poi(\lambda)$. Calculate $\mathbb{E}[X | X \geq 1]$.
I never done something like this, and I could not find any similar task, so I hope you can help.
What I know: The Conditional Expectation is defined as $\mathrm{E}(Y | B)=\frac{\mathrm{E}\left(1_{B} \cdot Y\right)}{P(B)}$
and the Poisson Expectation is defined as $ \mathrm{E}(X)=\sum_{k=0}^{\infty} k \frac{\lambda^{k}}{k !} \mathrm{e}^{-\lambda} = \lambda$
So now I have to combine those two definitions or?
Hint: $E(X|X\geq 1)=\frac 1 {P(X\geq 1)} \sum\limits_{k=1}^{\infty} ke^{-\lambda} \frac {\lambda^{k}} {k!}$ and $P(X\geq 1)=1-P(X=0)=1-e^{-\lambda}$.
Note that $\sum\limits_{k=1}^{\infty} k\lambda^{k}/k! =\lambda \sum\limits_{k=1}^{\infty} \lambda^{k-1}/(k-1)!=\lambda \sum\limits_{k=0}^{\infty} \lambda^{k}/k!=\lambda e^{\lambda}$.