I have found that if I have a $Y \sim \mathrm{Poi}(\lambda)$ and $Z=Y \mid Y>0$ then I say $$f_Z(k)=g(k)=Pr(Y=k\mid k>0)=\frac{\lambda^k}{k! (e^\lambda-1)}$$
Now I am trying to compute expectation, that by definiyion would be $\mathbb{E}[Z]=\sum\limits_{k=2}^\infty k g(k)$.
I am getting $f_Z(k)=k\lambda^k/k!(e^\lambda-1)$ but I know the answer should be $e^\lambda\lambda/(e^\lambda-1)$. I am not really sure how to get here.
Best Answer
You must to compute the sum
$$\sum _{k=1}^{\infty }{\frac {k{\lambda}^{k}}{k!}} \left( {{\rm e}^{ \lambda}}-1 \right) ^{-1} $$
and then you obtain
$$\sum _{k=1}^{\infty }{\frac {k{\lambda}^{k}}{k!}} \left( {{\rm e}^{ \lambda}}-1 \right) ^{-1}=\sum _{k=1}^{\infty }{\frac {{\lambda}^{k}}{(k-1)!}} \left( {{\rm e}^{ \lambda}}-1 \right) ^{-1}= \sum _{k=0}^{\infty }{\frac {{\lambda}^{k+1 }}{k!}} \left( {{\rm e}^{ \lambda}}-1 \right) ^{-1}= \\ \lambda(\sum _{k=0}^{\infty }{\frac {{\lambda}^{k }}{k!}} )\left( {{\rm e}^{ \lambda}}-1 \right) ^{-1}=\lambda{\rm e}^{ \lambda}\left( {{\rm e}^{ \lambda}}-1 \right) ^{-1} $$