Let $(X_i)$ be a collection of i.i.d. r.v. in $L^1$ and $N\sim \mathrm{Poi}(1)$ independent from $(X_i)$. We define $Z=\sum_{i=1}^N X_i$. I need to compute the distribution function of $\Bbb{E}(Z|N)$.
From the lecture I know that $\Bbb{E}(Z|N=n)=n\Bbb{E}(X_1)$. Now I know that I need to compute $\Bbb{P}(\Bbb{E}(Z|N)\leq x)$ for all $x\in \Bbb{R}$. But somehow I don't know where to start. I know that by definition $\Bbb{E}(X|Y=y)=\frac{\Bbb{E}(X1_{Y=y})}{\Bbb{P}(Y=y)}$ therefore I thought about doing a partition in $\Bbb{E}(Z|N)$ to get $\Bbb{E}(Z|N)=\sum_k \Bbb{E}(Z|N)1_{N=k}=\sum_k\Bbb{E}(Z|N=k)1_{N=k}$. But I do not get something nice. Can maybe someone help me?
Thanks a lot
You're on the right track.
$\Bbb{E}(Z|N)=\sum_k \Bbb{E}(Z|N)1_{N=k}=\sum_k\Bbb{E}(Z|N=k)1_{N=k} = =\sum_kk \Bbb{E}(X_1)1_{N=k} = \Bbb{E}(X_1)\sum_kk 1_{N=k} = \Bbb{E}(X_1) N$.
So in the end $\Bbb{P}(\Bbb{E}(Z|N)\leq x) = \Bbb{P}(N\Bbb{E}(X_1)\leq x)$ and you can finish from here using the known distribution function of $N$.