$X \sim$ Geometric$(q)$ and $Y \sim$ Binomial$(X,p)$

Question

The question is: how to determine $\mathbb{E}[Y]$ using the partition theorem?

The partition theorem says that if we have a partition ${\{B_1,B_2,...\}}$ of our outcomespace $\Omega$ with $\Bbb{P}[B_i]>0$, then $\Bbb{P}[A] = \sum_{i}\Bbb{P}(A\mid B_i)*\Bbb{P}(B_i)~\text{for}~A \in \mathcal{F}.$

First I noted that $\Bbb{P}(X=k)=q(1-q)^{k-1}$ and by definition we have $\Bbb{E}[Y]=\sum_{y\in Y[\Omega]} y*\Bbb{P}(Y=y).$ When I try to write out this definition with the dependence of $Y$ on $X$ I stumble into bad notation and do not see a way trough. Can you help me solve it?

Answer 1

Using the tower property of conditional expectation, \begin{align*} E(Y) &= E(E(Y \mid X)) \\ &= E(\text{Binom}(X,p) \mid X) \\ &= E(Xp) \\ &= pE(\text{Geom}(q)) = p\cdot \frac{1}{q} \end{align*}