Conditional expectation of conditional expectation

Question

I have a question about conditional expectation. I have always problem with that... It is a step of a proof that I just don't get... I appreciate any help! I have the random variable $$B=S+ \sum\limits_{j=1}^{Z} B_j.$$ The two terms on the rightside are NOT independent. Now we condition on $S$: $$M_B(s)=\mathbb{E}[e^{sB}]=\mathbb{E}[\ \mathbb{E}[ \ e^{sB} \vert S] \ ].$$ Then the next step is to calculate $\mathbb{E}[ \ e^{sB} \vert S]$. Here comes the problem up. We know that conditional on Z, the $B_i$'s are independent. But then I have conditional expectation of conditional expectation... I dont understand how to compute this. Can anybody help me? $$\mathbb{E}[ \ e^{sB} \vert S]=\mathbb{E}[ \mathbb{E}[ \ e^{sB} \vert S] \vert Z].$$ I mean, in words it is the expected mean value of $e^{sB}$ given $S$ and $Z$. Can I just do this: $$\mathbb{E}[ \ e^{sB} \vert S]=\mathbb{E}[ \mathbb{E}[ \ e^{sB} \vert S, Z]] ?$$

Answer 1

Yes, you've got the right idea. I'm assuming you intend the $B_i$ to be iid

$$ \begin{aligned}B^*(s) &= \mathbb E[ e^{sB} ]\\ &= \mathbb E[\mathbb E[e^{sB}|S]]\\ &= \mathbb E[\mathbb E[ \mathbb E[e^{s(S+\sum_{j=1}^Z B_j)}|S,Z]|S]]\\ &= \mathbb E[ e^{sS} \mathbb E[ \mathbb E[ e^{sB}|S,Z]^Z|S]]\\ &= \mathbb E[e^{sS}G(B^*(s))] \end{aligned}$$

where $G$ is the probability generating function of $Z$ given $S$. (Using $S$ for a random variable and $s$ for the dummy variable is a notation choice that might be worth trying to avoid in the future.)

[Edit: corrected typo involving sign]