Conditional expectation on Bernoulli Process If we have a Bernoulli Process with success probability p, and if X equals the number of successes in first trial, Y the number of successes in first two trials, and Z the number of successes in the first three trials,
how can I find E(XZ | Y)?

I know that X and Z must be conditionally independent given Y but I'm not sure how I can use it here. I couldn't just take E(X | Y)E(Z | Y) can I? 

Thanks everyone in advance.
Let $\{X_n:n=1,2,\ldots\}$ be a sequence of i.i.d. $\operatorname{Ber}(p)$ random variables and $\{S_n:n=0,1,2,\ldots\}$ the corresponding Bernoulli process (i.e. $S_0=0$ and $S_n=\sum_{j=1}^n X_j$ for $j&gt;0$). Let $\{\mathcal F_n:n=0,1,\ldots\}$ be the natural filtration of $\{S_n\}$. Then for any $n\geqslant0$ and bounded measurable $f$ we have we have 
\begin{align}
\mathbb E\left[f(S_{n+1})\mid \mathcal F_n\right] &amp;= \mathbb E\left[f(S_{n+1})\mathsf 1_{\{X_{n+1}=0\}} \mid\mathcal F_n\right] + \mathbb E\left[f(S_{n+1}\mathsf 1_{\{X_{n+1}=1\}}\mid\mathcal F_n\right]\\
&amp;= \mathbb E\left[f(S_n)\mid\mathcal F_n\right]\mathbb P(X_{n+1}=0) + \mathbb =E\left[f(S_n+1)\mid\mathcal F_n\right]\mathbb P(X_{n+1}=1)\\
&amp;= p\left(f(S_n)+f(S_n+1) \right)\\
&amp;= g(S_n),
\end{align}
where $g$ is the (measurable) map $x\mapsto p\left(f(x)+f(x+1)\right)$. Therefore $\{S_n\}$ is a Markov process, and in particular, 
$$
\mathbb E[S_1S_3\mid \mathcal F_2] = S_1\mathbb E[S_3\mid\mathcal F_2] = S_1\left(S_2+p\right).
$$
If we are conditioning on (the $\sigma$-algebra generated by) $S_2$ as opposed to $\mathcal F_2$, then the computation is a bit different:
\begin{align}
\mathbb E[S_1S_3\mid S_2] &amp;= \mathbb E[X_1S_2 + X_1X_3\mid S_2]\\
&amp;= S_2\mathbb E[X_1\mid S_2] + \mathbb E[X_3]\mathbb E[X_1\mid S_2]\\
&amp;= S_2\cdot \frac12 S_2 + p\cdot \frac12 S_2\\
&amp;= \frac12 S_2\left(p+S_2\right).
\end{align}
I know that X and Z must be conditionally independent given Y


You don't know that, because they are not. &nbsp; Consider the probability of having one success in the fist trial given that you have two successes in the first two trials.$$\mathsf P(X=1\mid Y=2) =1\quad\neq\quad p=\mathsf P(X=1)$$

Likewise the probability of having some successes in the first three trials is dependent on how many successes are given in the first two trials.



Let $A,B,C$ be the indicators for success in the first three trials.  These are independent and identically distributed.

Then $X=A\,, Y=A+B\,, Z=A+B+C$

$\mathsf E(XZ\mid Y) = \mathsf E(A^2+AB+AC\mid A+B) \\ = \mathsf E(A^2\mid A+B)+\mathsf E(AB\mid A+B)+\mathsf E(A\mid A+B)\mathsf E(C)$

Now, clearly $\mathsf E(C)=p$

And $\mathsf E(A\mid A+B) =\frac{A+B}{2}$ due to symmetry. Because $\mathsf E(A+B\mid A+B) = \underbrace{A+B}_Y$

Simmilarly $\mathsf E((A+B)^2\mid A+B)=(A+B)^2$, so...