Conditional expectation: $E[X_1 X_2\mid X_1 + X_2 X_3]$ Random variables $X_1, X_2, X_3$ are independent and $P(X_i=1)=P(X_i=-1)=\frac{1}{2}$ for $i=1,2,3$.
Compute:


*

*$E[X_1 \mid X_1 + X_2 + X_3]$

*$E[X_1 X_2 \mid X_1 + X_2 X_3]$
My answer to question 1 is $E[X_1 \mid X_1 + X_2 + X_3] = \dfrac{X_1+X_2+X_3}{3}$.
What about second question. Any idea? 
 A: For a visual guide, consider :
$\begin{array}{|r:r:r|r:r|}\hline X_1 & X_2 & X_3 & X_1+X_2X_3 & X_1X_2 \\ \hline
 1 & 1 & 1 & 2 & 1 \\
 1 & -1 & -1 & 2 & -1 \\ \hdashline
 1 & 1 & -1 & 0 & 1 \\
 1 & -1 & 1 & 0 & -1 \\
 -1 & 1 & 1 & 0 & -1 \\ 
 -1 & -1 & -1 & 0 & 1 \\ \hdashline
 -1 & 1 & -1 & -2 & -1 \\ 
 -1 & -1 & 1 & -2 & 1 \\ \hline
\end{array}$
A: The first one is easy:
$$
\operatorname{E}(X_1 \mid X_1+X_2+X_3) + \operatorname{E}(X_2 \mid X_1+X_2+X_3) + \operatorname{E}(X_3 \mid X_1+X_2+X_3)
$$
$$
= \operatorname{E}(X_1+X_2+X_3 \mid X_1+X_2+X_3) = X_1+X_2+X_3
$$
and all three terms on the first line are equal to each other, so each must be one-third of that last expression.
A: Since the $X_i$s are discrete with dichotomous support, it is actually very easy to write out all of the elementary outcomes and tabulate the expectation.  Let $\boldsymbol X = (X_1, X_2, X_3) \in \{-1,1\}^3$, and compute for each triplet the values $$g(\boldsymbol X) = X_1 X_2,$$ and $$h(\boldsymbol X) = X_1 + X_2 X_3.$$  Then since each elementary outcome has equal probability, the expectation of $g$ given $h$ is simply the arithmetic mean of $g$ calculated over those triplets yielding a given $h$.  For instance, there are two outcomes that give $h(\boldsymbol X) = 2$:  $\boldsymbol X \in \{(1,-1,-1), (1,1,1)\}$.  Then $g$ for these triplets are $-1$ and $1$, hence $$\operatorname{E}[g(\boldsymbol X) \mid h(\boldsymbol X) = 2] = 0.$$  Do it for the rest of the possible outcomes of $h(\boldsymbol X)$ and you're done.
A: Using this table i have 3 cases:
$E[g(X)|h(X)=2]=1*\frac{1}{2}+(-1)*\frac{1}{2}=0$
$E[g(X)|h(X)=0]=1*\frac{1}{4}+1*\frac{1}{4}+(-1)*\frac{1}{4}+(-1)*\frac{1}{4}=0$
$E[g(X)|h(X)=-2]=1*\frac{1}{2}+(-1)*\frac{1}{2}=0$
so summing up: $E[X_{1}X_{2}|X_{1}+X_{2}X_{3}]=0$
Could you check this? 
And the alternative solution:
$X_{1}+X_{2}X_{3}=X_{1}+(-X_{2})(-X_{3})$
so 
$E[X_{1}X_{2}|X_{1}+X_{2}X_{3}]=E[-X_{1}X_{2}|X_{1}+X_{2}X_{3}]$
and then
$2E[X_{1}X_{2}|X_{1}+X_{2}X_{3}]=0$
so finally
$E[X_{1}X_{2}|X_{1}+X_{2}X_{3}]=0$
Is this correct?
