Conditional expectation: $E[X_1 X_2\mid X_1 + X_2 X_3]$

Question

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:

1. $E[X_1 \mid X_1 + X_2 + X_3]$

2. $E[X_1 X_2 \mid X_1 + X_2 X_3]$

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?

Answer 1

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.

Answer 2

For a visual guide, consider : $\begin{array}{|r:r:r|r:r|}\hline X_1 & X_2 & X_3 & X_1+X_2X_3 & X_1X_2 \\ \hline 1 & 1 & 1 & 2 & 1 \\ 1 & -1 & -1 & 2 & -1 \\ \hdashline 1 & 1 & -1 & 0 & 1 \\ 1 & -1 & 1 & 0 & -1 \\ -1 & 1 & 1 & 0 & -1 \\ -1 & -1 & -1 & 0 & 1 \\ \hdashline -1 & 1 & -1 & -2 & -1 \\ -1 & -1 & 1 & -2 & 1 \\ \hline \end{array}$

Answer 3

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.

Answer 4

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?