Let $\Omega$ be a probability space with elements:
$\Omega = \{0,1,2,3\}$
Let $\mathbb{P}$ be a probability measure such that:
$\mathbb{P}(0) = \frac{1}{6}, \mathbb{P}(1) = \frac{1}{3}, \mathbb{P}(2) = \frac{1}{4}, \mathbb{P}(3) = \frac{1}{4}$
Finally, let X and Y be two random variables defined by:
$X(0) = 1, X(1) = 1, X(2) = -1, X(3) = -1$
$Y(0) = 1, Y(1) = -1, Y(2) = 1, Y(3) = -1$
Find $\mathbb{E}[Y|X]$.
Attempt At Solution
By the conditional expectation formula, we have:
$\mathbb{E}[Y|X] = \sum_{\omega \in \Omega} \frac{\mathbb{E}[Y\mathbb{1}_{X= \omega}]}{P(X= \omega)}$$\mathbb{1}_{X=\omega}$
But I'm not sure how to compute this sum?
Here, $X$ takes its values in $\{-1,+1\}$, so the sum in your formula should be on $\omega \in\{-1,+1\}$ (and $\omega$ is a very bad choice of notation).
Actually, the idea behind the formula is that $E[Y\mid X]=f(X)$ where $f(x) = E[Y \mid X=x]$. Since $X$ takes its values in $\{-1,+1\}$, you just have to compute $f(1)$ and $f(-1)$, that is:
$$ f(1) = E[Y \mid X=1] =\frac{E[Y1_{X=1}]}{P(X=1)},\qquad f(-1)=\frac{E[Y1_{X=-1}]}{P(X=-1)} $$
Therefore, the correct formula is: $$ E[Y\mid X] = \frac{E[Y1_{X=1}]}{P(X=1)}1_{X=1} + \frac{E[Y1_{X=-1}]}{P(X=-1)}1_{X=-1} $$
Here is a model of the computations required to get $f(1)$. I leave the rest to you. $$ P(X=1) = P(0)+P(1)=\frac{1}{6}+\frac{1}{3} = \frac{1}{2}, $$ and $$ E[Y1_{X=1}] = 1\times P(0) + (-1) \times P(1) + 0\times P(2) + 0 \times P(3) = -\frac{1}{6} $$