Conditional Expectation of Two Random Variable I'm new to conditional expectations and having trouble with the following problem: 


*

*For $\Omega = \{ a, b, c, d \}$, with $P(a) = 4P(b) = 2P(c) = 3P(d)$. 


Define the random variables:
$X(ω) = \begin{cases}~~~5 &:&\omega \in \{ a, b \}\\ −2&:& ω \in \{ c, d \}\end{cases}$
$Y (ω) = \begin{cases}~~~3&:& ω ∈ \{ a, c \}\\~~~ 4&:& ω ∈ \{ b, d \}\end{cases}$
I need to find $E[X|Y]$ and I've gotten this far:

(Also just realized it should say $y∈\{3,4\}$ not $y∈\{-2,5\}$.)
Not sure where to go from here, as I've only done problems where the probabilities were defined as, for example, $P(a)$ = $1/6$, etc.
 A: Firstly, you have $\mathsf P(a)=4\mathsf P(b)=2\mathsf P(c)=3\mathsf P(d)$, so you don't need to evaluate them since there will be cancelation.
Secondly, $\mathsf E(X\mid Y)$ is a random variable, measured over $Y$, so don't forget to leave in the indicator functions.. $$\begin{align}\mathsf E(X\mid Y)&=\begin{cases}\mathsf E(X\mid Y=3)&:&Y=3\\\mathsf E(X\mid Y=4)&:&Y=4\end{cases}
\\[2ex]&=\begin{cases}\dfrac{X(a)~\mathsf P(a)+X(c)~\mathsf P(c)}{\mathsf P(a)+\mathsf P(c)}&:&Y=3\\[2ex]\dfrac{X(b)~ \mathsf P(b)+X(d)~\mathsf P(d)}{\mathsf P(b)+\mathsf P(d)}&:& Y=4\end{cases}
\\[2ex]&=\dfrac{5~\mathsf P(a)-2~\mathsf P(c)}{\mathsf P(a)+\mathsf P(c)}\mathbf 1_{Y=3}+\dfrac{-2~ \mathsf P(b)+5~\mathsf P(d)}{\mathsf P(b)+\mathsf P(d)}\mathbf 1_{Y=4}
\\[2ex]&=\dfrac{10-2}{2+1}\mathbf 1_{Y=3}+\dfrac{-2~ \mathsf P(b)+5~\mathsf P(d)}{\mathsf P(b)+\mathsf P(d)}\mathbf 1_{Y=4}&&\text{since }\mathsf P(a)=2\mathsf P(c)
\\[2ex]&=\dfrac{8}{3}\mathbf 1_{Y=3}+\dfrac{-2~ \mathsf P(b)+5~\mathsf P(d)}{\mathsf P(b)+\mathsf P(d)}\mathbf 1_{Y=4}
\end{align}$$
And I'll leave the rest to you.
