If $X$ and $Y$ are discrete random variables then $E(X \mid Y)=E(X \mid Y^3)$ Suppose $X$ and $Y$ are discrete random variables. Show that $$E(X \mid Y)=E(X \mid Y^3).$$
The conditional expected value of a discrete random variable is expressed as
$$E(X \mid Y)=\sum xp_{X \mid Y}(x \mid y),$$
where$$p_{X \mid Y}(x \mid y)=\frac{p_{X,Y}(x,y)}{p_Y(y)}.$$ 
Similarly, you can say that 
$$E(X \mid Y^3)=\sum x p_{X \mid Y^3}(x \mid y^3),$$
where$$p_{X \mid Y^3}(x \mid y^3)=\frac{p_{X,Y^3}(x,y^3)}{p_{Y^3}(y^3)}.$$ 
The goal is to show that 
$$\sum xp_{X \mid Y}(x \mid y)=\sum xp_{X \mid Y^3}(x \mid y^3).$$
From here I don't really know how to show that the two are equal, some help would be appreciated. 
 A: Provide a generic approach. Strictly speaking $\mathbb E[X|Y]$ is just a notation, what it really means is $\mathbb E[X|\sigma(Y)]$, where $\sigma(Y)$ is the $\sigma$-field generated by $Y$ (see its definition below).
Proof. We need to show that $\sigma(Y)=\sigma(Y^3)$.
Let $f(x)=x^3$, $x \in \mathbb R$, then $Y^3=f \circ Y$
$$\sigma(Y)=\{Y^{-1}(A)|A \in \mathcal B(\mathbb R)\}=\{Y^{-1}\circ f^{-1}\circ f(A)|A \in \mathcal B(\mathbb R)\}$$
$$\sigma(Y^3)=\{(f\circ Y)^{-1}(A)|A \in \mathcal B(\mathbb R)\}=\{Y^{-1} \circ f^{-1}(A)|A \in \mathcal B(\mathbb R)\}$$
Now notice that both $f(x)=x^3$ and $f^{-1}(x)=\sqrt[3]{x}$ are monotone functions, and thus both of them are Borel-measurable (This is by Prop 5.10 of Real Analysis for Graduate Students Richard F. Bass).
Thus the above two sigma algebras are the same, because: $f$ Borel-measurable, we get $\sigma(Y^3)\subset \sigma(Y)$; $f^{-1}$ Borel-measurable, we get $\sigma(Y) \subset \sigma(Y^3)$.
A: Let $S$ denote the sample space. Notice that both $\mathbb{E}(X|Y):S\rightarrow \mathbb{R}$ and $\mathbb{E}(X|Y^3):S\rightarrow \mathbb{R}$ are random variables, and the following holds:
$\mathbb{E}(X|Y)(s) = \mathbb{E}(X|Y = Y(s)) = \mathbb{E}(X|Y^3 = (Y(s))^3) = \mathbb{E}(X|Y^3 = Y^3(s))= \mathbb{E}(X|Y^3)(s)$
for all $s\in S$.
Therefore, $\mathbb{E}(X|Y)= \mathbb{E}(X|Y^3)$.
