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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.

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  • $\begingroup$ I think $p_{X|Y}(x|y)=\frac{p_{X,Y}(x,y)}{p_Y(y)}$ $\endgroup$
    – GuPe
    Apr 23, 2017 at 4:15
  • $\begingroup$ @GuachoPerez you are right, yes, will edit now $\endgroup$
    – user435438
    Apr 23, 2017 at 4:19
  • $\begingroup$ The edited notation is still wrong. Sticking to your format, you should have something like $p_{X | Y^3}(x | y^3)$ for the conditional and $p_{Y^3}(y^3)$ for the marginal. You need to add a line saying that the goal is to show $\sum xp_{X|Y^3}(x|y^3) = \sum xp_{X|Y}(x|y)$ $\endgroup$
    – Lee David Chung Lin
    Apr 23, 2017 at 4:25
  • $\begingroup$ Use the fact that $Y^3$ is an injective transformation of $Y$. $\endgroup$
    – mlc
    Apr 23, 2017 at 4:32
  • $\begingroup$ Actually, $\sum\limits_xxp_{X|Y}(x|y)$ would be $E(X|Y=y)$ rather than $E(X|Y)$. $\endgroup$
    – Did
    Apr 23, 2017 at 9:23

2 Answers 2

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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 $\sigma(Y) \subset \sigma(Y^3)$.

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  • $\begingroup$ Intuitively, can't we say that because this is a one-to-one mapping between y and $y^3$, the information contained in $y^3$ is the same as that in $y$, and the expectation is therefore equal? $\endgroup$
    – Kwame Brown
    Nov 19, 2022 at 13:43
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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, $\mathbb{E}(X|Y)= \mathbb{E}(X|Y^3)$.

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