If $X\sim \text{Uniform}[-1,1]$ and $Y = X^2$, what is $E[X\mid Y]$?

Question

Given that $X$ is $\text{Uniform}(-1,1)$ continuous, $Y = X^2$, what is the expected value of $X$ given $Y$? I was asked to calculate the MMSE of $X$ given observing $Y$, which is $E[X\mid Y]$. The answer is $0$. But I don't know how to reason it.

How I calculated $E[Y|X]$ is $E[Y|X] = E[Y=X^2|X=x] = X^2$ because in this case, I think conditioning X, X^2 is a constant, so I get the value of E[Y|X] to be X^2. If I did it the same way for E[X|Y] = E[X = + $\sqrt{y}$ or X = - $\sqrt{y} | Y = y]$, I would think this equal to $\sqrt{Y}$ instead of 0 using the same logic that knowing value of Y, $x^2$ is a constant?

Can someone help me out to see what mistake I made? Thank you!

Answer 1

for $y\in (0,1)$,

$p_{X|Y}(x|y)=\displaystyle\frac{f_{X,Y}(x,y)}{f_{X,Y}(\sqrt{y},y)+f_{X,Y}(-\sqrt{y},y)}=\begin{cases}\frac{1}{2} & \text{if } x=\sqrt{y} \\ \frac{1}{2} & \text{if } x=-\sqrt{y}\\ 0 & \text{otherwise} \end{cases}$

and $p_{X|Y}(0|0)=1$.

Therefore, for $y\in (0,1)$, $\mathbb{E}(X|Y=y)=\frac{1}{2}(\sqrt{y})+\frac{1}{2}(-\sqrt{y})=0$ and $\mathbb{E}(X|Y=0)=0$. Consequently, $\mathbb{E}(X|Y)=0$.

Answer 2

The short answer is that it makes no sense to write $Y=X^2$ in an expectation, as you take the expectation of a random variable, not an equation and not a logical statement. The difference is that your first attempt can be rewritten as $$\mathbb{E}[Y\mid X] = \mathbb{E}[X^2\mid X] = X^2$$ (Note I've also not written $X=x$. That notation would essentially refer to the function $x\mapsto x^2$ rather than the random variable $X^2$)

To do the same thing for $\mathbb{E}[X|Y]$, we'd need to use indicators as stand-ins for the "or" appearing in your attempt, so you have the following: $$\mathbb{E}[X\mid Y] = \mathbb{E}\left[-\sqrt{Y}1_{X<0} + \sqrt{Y}1_{X>0}\mid Y\right] = \sqrt{Y}\left(\mathbb{P}(X>0\mid Y) - \mathbb{P}(X<0 \mid Y)\right)$$

Intuitively, knowing $Y$ tells us nothing about whether $X<0$ or $X>0$, so we'd have $$\mathbb{P}(X>0 \mid Y) = \mathbb{P}(X>0) = 1/2 = \mathbb{P}(X<0) = \mathbb{P}(X<0 \mid Y)$$ so their difference is zero.

More formally, if $0 < a < b$, then $$\mathbb{P}(X > 0 \mid a \leq Y < b) = \frac{\mathbb{P}(\sqrt{a} \leq X < \sqrt{b})}{\mathbb{P}(-\sqrt{b} < X \leq -\sqrt{a} \text{ or } \sqrt{a} \leq X < \sqrt{b})} = \frac12$$ so that $\mathbb{P}(X > 0 \mid Y) = \frac12$, and similarly $\mathbb{P}(X < 0 \mid Y) = \frac12$.

Answer 3

In this case, since $X$ is symmetric about $0$, both $X$ and $-X$ have the same distribution:

$$X \stackrel{d}= -X$$

So,

$$E\left[X\mid X^2\right]=E\left[-X\mid (-X)^2\right]=-E\left[X\mid X^2\right]$$

Hence, with probability $1$,

$$E\left[X\mid X^2\right]=0$$