Conditional Expectation of X given X^2

I also obtained the same result as @sds:

Notice that $\sigma(X^{2}) = \sigma(|X|)$. So by definition, for any $A \in \sigma(|X|)$ we have

$$ \int_{A} \Bbb{E} [X | X^{2}] \, d\Bbb{P} = \int_{A} X \, d\Bbb{P}.$$

On the other hand, since $\Bbb{E}[X | X^{2}]$ is $\sigma(|X|)$-measurable, there exists a Borel-measurable function $g : \Bbb{R} \to \Bbb{R}$ such that $\Bbb{E}[X|X^{2}] = g(|X|)$, i.e., $g(|X|)$ is a version of this conditional expectation. So if $X$ has density $f(x)$, then the density of $|X|$ is

$$ \frac{d}{dx} \Bbb{P}(|X| \leq x) = \frac{d}{dx} (F(x) - F(-x)) = f(x) + f(-x), \quad x \geq 0 $$

and for any event $A = \{ |X| \leq a \}$ for $a > 0$,

$$\int_{0}^{a} g(x) (f(x) + f(-x)) \, dx = \int_{A} g(|X|) \, d\Bbb{P} = \int_{A} X \, d\Bbb{P} = \int_{-a}^{a} x f(x) \, dx. $$

Differentiation both sides with respect to $a$, we have

$$ g(x) = x \frac{f(x) - f(-x)}{f(x) + f(-x)} \quad \text{a.s.} $$

Therefore it follows that

$$ \Bbb{E}(X | X^{2}) = g(|X|) = |X| \frac{f(|X|) - f(-|X|)}{f(|X|) + f(-|X|)}. $$

By definition, $$\begin{align} E[X|X^2=u^2] &= \sum_x xP(X=x|X^2=u^2) \\ & = u\frac{P(X=u \& X^2=u^2)}{P(X^2=u^2)}-u\frac{P(X=-u \& X^2=u^2)}{P(X^2=u^2)} \\ & =u\frac{P(X=u)-P(X=-u)}{P(X=u)+P(X=-u)} \end{align}$$

If $X$ has a density $f$, then

$$ E[X|X^2=u^2] = u\frac{f(u)-f(-u)}{f(u)+f(-u)} $$