Conditional distribution of continuous random variable $X$ given $|X|$? If a continuous random variable $X$ has a symmetric distribution around $0$, what is the conditional distribution of $X$ given $|X|$?
 A: For a rigorous answer one needs to rely on a precise definition of conditional distribution. Recall that, for any random variables $X$ and $Y$, one calls conditional distribution of $X$ conditionally on $Y$, any family $(\mu_y)_y$ of probability measures on the state space of $X$, indexed by the state space of $Y$, such that, for every bounded measurable function $u$,
$$
\mathbb E(u(X)\mid Y)=L_u(Y)\ \text{almost surely},\quad\text{where}\quad L_u(y)=\int u(z)\mathrm d\mu_y(z).
$$
In other words, one asks that, for every bounded measurable function $v$,
$$
\mathbb E(u(X)v(Y))=\mathbb E(L_u(Y)v(Y)).
$$

In the case at hand, $Y=|X|$ has density $g(y)=f_X(y)+f_X(-y)$ on $y\gt0$, hence
$$
\mathbb E(L_u(Y)v(Y))=\int_{y\gt0}L_u(y)v(y)g(y)\mathrm dy.
$$
On the other hand,
$$
\mathbb E(u(X)v(Y))=\int u(x)v(|x|)f_X(x)\mathrm dx=\int_{y\gt0} (u(y)f_X(y)+u(-y)f_X(-y))v(y)\mathrm dy.
$$
The only way these can coincide for every function $v$ is that, for Lebesgue almost every $y$,
$$
L_u(y)g(y)=u(y)f_X(y)+u(-y)f_X(-y),
$$
that is,
$$
L_u(y)=\frac{u(y)f_X(y)+u(-y)f_X(-y)}{f_X(y)+f_X(-y)}.
$$
Finally, the only way $L_u$ may be as above is when the family $(\mu_y)_y$ is such that, $\mathbb P_Y(\mathrm dy)$ almost surely,
$$
\mu_y(\mathrm dx)=\frac{f_X(y)\delta_y(\mathrm dx)+f_X(-y)\delta_{-y}(\mathrm dx)}{f_X(y)+f_X(-y)}.
$$
In particular, if the distribution of $X$ is symmetric, $f_X(y)=f_X(-y)$ hence, as you guessed,
$$
\mu_y(\mathrm dx)=\frac12(\delta_y(\mathrm dx)+\delta_{-y}(\mathrm dx)).
$$

A shortcut in the symmetric case is to note that $(X,|X|)$ and $(-X,|X|)$ coincide in distribution, hence
$$
\mathbb E(u(X)\mid |X|)=\mathbb E(u(-X)\mid |X|)=\mathbb E(L_u(|X|)\mid |X|)=L_u(|X|),
$$
where $L_u(y)=\frac12(u(y)+u(-y))$ for $y\gt0$. This yields directly that the measure $\mu_y$ is uniform on $\{-y,y\}$.
A: Let $f_X(x)$ be the density function.  The event that $|X|=x$ is the same as $X=\text{either }x\text{ or }-x$.  So you have
$$
\Pr(X=x\mid |X|=x)=\frac{f_X(x)}{f_X(x)+f_X(-x)},
$$
and $\Pr(X=-x\mid |X|=x)$ is complementary to that.
But you say it's symmetric about $0$, so that probability is $1/2$.
