Conditional distribution of $X$ given $|X|$ If a continuous random variable $X$ has a symmetric distribution around 0, what is the conditional distribution of $X$ given $|X|$? 
 A: Formally, your goal is to find, for each Borel set $A\in\mathscr{R}$, an integrable function $g_A$ (the conditional distribution of $X$ given $|X|$), whose value at $t$ we denote by
$$
  g_A(t)=:\mathrm{Pr}\left(X\in A\mid |X|=t\right) \, ,$$
satisfying
$$
  P(X\in A, |X|\in B) = \int_B g_A(t) \, d\mu_{|X|}(t) = \int_B \mathrm{Pr}\left(X\in A\mid |X|=t\right) \, d\mu_{|X|}(t) \, , \qquad (*)
$$
for every $B\in\mathscr{R}$, where $\mu_{|X|}$ is the distribution of $|X|$, defined by $\mu_{|X|}(B)=P\left( |X|\in B\right)$, and
$$
  \mathrm{Pr}\left(X\in {\small\bullet}\mid |X|=t\right)
$$
must be a probability measure over $(\mathbb{R},\mathscr{R})$ for almost every $t$ $[\mu_{|X|}]$ (if you can satisfy this for every $t$, then you have a regular version of the conditional distribution).
You may guess a candidate $g_A$ thinking as follows. 
If $t<0$, you don't have to worry: since $\mu_{|X|}$ puts zero mass on negative values, you can define $g_A$ arbitrarily, say, $g_A(t)=1/137.035999$.
If $t=0$, you know that $|X|=0$ if and only if $X=0$. So, it is reasonable to try $g_{\{0\}}(0)=1$.
If $t>0$, you know that $|X|=t$ if and only if $X=t$ or $X=-t$. Since the distribution of $X$ is symmetric around the origin, it is reasonable to guess that $g_{\{t\}}(t)=g_{\{-t\}}(t)=1/2$.
To write your formal proof, you have to show that this candidate $g_A$ satisfies the definition $(*)$.
Please, give it a try.
