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If a continuous random variable $X$ has a symmetric distribution around 0, what is the conditional distribution of $X$ given $|X|$?

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    $\begingroup$ For the intuition: Suppose I tell you $|X| = 3$. What can you conclude about the value of $X$, without considering anything about probability? Ok, now, what does symmetry buy you? $\endgroup$
    – cardinal
    Commented Oct 3, 2012 at 13:42
  • $\begingroup$ I guess the answer is $\frac{F_x(x)}{F_X(x)+ F_X(-x)}$, but I need a formal proof. $\endgroup$
    – ie86
    Commented Oct 3, 2012 at 14:06
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    $\begingroup$ Unfortunately, your conjectured answer is incorrect. You might play with a couple simple examples using my first comment as direction. To be better able to assist you, it would help if you could edit your question to include what you've thought about, where you are stuck and what level this problem is pitched at. (You say "formal proof", but that is hard to interpret. For example, we could give you a measure-theoretic answer to your question, but that may or may not be what you're looking for.) $\endgroup$
    – cardinal
    Commented Oct 3, 2012 at 14:26
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    $\begingroup$ A hint: what values can $X$ take given $|X| = c$? A second hint: what are the probabilities of those values? $\endgroup$
    – jbowman
    Commented Oct 3, 2012 at 16:59
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    $\begingroup$ Just to note that this question was also posted at MathSE. Link: math.stackexchange.com/questions/206586/… $\endgroup$
    – johnny
    Commented Oct 4, 2012 at 6:45

1 Answer 1

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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.

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