# Probability that the absolute value of one random variable is less than the absolute value of another

I'm having trouble wrapping my brain around this idea.

Suppose we have $X$ and $X'$ which are IID continuous random variables with conditional cdfs:

$F_{+}(x) = P(X \leq x | \epsilon = 1)$

$F_{-}(x) = P(-X \leq x | \epsilon = -1)$

Where $\epsilon = sign(X)$.

Suppose further that $P(\epsilon = 1) = p$.

The probability I am looking to compute is:

$P(X' < -X | \epsilon = 1 \wedge \epsilon' = -1)$

I want to know in terms of $F_+(x)$ and $F_{-}(x)$ (or possibly their derivatives) and p what this quantity might be. My instinct was to say 1/2 but I know that isn't necessarily right. I've basically reached a gap in understanding regarding the signed rank test statistic and understanding this quantity would really help to get me over the hump. Thanks a lot for any advice..

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The probability to be computed is $$\iint\mathbf 1_{0\leqslant x\lt y}\mathrm dF_+(x)\mathrm dF_-(y)=\int_0^{+\infty}F_+(y)\mathrm dF_-(y)=1-\int_0^{+\infty}F_-(x)\mathrm dF_+(x).$$