Question:
Suppose we have three uniform $[0,1]$ random variables $X,Y,Z$. What is $P(|X-Y|\leq r_1 | |X-Z| = r_2)$?
Things I know already / have tried:
I have that the CDF of $|X-Y|$ is $P(|X-Y|<r) = 2r-r^2$.
Also, $$ P(|X-Y|=r | X=x) = \begin{cases} 2 & 0\leq r \leq \max\{x, 1-x\} \\ 1 & x < r \leq 1-max\{x, 1-x\} \end{cases} $$ Would the CDF I'm looking for be the CDF of the above PDF? My reasoning is that the $|X-Y|$ is independent of $Z$, so conditioning on $|X-Z|$ would be the same as conditioning on $X$, but I'm not sure if this is true, because conditioning on $X$ is not the same as conditioning on $|X-Z|$, since there are infinitely many possible values of $X$ and $Z$ for a given $|X-Z|$. Other than this, I'm not really sure where to go next.
Any help would be greatly appreciated, Thanks :)
The trick to use here should be this one: conditionally on $X$, the two random variables $|X-Y|$ and $|X-Z|$ are iid.
Consider the notation $|X-Y|=\overline{XY}$.
Let's face the problem in terms of densities. Using the law of total probability in both numerator and denominator $$f_{\overline{XY}\vert\overline{XZ}=r_2}(r_1\vert r_2) = \frac{f_{\overline{XY},\,\overline{XZ}}(r_1, r_2)}{f_{\overline{XZ}}(r_2)} = \frac{\int{f_{\overline{XY},\,\overline{XZ}\vert{X}=u}(r_1, r_2|u)}f_X(u)\,du}{\int{f_{\overline{XZ}\vert X=u}(r_2\vert u)f_X(u)\,du}} = \frac{\int_0^1{f_{\overline{XY},\,\overline{XZ}\vert{X}=u}(r_1, r_2|u)}\,du}{\int_0^1{f_{\overline{XZ}\vert X=u}(r_2\vert u)\,du}}$$ Now we use the fact that, conditionally on $X$, the two random variables $\overline{XY}$ and $\overline{XZ}$ are iid. Thus $$ \dots = \frac{\int_0^1{f_{\overline{XY}\vert{X}=u}(r_1|u)f_{\overline{XZ}\vert{X}=u}(r_2|u)}\,du}{\int_0^1{f_{\overline{XZ}\vert X=u}(r_2\vert u)\,du}} = \frac{\int_0^1{g(r_1,u)\,g(r_2,u)\,du}}{\int_0^1{g(r_2,u)\,du}},$$ where $g(r,u)=f_{\overline{XY}\vert X=u}(r\vert u)$. With a bit of computations we get these two formulation for $g$ $$g(r,u) = \mathcal{I}_{(0,u)}(r) + \mathcal{I}_{(0,1-u)}(r) = \mathcal{I}_{(r,1)}(u) + \mathcal{I}_{(0,1-r)}(u)$$ Then the integral in the denominator is easy: $${\int_0^1{g(r_2,u)\,du}} = \int_0^1{\bigl[\mathcal{I}_{(r_2,1)}(u) + \mathcal{I}_{(0,1-r_2)}(u)\bigr]\,du} = \int_{r_2}^1 du + \int_0^{1-r_2}du = 2-2r_2$$ The integral in the numerator is trickier, but I think we'll figure it out from here.