I'm trying to learn about the sum of $ii$ random variables, conditional on a few of them being larger than the others.
Suppose I have four random variables, all drawn iid from the same absolutely continuous distribution $F$. Call them: $\theta^{A}_{1}, \theta^{A}_{2}, \theta^{B}_{1}$ and $\theta^{B}_{2}$.
I'm trying to derive the probability the sum of the 1st and 3rd are greater than the sum of the 2nd/4th, conditional on 1 being greater than two. That is:
$$\Pr(\theta^{A}_{1}+\theta^{B}_{2}>\theta^{A}_{2}+\theta^{B}_{1} | \theta^{A}_{1}>\theta^{A}_{2})$$
Everything in the summations is $iid$, although of course it's not $iid$ once we condition on $\theta^{A}_{1}>\theta^{A}_{2}$.
I'm not sure how to proceed, including what general toolkit to use. I could use Bayes theorem, but this seems like it would create another conditional probability in the numerator.
Let it be:
$A=\Pr(\theta^{A}_{1}+\theta^{B}_{2}>\theta^{A}_{2}+\theta^{B}_{1})$
$B=\Pr(\theta^{A}_{1}+\theta^{B}_{2}>\theta^{A}_{2}+\theta^{B}_{1}\cap\theta^{A}_{1}>\theta^{A}_{2})$
$C=\Pr(\theta^{A}_{1}+\theta^{B}_{2}>\theta^{A}_{2}+\theta^{B}_{1}\cap\theta^{B}_{2}>\theta^{B}_{1})$
$D=\Pr(\theta^{A}_{1}+\theta^{B}_{2}>\theta^{A}_{2}+\theta^{B}_{1}\cap\theta^{A}_{1}>\theta^{A}_{2}\cap\theta^{B}_{2}>\theta^{B}_{1})$
Evidently:
$A=B+C-D,\quad B=C,\quad A=1/2,\quad D=1/4$
Then:
$B=C=3/8$
And:
$\Pr(\theta^{A}_{1}+\theta^{B}_{2}>\theta^{A}_{2}+\theta^{B}_{1}\mid\theta^{A}_{1}>\theta^{A}_{2})=\frac{3/8}{1/2}=3/4$