Let two independent random variables $X$ and $Y$ have the continuous distribution functions which are symmetrical with equal medians $\texttt{med}(X) = \texttt{med}(Y) = m$. Prove that
$$\Pr(X<Y) = \Pr(X>Y) = \tfrac{1}{2}$$
Some thoughts: without loss of generality, we assume $m=0$. Because it's continuous distributed, either $\ge,\le$ or $>,<$ does not matter. Now we have two independent, symmetrically distributed continuous random variables, which means for any $\delta$:
$\begin{eqnarray} \Pr(X<-\delta) &= \Pr(X>\delta), \\ \Pr(X>-\delta) &= \Pr(X<\delta), \\ \Pr(X< \delta) + \Pr(X>\delta) &= 1, \\ \Pr(X<-\delta) + \Pr(X>-\delta) &= 1 \end{eqnarray}$
We also know $$\Pr(X<Y) + \Pr(X>Y) =1$$ and we only need to show $$\Pr(X<Y) = \Pr(X>Y)$$.
This approach seems stuck. If I took the integration approach described in this post Probability that a Random Variable is less than or equal to another, I will end up with
$$\begin{align} \Pr(X<Y) = E_Y[F_X(y)],\\ \Pr(Y<X) = E_X[F_Y(x)] \end{align}$$
It would work if $X$ and $Y$ shared the same distribution functions. The condition specifies that $X$ and $Y$ do NOT share the same distribution necessarily. This direction also goes no further.
as This post Does equal probability of rank imply two random variables symmetric around median? pointed out that either symmetric or identical distributions for both $X$ and $Y$ are necessary for the claimed statement to be true.
Any insights are welcome, and appreciate any help in advance.
