What can we conclude from the property $P(XY)=0.5$ of two random variables $X$ and $Y$? This is a similar question to Does the property $P(X<Y)=P(X>Y)=0.5$ imply equal expectations?
I have a series of short questions related to: the medians of $X$ and $Y$, the means, and the fact that $P(X>Y)=P(X<Y)=0.5$. Let us introduce the following properties:
\begin{align*}
P_{\mbox{prob}}: \quad P(X>Y)&=P(X<Y)=0.5,\\
P_{\mbox{mean}}: \hspace{0.9cm}\quad E[X]&=E[Y],\\
P_{\mbox{median}}: \quad med[X]&=med[Y].
\end{align*}
By the post in Does the property $P(X<Y)=P(X>Y)=0.5$ imply equal expectations? we know that $P_{\mbox{prob}} \nRightarrow P_{\mbox{mean}}$, and obviously we now that $P_{\mbox{mean}}$ and $P_{\mbox{median}}$ do not imply each other.
My questions are:

*

*$P_{\mbox{prob}} \Rightarrow P_{\mbox{median}}$?

*$P_{\mbox{mean}} \Rightarrow P_{\mbox{prob}}$?

*$P_{\mbox{median}} \Rightarrow P_{\mbox{prob}}$?

I am not sure if the continuous and discrete cases differ here. The continuous case is maybe more interesting.

I tried $X$ uniformly supported on $[a,b]$ and $Y$ on $[c,d]$ such that $a+b=c+d$ so that means are equal. But then I get $P(X<Y)=P(X>Y)$ which is then not a counterexample for question 2, or 1.
 A: All implications are wrong. I will here give counterexamples to each statement. All examples use continuous distributions.
Statement 1: Consider $X \sim \operatorname{Uniform}(0,10)$ and let $\delta$ be independent of $X$ with $\mathbb{P}(\delta = 1) = \frac{1}{2}=\mathbb{P}(\delta = 0)$. Define $Y$ as
$$Y= \begin{cases} X + 2 & \text{ if $\delta = 1$} \\
                     X-1 & \text{ if $\delta = 0$} \end{cases}.$$
Then we clearly have $\mathbb{P}(Y>X)=\mathbb{P}(Y<X)=\frac{1}{2}$ and $\operatorname{Median}(X)=5$, however using that the density of $Y$ must be
$$f_Y(y) = \frac{1}{2}(f_X(y+1)+f_X(y-2))= \begin{cases} \frac{1}{20} & y \in [-1,2] \cup (9,12] \\ \frac{1}{10} & y \in (2,9] \end{cases}$$
we can see that $\operatorname{Median}(Y)= 5.5 > 5=\operatorname{Median}(X)$.
Statement 2:
Consider $X \sim \operatorname{Uniform}(0,2)$ and $Y \sim \operatorname{Exponential}(1)$ to be independent. Then $\mathbb{E}[X] = 1 = \mathbb{E}[Y]$, however
$$P(Y>X) = \frac12 \int_0^2 P(Y>x) \: dx = \frac12 \int_0^2 e^{-x}dx = \frac{1-e^{-2}}{2} \neq \frac{1}{2}$$
Statement 3:
Consider $X \sim \operatorname{Uniform}(0,\log(4))$ and $Y \sim \operatorname{Exponential}(1)$ to be independent. Then $\operatorname{Median}[X] = \log(2) = \operatorname{Median}[Y]$, however
$$P(Y>X) = \frac{1}{\log(4)} \int_0^{\log(4)} P(Y>x) \: dx = \frac{1}{\log(4)} \int_0^{\log(4)} e^{-x}dx = \frac{1-4^{-1}}{\log(4)} \neq \frac{1}{2}$$
