Medians and means of $X, Y$ and $Z$, when it is known that $P(X > Y ) = P(Y > Z) = 2/3$. Suppose we have three random variables $X, Y$ and $Z$, it is known that $$P(X >
Y ) = P(Y > Z) = \dfrac23$$ Then,


*

*$P(Z > X) > 0.5$.

*$X-Z$ has positive mean.

*The median of $X$ is larger than the median of $Z$.

*The median of $Y$ is smaller than that of $X$.

*None of the above is correct.


I guess the answer is $2.$ since $X$ is greater than $Z$, either $X$ and $Z$ are both positive, both negative or $1$ positive $1$ negative, $X-Z$ is positive, is it a possible reason?
 A: Incomplete answer that excludes option2.
In this situation the mean of $X-Z$ is (if
it exists) not necessarily positive. Counterexample with independent discrete random variables:
$\begin{array}{c|cccc}
 & -60 & -3 & 0 & 3\\
\hline X & \frac{1}{3} &  &  & \frac{2}{3}\\
Y &  &  & 1\\
Z &  & \frac{2}{3} &  & \frac{1}{3}\end{array}$
Actually to exclude option2 it is allready enough to bring forward that the mean does not necessarily exists.
A: Counterexamples. Let 


*

*$P(X=-100)=\frac13$ and $P(X=4)=\frac23$ 

*$P(Y=1)=\frac13$ and $P(Y=3)=\frac23$

*$P(Z=1)=\frac13$ and $P(Z=2)=\frac23$


Obviously $$P(X>Y)=P(X=4)=\frac23 \quad  \text{and} \quad P(Y>Z)=P(Y=3)=\frac23$$ so these random variables do satisfy the given condition. But 


*

*Obviously $P(Z>X)=P(X=-100)=\frac13<0.5$ so 1. is false.

*$E[X-Z]=E[X]-E[Z]=-\frac{92}3-\frac53<0$ so 2. is false. 

*and 4. Now for the medians. The median of $X$, say $m_X$ is defined as a point such that $$P(X\le m_X)\le \frac12 \quad \text{and}\quad P(X\ge m_X)\ge \frac12$$ So, here things go really wrong, because there are many many medians for $X,Y$ and $Z$. For example, if we take $m_X=1$ then indeed $$P(X\le 1)=P(X=-100)=\frac13\le \frac12 \quad \text{and}\quad P(X\ge 1)=P(X=4)=\frac23 \ge \frac12 $$ so $m_X=1$ is a median for $X$. Repeat with any value between $-100$ and $4$ (included) to see that all of these values are medians for $X$. Similarly all values between $1$ and $3$ are medians for $Y$ and all values between $1$ and $2$ are medians for $Z$. So, choices $3.$ and $4.$ are inconclusive and therefore wrong. 


Go with 5.
