If the median of $X$ is greater than the median of $Y$ then $\Bbb P(X>Y) \geq \frac14$

Question

The median value of a random variable $X$ is $m$ if $\Bbb P (X \geq m) \geq \frac12$ and $\Bbb P (X \leq m) \geq \frac12$. Let $X$ and $Y$ be independent real random variables. Show that if the median value of $X$ is larger than the median value of $Y$, then $\Bbb P(X>Y) \geq \frac14$.

Answer 1

Note $m_X$ and $m_Y$ the medians. One subset of the event $X>Y$ is $X>m_x>m_Y > Y$. By independance, this event is the reunion of $Y<m_Y$ and $X>m_X$. Now we have, since $m_X>m_Y$:

$$P(X>m_x>m_Y>Y) = P(X>m_X) P(Y<m_Y). $$

Both of the terms equals $1/2$.

Then, $$P(X>m_x>m_Y>Y) = 1/4.$$

Since $X>m_x>m_Y>Y$ is a subset of $X>Y$, you have the desire inequality.