For a combined order statistics involving 2 sample, How would you find the probability (0<MedX<MedY)?

Question

Suppose that there are two samples $X_1$,$X_2$,...,$X_n$ and $Y_1$,$Y_2$,...,$Y_n$. An order Statistics is obtained by combining the two samples and arranging them in increasing order. Then how would you find the $$P(0<\text{Median of $X$ sample}<\text{Median of $Y$ Sample})?$$

Answer 1

we see on https://en.wikipedia.org/wiki/Order_statistic that an even number of i.i.d. random variables all with p.d.f. $f$ and c.d.f. $F$ has median with density

$$ f_{M}(x)={\frac {n}{2}}{n \choose n/2} [F(x)]^{n/2-1}[1-F(x)]^{n/2}f(x)$$

the two medians inherit independence from the variables they depend on and so

$$\mathbb{P}(0 < M_X < M_Y)=\int_{0}^{\infty} dy \int_{0}^{y} dx \ f_{M_X}(x) f_{M_Y}(y) $$