We are given a random sample of size $n$ and let the corresponding order statistics are as follows: $$Y_1 \leq Y_2 \leq Y_3.......Y_{n-1}\leq Y_n$$
The distribution of the sample median for $n$ even is required, so if $n=2k$ the sample median is given as : $\dfrac{Y_k + Y_{k+1}}{2}$
Also, the joint distribution of $Y_k,Y_{k+1}$ is given as follows (only defined for $x<y)$: $$f_{Y_k , Y_{k+1}}(x,y)=\frac{(2k)!}{((k-1)!)^2}[F(x)(1-F(x))]^{k-1}f(x)f(y)$$
Let $U=\dfrac{Y_k + Y_{k+1}}{2}$, starting off as follows:
$$ \mathbb{P}[U\leq u] = \mathbb{P}\left[\frac{Y_k + Y_{k+1}}{2}\leq u\right] = \mathbb{P}\left[Y_k + Y_{k+1}\leq 2u\right] $$
Thus, we are required to integrate the function $f(x,y)$ over the given shaded region:
Thus, the integral looks like this: $$\mathbb{P}[U\leq u] = \int_{- \infty}^{u} \int_{x}^{2u-x}f(x,y)dydx.$$
Is that expression okay ?
Because if I proceed further I get the result as $0$, whereas if I don't consider the condition $x<y$ the result is non-zero, but from the definition of the joint distribution of order statistics that condition is necessary, isn't it ?