Median of U-Statistics This is an exercise that we had before in course of Mathematical Statistics, but it turned out to be very difficult for me to solve.

Using hint:
$$\text{Med}(X_1,...,X_m)=\left\{\begin{matrix}
X_{\frac{(m+1)}{2}} & \text{for} & m-odd\\ 
\frac{X_{\frac{(m)}{2}} + X_{\frac{(m+1)}{2}}}{2}& \text{for} & m-even 
\end{matrix}\right.$$
Any help how to continue would be helpful for me.
 A: nice question and a bit tricky indeed.
Instead of solving the problem directly for general $n$ and $m$, it might be good to fix some small values, say $n=4$ and $m=3$, and write out the U-statistic explicitly.
Then you'll see that you get a linear combination of the order statistics and hence you can collect terms to get the weights $\omega_i$.
Let's fix (for example) $n=4$ and $m=3$, then there are the $4 = \binom{4}{3}$ index combinations such that $1 \leq i_1 < i_2 < i_3 \leq 4$, namely $(i_1,i_2,i_3) \in \{(1,2,3),(1,2,4),(1,3,4),(2,3,4)\}$.
I'll introduce the index $k \in \{1,\ldots,\binom{4}{3}\}$ to count the index sets and denote by $Y^k$ the corresponding selections of random variables, i.e.,
\begin{align}
Y^1 &= (Y^1_1,Y^1_2,Y^1_3) = (X_1,X_2,X_3),\\
Y^2 &= (Y^2_1,Y^2_2,Y^2_3) = (X_1,X_2,X_4),\\
Y^3 &= (Y^3_1,Y^3_2,Y^3_3) = (X_1,X_3,X_4),\\
Y^4 &= (Y^4_1,Y^4_2,Y^4_3) = (X_2,X_3,X_4).
\end{align}
For the $U$-statistic we now have
\begin{align}
\check{g}(X_1,X_2,X_3,X_4) &= \binom{4}{3}^{-1}\sum_{1 \leq i_1 < i_2 < i_3 \leq 4} \mbox{Med}(X_{i_1},X_{i_2},X_{i3})\\
&= \frac{1}{4} \sum_{k=1}^4 \mbox{Med}(Y^k_{1},Y^k_{2},Y^k_{3})\\
&= \frac{1}{4} \sum_{k=1}^4 Y^k_{(2)}.
\end{align}
Now for every $k$ the order statistic $Y^k_{(2)}$ corresponds to some order statistic $X_{(j)}$ of $(X_1,X_2,X_3,X_4)$ simply because $Y^k$ is a subset of $(X_1,X_2,X_3,X_4)$.
Collecting terms together gives then indeed
\begin{align}
\check{g}(X_1,X_2,X_3,X_4) &= \sum_{i=1}^n \omega_i  X_{(i)}
\end{align}
for some $\omega_i \geq 0$.
The tricky part (in my opinion) is however to realize that (for example taking $k=1$) the second order statistic of $(X_1,X_2,X_3)$ is not always equal to the second order statistic of $(X_1,X_2,X_3,X_4)$, or put differently $Y^1_{(2)}$ is not necessarily $X_{(2)}$.
Hence figuring out a formula for $\omega_i$ is tricky (at least for me), but the representation in terms of an $L$-statistic makes sense.
I hope this helps to get an idea of how to deal with the $m$ even case and then also with general $n$ and $m$.
