Inequality of difference of medians of two datasets.

Question

For $\{x_k\}_{k=1}^m$ and $\{y_k\}_{k=1}^m\in\mathbb{R}$ with $x_k-y_k=d_k$ define $x=\lim_{\delta\rightarrow 0}\left(\arg\min \frac{1}{m}\sum_{k=1}^m|x_k-x|+\delta x^2\right)$ and $y=\lim_{\delta\rightarrow 0}\left(\arg\min \frac{1}{m}\sum_{k=1}^m|y_k-y|+\delta y^2\right).$ Is there any way to get the inequality of this form $|x-y|\leq C|\frac1m\sum_{k=1}^md_k|?$ I have tried in several ways, but had problem with $|x_k-x|$ and $|y_k-y|$ terms. All I could come up with $|x-y|\leq \frac C m\sum_{k=1}^m|d_k|.$ Any help would be appreciated. Thanks in advance!

Answer 1

Unfortunately, there is no such way because otherwise if $\sum d_k=0$ then we should have $x=y$. This obviously fails, even in case $\sum x_k=\sum y_k=0$. For instance, put $k=4$, $x_1=x_2=1$, $x_3=x_4=-1$, $y_1=y_2=y_3=1$, $y_4=-3$. Then $x=0$, but $y=1$.