Let $\xi$ be a random variable with the median $m_ξ$. Is it true that $m_{\varepsilon\xi} = \varepsilon m_\xi, \forall \varepsilon \in \mathbb{R}$?
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@StefanHansen if not then a counter-example might be helpful – Henry Feb 24 '20 at 08:32
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@StefanHansen I disagree (see my answer). – drhab Feb 24 '20 at 09:10
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In general $m_{\xi}$ serves as median of $\xi$ iff: $$P\left(\xi\leq m_{\xi}\right)\geq0.5\wedge P\left(\xi\geq m_{\xi}\right)\geq0.5\tag1$$
Sidenote: a median is not necessarily unique. See also this question and its answer.
So the question can rephrased as: does $(1)$ imply that: $$P\left(\epsilon\xi\leq\epsilon m_{\xi}\right)\geq0.5\wedge P\left(\epsilon\xi\geq\epsilon m_{\xi}\right)\geq0.5\text{ for every }\epsilon\in\mathbb{R}$$?
The answer on this is: yes (also if $\epsilon\leq0$).
drhab
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1If $\epsilon>0$ then $\epsilon\xi\leq\epsilon m_{\xi}\iff\xi\leq m_{\xi}$ and $\epsilon\xi\geq\epsilon m_{\xi}\iff\xi\geq m_{\xi}$ (just divide by $\epsilon$) so we are dealing with the same events. If $\epsilon<0$ then $\epsilon\xi\leq\epsilon m_{\xi}\iff\xi\geq m_{\xi}$ and $\epsilon\xi\geq\epsilon m_{\xi}\iff\xi\leq m_{\xi}$ (again divide by $\epsilon$ but switch $\geq$ into $\leq$ and vice versa) and again we are dealing with the same events. If $\epsilon=0$ then $P(0\leq0)=P(0\geq0)=1\geq0.5$ where $\epsilon\xi$ is degenerated at $0$ and has $0$ as median. – drhab Feb 24 '20 at 19:19