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Given a finite set $S \subset \mathbb{Z}$, we can easily obtain its median $median(S)$. Given a set of sets $T = \{ S_1, S_2, \cdots, S_n \}$. where $S_i \cap S_j = \emptyset, \forall S_i, S_j \in T $

We can define two "median", $M_1 = median(\cup_{S \in T})$ and $M_2 = median(\{ median(S)|S \subset T \})$.

Would $M_1 == M_2$ ?

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  • $\begingroup$ How are you defining median? Is this the typical median from statistics where you can order the elements of the set? $\endgroup$ – Jack Neubecker Mar 22 at 3:20
  • $\begingroup$ yes, and i just find a counter-example :( $\endgroup$ – peng yu Mar 22 at 3:22
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The answer is No.

for example, $S_1 = (1, 2, 4)$, $S_2 = (3, 10, 11)$.

$M_1 = 3.5$, while $M_2 = 6$

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  • $\begingroup$ This would be more convincing if you also had an $S_3$ such as $(5, 20,23)$ to avoid the median of even numbers of terms issue $\endgroup$ – Henry Mar 22 at 3:33
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If the definition of median is as from statistics (no other definition is given), here is a counter example. Let $A = {0,1}$ and $B ={1, 10}$. Then the median of $A$ is $0.5$ and the median of $B$ is $5.5$; the median of these values is $3$. Whereas the median of the union is $1$.

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