Is there something similar to "mean of squares is greater than the square of mean" for medians?

Question

The mean statement is derived from the variance, but is there anything like that for medians and is there a good proof?

Answer 1

The inequality is true, as Especially Lime demonstrates.

In fact, the inequality is even an equality in many cases. If you apply any function $f$ on a list of numbers $x_1\le x_2\le \dots \le x_n,$ and that function is increasing (at least on the set of numbers you apply it to), then $f(x_1)\le f(x_2)\le \dots \le f(x_n),$ and thus $$ \begin{align} \min\{f(x_1),\dots,f(x_n)\} &= f(\min\{x_1,\dots,x_n\}), \\ \max\{f(x_1),\dots,f(x_n)\} &= f(\max\{x_1,\dots,x_n\}), \\ \text{median}\{f(x_1),\dots,f(x_n)\} &= f(\text{median}\{x_1,\dots,x_n\}) \quad (\text{if $n$ is odd}). \end{align}. $$

Answer 2

Yes, the median of the squares is greater than or equal to the square of the median. You can prove this by induction on the number of elements. It is trivially true for one element, and for two elements the median is the mean, so it is true by the mean inequality.

For $n$ elements where $n\geq3$, assume it's true for $n-2$ and remove the largest and smallest element. This does not change the median, or the square of the median. What does it do to the squares? One of the elements removed must have the largest square (the largest-magnitude element is either the top and positive or the bottom and negative). We also remove some other square. If the other square we remove was smaller than the median of the squares, we haven't changed the median of the squares (we've removed something larger and something smaller, so the middle hasn't moved). If the other square we remove was also larger than or equal to median of the squares, this may change which square is the median for the remaining numbers, but the new median of the squares can only be smaller. So the median of the squares of the original $n$ elements is at least the median of the squares of the remaining $n-2$ elements, which by assumption is at least the square of the median of the remaining numbers, which is equal to the square of the median of the original $n$ numbers.