Global minimum is attained at the median

Question

I have the following homework problem.

Let $(x_1,x_2,\ldots,x_n)$ be an increasing sample of size $n$. Show that the function $$f: x\mapsto \sum_{k=1}^n |x_k-x|$$ obtains a global minimum at the median $$x_{1/2}=\begin{cases} x_{(n+1)/2},\qquad & n \text{ odd} \\ 1/2 \left(x_{n/2}+x_{n/2+1}\right),\qquad &n \text{ even}\end{cases}.$$

My attempt

Firstly, since $f$ is not differentiable at $x_1,\ldots,x_n$, so these are all critical points. Also, $$f'(x)=2k-n,\qquad x\in (x_k,x_{k+1}), k=0,1,\ldots,n.$$ Thus, if $n$ is even, then all the points in the interval $(x_{n/2}, x_{n/2+1})$ are also critical points.

In either case, $x_{1/2}$ is a critical point of $f$. Moreover, $f'$ changes sign from negative to positive at $x_{1/2}$. Thus, $x_{1/2}$ is a local minimum of $f$.

How do I show it is also the global minimum?

Your cost function $f(x)$ is a weakly decreasing function of $x$ when $x$ is below the median and weakly increasing when $x$ is above the median, so the median is a global minimum — Henry, Jan 18 '23 at 14:59
Please clarify what How do I should it is also the global minimum? means. — Rodrigo de Azevedo, Jan 18 '23 at 16:35

score 0 · Answer 1 · answered Jan 18 '23 at 13:45

0

Since $f'(x) = 0$ for all $x \in (x_{n/2},x_{n/2+1})$, $f$ must be constant in that region. That gives you enough information for comparing the critical points. Hope that helps

answered Jan 18 '23 at 13:45

StiftungWarentest

877

Global minimum is attained at the median

1 Answers1