Find the minimum value of $f(x)=\sum_{i=1}^n |x-a_i|$.

Question

Let $a_1 \lt a_2 \lt \dots \lt a_n$.

My guess is the minimum occurs at the middle point. However, I don't know how to show this since I can't use calculus here. What kinds of idea should I use? I would greatly appreciate any help.

See this answer for the specific case $n=4$; it generalizes easily, though you do have to distinguish odd $n$ from even $n$. — Brian M. Scott, Aug 18 '15 at 18:16
@BrianM.Scott I get the idea intuitively that the function is decreasing as we move to the middle then increases as we move further away. But I don't know how to show it mathematically... — nomadicmathematician, Aug 18 '15 at 18:24
@takecare: Use the idea in that answer: pair up the points $a_k$, starting at the ends, and notice that as you move right away from $a_k$, you move towards $a_{n+1-k}$ by the same amount, so that $|x-a_k|+|x-a_{n+1-k}|$ remains constant. — Brian M. Scott, Aug 18 '15 at 18:26
One way to see it is to look at the value of $\vert x - a_j \vert$ on each interval $[a_i,a_{i +1}]$ in order to get rid of the absolute values. — mathcounterexamples.net, Aug 18 '15 at 18:27

score 0 · Answer 1 · answered Aug 18 '15 at 18:43

Okay how is this answer?

Let $x$ be in any interval $[a_{j-1},a_j]$. Now let $y$ be in $[a_j,a_{j+1}]$, and $|x-a_j|=|y-a_j|$. Then $|y-a_i|=|x-a_i|+|y-x|$ if $i\le j-1$, and $|y-a_i|=|x-a_i|-|y-x|$ if $i\ge j-1$.

Hence, $f(y)=f(x)+|y-x|\{(j-1)+(n-j)\}=f(x)+|y-x|\{(2j-1)-n\}$.

Now it is clear that as $x$ moves to the middle $a_i$, $f$ decrease, then increases as it moves further away.

score 0 · Answer 2 · answered Aug 18 '15 at 18:54

Suppose $a_1<a_2<\dots<a_n$. Then on each interval $\;(a_k,a_{k+1})$,

$$ f(x)=\sum_{i=1}^k(a_i-x)+\sum_{i=k+1}^n(x-a_i)=(n-2k)x+\sum_{i=1}^ka_i-\sum_{i=k+1}^n a_i $$ Thus, $f(x)$ is decreasing on $\;(a_k,a_{k+1})$ as long as $n-2k>0$, i.e. $k<\dfrac n2$, increasing as soon as $k>\dfrac n2$.

We see there are two cases:

If $n$ is odd; $n=2p+1$, there is a unique minimum, attained at the middle point $a_{p+1}$, and its value is: $$f(a_{n+1})= -a_1-\dots-a_n+a_{n+2}+\dots+a_{2p+1}.$$
If $n$ is even: $n=2p$, the function attains its minimum on the middle interval $\;[a_p,a_{p+1}]$ and its value is: $$-a_1-\dots-a_p+a_{p+1}+\dots+a_{2p}.$$

Augustin · Answer 3 · 2015-08-18T19:32:57.267

Your intuition is good, the answer is the median of $(a_i)_{1\leq i\leq n}$.

Let $f_1(x)=\sum_{i=2}^{n-1}\vert x-a_i\vert$. Then you have, for all $x$:

$$f(x)=\vert x-a_1\vert + f_1(x)+\vert x - a_n\vert$$

Considering this equation, you can show that $f_1$ et $f$ have the same minimum, noticing that:

the minimum is not outside $[a_1,a_n]$
$\forall x\in\lbrack a_1,a_n\rbrack$, $f(x)=f_1(x)+(a_n-a_1)$

Repeating this, you can "forget" extremal points at each step. At the end you only have one or two points left. Then you just have to consider the two cases: $n$ even and $n$ odd.

score 0 · Answer 4 · answered Aug 18 '15 at 19:21

My guess is the minimum occurs at the middle point

The middle (median) interval. There are two median points if $n$ is even and $f$ is constant and minimum on the interval between them. If $n$ is odd the interval reduces to a point.

This was the starting point of Median Regression. Wikipedia attributes the idea to Boskovic in the 1700's, which would make it one of the first examples of regression methods.

Find the minimum value of $f(x)=\sum_{i=1}^n |x-a_i|$.

4 Answers4