# The Median Minimizes the Sum of Absolute Deviations (The ${L}_{1}$ Norm)

Suppose we have a set $S$ of real numbers. Show that $$\sum_{s\in S}|s-x|$$ is minimal if $x$ is equal to the median.

This is a sample exam question of one of the exams that I need to take and I don't know how to proceed.

• Please replace THE median by ANY median. – Did Feb 25 '12 at 18:47

Introduction: The solution below is essentially the same as the solution given by Brian M. Scott, but it will take a lot longer. You are expected to assume that $S$ is a finite set. with say $k$ elements. Line them up in order, as $s_1<s_2<\cdots <s_k$.

The situation is a little different when $k$ is odd than when $k$ is even. In particular, if $k$ is even there are (depending on the exact definition of median) many medians. We tell the story first for $k$ odd.
Recall that $|x-s_i|$ is the distance between $x$ and $s_i$, so we are trying to minimize the sum of the distances. For example, we have $k$ people who live at various points on the $x$-axis. We want to find the point(s) $x$ such that the sum of the travel distances of the $k$ people to $x$ is a minimum.

The story: Imagine that the $s_i$ are points on the $x$-axis. For clarity, take $k=7$. Start from well to the left of all the $s_i$, and take a tiny step, say of length $\epsilon$, to the right. Then you have gotten $\epsilon$ closer to every one of the $s_i$, so the sum of the distances has decreased by $7\epsilon$.

Keep taking tiny steps to the right, each time getting a decrease of $7\epsilon$. This continues until you hit $s_1$. If you now take a tiny step to the right, then your distance from $s_1$ increases by $\epsilon$, and your distance from each of the remaining $s_i$ decreases by $\epsilon$. What has happened to the sum of the distances? There is a decrease of $6\epsilon$, and an increase of $\epsilon$, for a net decrease of $5\epsilon$ in the sum.

This continues until you hit $s_2$. Now, when you take a tiny step to the right, your distance from each of $s_1$ and $s_2$ increases by $\epsilon$, and your distance from each of the five others decreases by $\epsilon$, for a
net decrease of $3\epsilon$.

This continues until you hit $s_3$. The next tiny step gives an increase of $3\epsilon$, and a decrease of $4\epsilon$, for a net decrease of $\epsilon$.

This continues until you hit $s_4$. The next little step brings a total increase of $4\epsilon$, and a total decrease of $3\epsilon$, for an increase of $\epsilon$. Things get even worse when you travel further to the right. So the minimum sum of distances is reached at $s_4$, the median.

The situation is quite similar if $k$ is even, say $k=6$. As you travel to the right, there is a net decrease at every step, until you hit $s_3$. When you are between $s_3$ and $s_4$, a tiny step of $\epsilon$ increases your distance from each of $s_1$, $s_2$, and $s_3$ by $\epsilon$. But it decreases your distance from each of the three others, for no net gain. Thus any $x$ in the interval from $s_3$ to $s_4$, including the endpoints, minimizes the sum of the distances. In the even case, I prefer to say that any point between the two "middle" points is a median. So the conclusion is that the points that minimize the sum are the medians. But some people prefer to define the median in the even case to be the average of the two "middle" points. Then the median does minimize the sum of the distances, but some other points also do.

We're basically after: $$\arg \min_{x} \sum_{i = 1}^{N} \left| {s}_{i} - x \right|$$

One should notice that $\frac{\mathrm{d} \left | x \right | }{\mathrm{d} x} = \operatorname{sign} \left( x \right)$ (Being more rigorous would say it is a Sub Gradient of the non smooth ${L}_{1}$ Norm function).
Hence, deriving the sum above yields $\sum_{i = 1}^{N} \operatorname{sign} \left( {s}_{i} - x \right)$.
This equals to zero only when the number of positive items equals the number of negative which happens when $x = \operatorname{median} \left\{ {s}_{1}, {s}_{2}, \cdots, {s}_{N} \right\}$.

One should notice that the median of a discrete group is not uniquely defined.
Moreover, it is not necessarily an item within the group.

• Using derivatives here is overkill; the problem can be done by more elementary methods. $\qquad$ – Michael Hardy Oct 28 '16 at 17:06
• @MichaelHardy actually among all answers I find this one to be the simplest, rather than the "walk to the left and then to the right of the real line". – gented Feb 21 '19 at 23:21
• To the point. :) – dksahuji Apr 12 '19 at 8:26
• The derivative of |x| is x/|x| as proven here : math.stackexchange.com/questions/83861/… – clyton dantis Apr 15 '19 at 3:07
• @clytondantis, The function $\frac{x}{ \left| x \right| }$ is one of the definitions of $\operatorname{sign} \left( \cdot \right)$. – Royi Apr 15 '19 at 4:04

Suppose that the set $$S$$ has $$n$$ elements, $$s_1. If $$x, then $$f(x)=\sum_{s\in S}|s-x|=\sum_{s\in S}(s-x)=\sum_{k=1}^n(s_k-x)\;.\tag{1}$$ As $$x$$ increases, each term of $$(1)$$ decreases until $$x$$ reaches $$s_1$$, therefore $$f(s_1) for all $$x.

Now suppose that $$s_k\le x\le x+d\le s_{k+1}$$. Then

\begin{align*}f(x+d)&=\sum_{i=1}^k\Big(x+d-s_i\Big)+\sum_{i=k+1}^n\Big(s_i-(x+d)\Big)\\ &=dk+\sum_{i=1}^k(x-s_i)-d(n-k)+\sum_{i=k+1}^n(s_i-x)\\ &=d(2k-n)+\sum_{i=1}^k(x-s_i)+\sum_{i=k+1}^n(s_i-x)\\ &=d(2k-n)+f(x)\;, \end{align*}

so $$f(x+d)-f(x)=d(2k-n)$$. This is negative if $$2k, zero if $$2k=n$$, and positive if $$2k>n$$. Thus, on the interval $$[s_k,s_{k+1}]$$

$$f(x)\text{ is }\begin{cases} \text{decreasing},&\text{if }2kn\;. \end{cases}$$

From here it shouldn’t be too hard to show that $$f(x)$$ is minimal when $$x$$ is the median of $$S$$.

• But a small typo. In "As x increases, each term of (1) decreases until x reaches x_1, so" You intended to say s_1 instead of x_1. – Neo M Hacker Sep 28 '17 at 3:23

You want the median of $n$ numbers. Say $x$ is bigger than $12$ of them and smaller than $8$ of them (so $n=20$). If $x$ increases, it's getting closer to $8$ of the numbers and farther from $12$ of them, so the sum of the distances gets greater. And if $x$ decreases, it's getting closer to $12$ of them and farther from $8$ of them, so the sum of the distances gets smaller.

A similar thing happens if $x$ is smaller than more of the $n$ numbers than $x$ is bigger than.

But if $x$ is smaller than $10$ of them and bigger than $10$ of them, then when $x$ moves, it's getting farther from $10$ of them and closer to just as many of them, so the sum of the distances is not changing.

So the sum of the distances is smallest when the number of data points less than $x$ is the same as the number of data points bigger than $x$.

Starting with $$f(x)=\sum_{i=1}^n |s_i-x|$$

Assume we rearranged our terms such that $s_1<s_2<\cdots<s_n$

We first proceed by making the following observation $$\sum_{i=1}^n |s_i-x| = \sum_{i=2}^{n-1} |s_i-x| +(s_n -s_1) \quad \text{when} \quad x \in [s_1,s_n]$$

Now suppose that $n$ is odd, then by applying the above identity repeatedly we get $$f(x)=\sum_{i=1}^n |s_i-x|=|s_{\frac{n+1}2}-x|+(s_n -s_1)+(s_{n-1}-s_2)+\cdots+(s_{\frac{n+3}2}-s_{\frac{n-1}2})$$ or in other words $$f(x)=|s_{\frac{n+1}{2}}-x|+\text{constant}$$

This is just the absolute value function with its vertex being at $(s_{\frac{n+1}{2}},\text{constant})$, the minimum of the absolute value function occurs at its vertex, therefore $s_{\frac{n+1}{2}}$(median) minmizes $f(x)$.

Now suppose $n$ is even, again by using our identity, we have $$f(x)=\sum_{i=1}^n |s_i-x|=|s_{\frac{n}2}-x|+|s_{\frac{n+2}2}-x| + \text{constant}$$

Where the minimum occurs at $f'(x)=0$(or when not defined), therefore by differentiating and setting $f'(x)$ to zero we get $$\dfrac{|s_{\frac{n}{2}}-x|}{s_{\frac{n}{2}}-x}+\dfrac{|s_{\frac{n+2}{2}}-x|}{s_{\frac{n+2}{2}}-x}=0$$

Observe that $s:=\dfrac{s_{\frac{n+2}{2}}+s_{\frac{n}{2}}}{2}$(median) satisfies the above equation, since $s$ is halfway between $s_{\frac{n}{2}}$ and $s_{\frac{n+2}{2}}$ $$s_{\frac{n}{2}}-s=-(s_{\frac{n+2}{2}}-s)$$ that is by setting $x=s$ we get $$\dfrac{|s_{\frac{n}{2}}-s|}{s_{\frac{n}{2}}-s}+\dfrac{|s_{\frac{n}{2}}-s|}{-(s_{\frac{n}{2}}-s)}=0$$

Therefore $s$ is a minimum.

• I think some theory about minimum being where $f'(x) = 0$ for non differentiable functions is needed here – Guerlando OCs Aug 27 '18 at 0:56

Consider two $$x_i$$'s $$x_1$$ and $$x_2$$,

For $$x_1\leq a\leq x_2$$, $$\sum_{i=1}^{2}|x_i-a|=|x_1-a|+|x_2-a|=a-x_1+x_2-a=x_2-x_1$$

For $$a\lt x_1$$, $$\sum_{i=1}^{2}|x_i-a|=x_1-a+x_2-a=x_1+x_2-2a\gt x_1+x_2-2x_1=x_2-x_1$$

For $$a\gt x_2$$,$$\sum_{i=1}^{2}|x_i-a|=-x_1+a-x_2+a=-x_1-x_2+2a\gt -x_1-x_2+2x_2=x_2-x_1$$

$$\implies$$for any two $$x_i$$'s the sum of the absolute values of the deviations is minimum when $$x_1\leq a\leq x_2$$ or $$a\in[x_1,x_2]$$.

When $$n$$ is odd, $$\sum_{i=1}^n|x_i-a|=|x_1-a|+|x_2-a|+\cdots+\left|x_{\tfrac{n-1}{2}}-a\right| + \left|x_{\tfrac{n+1}{2}}-a\right|+\left|x_{\tfrac{n+3}{2}}-a|+\cdots+|x_{n-1}-a\right|+|x_n-a|$$ consider the intervals $$[x_1,x_n], [x_2,x_{n-1}], [x_3,x_{n-2}], \ldots, \left[x_{\tfrac{n-1}{2}}, x_{\tfrac{n+3}{2}}\right]$$. If $$a$$ is a member of all these intervals. i.e, $$\left[x_{\tfrac{n-1}{2}},x_{\tfrac{n+3}{2}}\right],$$

using the above theorem, we can say that all the terms in the sum except $$\left|x_{\tfrac{n+1}{2}}-a\right|$$ are minimized. So $$\sum_{i=1}^n|x_i-a|=(x_n-x_1)+(x_{n-1}-x_2)+(x_{n-2}-x_3)+\cdots + \left(x_{\tfrac{n+3}{2}}-x_{\tfrac{n-1}{2}}\right) + \left|x_{\tfrac{n+1}{2}}-a\right| = \left|x_{\tfrac{n+1}{2}}-a \right|+\text{costant}$$ Now since the derivative of modulus function is signum function, $$f'(a)=\operatorname{sgn}\left(x_{\tfrac{n+1}{2}}-a\right)=0$$ for $$a=x_{\tfrac{n+1}{2}}=\text{Median}$$

$$\implies$$ When $$n$$ is odd,the median minimizes the sum of absolute values of the deviations.

When $$n$$ is even, $$\sum_{i=1}^n|x_i-a|=|x_1-a|+|x_2-a|+\cdots+|x_{\tfrac{n}{2}}-a|+|x_{\tfrac{n}{2}+1}-a|+\cdots+|x_{n-1}-a|+|x_n-a|\\$$ If $$a$$ is a member of all the intervals $$[x_1,x_n], [x_2,x_{n-1}], [x_3,x_{n-2}], \ldots, \left[x_{\tfrac{n}{2}},x_{\tfrac{n}{2}+1}\right]$$, i.e, $$a\in\left[x_{\tfrac{n}{2}},x_{\tfrac{n}{2}+1}\right]$$,

$$\sum_{i=1}^n|x_i-a|=(x_n-x_1)+(x_{n-1}-x_2)+(x_{n-2}-x_3)+\cdots + \left(x_{\tfrac{n}{2}+1}-x_{\tfrac{n}{2}}\right)$$

$$\implies$$ When $$n$$ is even, any number in the interval $$[x_{\tfrac{n}{2}},x_{\tfrac{n}{2}+1}]$$, i.e, including the median, minimizes the sum of absolute values of the deviations. For example consider the series:$$2, 4, 5, 10$$, median, $$M=4.5$$.

$$\sum_{i=1}^4|x_i-M|=2.5+0.5+0.5+5.5=9$$ If you take any other value in the interval $$\left[x_{\tfrac{n}{2}},x_{\tfrac{n}{2} + 1} \right] =[4,5]$$, say $$4.1$$ $$\sum_{i=1}^4|x_i-4.1|=2.1+0.1+0.9+5.9=9$$ For any value outside the interval $$\left[x_{\tfrac{n}{2}},x_{\tfrac{n}{2}+1}\right]=[4,5]$$, say $$5.2$$ $$\sum_{i=1}^4|x_i-5.2|=3.2+1.2+0.2+4.8=9.4$$

Consider two real numbers $a<b$. Then the objective becomes

$$dist(a,b) = |x-a|+|x-b|$$

This expression is minimum when $a\leq x \leq b$. It can be proved by calculating the objective on 3 cases ($x<a, a\leq x\leq b, x>b$).

Now consider the general case where $S$ has $n$ elements. Sort them in increasing order as $S_1, S_2, \ldots, S_n$.

Pair the smallest and the largest numbers. As explained above, $dist(S_1,S_n)$ is minimum when $S_1\leq x\leq S_n$. Remove these two elements from the list and continue this procedure until there is at most one element left in the set.

If there is an element $S_i$ left, then $x=S_i$ minimizes $dist(x-S_i)$. It also lies between all the pairs.

In the case of even elements, finally the sequence will be empty. As in the case above, median lies between all the pairs.

Suppose $S$ is finite (with cardinal $s$), without repetitions, and ordered. Then the sum of absolute values is continuous (sum of continuous functions), and piecewise linear (hence differentiable), with left-most slope $-s$. By induction, the slope increases by 2 for each interval from left to right, with right-most slope $+s$. Hence the piece-wise slope first reaches either $-1$ or $0$ at index $\left\lfloor \frac{s+1}{2}\right\rfloor$, and $0$ or $+1$ at index $\left\lceil \frac{s+1}{2}\right\rceil$.

Hence the function attains its minima in the interval $\left[\left\lfloor \frac{s+1}{2}\right\rfloor, \left\lceil \frac{s+1}{2}\right\rceil\right]$, which reduces to a singleton when $s$ is odd.

The notion of median for continuous functions is detailed in Sunny Garlang Noah, The Median of a Continuous Function, Real Anal. Exchange, 2007