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.

**Answer**

**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.

**Attribution***Source : Link , Question Author : hattenn , Answer Author : Michael Hardy*