Why does the median minimize $E(|X-c|)$?

Suppose $X$ is a real-valued random variable and let $P_X$ denote the distribution of $X$. Then $$ E(|X-c|) = \int_\mathbb{R} |x-c| dP_X(x). $$ The medians of $X$ are defined as any number $m \in \mathbb{R}$ such that $P(X \leq m) \geq \frac{1}{2}$ and $P(X \geq m) \geq \frac{1}{2}$.

Why do the medians solve $$ \min_{c \in \mathbb{R}} E(|X-c|) \, ? $$

Solution 1:

For every real valued random variable $X$, $$ \mathrm E(|X-c|)=\int_{-\infty}^c\mathrm P(X\leqslant t)\,\mathrm dt+\int_c^{+\infty}\mathrm P(X\geqslant t)\,\mathrm dt $$ hence the function $u:c\mapsto \mathrm E(|X-c|)$ is differentiable almost everywhere and, where $u'(c)$ exists, $u'(c)=\mathrm P(X\leqslant c)-\mathrm P(X\geqslant c)$. Hence $u'(c)\leqslant0$ if $c$ is smaller than every median, $u'(c)=0$ if $c$ is a median, and $u'(c)\geqslant0$ if $c$ is greater than every median.

The formula for $\mathrm E(|X-c|)$ is the integrated version of the relations $$(x-y)^+=\int_y^{+\infty}[t\leqslant x]\,\mathrm dt$$ and $|x-c|=((-x)-(-c))^++(x-c)^+$, which yield, for every $x$ and $c$, $$ |x-c|=\int_{-\infty}^c[x\leqslant t]\,\mathrm dt+\int_c^{+\infty}[x\geqslant t]\,\mathrm dt $$

Solution 2:

Let $f$ be the pdf and let $J(c) = E(|X-c|)$. We want to maximize $J(c)$. Note that $E(|X-c|) = \int_{\mathbb{R}} |x-c| f(x) dx = \int_{-\infty}^{c} (c-x) f(x) dx + \int_c^{\infty} (x-c) f(x) dx.$

To find the maximum, set $\frac{dJ}{dc} = 0$. Hence, we get that, $$\begin{align} \frac{dJ}{dc} & = (c-x)f(x) | _{x=c} + \int_{-\infty}^{c} f(x) dx + (x-c)f(x) | _{x=c} - \int_c^{\infty} f(x) dx\\ & = \int_{-\infty}^{c} f(x) dx - \int_c^{\infty} f(x) dx = 0 \end{align} $$

Hence, we get that $c$ is such that $$\int_{-\infty}^{c} f(x) dx = \int_c^{\infty} f(x) dx$$ i.e. $$P(X \leq c) = P(X > c).$$

However, we also know that $P(X \leq c) + P(X > c) = 1$. Hence, we get that $$P(X \leq c) = P(X > c) = \frac12.$$

EDIT

When $X$ doesn't have a density, all you need to do is to make use of integration by parts. We get that $$\displaystyle \int_{-\infty}^{c} (c-x) dP(x) = \lim_{y \rightarrow -\infty} (c-y) P(y) + \displaystyle \int_{c}^{\infty} P(x) dx.$$ Similarly, we also get that $$\displaystyle \int_{c}^{\infty} (x-c) dP(x) = \lim_{y \rightarrow \infty} (y-c) P(y) - \displaystyle \int_{c}^{\infty} P(x) dx.$$

Solution 3:

Let $m$ be any median of $X$. Wlog, we can take $m=0$ (consider $X':=X-m$). The aim is to show $E|X-c|\ge E|X|$.

Consider the case $c\ge 0$. It is straightforward to check that $|X-c|-|X|=c$ when $X\le0$, and $|X-c|-|X|\ge -c$ when $X>0$. It follows that $$ (|X-c|-|X|)\,I(X\le0)=c\,I(X\le0)\tag1 $$ and $$(|X-c|-|X|)\,I(X>0)\ge-c\,I(X>0).\tag2 $$ Adding (1) and (2) and taking expectation yields $$ E(|X-c|-|X|)\ge c\left[P(X\le0)-P(X>0)\right].\tag3 $$ The RHS of (3) equals $c\,[2P(X\le0)-1]$, which is non-negative since $c\ge0$ and zero is a median of $X$. The case $c\le0$ is reduced to the previous one by considering $X':=-X$ and $c':=-c$.