Krdytkyu

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
$$
min_{c in mathbb{R}} E(|X-c|) , ?
$$

For every real valued random variable $X$,
$$
mathrm E(|X-c|)=int_{-infty}^cmathrm P(Xleqslant t),mathrm dt+int_c^{+infty}mathrm P(Xgeqslant t),mathrm dt
$$
hence the function $u:cmapsto mathrm E(|X-c|)$ is differentiable almost everywhere and, where $u'(c)$ exists, $u'(c)=mathrm P(Xleqslant c)-mathrm P(Xgeqslant c)$. 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 $$(x-y)^+=int_y^{+infty}[tleqslant x],mathrm dt$$ and $|x-c|=((-x)-(-c))^++(x-c)^+$, which yield, for every $x$ and $c$,
$$
|x-c|=int_{-infty}^c[xleqslant t],mathrm dt+int_c^{+infty}[xgeqslant t],mathrm dt
$$

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 $$displaystyle int_{c}^{infty} (x-c) dP(x) = lim_{y rightarrow infty} (y-c) P(y) - displaystyle int_{c}^{infty} P(x) dx.$$

The following intends to complement Did's answer.

Claim

Denote by $M$ be the set of $X$'s medians. Then

1. $M = [m_1, m_2]$ for some $m_1, m_2 in mathbb{R}$, such that $m_1 leq m_2$.




2. For every $m in M$ and for every $x in mathbb{R}$ we have
   $$
   Eleft(|X-m|right) leq Eleft(|X-x|right).
   $$
   (In particular, $mmapsto Eleft(|X-m|right)$ is constant on $M$.)

Part 2's proof builds on Did's answer.

Proof

1. 
   

   It is known that $M neq emptyset$. Define
   $$
   begin{align}
   M_1 &:= left{tinmathbb{R} |!: F_X(t) geq frac{1}{2}right}, \
   M_2 &:= left{tinmathbb{R} |!: P(X<t) leq frac{1}{2}right}.
   end{align}
   $$
   Then $M = M_1 cap M_2$. It therefore suffices to show that $M_1 = [m_1, infty)$ and that $M_2 = (-infty, m_2]$, for some $m_1, m_2 in mathbb{R}$.

   
   
   

   Since $lim_{trightarrow-infty}F_X(t) = 0$, $M_1$ is bounded from below. Since $lim_{trightarrowinfty}F_X(t) = 1$, $M_1$ is an interval that extends to infinity. Hence $M_1 = (m_1,infty)$ or $M_1 = [m_1,infty)$, for some $m_1 in mathbb{R}$. It follows from $F_X$'s right-continuity that $m_1 in M_1$. An analogous argument shows that $M_2 = (-infty,m_2]$ (just verify that $tmapsto P(X<t)$ is left-continuous).

   
   


2. 
   

   Define a function $f:mathbb{R}rightarrowmathbb{R}$ as follows. For every $c in mathbb{R}$, set
   $$
   f(c) := Eleft(|X-c|right).
   $$

   
   
   

   We will begin by showing that $f$ is convex. Let $a, b in mathbb{R}$, and let $t in (0,1)$. Then
   $$
   begin{align}
   fleft(ta+(1-t)bright) &= Eleft(left|X-left(ta+(1-t)bright)right|right) \
   &= Eleft(left|left(tX-taright)+left((1-t)X-(1-t)bright)right|right) \
   &leq Eleft(left|left(tX-taright)right|+left|left((1-t)X-(1-t)bright)right|right) \
   &=Eleft(left|left(tX-taright)right|right)+Eleft(left|left((1-t)X-(1-t)bright)right|right) \
   &= t Eleft(|X-a|right) + (1-t) Eleft(|X-b|right) \
   &= t f(a) + (1-t) f(b).
   end{align}
   $$

   
   
   

   Since $f$ is convex, then, by Theorem 7.40 of [1] (p. 157), there exists a set $A subseteq mathbb{R}$ such that $mathbb{R}setminus A$ is countable, and such that $f$ is finitely differentiable on $A$. Moreover, letting $m in M$, and letting $x in (-infty, m_1)$, Theorem 7.43 of [1] (p. 158) yields that $f'$ is Lebesgue-integrable on $[x,m] cap A$, and that
   $$
   f(m) - f(x) = int_{[x,m]cap A} f' dlambda.
   $$

   
   
   

   Applying Did's answer, we find that $f'leq 0$ on $[x,m]cap A$. Hence $f(m) leq f(x)$. Similar considerations show that, for every $x in (m_2,infty)$, $f(m) leq f(x)$, and also that $f(m) = f(m_1)$ (implying that $f$ is constant on $M$, since $m$ was chosen arbitrarily in $M$).

   
   
   

   (The argument of the last paragraph was suggested to me by copper.hat in their answer to a related question of mine.)

   
   

Q.E.D.

References

[1] Richard L. Wheeden and Antoni Zygmund. Measure and Integral: An Introduction to Real Analysis. 2nd Ed. 2015. CRC Press. ISBN: 978-1-4987-0290-4.

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 $cge 0$. It is straightforward to check that $|X-c|-|X|=c$ when $Xle0$, and $|X-c|-|X|ge -c$ when $X>0$.

It follows that
$$
Eleft[(|X-c|-|X|)I(Xle0)right]=cP(Xle0)tag1
$$
and
$$Eleft[(|X-c|-|X|)I(X>0)right]ge-cP(X>0).tag2
$$
Adding (1) and (2) yields
$$
E(|X-c|-|X|)ge cleft[P(Xle0)-P(X>0)right].tag3
$$
The RHS of (3) equals $c[2P(Xle0)-1]$, which is non-negative since $cge0$ and zero is a median of $X$. The case $cle0$ is reduced to the previous one by considering $X':=-X$ and $c':=-c$.