Median of distribution

Question

The median of real-valued random variable $X$ is defined as a number: $P(X<m) \leq \frac{1}{2} \leq P(X \leq m)$. I need to prove that for every $b \in \mathbb{R}$ and non-decreasing function $f: [0,\infty] \to \mathbb{R}$ is: $f(|m-b|) \leq 2E(f(|X-b|))$.

Thank you for any help.

Answer 1

$E(f(|X-b|))=E(f(|X-b|) | m\leq X)P(m\leq X)+E(f(|X-b|)|X\leq m)P(X\leq m)$

Since $f$ is a non decreasing function, $E(f(|X-b|)|X\leq m) \geq 0$ this gives $E(f(|X-b|))\geq E(f(|X-b|) | m\leq X)P(m\leq X)$

From the definition of the median $P(m\leq X) \geq \frac{1}{2}$ so you get

$E(f(|X-b|))\geq \frac{1}{2} E(f(|X-b|) | m\leq X)$

Now we use the fact that f is non decreasing a second time to get $E(f(|X-b|) | m\leq X) \geq E(f(|m-b|) | m\leq X) = f(|m-b|) $

So you get $E(f(|X-b|))\geq \frac{1}{2} f(|m-b|) $