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The median of real-valued random variable $X$ is defined as a number: $P(X

Thank you for any help.

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    As stated, it is not true if $f$ can be negative: consider the constant function $f(x)=-1$. Did you intend $f: \mathbb{R} \to [0,\infty]$ rather then $f: [0,\infty] \to \mathbb{R}$ ? – 2017-09-22

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

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|) $

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    Thank you so much. But how we know that: $E(f(|X-b|)) = E(f(|X-b|)m \leq X)P(m \leq X) + E(f(|X-b|)X \leq m)P(X \leq m)$? – 2017-02-16
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    $E(f(|X-b|))=E(f(|X-b|) 1_{X \geq m})+E(f(|X-b|)1_{m \geq X})$ – 2017-02-16