Differentiability of the expectation of an arbitrary random variable

Question

I was reading an answer to a probability theory question (linked below), and I found a claim that I wasn't able to justify.

Let $X$ be an arbitrary random variable and let $c\in \mathbb{R}$, then we he have that the function $c\mapsto E[|X-c|]$ is differentiable almost everywhere, where

$E[|X-c|] = \int_{-\infty}^cP(X\le t) dt + \int_{c}^{\infty}P(X\ge t) dt$.

My guess is that this follows from the right continuity of the CDF, but I wasn't sure if we also needed left continuity to apply the fundamental theorem of calculus.

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

Answer 1

This is a somewhat pedantic technical detail, but this will not hold for arbitrary $X$-- you need $E[|X|] < \infty$ (the derivative does not exist if $E[|X|] = \infty$). In this case both integrands in the formula $$E[|X-c|] = \int_{-\infty}^cP(X\le t) dt + \int_{c}^{\infty}P(X\ge t) dt$$ are $L^1$ functions, and the a.e. differentiability follows most easily from the Lebesgue differentiation theorem.