Assuming $\mu$ is the median of random variable $\xi$, prove, that,
$$\mathbb{E}~|\xi - \mu| \leq \mathbb{E}~|\xi - \mathbb{E}\xi| $$
The case when $\mu = \mathbb{E}\xi $ is trivial, but I have no idea what to do when they are not equal.
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1Maybe it will help: http://math.stackexchange.com/questions/1180801/the-minimum-of-the-sum – Mesmerized student Oct 04 '15 at 15:05
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Actually, $\mathbb{E}~|\xi - \mu| \leq \mathbb{E}~|\xi - x| $ for every $x$, as differentiating the RHS shows. – Did Oct 04 '15 at 15:11
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Thanks for posting such a useful link. It helped me very much – xolodec Oct 04 '15 at 15:40
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(I'm going to use $X$ instead of $\xi$, just to make it easier to write and read.) Look at $ g(m):=\mathbb{E}\lvert X-m \rvert$. If the distribution function of $X$ is $F$, this is $$ \int_{X \leqslant m} (m-x) \, dF(x) + \int_{X>m} (x-m) \, dF(x). $$
Integrate by parts: $$ \begin{align} \int_{X \leqslant m} (m-x) \, dF(x) &= \left[ (m-x)F(x) \right]_{-\infty}^m+\int_{X \leqslant m} F(x) \, dx \\ &= \int_{X \leqslant m} F(x) \, dx \\ \int_{X>m} (x-m) \, dF(x) &= [-(x-m)(1-F(x))]_{m}^{\infty}+\int_{X > m} (1-F(x)) \, dx \\ &= \int_{X > m} (1-F(x)) \, dx \end{align} $$
Differentiating with respect to $m$ gives $$ g'(m) = 2F(m)-1. $$ This is
- Zero when $F(m)=1/2$,
- Nondecreasing in $m$.
It follows that $g(m)$ has a minimum exactly when $F(m)=1/2$, which is the definition of the median. Hence $g(\mu) \leqslant g(m)$ for any other $m$, including $\mathbb{E}X$, which is exactly the inequality you want.
Chappers
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@Chappers: I think a little bit more care is needed here. It is not given in the question that the CDF of $X$ is continuous so it may happen that $\frac{1}{2}$ is not in the image of CDF $F$. – MerylStreep Oct 05 '15 at 01:23
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@MerylStreep I'm not sure how one would define the median, were that not the case. – Chappers Oct 05 '15 at 01:29
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@Chappers: Usually the median is defined as $\inf{x: F(x) \geq \frac{1}{2}}$. – MerylStreep Oct 05 '15 at 01:30
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I think fundamentally the point is that $g'(m)$ is nondecreasing, so $g$ is convex. Therefore, it has a unique minimum, since $g'(\pm \infty)=\pm 1$. – Chappers Oct 05 '15 at 01:42
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