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Let $X$ be a real random variable. It is clear that any median $m\in\mathbb{R}$ of $X$ satisfies that $$\text{E}[|X - m|] = \min_{x\in\mathbb{R}}\text{E}[|X - x|]$$.

My question is the following. If $b\in\mathbb{R}$ is a number such that $\text{E}[|X - b|] = \min_{x\in\mathbb{R}}\text{E}[|X - x|]$, is $x$ a median of $X$?

Thanks in advance.

user13761697
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1 Answers1

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Yes, if you define a median as a value $m$ where $\mathbb P(X<m) \le \frac12$ and $\mathbb P(X\le m) \ge \frac12$. Your $b$ will have this property.

Henry
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  • And could you give a proof? I'm not able to see why it is connected the minimum of this function and the definition of the median. – user13761697 Dec 18 '20 at 11:14
  • https://math.stackexchange.com/questions/85448/why-does-the-median-minimize-ex-c has a proof and https://math.stackexchange.com/questions/85696/does-a-median-always-exist-for-a-random-variable shows there will be a median. Essentially any point with $\text{E}[|X - b|] = \min_{x\in\mathbb{R}}\text{E}[|X - x|]$ cannot have $\mathbb P(X<b) \gt \frac12$ or $\mathbb P(X\le b) \lt \frac12$ as you would then get $\text{E}[|X - b|]> \text{E}[|X - m|]$ – Henry Dec 18 '20 at 11:16
  • https://math.stackexchange.com/questions/85448/why-does-the-median-minimize-ex-c proves it in the points where the derivative exists, that is, where the cummulative distribution function is continuous. But what about on the other points? – user13761697 Dec 18 '20 at 16:25
  • The cumulative distribution function has a derivative almost everywhere, in particular somewhere between any point that is not a median and the nearest point that is a median. That is enough. – Henry Dec 18 '20 at 16:34