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Let $X$ be a random variable with support ${1,2,3,5,15,25,50}$, each point of which has the same probability $\frac{1}{7}$. Argue that $c=5$ is the value that minimizes $h(c)=E(|X-c|)$. Compare $c$ with the value of $b$ that minimizes $g(b)=E[(X-b)^{2}]$.

I am over here: $$h(c)=E(|x-c|)=(|1-c|+|2-c|+|3-c|+|5-c|+|15-c|+|25-c|+|50-c|)\times\frac{1}{7}$$ My main concern is I am not sure using absolute values to argue that 5 is the value that minimizes the equation.

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    $\begingroup$ Do you know any calculus? You can think about it in terms of the derivative with respect to $c$. $\endgroup$
    – Tyberius
    Jan 21 '20 at 1:20
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    $\begingroup$ I don't have time to write an answer, but it's worth graphing some functions like $|x|$ and $|x|+|x-1|$ or $|x|+|x-1|+|x-3|$ - just smaller sums of absolute values - and seeing that these are just a few line segments pasted end to end - and that you can characterize the slope of that segment by counting how many subtracted terms are greater/less than the current one. $\endgroup$ Jan 21 '20 at 1:41
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It is a fact, that the median minimizes the sum of absolute deviations. See here, for example, for explanations and proofs.

Since $5$ is the median of the given data set and the weights are all equal to $\frac 17$, it minimizes $E(X-c)$.

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  • $\begingroup$ This also hints at how you are supposed to approach the 2nd part of the question of how to minimize the sum of squared deviations. $\endgroup$
    – Tyberius
    Jan 21 '20 at 1:54
  • $\begingroup$ @Tyberius : Exactly. In my opinion, this problem aims at the learning outcome to realize that the median and the mean are minimizers of measures of distance. $\endgroup$ Jan 21 '20 at 2:07

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