$\displaystyle S_n(x)=\frac 1n\sum_{i=1}^n|x_i-x| \Longleftrightarrow f(x)=nS_n(x)=\sum_{i=1}^n|x_i-x|$Prove that it has the minimum when $\displaystyle x$ is the median.

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- Feb 5th 2008, 10:04 AMjames_bondAverage Absolute Deviation (find the minimum)$\displaystyle S_n(x)=\frac 1n\sum_{i=1}^n|x_i-x| \Longleftrightarrow f(x)=nS_n(x)=\sum_{i=1}^n|x_i-x|$Prove that it has the minimum when $\displaystyle x$ is the median.
- Jun 3rd 2010, 07:24 AMjames_bond
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