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Ho do we prove that the standard deviation is greater than or equal to the mean deviation about the arithmetic mean ?

$$ \sqrt\frac{\sum_{i=1}^{n}(x_i-\bar{x})^2}{n}\geq\frac{\sum_{i=1}^{n}|x_i-\bar{x}|}{n} $$

and under what conditions we get the equality ?

I think i understand that it is because of the squaring in standard deviation which tends to give more weightage to the data far from the central tendency.

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    $\begingroup$ The Cauchy–Schwarz inequality will do. $\endgroup$ – Chappers Jul 20 '17 at 18:18
  • $\begingroup$ @Chappers could u help me how to proceed. i'm unable to find where to start inorder to prove it. $\endgroup$ – ss1729 Jul 20 '17 at 18:41
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Let $v=\left|\vec{x} - \bar{x}\vec{1}\right|$, where $|\cdot|$ is component-wise. Then: \begin{align} \frac{1}{n^2}\left( \sum_i |x_i - \bar{x}| \right)^2 &= \frac{ \left(v\cdot\vec{1}\right)^2}{n^2}\\ &\leq \frac{(\vec{1}\cdot\vec{1})(v\cdot v)}{n^2}\\ &= \frac{1}{n}\sum_i|x_i - \bar{x}|^2 \end{align} where we used the CS inequality for the second step. Now take the root of the first and last terms: $$ \sqrt{\frac{1}{n}\sum_i|x_i - \bar{x}|^2\;} \,\geq \frac{1}{n}\sum_i |x_i - \bar{x}| $$

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  • $\begingroup$ How does $\left|\vec{x} - \bar{x}\vec{1}\right|= \sum_i \left|x_i - \bar{x} \right|$ ? $\endgroup$ – ThePassenger Jul 21 '17 at 8:22
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The Cauchy Schwarz inequality

$$\bigg(\sum_{i=1}^{n}a_{i}^2\bigg).\bigg(\sum_{i=1}^{n}b_{i}^2\bigg)\ge \bigg(\sum_{i=1}^{n}a_ib_i\bigg)^2 $$

taking $a_i=|x_i-\bar{x}|$ and $b_i=1/n$,

$$ \bigg(\sum_{i=1}^{n}|x_i-\bar{x}|^2\bigg).\bigg(\sum_{i=1}^{n}\tfrac{1}{n^2}\bigg)\ge \bigg(\sum_{i=1}^{n}|x_i-\bar{x}|.\tfrac{1}{n}\bigg)^2\\ \bigg(\sum_{i=1}^{n}(x_i-\bar{x})^2\bigg).\bigg(n.\tfrac{1}{n^2}\bigg)\ge \bigg(\frac{\sum_{i=1}^{n}|x_i-\bar{x}|}{{n}}\bigg)^2\\ \frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n}\ge \bigg(\frac{\sum_{i=1}^n|x_i-\bar{x}|}{n}\bigg)^2\\ \sqrt\frac{\sum_{i=1}^n(x_i-\bar{x})^2}{n}\ge\frac{\sum_{i=1}^n|x_i-\bar{x}|}{n}\\ S.D\ge M.D $$

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