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.
Best Answer
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: $$ \sqrt{\frac{1}{n}\sum_i|x_i - \bar{x}|^2\;} \,\geq \frac{1}{n}\sum_i |x_i - \bar{x}| $$