Why is Standard Deviation not defined as the expected value of the distances of points from mean?

Question

Variance is defined as $$V(X) = \sum (x-\mu)^2 .p(x)$$ And standard deviation is $\sigma_X = \sqrt{V(x)}$

But I feel it makes more sense to define $\sigma_X$ as $\sum( |x-\mu|.p(x))$ instead because the mod takes care of negative distances and multiplication by p(x) would give us expected value of the deviation we should expect. Then why is SD defined the way it is?

One reason is that the density function for the normal distribution is easily defined in terms of standard deviation: ${\frac {1}{\sigma {\sqrt {2\pi }}}}\exp({-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}})$, but not the mean absolute deviation, and the normal distribution is very important because of the central limit theorem. — Jair Taylor, Jan 28 '20 at 11:15

score 2 · Answer 1 · answered Jan 28 '20 at 11:14

Standard deviation has a natural geometricc interpretation as the $L^2$ distance of a random variable from its mean. What you're describing is an example of absolute deviation, and would be $L^1$ instead. The $L^2$ case for deviation and related issues is more tractable, is amenable to more incisive theorems (such as this), and admits the only geometric interpretation on which our random variables form a Hilbert space, which has all sorts of benefits when we linearly transform them to analyze latent variables (e.g. viz here and here). In particular, the concern here is covariance as a generalization of variance.

Why is Standard Deviation not defined as the expected value of the distances of points from mean?

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