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I have a beautiful function
$$f(x)= \frac{\sum_0^n|x-E(x)|}{n} $$ which calculates the average differences between all values and the mean value
The results are very simple to understand
I m trying to understand the motivation behind creating the variance
$$f(x)= \frac{\sum_0^n(x-E(x))^2}{n} $$ What is this squaring good for?
Tnx

Michael Hardy
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Chris Border
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  • My professor has said that one motivation is differentiability. The squaring is nicer than taking absolute value in that sense. – SOULed_Outt Apr 13 '18 at 15:21
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    Maybe https://math.stackexchange.com/questions/700160/intuition-behind-variance-forumla/700231#700231 is helpful. – Michael Hoppe Apr 13 '18 at 15:33
  • Also, the squaring lends greater weight to the most deviant values. Some times you want that, since if you're after an estimate, it is often more important that the estimate is rarely very wrong, than that it is often entirely correct. – Arthur Apr 13 '18 at 15:33
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    The Euclidean norm coincides with our geometric intuition – user251257 Apr 13 '18 at 15:39
  • Another observation which may help convince you: $ f_2(x)= \frac{\Sigma_1^n(x-\hat{x}))^2}{n} $ is minimized when $\hat{x} = E(x)$, i.e. when centered at the mean, but $f_1(x)= \frac{\Sigma_1^n|x-\hat{x}|}{n}$ is minimized when $\hat{x} = Median(x)$, i.e. when centered at the median, and in fact, if the number of samples is even, $\hat{x}$ can be any value between the center 2 values and it wouldnt affect $f_1$ value. Would that perhaps change your vote as to which one is "better"? :) – antkam Apr 13 '18 at 20:28

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The main motivation for using the square root of the mean squared deviation instead of using the mean absolute deviation is that it makes possible this identity: $$ \operatorname{var}(X_1+\cdots+X_n) = \operatorname{var}(X_1) + \cdots + \operatorname{var}(X_n) \text{ if } X_1,\ldots,X_n \text{ are independent.} $$

Abraham de Moivre used this in the 18th century when he showed that the probability mass function of the number of "heads" that appear when a coin is tossed 1800 times is well approximated by the integral of a Gaussian function over a bounded interval.

Michael Hardy
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