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I was reading about variance from Head First Statistics :

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And then -

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Q. I find the reasoning a little absurd. Wouldn't just taking the absolute distance suffice if cancelling out of the terms was the reason ? Why do squares to make it positive and complicate the calculations further (In terms of computations on a computer, square would be costlier than subtraction, right?) ?

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  • $\begingroup$ The variance emphasizes the largest difference, and the mechanism to do that is squaring each term. $\endgroup$ – abiessu Dec 27 '14 at 14:59
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    $\begingroup$ This question is arguably better suited for the statistics stackexchange. The following statistics stackexchange question offers a discussion of the choice of standard deviation vs. the statistic you're proposing (which is called the mean absolute error): stats.stackexchange.com/questions/118/… $\endgroup$ – user78270 Dec 27 '14 at 15:45
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You can take the average absolute deviation from the mean as a measure of dispersion if you wish.

There are various reasons why the variance and its square root (the standard deviation) are often used, including:

  • the average square of the deviation is minimised when taken about the mean, while the average absolute deviation is minimised when taken about the median, so the variance is in this sense a natural partner of the mean

  • for a random variable with a normal (Gaussian) distribution, the mean and variance together are sufficient statistics, while the mean and average absolute distribution are not

  • the average square of the deviation is easily differentiable while the average absolute deviation, in the same way as $x^2$ is differentiable for all $x$ while $|x|$ is not

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  • $\begingroup$ could you rephrase the first item, please? $\endgroup$ – Guillermo Mosse Apr 4 at 14:07
  • $\begingroup$ @GuillermoMosse Where are you having difficulties? Deviation is the difference between values observed and a particular given value such as the mean or median or something else. You can if you wish square this or take the absolute value if you want a non-negative number, and you can average the results; you may want to minimise this average by changing the particular given value $\endgroup$ – Henry Apr 4 at 17:48
  • $\begingroup$ I'm having trouble with the words "taken about", particularly. Moreover, I don't understand why you say that the deviation is minimised. I understand the deviation as something you just calculate, not something you minimize. Or are you talking about estimators? I am sorry for mi ignorance :-). $\endgroup$ – Guillermo Mosse Apr 4 at 17:58
  • $\begingroup$ I am saying that the sum of [squares/absolute value] of the deviations varies depending on the particular given value you are measuring deviation from. Since this sum is non-negative, there is a value which minimises the sum. That minimising value is the mean when minimising the sum of squares and is the median when minimising the sum of absolute values $\endgroup$ – Henry Apr 4 at 20:02
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Actually the real "absolute distance" is the standard deviation (i.e $\sqrt{\text{Var}(X)}$) ! I'll make the exemple in the case where the expected value is $0$. The reason is that $$\sqrt{x_1^2+...+x_n^2}\leq |x_1|+...+|x_n|$$ therefore $\sqrt{x_1^2+...+x_n^2}$ gives a better approximation than $|x_1|+...+|x_n|$.

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  • $\begingroup$ "Real" in which sense? "Better" for which criterion? Not sure this answer helps. $\endgroup$ – Did Dec 27 '14 at 19:41
  • $\begingroup$ I din't get the point you are trying to put at all. $\endgroup$ – Amit Tomar Dec 27 '14 at 20:04

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