Why do we use squared deviations to compute the SD, given that it amplifies the effect of outliers? [duplicate]

Question

Suppose I have the following hypothetical data:

• One thousand times value 15 (i.e., 15 occurs 1000 times)
• and a single outlier value - 115 (i.e., 115 occurs just once - an outlier)

Thus the mean is: $$((15 \times 1000) + 115)/1001 \Rightarrow (15000 + 115) / 1001 \Rightarrow 15.1$$

The standard deviation is: $3.16$.

Whereas, $Σ|x - x̄| / N$ (i.e., using the absolute deviation instead of squaring) I get: $0.20$.

I think the later value is more appropriate, but I have read in many places that one of the reasons we take the square for calculating the SD is because we want to give more weight to outliers. But I have always thought that outliers are supposed to be ignored!

What is the benefit of using the standard deviation formula (with squared deviations) which amplifies the effect of outliers (as compared to using absolute deviations)?

Answer 1

I have read in many places that one of the reasons we take the square for calculating the SD is because we want to give more weight to outliers.

This is not correct. The use of the square in calculating variance is almost never because of giving more weight to outliers

It might, however, be an intuitive way to think of it. I believe people more often say that the square gives more weight to values further away from the mean. But this is not to be considered outliers.

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The type of cost function to apply to the difference, squaring, absolute difference or something else, should depend on the type of distribution that one assumes/expects/uses for the error distribution.

Due to how errors are often created (as a sum of many little incremental errors), the errors often follow a normal distribution or something resembling. That is the more common reason for using the square

• The sample standard deviation $s = \sqrt{\sum (x_i-\bar{x})^2/(n-1)}$, based on the squared difference, is an efficient estimator for the parameter $\sigma$ in the normal distribution (and also the minimum variance unbiased estimator).

  See Fisher's 1920 article A Mathematical Examination of the Methods of Determining the Accuracy of an Observation by the Mean Error, and by the Mean Square Error.

• The squared difference, or variance, as a measure of te deviation has also the advantage that it is additive. $Var(X_1+X_2) = Var(X_1) + Var(X_2)$.

  See Fisher's 1918 article The Correlation Between Relatives on the Supposition of Mendelian Inheritance

So in your hypothetical case, with the thousand values of 15 and one of 115, the use of the square might be very well inappropriate. But... it depends a lot on what the data represents and what kind of statistical model would be appropriate to model it. The question is not clear about that.