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To measure spread we use Variance or Standard Deviation. Variance and hence Standard deviation uses mean to find out the spread.

Recently came across the MAD (median absolute deviation). http://en.wikipedia.org/wiki/Median_absolute_deviation

Why is Median absolute deviation not as popular as Standard Deviation, although it looks more robust (immune to outlier)? In other words why are artifacts which measure the SD from the median are not as popular as the artifacts which measure it from Mean?

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    $\begingroup$ See stats.stackexchange.com/questions/147001/… and stats.stackexchange.com/questions/130939/… $\endgroup$
    – Tim
    May 4, 2015 at 22:20
  • $\begingroup$ @Tim: In case the square form have better closed properties, Why is the squared deviation (variance) from median not popular? It will make the math consistent and make the moment immune to outlier also! $\endgroup$
    – David
    May 4, 2015 at 22:31
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    $\begingroup$ Many of the comments in posts about using variance rather than mean absolute deviation from the mean (e.g. here) apply also to median absolute deviation from the median. Then on top of that, generally speaking properties of medians are not as nice as those of means. For example in general $\text{med}(X+Y)\neq$ $\text{med}(X)+\text{med}(Y)$; things like this makes it much less convenient to work with. $\endgroup$
    – Glen_b
    May 4, 2015 at 23:11

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Robustness to outliers is a double-edged sword: Sometimes we want to estimate things in a way that is robust to outliers, which means that we do not mind getting large outliers. At other times we want to avoid large outliers, so we want to estimate things in a way that is not robust to outliers. Similarly, with measures of spread, sometimes we want something that is robust to outliers, so that large outliers do not increase the measure. At other times we want our measure of spread to reflect the presence of large outliers by manifesting in a larger value.

In decision-theory, issues like this are dealt with by specifying a penalty/loss function which penalises you for your error in estimation of a quantity. Two common loss functions are absolute-error loss and squared-error loss (shown in the following plots, taken from this answer by Jean-Paul).

Absolute-error and squared-error penalty functions

Absolute-error loss penalises you according to the absolute deviation of your estimate from the true value. This form of loss function leads to estimation using medians. This form of loss function is robust to outliers in the sense that outliers contribute a penalty that is proportionate to their size. Measures of spread in this context reflect the expected loss of a particular estimate of central location, with the expected loss being a weighted sum of absolute deviations from the estimated central location.

Squared-error loss penalises you according to the squared deviation of your estimate from the true value. This form of loss function leads to estimation using means. This form of loss function is sensitive to outliers in the sense that outliers contribute a penalty that is proportionate to their squared deviation - this magnifies the effect of large outliers. Measures of spread in this context reflect the expected loss of a particular estimate of central location, with the expected loss being a weighted sum of squared deviations from the estimated central location.


In regard to the choice between median absolute deviation and standard deviation these same considerations apply. The former measure is a measure of spread that represents expected absolute-error loss, and is more robust to outliers. In this case, outliers do not manifest in large increases in the measure of spread. The latter is a measure of spread that represents expected squared-error loss, and is more sensitive to outliers. In this case, the outliers will manifest in large increases in the measure of spread.

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  • $\begingroup$ Do you have any examples were we want our measurement of spread to be sensitive to outliers on purpose? $\endgroup$ Nov 24, 2021 at 9:58
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    $\begingroup$ A typical example would be when you have a decision problem using a loss function that has high loss for extreme deviations from the expected outcome. In that case you would want your measurement of risk to be sensitive to outliers. $\endgroup$
    – Ben
    Nov 24, 2021 at 10:05
  • $\begingroup$ Is there a similar way to extend this reasoning towards measuring the skew? What is the "absolute-error" version of the skew? Similarly with kurtosis... $\endgroup$ Jun 6, 2022 at 6:03

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