Or to say in the formula MD=$\frac{1}{N}\sum_{i=1}^{N}|x_i-\overline{x}|$ why can't we have median or mode instead of mean($\overline{x}$) and as a result speak about additional possible measures of deviation?
1 Answer
Yes we can, there is, for example, median absolute deviation (MAD):
$$ \operatorname{MAD}(X) = \operatorname{median}\left(\ \left| X_{i} - \operatorname{median} (X) \right|\ \right)$$
It has even some nice properties, e.g. $1.4826 \times \operatorname{MAD}(X)$ is estimator for standard deviation of normally distributed $X$. You can read more about it in the paper by Rousseeuw and Croux (1993).
We often look at much more then mean of errors, e.g. quantiles of errors, like in standard output of summary(lm)
in R:
Call:
lm(formula = mpg ~ ., data = mtcars)
Residuals:
Min 1Q Median 3Q Max
-3.4506 -1.6044 -0.1196 1.2193 4.6271
We also look at distributions of errors by plotting residuals and absolute errors etc.
It is not that commonly used because mean is more sensitive to outliers then median, and in most cases this is a desired property when talking about error measures.
Rousseeuw, P.J. and Croux, C. (1993). Alternatives to the median absolute deviation. Journal of the American Statistical Association. 88 (424): 1273–1283.
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$\begingroup$ by "on the lines" I meant something mathematically similar to mean deviation as written in the question details section. Here, median operation is used in the median absolute deviation and not the mean operation. $\endgroup$– ankitNov 25, 2016 at 12:42
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$\begingroup$ @ankit check the last paragraph of my answer. Moreover, median of deviations from median has some nice properties that are absent in case of median deviation of mean -- why would you want to use it? $\endgroup$– Tim ♦Nov 25, 2016 at 12:49
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$\begingroup$ On the page(en.wikipedia.org/wiki/…) I've discovered the same mathematical thing being performed over the examples as requested in the question. Are they called mean absolute deviation around median and mean absolute deviation around mode respectively? If yes then does it means that mean absolute deviation is of three types? KINDLY HELP ME TIM $\endgroup$– ankitNov 25, 2016 at 12:58
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2$\begingroup$ @ankit I guess you already answered yourself, didn't you? $\endgroup$– Tim ♦Nov 25, 2016 at 13:29
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