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Twinbee
Jun5-09, 11:24 AM
What are the advantages of using the absolute mean deviation over the standard deviation. Is it possible to show a simple example where the former is more (or less) appropriate?

Also, related to the mean deviation is my own variation. Does it have a name? Instead of calculating the absolute differences from the mean for each number, my technique would instead find the average of all the absolute differences for each number against each other number.

So for example: given the numbers 3,7,7,19
Average is: 9
Absolute Mean deviation is: 5
My 'special' deviation is: 6

This is found thusly:

(|3-3| + |7-3| + |7-3| + |19-3| +
|3-7| + |7-7| + |7-7| + |19-7| +
|3-7| + |7-7| + |7-7| + |19-7| +
|3-19| + |7-19| + |7-19| + |19-19| ) / 16

= 6

As you can see, everything is compared against everything else. What do people here think? One could also remove the 3-3, 7-7, 7-7, and 19-19 bits, and then divide by 12 for a similar variation (results in 8 by the way).

Could this method be usefully applied in stats?

statdad
Jun5-09, 12:49 PM
Your variation is essentially the Gini Mean difference, if I understand your explanation correctly.

The Mean Absolute Deviation (MAD), which is


\frac 1 n \sum |x_i - \bar x |


was proposed as an estimate of variation, but in the case of normally distributed data it is neither unbiased nor particularly efficient, compared to the usual estimates.

Note that there are other, better (more robust) measures of variability. The median absolute deviation (another MAD) is


MAD = \text{median}_i \left( | x_i - \text{median}_j (x_j)|\right)


(The same name is also given to this estimate: 1.4826 MAD - the multiplication by 1.4826 makes this unbiased for \sigma in the normal distribution case. Here also MAD refers to the median absolute deviation.)

Twinbee
Jun6-09, 02:24 PM
Ah, the Gini Mean Difference looks like the one. I wonder what applications it should be used for over SD or MD.

was proposed as an estimate of variation, but in the case of normally distributed data it is neither unbiased nor particularly efficient, compared to the usual estimates.
Interesting you say that it's biased. Doesn't it depend on the distribution? In that sense, the standard deviation would appear biased for evenly distributed data (non-normal).

There's this page which sings the praises of the MD over the SD and says it should be used in most cases of 'real data' where even slight errors may creep in. One other advantage apparently is that when outliers (long tail data) are squared, this creates bias, and the MD avoids this. Of course, I'm not sure how much all this is true, but here's the page:

http://www.leeds.ac.uk/educol/documents/00003759.htm

Thanks for the reply.