Standard deviation versus absolute mean deviation

[Twinbee]

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]

Your variation is essentially the Gini Mean difference, if I understand your explanation correctly.

The Mean Absolute Deviation (MAD), which is

<br /> \frac 1 n \sum |x_i - \bar x |<br />

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

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

(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]

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.

 

[Twinbee]

An old thread, but a goodie. My first post did indeed describe the Gini mean difference, but I described two different versions which I'll call GiniA and GiniB:

GiniA(3,7,7,19) = 8
GiniB(3,7,7,19) = 6

For the above values, 8 is the correct answer according to the standard Wikipedia definition. However, I think a 'better?' value may be 6 as GiniB shows, since it includes the missing 4 differences (3-3, 7-7, 7-7 and 19-19) and divides the result by n*n instead of n*(n-1). This is demonstrated in my first post as well (16 sums divided by 16 instead of by 12).

Is there any reason to believe GiniB could be useful? It seems natural to divide by 16, since one could say that each value is an 'error of itself' (i.e. no error) as well as errors of each other.