Talk:Mean difference
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I think the formula is wrong - it should be 1/(n*(n-1)), not 1/n^2 ....
PBoyd (talk) 13:40, 21 April 2008 (UTC) I agree, I think that that would exclude the deviations of the observations from themselves (which, I hope, are always zero).
Upon further reflection ... the number of pairs and the number of differences should be n!(n-1)!/2. Each paired difference should be calculated only once (except that an observation should not be differenced from itself ... |x-x|). However, if each difference is added twice (|x-y| and |y-x|), then the denominator is n!(n-1)! [=2*(n!(n-1)!/2)]. Since |x-x|=0, it doesn't matter whether or not you include them in the numerator. I will make the change. PBoyd (talk) 12:55, 2 May 2008 (UTC)
