On 31 Jan 2004 17:20:20 -0800, [EMAIL PROTECTED] (Michael Stembera) wrote: >I would like to know if there is a solution for computing MD (mean >deviation), which would allow me to compute it in one pass. i.e., w/o >having to compute <x> the mean first. > >MD = 1/(N-1) * Sum(i=1 to N)||(xi - <x>)||
I doubt it: the average absolute deviation about a point typically needs the point to be known first. More seriously, why are you dividing by 1/(N-1)? It is not obvious to me that this will produce an unbiased estimate of the population average absolute deviation about the population mean - I would guess that there is no simple and general expression for producing an unbiased estimate (as is also the case with the standard deviation). It is also worth noting that the average absolute deviation about the median is less than or equal to the average absolute deviation about the mean. If you are insistent on using such a measurement of deviation, you might ask yourself why you are not interested in the median. . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
