"Robert J. MacG. Dawson" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED]... > > > "Wuensch, Karl L" wrote: > > > > How about simply using the M.A.D.? No, not the mad spouse who noticed she > > was getting short-shrimped, rather the mean absolute deviation of individual > > shrimp from the mean of all shrimp. > > I thought of this, but IIRC the mean absolute deviation (whether from > the mean or from the median - see below) does *not* relate to the mean > absolute difference over all pairs in the same way that the mean squared > deviation from the mean [variance or, backtransformed, SD] relates to > the mean squared difference over all pairs. Please correct me if I'm > wrong here... > > Why from the median? Well, the median is the value from which MAD is > minimized in the same way that the mean is the value from which squared > deviation is minimized, and the mode is the value from which "boolean > deviation" (code 1 if they differ, 0 if they are the same) is minimized. > > -Robert Dawson > >
To answer thew original problem, the expected value of the difference can be calculated: Let wk, k = 1, 2, ..., 2n, be the ordered weights of the shrimp (from smallest to largest). Then the expected weight of the heavier shrimp - lighter shrimp is (if I didn't make an error): sum ( from 1 to 2n) { wk ( 2k - 2n - 1) 2n / [2n (2n-1)] } This can be calculated by looking at all (2n)! orderings in which the shrimp can be removed from the bowl. Roman Mureika ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at http://jse.stat.ncsu.edu/ =================================================================