"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
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