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Jon Dodd wrote:
> ...
> Dear Harald,
> 
> Thankyou for your reply. I don't think you understood my problem correctly or
> perhaps I misinterpreted your reply.


I have not read your original question carefully, so I my answer was
useless, sorry!


Now my second attempt:

1. the normal probability density function (normPDF) for A (and B) is 
  normPDF[z] := 1/Sqrt[2 Pi sd^2] * Exp[(-1/2)* (z-m)^2/sd^2]

2. the probability for any A being greater then B (A and B from the same
normPDF) is:
  Integrate[normPDF[z], with z running from -Infinity to B]
This gives the cumulative probability density function (normCDF) for 
A > B:
normCDF[z] := 1/2 * (1-Erf[(x-m)/(Sqrt[2]*sd)]) with Erf being the
so-called "error function" (see any statistic book for details).
Some values: 
  normCDF[-Infinity] = 0
  normCDF[0] = 0.5
  normCDF[+Infinity] = 1
Essentially, normCDF rises from 0 to 1 within (m - 4*sd) and (m + 4*sd). 

3. the conditional probability density function X(A|A>B) (PDF for A
given A>B) is just the product:
  condPDF[z] := normPDF[z]*normCDF[z]
( that for A<B is simply  normPDF[z]*(1-normCDF[z]) )

Now the problem: 
  if you assume the mean of a density distribution to be the "value of
maximum likelihood" i.e. the maximum of the function, then you have to
differentiate condPDF[z] for z (no problem) and solve the result for its
root: Derivative[condPDF[z] for z] = 0 .
The latter makes the problem if you look for an analytical result,
rather than a numerical.

I tried it with Mathematica - an algebra software (www.wolfram.com) -
but that failed for the analytical solution.

Regards,
Harald


-- 
Harald Giese
Email: giese@dkrz.de
Phone: +49 (0)40 4123 5796; Fax: +49 (0)40 5605724
Institut fuer Meereskunde der Universitaet Hamburg
(Institute of Oceanography of the University of Hamburg)
Troplowitzstrasse 7, D-22529 Hamburg