From ghill@alizarin.physics.wisc.edu Wed Aug  9 22:47:30 2000
Date: Sat, 5 Aug 2000 10:40:25 -0500 (CDT)
From: Gary Hill <ghill@alizarin.physics.wisc.edu>
To: amanda-c // mailing_list
Subject: More on the Explanation of error treatment difference Ralf/Ty 



      Hello Jan, and others. By popular demand, here's the explanation of
the error treatment for weighted MC. This shows how the formula Ralf used
is a special case of the more general form of the "basic theorem of MC
integration". For more details, see my note on this subject listed below,
or any basic statistics book. In particular, the section in "Numerical
Recipes in C", Second Edition, on Importance Sampling is very good (page
316). The "basic theorem of MC integration" (un-weighted treatment) is
found on page 305, equation 7.6.1. The generalisation to weighted MC is on
page 316, eqn 7.8.3.

      Enjoy!

      Gary


---------- Forwarded message ----------
Date: Fri, 4 Aug 2000 09:57:57 -0500 (CDT)
From: Gary Hill <ghill@alizarin.physics.wisc.edu>
To: amanda-i // mailing list
cc: Ty DeYoung <deyoung@alizarin.physics.wisc.edu>
Subject: Explanation of error treatment difference Ralf/Ty
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   To all interested in weighted MC error analysis -

   I have found the difference in the Ralf/Ty error treatments. 

   Ralf indeed has done the treatment correctly -

   The variance in a weighted MC using N generated primary events is given
by (see eg http://alizarin.physics.wisc.edu/ghill/is/is.ps)

     sig^2 = (sum (w_i^2))/N -   ( (sum w_i)/N )^2

    By the central limit theorem, the error in the mean (sum w_i)/N is
then sqrt ( sig^2/N )

    Then error in the total weight is then 

     sum w_i    +/-    N sqrt ( sig^2/N ) 

     =  sum w_i    +/-  sqrt(sum (w_i^2)))    if N is large (killing the
                                                              second
                                                        term in sig^2)

      as used by Ralf.


     What Ty did was divide the MC up into 8 (not 4 as Ralf states)

((((====> Ralf adds: 
	Ty agreed that I was right ... (his mail from 08/05/2000)
)))))

independent subsets and compute 8 means, then find the variance of those
means directly. However, this procedure finds the variance of a simulation
of size N/8 and not of size N as we actually have.  


      I hope this has made the difference clear.  


      Gary