From help-octave-request at bevo dot che dot wisc dot edu Tue Nov 12 11:46:30 2002
Subject: weighted var, etc. (was "var(x) and std(x)")
From: Mike Miller <mbmiller at taxa dot epi dot umn dot edu>
To: Francesco Potorti` <F dot Potorti at cnuce dot cnr dot it>
cc: Help-Octave List <help-octave at bevo dot che dot wisc dot edu>
Date: Tue, 12 Nov 2002 11:46:12 -0600 (CST)

On Tue, 12 Nov 2002, Francesco Potorti` wrote:

> If you compute the mean from the samples, this is correct.  If you know
> the mean a priori, then you should divide by n.  The code you cite
> computes the mean from the samples, so that's okay.


That's true.

A related issue (question):  Is there an Octave procedure for computing
weighted means and variances?  If we have a vector x of values and a
corresponding (same size) vector p of probabilities, the mean of x is p'*x
and the variance is p'*(x-p'*x).  If length(x)=n, and p=ones(n,1)/n, then
p'*(x-p'*x) = ((n-1)/n)*var(x).

If appropriate functions don't already exist, I would recommend the
following names: wmean, wvar, wstd and wcov.  I would recommend that they
work like this:

y=wmean(x,p)  <--where p is a matrix the same size as x
y=wmean(x,p)  <--where p is column vector with length(p)=size(x,1)
y=wmean(x)    <--where p is set to p=ones(size(x,1),1)/size(x,1)

wmean would compute weighted means of the columns of matrix x using the
weights given in matrix (or vector) p.  The same system would be used for
wvar, etc.  This approach has the nice side effect that wvar(x) provides
var(x)*(n-1)/n saving the user a few keystrokes.  wmean(x)=mean(x).

Has this already been done?

Mike



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