Re: var(x) and std(x)

[Paul Kienzle]

Oops.  What about complex numbers?  

It ought to be:

    var(x+iy) = var(x) + var(y)

which is equivalent to:

    y = abs ( (x - mean(x)) * (x - mean(x))' ) / (n-1) 

The abs() is necessary because machine precision won't necessarily 
yield a real value from x * conj(x).

However one web page gives the following formula:

    var(x+iy) = 1/2 * (var(x) + var(y))

and says that the factor of 1/2 is controversial.  Anyone have
an authoritative definition?

Paul Kienzle
address@hidden

On Tue, Nov 12, 2002 at 01:03:13PM -0600, John W. Eaton wrote:
> On 12-Nov-2002, Paul Kienzle <address@hidden> wrote:
> 
> | Index: var.m
> | ===================================================================
> | RCS file: /cvs/octave/scripts/statistics/base/var.m,v
> | retrieving revision 1.6
> | diff -c -r1.6 var.m
> | *** var.m   2002/09/26 22:48:24     1.6
> | --- var.m   2002/11/12 18:30:44
> | ***************
> | *** 40,48 ****
> |       y = 0;
> |     elseif ((nr == 1) || (nc == 1))
> |       n = length (x);
> | !     y = (sumsq (x) - abs(sum(x))^2 / n) / (n - 1);
> |     else
> | !     y = (sumsq (x) - abs(sum(x)).^2 / nr) / (nr - 1);
> |     endif
> |   
> |   endfunction
> | --- 40,48 ----
> |       y = 0;
> |     elseif ((nr == 1) || (nc == 1))
> |       n = length (x);
> | !     y = sumsq (x - sum (x) / n) / (n - 1);
> |     else
> | !     y = sumsq (x - ones (nr, 1) * (sum (x) / nr) ) / (nr - 1);
> |     endif
> |   
> |   endfunction
> 
> I applied this patch.
> 
> Thanks,
> 
> jwe
> 
> 
> 
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