Re: var(x) and std(x)

[Dave Cottingham]

I see several people have already explained the n - 1, but I have
another concern.  The method cited here is prone to bad roundoff error
in cases where sqrt(variance) is much smaller than the mean.  A better
method would be

   y = sumsq(x - sum(x) / n) / (n - 1);

 - Dave Cottingham

> Hi,
>   I have a doubt regarding the implementation of variance function var(x) in 
> octave.It finds the variance as follows
> 
> function y = var(x)
> 
>   if (nargin != 1)
>     usage ("var (x)");
>   endif
> 
>   [nr, nc] = size (x);
>   if (nr == 0 || nc == 0)
>     error ("var: x must not be empty");
>   elseif ((nr == 1) && (nc == 1))
>     y = 0;
>   elseif ((nr == 1) || (nc == 1))
>     n = length (x);
>     y = (sumsq (x) - sum(x)^2 / n) / (n-1);
>   else
>     y = (sumsq (x) - sum(x).^2 / nr) / (nr-1);
>   endif
> 
> endfunction
> 
>    It gives variance after averaging over n-1 samples, rather than averaging 
> over n samples. for example, if we calculate the variance of 
> x = [1,2,3] , it returns the value 1, instead of 2/3..This is because it
> finds 2/2 and not 2/3..The same is the case with std(x) also. 
> Is it an error or is there any explanation for that?
> Please correct me if I am wrong.
> 
> Thanks and Regards
> Anand
> 
> 



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