
 
 Peter Dalgaard BSA <p.dalgaard@biostat.ku.dk> wrote:
Rolf Turner writes:

> Peter Dalgaard writes:
> 
> > Shutnik writes:
> > 
> > > Hello,
> > > I have the normal random variables y(t)~N(mu, sigma.sq) and want
> > > to decompose them into n normal variables:
> > > 
> > > y(t) = x(t,1) + … + x(t,n) 
> 
> I presume this means y(t) = x(t,1) + ... + x(t,n) (R.T.)
> 
> > > 
> > > x(t,i)~N(mu, sigma.sq/n) 
> 
> I presume you want x(t,i)~N(mu/n, sigma.sq/n),
> elsewise the question doesn't make sense.
> 
> I also presume you want the x(t,i) to be independent,
> elsewise the question is trivial. (R.T.)
> 
> > > 
> > > The problem is not as simple as can appear. All my experiments
> > > didnt give me anything so far. Are there any tools to do this?
> > > 
> > 
> > This should work, provided I understand the problem correctly:
> > 
> > x <- rnorm(n,sd=sqrt(sigma.sq/n))
> > x <- x - mean(x) + y/n
> 
> I don't think it's that simple: By my calculations,
> 
> Var(x_i) = 2*sigma.sq/n - sigma.sq/n^2, not sigma.sq/n.

>Try again... Var(y/n) = sigma.sq/n^2, not sigma.sq/n so it >cancels the
>second term rather than doubling the first.

> I think the problem is actually fairly subtle (although I may
> be overlooking something simple). Something like a Gramm-Schmidt
> approach should work, but I can't quite suss it out.

>You just need to orhtogonalize against the vector (1,1,...,1), which
>is what I did, effectively. The residuals x-mean(x) are independent of
>mean(x) by the Fisher-Cochran theorem (if I remember the name
>correctly...) and y/n has same distribution as mean(x) so you can
>substitute y/n for mean(x) and paste things back together again.

>But where does the t in y(t) and x(t,i) enter in all of this??


In 



x <- x - mean(x) + y/n 

 when we add y/n to x - mean(x) the variance of x(i) is no longer sigma.sq/n while it is required:



y = x(1) + ... + x(n)

x(i)~N(0,sigma.sq/n)





The variance of x(i) is important because the property will be used is E(sum(x^2)) = sigma.sq.



Regards,

Shutnik





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