[R] R help
Peter Dalgaard BSA
p.dalgaard at biostat.ku.dk
Wed Apr 30 00:27:33 CEST 2003
Rolf Turner <rolf at math.unb.ca> writes:
> Peter Dalgaard writes:
> > Shutnik <shutnik_xx at yahoo.co.uk> 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) +
> 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??
O__ ---- Peter Dalgaard Blegdamsvej 3
c/ /'_ --- Dept. of Biostatistics 2200 Cph. N
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk) FAX: (+45) 35327907
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