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From: Obaid Mushtaq on 5 Aug 2010 07:00
Hi,

Thanks. I still think that the formula resembles to the linear one except for the scaling with square roots. The looped version was on purpose as I had to do this in C++ later on. I didn't know how to use 2 variables in a for loop in ML so I was doing something like that.

BR,

Obaid
From: Matt J on 5 Aug 2010 10:46
"Obaid Mushtaq" <obaidmushtaq(a)yahoo.com> wrote in message <i3e5k6$ju9$1(a)fred.mathworks.com>...

> Thanks. I still think that the formula resembles to the linear one except for the scaling with square roots.
=======

Yes, and it's pretty clear what motivated that. The formula we had from before

Ycov=T*Zcov*T.'

means that the variance vectors Var_Y=diag(Ycov)
and Var_Z=diag(Zcov) are specifically related by

Var_Y=T.^2 *Var_Z

So, if we select T=sqrt(S) where S is linear interpolation than

Var_Y=S*Var_Z

Meaning that interpolating the signal with sqrt(S) is the same as linearly interpolating
the variances Var_Z. Hence, if Var_Z is a constant Zvar than likewise Var_Y=Zvar and you've preserved variance.

Clearly, though, you can do this with any uniformity-preserving interpolator (e.g. B-splines)
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