Maybe Not? Re: Is it always true for Gaussian random vairables?

From: cheshirecat (quzhang_at_ecs.umass.edu)
Date: 11/15/04 

Date: Mon, 15 Nov 2004 02:27:08 +0000 (UTC)

It seems that the decomposition does not exist: When we assume such Z
exists which is independent of X and Y, then \sigma_z =
sqrt(1-(r*\sigma_y/\sigma_x)^2)*\sigma_y. However, this equation can
also be written as X = 1/(r*\sigma_y/\sigma_x) * (Y-Z), and we can see
that \sigma_z is not equal to the former one.

Am I right?