Re: Statistical Indep and Corr of Normal RVs
- From: "David Jones" <dajxxx@xxxxxxxxx>
- Date: Fri, 20 Oct 2006 16:20:18 +0100
Brenneman wrote:
Hi,Corr(x,y)
I have come across the fact that for 2 normal rvs, x and y,
= 0 does not imply x and y are independent.
It may depend somewhat on where you came across this. It is strictly
true, but it may be wrong depending on what is meant by "2 normal
rvs". See below. There is just a small possibility that where you
found this was actaually the way you were.
It seems to me howeverthe
that this contradicts the condition that x and y are independent iff
their pdfs separate. Here is the simple argument I am thinking of:
In a problem I am looking at, x and y are two normal rvs both with
mean zero and sdev=1. In this case, their joint pdf is given by the
standard bivariate normal pdf:
f(x,y) = 1/[sqrt(1-r^2)*2*pi] * exp[ - { x^2 -2*x*y*r +
y^2}/(2*(1-r^2))]
But r=0 implies this just reduces down to
f(x,y | r=0) = 1/[2*pi] * exp[ - { x^2 + y^2}/2]
which can be written as the individual pdfs for x and y.
Since stat independence is also equivalent to being able to write
joint pdfs of the two rvs and the product of their individual pdfs,indep
does zero correlation in this case imply stat indep? If this is the
case, what happens in those cases where r=0 but x and y are not
that causes this argument to break down?
The above looks true (I have not read it in detail), except that you
are assuming that "2 normal rvs" and "bivariate normal" are the same
thing. It is possible for two variables to have normal distributions
for the marginal distributions, but not be bivariate normal.
To find an example, let X be standard Normal and set Y=X except over a
central region -c<X<c, andset Y=-X over this region. Then Y is Normal.
I believe c can be chosen to give a correlation of zero. Yet Y is a
deterministic function of X, and so "fully" dependent.
David Jones
.
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