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From: DZ on 19 Jun 2005 20:49
Kenneth T. Onyee <kentonyee(a)hotmail.com> wrote:
> Why does Cov[X,Y] = 0 insufficient to imply that
> Cov[ X^2, Y^2 ] = 0?

The dependency need to have a linear component for the covariance to
be zero. Consider the following

X ~ N(0, 1)
E ~ N(0, 1/10)
Y = X^2 + E

That is, the functional form of dependency between X and Y is the
parabola with added normal noise from E. Therefore Cov(X, Y) = 0.

But if you look at the relation between Y and X^2 then it is entirely
linear apart from the noise from E. Therefore, the relation between
Y^2 and X^2 is described by the positive part of the parabola and the
Cov(X^2,Y^2) is also very high.
From: DZ on 19 Jun 2005 21:06
DZ wrote:
> Kenneth T. Onyee <kentonyee(a)hotmail.com> wrote:
>> Why does Cov[X,Y] = 0 insufficient to imply that
>> Cov[ X^2, Y^2 ] = 0?
>
> The dependency need to have a
.......................^^^^^^^^^

err... read "to LACK the" (linear component for the covariance to be
zero).


> linear component for the covariance to
> be zero. Consider the following
>
> X ~ N(0, 1)
> E ~ N(0, 1/10)
> Y = X^2 + E
>
> That is, the functional form of dependency between X and Y is the
> parabola with added normal noise from E. Therefore Cov(X, Y) = 0.
>
> But if you look at the relation between Y and X^2 then it is entirely
> linear apart from the noise from E. Therefore, the relation between
> Y^2 and X^2 is described by the positive part of the parabola and the
> Cov(X^2,Y^2) is also very high.
From: Robert Israel on 19 Jun 2005 21:12
In article <1119225653.199817.198360(a)g44g2000cwa.googlegroups.com>,
Kenneth T. Onyee <kentonyee(a)hotmail.com> wrote:
>Why does Cov[X,Y] = 0 insufficient to imply that
>Cov[ X^2, Y^2 ] = 0?

Cov[X,Y] = 0 is one scalar equation. Cov[X^2, Y^2] = 0 is a different
equation. There's no reason to think that either should imply the other,
and just about any nontrivial exploration (say with distribution
of X depending on a parameter) will provide counterexamples.

>What are necessary and sufficient conditions to guarantee that Cov[X^2,
>Y^2] = 0?

The definition: E[X^2 Y^2] - E[X^2] E[Y^2] = 0, or restatements thereof.

Robert Israel israel(a)math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada


From: kentonyee on 19 Jun 2005 21:46
> >What are necessary and sufficient conditions to guarantee that Cov[X^2,
> >Y^2] = 0?

> The definition: E[X^2 Y^2] - E[X^2] E[Y^2] = 0, or restatements thereof.

i was thinking of a less tautological relationship, like
Cov[X^2, Y^2] = 0 if, and only if, X is "independent" of Y.
(But I'm not sure if "independence" is necessary or sufficient...)

Onyee

From: Robert Israel on 19 Jun 2005 23:44
In article <1119231980.397473.240530(a)g43g2000cwa.googlegroups.com>,
<kentonyee(a)hotmail.com> wrote:
>> >What are necessary and sufficient conditions to guarantee that Cov[X^2,
>> >Y^2] = 0?
>
>> The definition: E[X^2 Y^2] - E[X^2] E[Y^2] = 0, or restatements thereof.
>
>i was thinking of a less tautological relationship, like
>Cov[X^2, Y^2] = 0 if, and only if, X is "independent" of Y.
>(But I'm not sure if "independence" is necessary or sufficient...)

Sufficient, of course. But not at all necessary. Again, counterexamples
are very easy to construct.

Robert Israel israel(a)math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

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