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From: Ron Baker, Pluralitas! on 19 Jun 2005 16:57

"Stephen J. Herschkorn" <sjherschko(a)netscape.net> wrote in message
news:42B5D31A.8010609(a)netscape.net...
> Ron Baker, Pluralitas! wrote:
>
>>"Stephen J. Herschkorn" <sjherschko(a)netscape.net> wrote in message
>>news:42B5BE28.9060609(a)netscape.net...
>>
>>>Ron Baker, Pluralitas! wrote:

<snip>

>>>>
>>>Well, first of all, correlation is defined by
>>>
>>>corr(x, y) = Cov(X, Y) / [sqrt(Var(X)) sqrt(Var(Y))],
>>>
>>
>>I see that defined as the correlation *coefficient*.
>>http://mathworld.wolfram.com/CorrelationCoefficient.html
>>
>>
>>>where Cov(X,Y) = E[(X- EX) (Y-EY)] = E[XY] - EX EY,
>>>
>>>so your definitions are quite nonstandard.
>>>
>>
>>See: Alberto Leon-Garcia, "Probability and Random Processes
>>for Electrical Engineering", second editition, page 233.
>>"... it is customary to call E[XY], the correlation of X
>>and Y."[...]
>>
>
> That is completely nonstandard. My specialty is applied probablity, and I
> have *never* seen correlation defined that way.

You have now and you are about to again.

> I challenge you to find another reference that defines it as such. My
> definition is the standard. See any standard probability text, e.g., by
> Ross, Billingsley, or Feller. Even mathworld does not define correlation
> as does your source.

Note that Mathworld calls it the _correlation coefficient_.

Second reference: Cooper & McGillem, "Probabilistic Methods
of Signal and System Analysis", page 83. "If two random
variables X and Y have possible values x and y, then the expected
value of their product is known as the correlation, defined
in (3-5) as E[XY] = .....". Then they go on to separately
define the "correlation coefficient or normalized covariance".

I suspect that 'corrolation coefficient' is the more formal
term and that in your field people aren't normally concerned
with the 'correlation' I've cited so rather than say 'correlation
coefficient' all the time they abbreviate to 'correlation' with
practically no ambiguity.
In engineering (and physics too, I suspect) we are
more often concerned with the 'correlation' I've cited
than the 'correlation coefficient'.

<snip>

>>>
>>
>>X and Y uncorrelated? (That isn't right unless
>>they are uncorrelated.) In which case it is also:
>> E[X^2]E[Y^2] - E[X]^2 E[Y]^2
>>
>
> The OP said off the bat that X and Y are uncorrelated. However, that
> does *not* imply that X^2 and Y^2 are uncorrelated. For an example:
>
> P{X = 0} = 1/2.
> X has conditional Uniform(-1,1) distribution given X != 0.
> Y has conditional Uniform(-1,1) distribution given X = 0.
> P(Y = 0 | X != 0) = 1.

Oops. My bad. I confused uncorrelated with
independent.

--
rb


From: Ron Baker, Pluralitas! on 19 Jun 2005 17:02

"Kenneth T. Onyee" <kentonyee(a)hotmail.com> wrote in message
news:1119201866.251158.217870(a)g43g2000cwa.googlegroups.com...
> Very interesting. What is the definition of "acor"?
> Kenneth
>


"Ron Baker, Pluralitas!" <stoshu(a)bellsouth.net.pa> wrote in message
news:Qc9te.21572$h86.19145(a)tornado.socal.rr.com...
>
> "Kenneth T. Onyee" <kentonyee(a)hotmail.com> wrote in message
> news:1119151127.412846.142900(a)g44g2000cwa.googlegroups.com...
>> Let X and Y be uncorrelated random variables.
>> By construction, this means Cov[X,Y] = 0.
>> Is there a simple expression for Var[XY]?
>
>
> var[XY] = E[ (XY - xyBar)^2 ]
> = E[ (XY)^2 - XYxyBar - XYxyBar + xyBar^2 ]
> = E[ (XY)^2 ] - 2*xyBar*E[XY] + xyBar^2
> = E[ (XY)^2 ] - xyBar^2
> = E[ (XY)^2 ] - E[XY]^2
> = acor(XY) - cor(X,Y)^2
>
> note that 'cor' is correlation, not correlation coefficient.
>
> If X and Y independent
>
> = acor(X) * acor(Y) - xBar^2 * yBar^2


As Stephen J. Herschkorn has pointed out independent
is not the same as uncorrelated, so the last line above
does not apply to your original question.

--
rb


From: Robert Israel on 19 Jun 2005 17:31
In article <1119213985.468998.45010(a)g44g2000cwa.googlegroups.com>,
Kenneth T. Onyee <kentonyee(a)hotmail.com> wrote:
>I am the OP. Let me re-state my problem. It is:
> Assume Cov[X,Y] = 0. What is Cov[X Y, X Y]?

>I am not interested in "correlation." I do not make any additional
>assumptions
>about independence -- only that Cov[X,Y]=0. Is there a simple, general
>formula for the Var[X Y] under these circumstances?

In case you missed it, the answer was that it's
Var[X Y] = E[(X Y)^2] - E[X Y]^2 = E[X^2 Y^2] - E[X]^2 E[Y]^2.
Well, you could also call it
Cov[X^2, Y^2] + Var[X] Var[Y] + E[X]^2 Var[Y] + Var[X] E[Y]^2
but I don't think I'd call that simpler.

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

From: Kenneth T. Onyee on 19 Jun 2005 20:00
Why does Cov[X,Y] = 0 insufficient to imply that
Cov[ X^2, Y^2 ] = 0?

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

From: Jerry Dallal on 19 Jun 2005 19:38
Kenneth T. Onyee wrote:
> Why does Cov[X,Y] = 0 insufficient to imply that
> Cov[ X^2, Y^2 ] = 0?

Let (X,Y) = (1,1), (-1,1), (1,-1), (-1,-1), (0,0)
each with probability 1/5 (draw picture)

Then, (X^2,Y^2)=(1,1) with probability 4/5
and (0,0) with probability 1/5

corr(X,Y)=0, while corr(X^2,Y^2)=1

(unless I've made a typo)
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