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sum of correlation

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roy

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Aug 31, 2002, 12:41:21 PM8/31/02
to
roy wrote:
>
> I have some vectors X,Y,Z, after some kind of normalization,
> I get the equation
> Corr(X,S)= p(Corr(X,Y)+ Corr(X,Z))
> where S=Y+Z, p is just a parameter.

We know that if there is strong relationship between two vectors,
the Correlation will be 1 or -1, otherwise 0.
However, if I can limit the correlation in [0,1], that means,
if there is strong relationship between two vectors,
the correlation is 1 (not 1 or -1), if there is no relationship, correlation
is 0.
Thus, I can get the conclusion that
Corr(X,S)=0 if and only if Corr(X,Y)=0 and Corr(X,Z)=0.
Corr(X,S)=0+c if and only if Corr(X,Y)=0+c and Corr(X,Z)=0+c.
( 0<c<<1 )

That is what I wanted. Because I don't know the value of Corr(X,Y)and
Corr(X,Z), I just want to decide their relationship from the value
of Corr(X,S).

Do you have some idea about that?
Thank you very much for your kind help.

Lynn Killingbeck
>Plus 1 then divided by 2


roy

Thank you very much.
Plus 1 then divided by 2 really makes the correlation vary in [0,1]
From Corr(X,S)= p(Corr(X,Y)+ Corr(X,Z))we get
(Corr(X,S)+2)/2= p( (Corr(X,Y)+1)/2 + (Corr(X,Z)+1)/2)
where S=Y+Z, p is just a positive parameter.
Thus, if left side=0, then each part of right side should be 0.
So Corr(X,Y)=-1 and Corr(X,Z)=-1. But if left side is large, we cannot
decide the value of each part of right side, may be both large, maybe
one
is 0, the other is very large.

But if after some transformation, when the left side=0, we know that
the
Corr(X,Y)=0 and Corr(X,Z)=0, then things become easier. We dont have
to think about Corr(X,Y)and Corr(X,Z) any more because they have no
relationship to each other, we can just throw them away.
If the left side is large, we will know there is strong relationship
between vectors in the right side. Then we can furtherly work on these
vectors.


--------------------------------------------------------------------------------

Greg Heath

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Sep 2, 2002, 5:20:11 AM9/2/02
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roy...@eyou.com (roy) wrote in message news:<eafd439e.0208...@posting.google.com>...

> roy wrote:
> >
> > I have some vectors X,Y,Z, after some kind of normalization,
> > I get the equation
> > Corr(X,S)= p(Corr(X,Y)+ Corr(X,Z))
> > where S=Y+Z, p is just a parameter.

What am I missing?

Problem #1:

Where did you get the above relationship? Corr is a function of
time-lag (or space shift). For real-valued time vectors the
"U"nnormalized correlation function is

UCxs(tau) = UCorr(X,S;tau) = INT(-oo,oo){ dt X(t)S(t+tau) }
= INT(-oo,oo){ dt S(t)X(t-tau) }
= UCsx(-tau).

Obviously

UCxs = UCxy + UCxz

and

UCss = UCyy + UCyz + UCzy + UCzz.

The corresponding normalized correlation coefficient is

Cxs = UCxs/sqrt(UCxx UCss).

Therefore

sqrt(UCxx UCss) Cxs = sqrt(UCxx UCyy) Cxy + sqrt(Uxx UCzz) Cxz

and

Cxs = sqrt(UCyy/UCss) Cxy + sqrt(UCzz/UCss) Cxz.

= p (Cxy + Cxz)

only if

UCzz = UCyy,

where the latter is an equation in tau.

Please clarify.

Thanks,

Greg

Brian

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Sep 5, 2002, 1:47:07 PM9/5/02
to
In article <eafd439e.0208...@posting.google.com>, roy...@eyou.com
says...

>
>roy wrote:
> >
> > I have some vectors X,Y,Z, after some kind of normalization,
> > I get the equation
> > Corr(X,S)= p(Corr(X,Y)+ Corr(X,Z))
> > where S=Y+Z, p is just a parameter.
>
> We know that if there is strong relationship between two vectors,
> the Correlation will be 1 or -1, otherwise 0.
> However, if I can limit the correlation in [0,1], that means,
> if there is strong relationship between two vectors,
> the correlation is 1 (not 1 or -1), if there is no relationship, correlation
> is 0.
> Thus, I can get the conclusion that
> Corr(X,S)=0 if and only if Corr(X,Y)=0 and Corr(X,Z)=0.
> Corr(X,S)=0+c if and only if Corr(X,Y)=0+c and Corr(X,Z)=0+c.
> ( 0<c<<1 )
>
> That is what I wanted. Because I don't know the value of Corr(X,Y)and
> Corr(X,Z), I just want to decide their relationship from the value
> of Corr(X,S).
>
> Do you have some idea about that?
> Thank you very much for your kind help.
>

You can find out many useful things using algebra, starting from the definition
of correlation,

Corr(X, Y) = Covariance(X,Y) / ((Stdev(X) * (Stdev(Y))

where

Covariance(X,Y) = Sum((Xi*mean(X))*(Yi*mean(Y)) / N

Stdev(X) = sqrt(Var(X))

Var(X) = Sum((Xi*mean(X))*(Xi*mean(X))) / N


It's fairly easy, for example, to find an equation for Cov(X,S) given Cov(X,Y)
and Cov(X,Z), if S = Y + Z. Each Si,and mean(S), can be expressed in terms of
Yi,Zi,mean(Y),and mean(Z).

I'm not sure what knowing just Corr(X,S), or Corr(X,Y) and Corr(X,Z), will tell
you, except in the case where Corr(X,Y) = Corr(X,Z) = 1.

Suppose, for example, that Corr(X,S) = 0. One possibility is that Corr(X,Y)
and Corr(X,Z) are both 0. Another is that S is the zero-vector (Cov(X,S) = 0),
but Corr(X,Y) = 1 and Corr(X,Z) = -1.

Note that Corr(X,S) will depend in part on Stdev(S). In this case, since S = Y
+ Z,

Var(S) = Var(Y) + Var(Z)
Stdev(S) = sqrt(Var(Y) + Var(Z))

This makes the following:

> Corr(X,S)=0+c if and only if Corr(X,Y)=0+c and Corr(X,Z)=0+c.
> ( 0<c<<1 )

highly dubious.


Brian


roy

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Sep 6, 2002, 12:58:26 PM9/6/02
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a...@c.com (Brian) wrote in message news:<vMMd9.196832$vg.34...@twister.nyroc.rr.com>...

Thank you very much,Brian and Greg.
Yes, the example


> Suppose, for example, that Corr(X,S) = 0. One possibility is that Corr(X,Y)
> and Corr(X,Z) are both 0. Another is that S is the zero-vector (Cov(X,S) = 0),
> but Corr(X,Y) = 1 and Corr(X,Z) = -1.

described the difficulty I met. What I think is, is it possible to transform
the vector or transform the equation of correlation so that we can

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