Correlation Proof

[statsq45]

Hi guys, wondering if you could help me with a proof that I am trying to figure out. Here it is...

Corr (A, B) = 0 and Corr (B, C) = 0. This implies that Corr (A, C) = 0, prove this is true? If not provide a counter example.

I started out by saying sigma(a,b)/(sigma(a)*sigma(b))=0 and did the same for corr(b,c) and then set them equal to each other. Then I solved for sigma(a) and then substituted that into the equation sigma(a,c)/(sigma(a)*sigma(c))=0 but I don't know where to go from there.

Am I doing this right, or am I missing something. Thanks for any responses!

I apologize for double posting in different forums.

[CowboyBear]

I apologize for double posting in different forums.

No worries. I have deleted the other thread. Don't worry too much about which forum you place a question in - they'll get seen regardless via the latest posts feature.

This isn't a question I can really help with (I'm about as much good with the mathematical side of statistics as I am at tap-dancing ), but intuitively I'm thinking of the case where C is just a linear transformation of A; therefore presumably:

Corr (A, B) = 0; Corr (B, C) = 0; Corr (A, C) = 1.

[statsq45]

Thanks for the response. By "corr" I mean correlation and the two variables signify the two data points that are supposed to be used to find the correlation. I don't know if a linear transformation would work here would it? Any other ideas from anyone?

[BGM]

Supporting CowBoyBear

If you are not sure the meaning of "C is just a linear transformation of A",
in fact it just mean something like

Of course it include , the identity itself.

So is it possible to put ?
If yes, then it is the simplest counter example you can give.

[statsq45]

Supporting CowBoyBear

If you are not sure the meaning of "C is just a linear transformation of A",
in fact it just mean something like

Of course it include , the identity itself.

So is it possible to put ?
If yes, then it is the simplest counter example you can give.

What do you mean by "So is it possible to put ?". Seems to me if it is a linear transformation, which it is, then it must be possible. Correct?

[BGM]

I mean if you allowed to put (or any linear transformation),
then it should be a valid counter example to your problem.

[ichbin]

It's not true. Here is a simple counter-example:

Suppose the points of the scatterplot of A vs. B lie in a perfect circle around the origin. Then Corr(A,B) = 0, because for any value of A, there are equally likely, equally large positive and negative values of B. Suppose furthermore than A = C, so Corr(A,C) = 1. The the points of the scatterplot of B vs. C will also lie on the same circle, so Corr(B,C) = 0. So we have Corr(A,B) = 0 and Corr(B,C) = 0 but Corr(A,C) != 0. Q.E.D.

This is, by the way, a classic illustration of how correlation does not necessarily measure association.

[Dason]

It's not true. Here is a simple counter-example:

Suppose the points of the scatterplot of A vs. B lie in a perfect circle around the origin. Then Corr(A,B) = 0, because for any value of A, there are equally likely, equally large positive and negative values of B. Suppose furthermore than A = C, so Corr(A,C) = 1. The the points of the scatterplot of B vs. C will also lie on the same circle, so Corr(B,C) = 0. So we have Corr(A,B) = 0 and Corr(B,C) = 0 but Corr(A,C) != 0. Q.E.D.

This is, by the way, a classic illustration of how correlation does not necessarily measure association.

Well correlation does measure association. Just not every type of association. If we're talking about pearson correlation then it's just measuring linear association.