I've read that centering two normal (or symmetrical) variables $X$ and $Z$ affects correlation of centered $X$ with interaction term $X\cdot Z$ in such way, that this correlation $cor(X-EX, X\cdot Z)$ is $0$. I am not sure ... (here I use the numerator of correlation, which is covariance)

When I'm doing my own calculations I get stuck here:

$cov(X,X\cdot Z) = E(X\cdot X \cdot Z) - EX\cdot E(X\cdot Z) = EX^2\cdot Z - EX\cdot EXZ$

because without any information about independence between $X$ and $Z$ it's over. Even knowing that these two variables are normal it gives me nothing. At least me :-) The independence between $X$ and $Z$ would give me only that

$cov (X, X\cdot Z) = EX^2EZ-EX\cdot EX\cdot EZ = EZ\cdot varX$

It's not $0$. But the book 'says' explicitly:

$cov(X\cdot Z,X) = var(X)\cdot EZ + cov(X,Z)\cdot EZ$

If $X$ and $Z$ are centered, then $EX$ and $EZ$ are both zero, and the covariance between $X$ and $XZ$ is zero as well. Thus the correlation between $X$ and $XZ$ is also zero. The same holds for the correlation between $Z$ and $XZ$

So did I missed something (and the book is right) or ... is my thinking correct?

The book is "Applied Multiple Regression/Correlation Analysis for the Behavioral Sciences" by Cohen, Cohen, Aiken, West.


cov(X,Z).EZ has the same units than X.Z²

var(X).EZ has the same units than X².Z

cov(X.Z, X) has the same units than X².Z

so the equation cov(X.Z, X) = var(X).EZ + cov(X,Z).EZ cannot be right.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.