# centering two variables $X$ and $Z$ makes $cov (X,XZ) = 0$

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.