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:

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

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

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

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

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

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.

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  • $\begingroup$ The book is clear and correct. Your formulas are murky because you do not use enough parentheses to show what you are taking expectations of, making it impossible to assess where you are running into problems. They do suggest you're playing fast and loose with the rules of expectation, which you might want to review. $\endgroup$ – whuber Jan 30 '15 at 16:19
  • $\begingroup$ I added parentheses. $\endgroup$ – Lil'Lobster Jan 30 '15 at 16:58
  • $\begingroup$ That use of parentheses implies you are thinking of "$X$" and "$X$" as being independent variables! It is not generally the case that $E(X\cdot X) = E(X)E(X)$, but that's exactly what the last equality in your first equation is asserting. $\endgroup$ – whuber Jan 30 '15 at 17:08
  • $\begingroup$ Yes, I agree completely that $X$ 'can't' be independent of itself. In the last equality I simply state that $X\cdot X$ is $X^2$, so $E(X\cdot X\cdot Z)=E(X^2\cdot Z)$. I use the formula $cov(A,B)=E(A\cdot B)-E(A)\cdot E(B)$ where $A$ is my $X$ and $B$ is my $X\cdot Z$. $\endgroup$ – Lil'Lobster Jan 30 '15 at 17:16
  • $\begingroup$ I may have misread part of it. I'll post an answer. $\endgroup$ – whuber Jan 30 '15 at 17:20

You are correct. As a simple counterexample, consider the variables $X$ and $Z$ whose joint distribution has probability $1/2$ on $(X,Z)=(0,-1)$ and probability $1/4$ on each of $(\pm 1, 1)$. Thus both $X$ and $Z$ are centered at $0$; indeed, even the mean of $XZ$ is $0$, as you can readily check. The covariance of $X$ and $XZ$ therefore is the expectation of their product

$$\text{Cov}(X, XZ) = \mathbb{E}(X\cdot XZ) = \frac{1}{4}\left(1\right) + \frac{1}{2}\left(0\right)+\frac{1}{4}\left(1\right) = \frac{1}{2} \ne 0,$$

contradicting the assertion.

The authors might have had a bivariate normal distribution in mind. That has a bivariate symmetry: after centering, the distribution of $(X,Z)$ is the same as the distribution of $(-X,-Z)$. This time $$\text{Cov}(X, XZ) = \mathbb{E}(X\cdot XZ) = \mathbb{E}(-X\cdot (-X)(-Z)) = - \mathbb{E}(X\cdot XZ) = -\text{Cov}(X,XZ),$$

which is possible only when the covariance is zero.

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  • $\begingroup$ Is it possible to relax the assumption of bivariate normal to just bivariate symmetry distribution? $\endgroup$ – Lil'Lobster Jan 30 '15 at 17:56
  • $\begingroup$ Yes: as you can see, that is the only assumption needed to show the covariance is zero. Normality was not used. (But I believe it is rare to make such a bivariate symmetry assumption: as soon as the variables are assumed to be non-normal, symmetry of the joint distribution typically becomes implausible.) $\endgroup$ – whuber Jan 30 '15 at 17:59

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