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Let $X, Y, Z$ be three random variables such that $Z$ and $X$ are independent, and $Z$ and $Y$ are independent. Let $\rho(X,Y)$ be the Pearson correlation coefficient between $X$ and $Y$. It seems to me that $\rho(XZ, YZ) \geq \rho(X, Y)$, but I can't prove it. I tried using the identity $\rho(X,Y) = \frac{Cov(X,Y)}{\sqrt{V(X)V(Y)}}$ and grinding through the algebra, to no avail.

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    $\begingroup$ Have you taken some sample X, Y, and Z and seen whether it's true for those variables? I think you will find it's not true, especially if they have mean zero. $\endgroup$
    – Acccumulation
    Commented Nov 8, 2017 at 0:44
  • $\begingroup$ @Accumulation It looks to me like the claim will be true if X and Y both have mean 0 (it would be equality in that case); the algebra is simple in that situation $\endgroup$
    – Glen_b
    Commented Nov 8, 2017 at 0:52
  • $\begingroup$ However, here's a clear counterexample in R by giving one of X and Y a zero mean and the other a large mean (relative to its sd), while Z has a moderate mean relative to its standard deviation: xi=rnorm(10000); y=.8*xi+.6*rnorm(10000); x=xi+10; z=rnorm(10000,1,1); cor(x,y); cor(x*z,y*z) $\endgroup$
    – Glen_b
    Commented Nov 8, 2017 at 1:10
  • $\begingroup$ As for the algebra, you'll need to make use of the result $\text{Var}(UV) = \text{Var}(U)\text{Var}(V)+E(U)^2 \text{Var}(V)+\text{Var}(U)E(V)^2$ for independent U and V. If you take all the variables as having variance 1 and Y as having mean 0, the calculations are considerably simplified. Taking the mean of Z to be 1 simplifies things further which makes it easier to see how to obtain counterexamples. $\endgroup$
    – Glen_b
    Commented Nov 8, 2017 at 1:22
  • $\begingroup$ The claim cannot generally be true for the simple reason that the inequality is reversed when either of $X,Y$ (but not both) is negated. But you will find that if you try to patch this up by considering the absolute value of the correlation, the inequality still doesn't hold. $\endgroup$
    – whuber ♦
    Commented Nov 8, 2017 at 15:07

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The claim is generally not true. Thank you @Glen_b for the following counterexample (in R): 

xi <- rnorm(10000)
y <- .8*xi+.6*rnorm(10000)
x <- xi+10
z <- rnorm(10000,1,1)
cor(x*z,y*z) - cor(x,y) > 0 # returns FALSE
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