# multiplying two random variables by a third increases their correlation?

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.

• 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. – Acccumulation Nov 8 '17 at 0:44
• @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 – Glen_b Nov 8 '17 at 0:52
• 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) – Glen_b Nov 8 '17 at 1:10
• 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. – Glen_b Nov 8 '17 at 1:22
• 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. – whuber Nov 8 '17 at 15:07

xi <- rnorm(10000)