Suppose we have three random variables X, Y, T.

X is positively correlated with T: we know $\rho_{XT}$ and it is greater than 0, let's say for the sake of argument 0.5.

Y is positively correlated with T: we know $\rho_{YT}$ and it is greater than 0; similarly about 0.5.

If we find that the product of X and Y, XY is positively correlated with T, does $\rho_{(XY)T}$ tell us anything new (like an interaction term in multivariate regression)? Or is high $\rho_{(XY)T}$ inevitable given $\rho_{XT}$ and $\rho_{YT}$?

Update: I identified a possibly duplicate question sum of correlation coefficients and correlation coefficient of product . Not quite a duplicate as the other question asks whether in the general case $\rho_{(XY)T} = \rho_{XT} + \rho_{YT}$ which it isn't. I am interested in the less specific question of whether high, positive $\rho_{(XY)T}$ is inevitable given high, positive $\rho_{XT}$ and $\rho_{YT}$, and whether $\rho_{(XY)T}$ tells us anything useful we didn't already know.

  • $\begingroup$ It may be inevitable if X and Y are restricted to be positive. Should I explore that case in my answer? $\endgroup$ – eric_kernfeld Dec 13 '17 at 16:48
  • $\begingroup$ Yes, please explore that - I didn't think it made a difference but your initial answer shows me it does. And in the case I'm dealing with all variables are positive. $\endgroup$ – Sideshow Bob Dec 13 '17 at 16:52
  • $\begingroup$ OK, I'll work on it. It's tricky. $\endgroup$ – eric_kernfeld Dec 13 '17 at 17:53

It is not inevitable. Suppose T is a shifted Bernoulli r.v. that takes the values -1 and 0 with probability 0.5. Suppose X=Y=T. Then XY is negatively correlated with T: as X, Y, T decrease to -1, XY is increasing.

EDIT: Suppose all the R.V.'s are positive. As a preliminary question, is it possible to have a negative correlation between X and Y? The answer is yes. For example, suppose X, Y, T each live in {1, 2}. Suppose T is again a 50-50 coin flip. Suppose that:

  • $X=Y=T$ with 20% probability,
  • $X\neq T$ but $Y=T$ with 40% probability
  • $X=T$ but $Y\neq T$ with 40% probability

This setup completely determines a joint distribution over $X,Y,T$ such that $Y\neq X$ with 80% probability, even though X and Y are each correlated with T (equaling it with 60% probability).

This possibility makes it hard for me to reason abstractly about XY and guarantee its behavior. It also opens up avenues for a counterexample: if X and Y are negatively correlated, then at least the motion of XY relative to Z could be dampened rather than amplified as one expects intuitively (or reversed, as when negative numbers are allowed).

In the example above, E[XY] is above 2, so XY is usually below its mean, even when T is above its mean. Unfortunately, there is one case where XYT is really big.

  • T, X, Y, (XY): % chance
  • 1, 1, 1 (1): 10
  • 1, 2, 1 (2): 20
  • 1, 1, 2 (2): 20
  • 2, 2, 2 (4): 10
  • 2, 1, 2 (2): 20
  • 2, 2, 1 (2): 20

My instinct is that this basic structure could be tweaked until cov[XY, T] is negative. Maybe I'll try it later today.

  • $\begingroup$ This has already clarified my thoughts a lot, thank you. The point you raise is that the actual values of x and y matter. In the general case you would suspect therefore that $\rho_{(XY)T}$ provides new information, even if we already know the other factors, and even if we know $\rho_{XY}$ as well? $\endgroup$ – Sideshow Bob Dec 13 '17 at 22:50
  • $\begingroup$ Yes, that is the idea. $\endgroup$ – eric_kernfeld Dec 14 '17 at 2:08

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