In other words, if the Pearson r between x and z, and between y and z, is zero, does it follow that x/y is uncorrelated with z?

  • $\begingroup$ Thanks for the edit. Yes, Pearson's $r$ is the correlation (or rather, we're discussing random variables, so we're talking about population correlation, $\rho$). If $\rho\neq 0$ then the variables are correlated, but the title and the body text still ask distinctly different questions. Please clarify which one is to be answered. I believe a quick, clear answer to the title question is possible (so I suggest modifying the body), but if you require an answer to the body question instead, modify the title. $\endgroup$
    – Glen_b
    Commented Nov 30, 2018 at 9:23
  • 2
    $\begingroup$ if we can't say it's uncorrelated that doesn't mean we couldn't say something. The first question could be answered in the negative simply by an example where the correlation is nonzero, but the second question (essentially, needing an answer to what things can we say about the correlation) would seem to require something approaching a characterization of the ways the correlation would or would not be zero; a considerably larger task $\endgroup$
    – Glen_b
    Commented Nov 30, 2018 at 9:29

1 Answer 1


The answer is no, here is a counter example:

$P(y = 1) = P(y = -1) = 0.5, ~ x = z y$ with $P(z = 1) = P(z = 2) = 0.5$ and $z$ independent of $y$. Since $z = x/y$, $z$ and $x/y$ are perfectly correlated. However, $z$ and $y$ are independent, therefore they are clearly uncorrelated. Finally: Cov(x,z) = $\mathbb{E} (xz) - \mathbb{E}(x) \mathbb{E}(z)$ with the expected value of $x$ being 0 and $\mathbb{E} (xz) = \mathbb{E} (yz^2) = \mathbb{E} (y) \mathbb{E} (z^2) = 0$ because the expected value of $y$ is again 0


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