Assumme a variable $x$ is positively correlated with an event $A$, by which I mean that $\rho_{x,1_{A}} > 0$, where $\rho$ is the correlation coefficient and $1_{A}$ is the indicator function of the event $A$. Is there a simple argument why from here follows $\rho_{x,1_{\neg A}} < 0$, i.e., $x$ is negatively correlated with the negation/complementary event of $A$? Moreover, is it true that $\rho_{x,1_{A}} = -\rho_{x,1_{\neg A}}$? Both statements appear to be highly intuitive, and I am rather confident that they are true. However, I do not have lots of experience with stochastics and do not see an immediate argument to make this plausible which is also brief enough to be put in a footnote, and I would prefer not to start a full proof based on the definition of the correlation coefficient etc.
1 Answer
Let $1$ be the constant function identically equal to $1$. Then $\operatorname{cov}(x,1) = 0$ because covariance with a constant is zero. Since $\operatorname{cov}(x,y+z)=\operatorname{cov}(x,y)+\operatorname{cov}(x,z)$ and $1 = 1_A+1_{\neg A}$, it follows that $\operatorname{cov}(x,1_A) = -\operatorname{cov}(x,1_{\neg A})$. Since $\operatorname{var}(a+bx) = b^2\operatorname{var}(x)$ and $1_A=1-1_{\neg A}$, it follows that $\operatorname{var}(1_A) = \operatorname{var}(1_{\neg A})$ and therefore their standard deviations are also equal. The desired result then follows from the definition of correlation as covariance divided by the product of the standard deviations.
Note that it is necessary in this argument to go via the covariance because correlation with a constant is undefined (it is of the form $0/0$).
- $\begingroup$ Thank you very much, Jayanth! I will check your argument later today and then assign the green check. $\endgroup$Florian Biermann– Florian Biermann2025-08-13 12:22:16 +00:00Commented Aug 13 at 12:22