Let $X$ be a random variable taking values on the real line. Let $R(X) = max\{0, X\}$. Is it true that the covariance $Cov[X, R(X)] \ge 0$ irrespective of the distribution of $X$? Many experiments, as well as the intuition seem to suggest that their covariance must be non-negative. Is it true? If so, how can I prove it?
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$\begingroup$ If $X$ is mean zero then I believe $\mathrm{Cov}(X,R(X)) = \int_\Omega X(\omega) R(X(\omega))d\mathbb{P}(\omega) = \int_{X>0} (X(\omega))^2 d\mathbb{P}(\omega) \geq 0$ $\endgroup$– rubikscube09Commented Mar 20, 2021 at 19:24
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$\begingroup$ No, it need not be zero mean. In fact, if the mean is negative, or zero, the non-negativeness emerges trivially from the definition : $ cov(X, R(X)) = E[X.R(X) ] - E[X]E[R(x)]$. Because $E[R(x)]$ is always non-negative. But I can't figure out the case for $E[X] > 0$. $\endgroup$– Trade PaulCommented Mar 20, 2021 at 19:34
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$${\rm Cov}\,(X,R(X))= \int_0^\infty x^2P(x)\,dx - \left(\int_{-\infty}^\infty xP(x)\,dx\right)\left(\int_{0}^\infty xP(x)\,dx\right)$$ $$=\int_0^\infty x^2P(x)\,dx - \left(\int_{0}^\infty xP(x)\,dx\right)^2- \left(\int_{-\infty}^0 xP(x)\,dx\right)\left(\int_{0}^\infty xP(x)\,dx\right)$$ $$={\rm Var}\,R(X)- \left(\int_{-\infty}^0 xP(x)\,dx\right)\left(\int_{0}^\infty xP(x)\,dx\right)\geq 0.$$ The variance is positive by definition and the second term is minus the product of a negative and a positive integral.
answered Mar 20, 2021 at 19:45