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Mar 20, 2021 at 20:20 vote accept Trade Paul
Mar 20, 2021 at 19:45 answer added Carlo Beenakker timeline score: 2
Mar 20, 2021 at 19:34 comment added Trade Paul 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 $E[X] > 0$.
Mar 20, 2021 at 19:24 comment added rubikscube09 If $X$ is mean zero then I believe $\mathrm{Cov}(X,R(X)) = \int_\Omega X(\omega) R(X(\omega))d\mathbb{P}(\omega) = \int_{X>0} (X(\omega))^2 d\mathbb{P}(\omega) \geq 0$
Mar 20, 2021 at 19:21 review First posts
Mar 20, 2021 at 19:55
Mar 20, 2021 at 19:16 history asked Trade Paul CC BY-SA 4.0