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Working on fairly unrelated stuff, I needed to prove the following, fairly easy results, and I wonder if anyone can provide references to the literature. Not being a probabilist I wouldn't know where to start looking.

I. Let $X$ and $Y$ be two integrable random variables, and let $X'$ and $Y'$ have the same distributions, respectively, and be indepdendent from $X$ and $Y$. Then $$ E[\min \{X,X'\} + \min \{Y,Y'\} - 2\min\{ X,Y'\}] \geq 0 $$ and equality holds iff $X$ and $Y$ have the same distribution

II. Let $X$ be a random variable, and let $\cal F$ be some sub-algebra. Let $Y$ have the same conditional distribution as $X$ over $\cal F$, indepndently from $X$ over $\cal F$, and let $Z$ have the same distribution as $X$, independently from $X$. Then $$ E[\min \{X,Y\}] \geq E[ \min \{X,Z\} ] $$ and equality holds iff $X$ is independent from $\cal F$.

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    $\begingroup$ I would suggest using functional notation for $\min,$ that is $\min(X, Y),$ otherwise the equations are very confusing. $\endgroup$
    – Igor Rivin
    Commented Sep 10, 2012 at 21:53

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If the cumulative distribution function of $X$ is $G$ while the probabiity density $g=G^\prime,$ then the probability density of $\min(X, X^\prime)$ is $2 g G.$ Similarly, if the CDF of $Y$ is $H,$ with density $h,$ the probability density of $\min(Y, Y^\prime)$ is $2 h H,$ and that of $\min(X, Y')$ is $g H + h G.$ So, your expectation is $2\int x (g G + h H - g H - h G) d x = \int x d(G-H)^2.$ If $G=H,$ the integral is $0,$ as you conjecture, but if $G\neq H$ the expectation is nonnegative if the random variable is nonnegative.

I don't quite understand the second question...

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  • $\begingroup$ Thank you, but I do know how to prove these things - my question is whether this exists in the literature. $\endgroup$
    – Itaï BEN YAACOV
    Commented Sep 11, 2012 at 7:04
  • $\begingroup$ To explain II, maybe a quick proof is best: You can reduce fairly easily to the case where $X$ is the indicator function of some event $A$. Let $f = P[A|{\cal F}]$. Then the inequality becomes $E[f^2] \geq E[f]^2$, i.e., $V(f) \geq 0$, and equality holds iff $V(f) = 0$, i.e., if $f$ is constantly $E[f] = P[A]$, i.e., if $A$ is independent from $\cal F$. $\endgroup$
    – Itaï BEN YAACOV
    Commented Sep 11, 2012 at 7:58
  • $\begingroup$ @Itai: re your first comment: I have a certain sense of deja-vu, so it probably does exist in the literature; looking it up is harder than proving it :( $\endgroup$
    – Igor Rivin
    Commented Sep 11, 2012 at 12:23

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