Let $X,Y$ be positive-valued, well-behaved random variables. Further, let $g(\cdot) \ge 0$ and $f(\cdot)\ge 0$ be two functions and $E(\cdot)$ denotes expectation operator. I am trying to prove the following relationship \begin{equation*} E(f(X))E(Y^2g(Y))+E(X^2f(X))E(g(Y)) \ge 2E(Xf(X))E(Yg(Y)). \end{equation*}

This is valid when both $X$ and $Y$ follow a Bernoulli distribution. I want to extend the assertion to general continuous random variables.

I have asked this question here, too. But it is found off-topic. One comment, though, suggests some hint: Divide by $Ef(X)Eg(Y)$. Treat $f(X)/Ef(X)$ and $g(Y)/Eg(Y)$ as Radon-Nikodym derivative. Bound from above the right side by sum of squares. Then apply Cauchy-Shwartz.

My problem is that I am not sure what Radon-Nikodym derivative is. When, I apply the hint what I get is the following: \begin{equation*} \frac{E(f(X))E(Y^2g(Y))} {E(f(X))E(g(Y))} + \frac {E(X^2f(X))E(g(Y))}{E(f(X))E(g(Y))} \ge \frac {2E(Xf(X))E(Yg(Y))} {E(f(X))E(g(Y))}, \end{equation*} which is equivalent to \begin{equation*} \frac{E(Y^2g(Y))} {E(g(Y))} + \frac {E(X^2f(X))}{E(f(X))} \ge \frac {2E(Xf(X))E(Yg(Y))} {E(f(X))E(g(Y))}. \end{equation*}

Now, there should be some $f(X)/Ef(X)$ to be treated as Radon-Nikodym derivative but I do not see any. Any help is much appreciated!

  • $\begingroup$ Are $Y$ and $X$ independent? $\endgroup$ – Alexander Vigodner Apr 22 '15 at 15:27
  • $\begingroup$ Yes, in my case $X$ and $Y$ are independent, but I guess the above result should be obtained for even dependent random variables. Thanks! $\endgroup$ – emper Apr 22 '15 at 15:56
  • $\begingroup$ Well I am not sure but for independent case it is easy to prove. See my answer in a few minutes $\endgroup$ – Alexander Vigodner Apr 22 '15 at 15:57

Assume $X$ and $Y$ are independent. Let's redefine $Y$ and $X$ to $\tilde Y$ and $\tilde X$ so that $$ E( \tilde Y^n )=E(Y ^n g(Y))/E(g(Y))\\ E( \tilde X^n )=E(X ^n f(X))/E(f(X))\\ $$ Since $g(\ )/E(g(Y))$ and $f(\ )/E(f(X))$ are densities with respect to the distributions of $Y$ and $X$ respectively, this is possible. Then $$ E((\tilde Y-\tilde X)^2)=E(\tilde Y^2)+E(\tilde X^2)-2E(\tilde Y)E(\tilde X)\ge 0 $$ Getting back to original $Y$ and $X$ you get your inequality.

PS: if $\mu_{Y,X}$ is the probability distribution of $(Y,X)$ then $$ \nu_Y(A)=\int_A g(y)E(g(Y))^{-1}d\mu_Y(y)\\ \nu_X(A)=\int_A f(x)E(f(X))^{-1}d\mu_X(x) $$ are the probability distributions of $\tilde Y$ and $\tilde X$ respectively.

REMARK. It was assumed that $E(f(X))>0$ and $E(g(Y))>0$. In the vanished case when $E(f(X))=0$ or $E(g(Y))=0$ the result follows immediately.

REMARK 2. I used the fact that $\tilde X,\tilde Y$ are independent assuming that $X,Y$ are independent. Indeed if $X,Y$ are independent then for any measurable $z(x,y)$ $$ \int_{A\times B} z(x,y)d\mu =\int_Ad\mu_x \int_B z(x,y)d\mu_y $$ But then $$ \int_{A\times B} z(x,y)g(y) f(x)d\mu =\int_A f(x)d\mu_x \int_B z(x,y) g(y)d\mu_y=\int_{A\times B } z(x,y) d\nu $$

  • $\begingroup$ Thanks! This is the exact thing I was looking for! Much appreciated! $\endgroup$ – emper Apr 23 '15 at 2:44
  • $\begingroup$ I tried to correct as many misprints as I could. Remain the independence considerations, which are not dealt with at present and should be reworded. For example one should assume (and one should show that one can assume) that $\tilde X$ and $\tilde Y$ are independent. $\endgroup$ – Did Apr 23 '15 at 6:47
  • $\begingroup$ Thanks Did. Your remark is important.Probably we can deduct it from independence of $X,Y$, however I don't see it immediately and may be it is even wrong. $\endgroup$ – Alexander Vigodner Apr 23 '15 at 15:23
  • $\begingroup$ Well, I think I proved that independence of $(X,Y)$ implies independence of $(\tilde X,\ tilde Y)$ $\endgroup$ – Alexander Vigodner Apr 23 '15 at 17:55

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