0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $$ \int_{A\times B} z(x,y)g(y) f(x)d\mu =\int_A f(x)d\mu_x \int_B z(x,y) g(y)d\mu_y=\int_{A\times B } z(x,y) d\nu $$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.