Covariance of functions of two random variables

Question

I am trying to show the following: Suppose that x and y follow identical and independent distribution. F(x, y) is an increasing function of x and y. Is there any way to show that the covariance between F(x, y) and F(y, x) is positive or not? Thank you so much!

Answer 1

Let $X,X_1,X_2,Y_1,Y_2,Y$ have the distribution of $X$ (same as $Y$) and are independent. Since $F(x,y)$ is increasing in both arguments, we have $$ [F(X_2,Y)-F(X_1,Y)][F(Y,X_2)-F(Y,X_1)]\ge 0 $$ as both multipliers are either simultaneously positive or simultaneously negative. Taking the expectation and using the fact that all the r.v. have the same distribution, we get $$ \mathbb{E} \left[F(X,Y)F(Y,X)\right]\ge \mathbb{E} \left[F(X_2,Y)F(Y,X_1)\right]. $$ Next, by the same token, $$ [F(X_2,Y_1)-F(X_1,Y_1)][F(Y_2,X_2)-F(Y_2,X_1)]\ge 0 $$ so $$ \mathbb{E} \left[F(X,Y_2)F(Y_1,X)\right]\ge \mathbb{E} \left[F(X_1,Y_1)F(Y_2,X_2)\right]. $$ However, $$ \mathbb{E} \left[F(X,Y_2)F(Y_1,X)\right]=\mathbb{E} \left[F(X_2,Y)F(Y,X_1)\right] $$ because all $X$s and $Y$s have the same distribution. ALso, trivially, $$ \mathbb{E} \left[F(X_1,Y_1)F(Y_2,X_2)\right]=\mathbb{E} \left[F(X,Y)\right] \mathbb{E} \left[F(Y,X)\right]=\left(\mathbb{E} \left[F(X,Y)\right]\right)^2 $$ Combining all above, we get $$ \mathbb{E} \left[F(X,Y)F(Y,X)\right]\ge \mathbb{E} \left[F(X,Y)\right] \ \mathbb{E} \left[F(Y,X)\right] $$ thus the covariance of $F(X,Y)$ and $F(Y,X)$ is indeed non-negative.