If the PDF satisfies $f(x_1,y_1)f(x_2,y_2)\leq f(x_1,y_2)f(x_2,y_1)$ then the CDF's satisfy $F(a,b)\leq F_X(a)F_Y(b)$.

Question

Let $(X,Y)$ have joint density function $f$ and joint distribution function $F$. Suppose that $$f(x_1,y_1)f(x_2,y_2)\leq f(x_1,y_2)f(x_2,y_1),$$ holds for $x_1\leq a\leq x_2$ and $y_1\leq b\leq y_2$. How to show that $$F(a,b)\leq F_X(a)F_Y(b)?$$ Here $F_X$ and $F_Y$ are the cumulative distribution functions of $X$ and $Y$ respectively.

My approach was this: $$ \begin{align}(F(a,b))^2 &= \left(\int_{-\infty}^a\int_{-\infty}^b f(x_1,y_1)dx_1dy_1\right)\left(\int_{-\infty}^a\int_{-\infty}^b f(x_2,y_2)dx_2dy_2\right) \\ &= \int_{-\infty}^a\int_{-\infty}^b\int_{-\infty}^a\int_{-\infty}^b f(x_1,y_1)f(x_2,y_2)dx_1dy_1dx_2dy_2 \\ &\leq\int_{-\infty}^a\int_{-\infty}^b\int_{-\infty}^a\int_{-\infty}^b f(x_1,y_2)f(x_2,y_1)dx_1dy_1dx_2dy_2. \\ \end{align} $$ But this leads to nowhere.

Answer 1

You have to use the hypothesis more specifically.

HINTS: $F_X(a) = \displaystyle \int_{-\infty}^a \int_{-\infty}^\infty f(x_1,y_2) dy_2\,dx_1$, $F_Y(b) = \displaystyle \int_{-\infty}^b \int_{-\infty}^\infty f(x_2,y_1) dx_2\,dy_1$. Consider \begin{multline*} \int_{-\infty}^b\int_a^\infty \int_{-\infty}^a \int_b^\infty f(x_1,y_2)f(x_2,y_1) dy_2\,dx_1\,dx_2\,dy_1 \\ = (F_X(a)-F(a,b))(F_Y(b)-F(a,b)). \end{multline*} But, changing the order of integration, this integral is greater than or equal to \begin{multline*} \int_b^\infty\int_a^\infty\int_{-\infty}^b \int_{-\infty}^a f(x_1,y_1)f(x_2,y_2) dx_1\,dy_1\,dx_2\,dy_2 \\= F(a,b)\big(1-(F_X(a)+F_Y(b)-F(a,b))\big). \end{multline*}