# Suppose that $f(x_1,y_1)f(x_2,y_2) \le f(x_1,y_2)f(x_2,y_1)$ holds for $x_1 \le a \le x_2$ and $y_1 \le b \le y_2$ .

Let $(X,Y)$ have the joint density function $f$ and joint distribution function $F$.Suppose that $f(x_1,y_1)f(x_2,y_2) \le f(x_1,y_2)f(x_2,y_1)$ holds for $x_1 \le a \le x_2$ and $y_1 \le b \le y_2$ .Show that $F(a,b) \le F_{x}(a)F_{y}(b)$ where $F_{x}$ and $F_{y}$ are marginal dfs. I tried by finding that RHS is $F_{X}(a)F_{Y}(b)=(F_{X}(x_1)+ \int_{x_1}^{a}f(x,y) dy)(F_{Y}(y_1)+ \int_{y_1}^{b}f(x,y) dx)$ But I don't know how to proceed next

• The condition which you have written for $x_1 \leq a \leq x_2$ and $y_1 \leq b \leq y_2$ does not seem to depend on $a,b$, it's written only in terms of $x_i,y_i$. Please clarify, because I like this question. – астон вілла олоф мэллбэрг Aug 31 '17 at 3:08
• The question itself says this.You can think of $a$ being a number between $x_1,x_2$ and $b$ a number between $y_1,y_2$. Nothing more is said in the original question too. – Legend Killer Aug 31 '17 at 3:23
• Okay, I have understood the question. – астон вілла олоф мэллбэрг Aug 31 '17 at 3:29

I liked your question, so $+1$.

Let's consider the inequality $f(x_1,y_1)f(x_2,y_2) \leq f(x_2,y_1)f(x_1,y_2)$.

Integrate both sides over corresponding intervals, as follows(involving four variables, so we have to integrate four times). It's all rigorously done, since Fubini's theorem holds. $$\int_{-\infty}^a\int_{-\infty}^b\int_{a}^{\infty}\int_{b}^\infty f(x_1,y_1) f(x_2,y_2) \leq \int_{-\infty}^a\int_{-\infty}^b \int_{a}^{\infty} \int_{b}^\infty f(x_2,y_1) f(x_1,y_2)$$

Your task : rearrange the integrals on the left and right hand sides so that you get the following inequality: $$P(x \leq a, y \leq b) P (x \geq a, y \geq b) \leq P(x \leq a, y \geq b)P(x \geq a, y \leq b)$$ Once this is done, merely for our convenience, we define: $$A := P(x \leq a, y \leq b) ; B := P (x \leq a, y \geq b); C:=P(x \geq a, y \leq b)$$ Once we have done the above, it's clear that the above statement says $A(1-A-B-C) \leq BC$.

Rearrange all the terms, to get $A \leq A^2 + AB+AC+BC = (A+B)(A+C)$.

The left hand side is $A = F(a,b)$, and the right hand side, if seen carefully, is: $$A+B = P(x \leq a) = F_x(a) ; A+C = P(y \leq b) = F_y(b)$$

Hence, the result follows.

Of course, if you do not get your "task", then do get back, I'm happy to help.

• I am thrilled by your solution! – Legend Killer Sep 2 '17 at 3:29
• I too feel nice, since this is the first bounty problem I solved. Took me forty five minutes. I would be honoured it you award me the bounty (whenever it is possible). – астон вілла олоф мэллбэрг Sep 2 '17 at 3:32
• I would like to know more from you...Is there any way I can contact you? – Legend Killer Sep 2 '17 at 8:21
• How would you like to contact me? Would you like an email id? By the way, I answer all kinds of questions, this is not really my expertise. – астон вілла олоф мэллбэрг Sep 2 '17 at 8:24
• yup sure, sure email id is perfect – Legend Killer Sep 2 '17 at 8:26