# Monotone coupling of random variables.

Let $$F_1,F_2$$ be two cumulative distribution functions. Let also the distrubutions have the property that $$F_1\leq F_2$$.

I want to prove that there exists two random variables $$X,Y$$ such that:

$$F_X=F_1,F_Y=F_2$$, but $$P(X.

Any tips on what to use to prove this result?

• You should use different symbols for the random variables as opposed to the distribution functions. In any case $X=Y$ works. – herb steinberg Oct 13 '18 at 22:01

On $$(0,1)$$ with Lebesgue measure define $$X(\omega) =\inf \{t: F_1(t) \geq \omega \}$$ and $$Y(\omega) =\inf \{t: F_2(t) \geq \omega \}$$. These random variables have the required properties. Hint: $$F_1(t) \geq \omega$$ iff $$X(\omega) \leq t$$ and $$F_2(t) \geq \omega$$ iff $$Y(\omega) \leq t$$.
• What if we aren't working in $(0,1)$ though? – Dole Oct 15 '18 at 0:12
• You are asked to prove that there exist random variables $X,Y$ satisfying certain condition. It is entirely up to you to choose the probability space on which they are defined. – Kavi Rama Murthy Oct 15 '18 at 0:30
• Any tip on how to prove the final part, that is $P(X<Y)=0$? – Dole Oct 15 '18 at 0:48
• If $X(\omega) <Y(\omega)$ then there exists $t$ such that $Y(\omega)>t$ and $F_1(t) \geq \omega$. [This follows by definition of $X(\omega)$]. But $F_1(t) \leq F_2(t)$ so $F_2(t) \geq \omega$. But this contradicts the definition of $Y(\omega)$. – Kavi Rama Murthy Oct 15 '18 at 4:33