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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 $P(X<Y)=0$.

Any tips on what to use to prove this result?

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  • $\begingroup$ You should use different symbols for the random variables as opposed to the distribution functions. In any case $X=Y$ works. $\endgroup$ – herb steinberg Oct 13 '18 at 22:01
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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 $Y(\omega) \leq t$.

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  • $\begingroup$ What if we aren't working in $(0,1)$ though? $\endgroup$ – Dole Oct 15 '18 at 0:12
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    $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Oct 15 '18 at 0:30
  • $\begingroup$ Any tip on how to prove the final part, that is $P(X<Y)=0$? $\endgroup$ – Dole Oct 15 '18 at 0:48
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    $\begingroup$ 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 $Y(\omega)$. $\endgroup$ – Kavi Rama Murthy Oct 15 '18 at 4:33

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