# Prove that a distribution function is stochastically smaller than another distribution function.

I'm trying to prove two statements:

1.) Suppose $$X,Y$$ are random variables on $$(\Omega,\mathcal F, P)$$ such that $$P(X \leq Y)=1$$. Show that $$F_Y(x) \leq F_X(x)$$ for every $$x \in \mathbb R$$. (Show that $$F_X$$ is stochastically smaller than $$F_Y$$.) Here $$F_X=P(X \leq x)$$ is the distribution function of X.

2.) Suppose $$F$$ is stochastically smaller than $$G$$. Show that there exist random variables $$X,Y$$ on some probability space $$(\Omega,\mathcal F, P)$$such that $$F_X = F, F_Y =G$$, and $$P(X \leq Y)=1$$.

Try at 1.): For any $$x \in \mathbb R$$, let $$A_1$$ and $$A_2$$ sets of $$\Omega$$ such that: $$A_1= \{\omega: X(\omega) \le x\}, A_2 = \{\omega: Y(\omega) \le x\}.$$ Now, $$P(X \le x)=P(\{\omega: X(\omega) \le x\}) = F_X(x)$$ and $$P(Y \le x)=P(\{\omega: Y(\omega) \le x\}) = F_Y(x)$$ by defintion. Then since $$P(X\leq Y)=1$$, we can write $$X(\omega)\leq Y(\omega)$$ for every $$\omega \in \Omega$$, thus $$X(\omega)\leq Y(\omega)\leq x$$. It follows that $$A_2\subset A_1$$, and $$F_Y(x)=P(Y\leq x)=P(A_2)\leq P(A_1)=P(X\leq x)=F_X(x)$$ by monotonicity of probability measures, thus $$F_X$$ is stochastically smaller than $$F_Y$$. As I was asking I actually found my proof to be very similar to the OP's proof here. I just wanted to check if the proofs were ok.

Try at 2.): Since $$F,G$$ are distribution functions, there exists a probability space $$(\Omega,\mathcal F, P)$$, and random variables $$X:\Omega \to \mathbb R, Y:\Omega \to \mathbb R$$ such that $$F=F_X,G=F_Y$$. But I'm not really sure where to go from this point on.

Any advice or help would be appreciated!

1) is OK. For 2) there is a standard construction. Consider the probability space $$(0,1)$$ with Borel sets and Lebesgue measure. Let $$X(\omega)=\inf \{t: F(t) \geq \omega \}$$ and $$Y(\omega)=\inf \{t: G(t) \geq \omega \}$$. It is trivial to check that $$X(\omega) \leq Y(\omega)$$ for all $$\omega$$. To show that the distribution of $$X$$ is $$F$$ show that the statements $$X(\omega) \leq t$$ and $$F(t) \geq \omega$$ are equivalent. [You will need right continuity of $$F$$ for this]. Taking probability of $$\{\omega: X(\omega) \leq t\}$$ you will see that $$X$$ indeed has distribution $$F$$. Similarly, $$Y$$ has distribution $$F$$.
• "To show that the distribution of $X$ is $F$ show that the statements $X(\omega) \leq t$ and $F(t) \geq \omega$ are equivalent." Why is this so? – Sank Sep 30 '18 at 1:59
• Also would it make more sense to use $X(\omega)$=$sup${$t:F(t)<\omega$}? – Sank Sep 30 '18 at 2:20