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 $F$.

  • $\begingroup$ "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? $\endgroup$ – Sank Sep 30 '18 at 1:59
  • $\begingroup$ Also would it make more sense to use $X(\omega)$=$sup${$t:F(t)<\omega$}? $\endgroup$ – Sank Sep 30 '18 at 2:20
  • $\begingroup$ I guess this would give you a left continuous function instead of a right continuous one. $\endgroup$ – Kabo Murphy Sep 30 '18 at 4:16
  • $\begingroup$ Would it matter which one I use? $\endgroup$ – Sank Sep 30 '18 at 5:18
  • $\begingroup$ It depends how distribution functions are defined. Most books define them to be right continuous. $\endgroup$ – Kabo Murphy Sep 30 '18 at 11:31

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