Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

When I have shown, for $s\le t$ and for two continuous stochastic process an inequality:

$$ X_s \le Y_t$$ P-a.s.

How can I deduce that this P-a.s. simultaneously for all rational $s\le t$ ? Thank you for your help

EDIT: According to Ilya's answer, I see that we have $$P(X_s\le Y_t\text{ simultaneously for all rationals }s\le t) = 1.$$ How could we use continuity of $X,Y$ to deduce $P(X_s\le Y_t,s\le t)=1$. Of course we take sequences of rational, however I mess up the details. So a detailed answer how to do this, would be appreciated.

share|improve this question
Hint: there are countably many pairs of rationals $s \le t$. –  Robert Israel May 7 '12 at 16:35
I have updated my answer, using outcome-wise approach. –  Ilya Jun 26 '12 at 15:44
I think, now it is very explicit. –  Ilya Jun 26 '12 at 16:15

2 Answers 2

up vote 1 down vote accepted

If I got you correct, you have $P(X_s\leq Y_t) = 1$ for all $s\leq t$. Recall that if $P(A_n) $ for all $n\in \mathbb N$ then also you have $P(\bigcap_n A_n) = 1$ since $$ P(\bigcap_n A_n) = 1-P(\bigcup_n A^c_n)\geq 1-\sum\limits_n P(A^c_n) =1. $$ Now take $s_n$ be the $n$-th rational number that is less or equal to $t$ and $A_n=\{X_{s_n}\leq Y_t\}$. You can even use continuity of $X,Y$ to show that $P(X_s\leq Y_t,s\leq t) = 1$.

To show the latter, consider two sets: $$ C = \{\omega\in \Omega: X_t(\omega),Y_t(\omega)\text{ are continuous }\} $$ $$ D = \{\omega\in \Omega:X_s(\omega)\leq Y_t(\omega) \text{ simultaneously for all rational }s\leq t\}. $$

It is given to us that $\mathsf P(C) = 1$ and we have proved above that $\mathsf P(D) = 1$. As a result, $$ \mathsf P(C\cap D) = 1-\mathsf P(C^c\cup D^c)\geq 1-\mathsf P(C^c)-\mathsf P(D^c) = 1. $$

Consider now any $\omega\in C\cap D$.

  1. It holds that for this $\omega$: $X_s(\omega)\leq Y_t(\omega)$ for all rational $s\leq t$. Since both $X,Y$ are continuous on $\omega$, it follows that $X_s(\omega)\leq Y_t(\omega)$ for all $s\leq t$. Indeed, if that would not be true, i.e. for some $s'\leq t$ you have $X_{s'}(\omega)>Y_{t}(\omega)$ then $X_s(\omega)>Y_t(\omega)$ in the neighborhood of $s'$ which cannot happen since there will be at least one rational number $s$.

  2. Thus, for any $\omega\in C\cap D$ the desired relation holds: $C\cap D \subseteq \{X_s\leq Y_t,s\leq t\}$, so that $$ 1 =\mathsf P(C\cap D) \leq \mathsf P\{X_s\leq Y_t,s\leq t\}. $$

share|improve this answer
@ Ilya: So I agree we have $P(X_s\le Y_t \mbox{ simultaneously for all rational $s\le t$})$. But how could I use continuity of $X,Y$ to deduce $P(X_s\le Y_t,s\le t)=1$? If you could this explain in a more detailed way, I will accept your answer. Thank you for your help. Sorry for the late response! –  user20869 Jun 25 '12 at 11:49
@hulik: You show that $\{X_s\leq Y_t, s\leq t, s\in\mathbb{Q}\}\subseteq \{X_s\leq Y_t,s\leq t\}$. Pick an $\omega$ in the left-hand set. Then $X_s(\omega)\leq Y_t(\omega)$ for all $s\leq t$ with $s\in\mathbb{Q}$. Now suppose that $u\leq t$ is arbitrary. Then we pick a sequence $(u_n)\subseteq \mathbb{Q}$ such that $u_n\to u$ for $n\to\infty$. Since $X_{u_n}(\omega)\leq Y_t(\omega)$ for all $n$, the continuity gives that $X_u(\omega)=\lim_{n\to\infty} X_{u_n}(\omega) \leq Y_t(\omega)$ and hence $\omega$ is in the right-hand set. –  Stefan Hansen Jun 25 '12 at 12:32
@ Stefan Hansen: If I understand you right, you want to show, that $\{X_s\le Y_t,s\le t,s,t\in \mathbb{Q}\}=\{X_s\le Y_t,s\le t\}$ and the first set has measure 1. But why do you take just the set $\{X_s\le Y_t,s\le t,s\in \mathbb{Q}\}$? Thanks for your help –  user20869 Jun 25 '12 at 15:20
@ Ilya: Thanks for your updated answer. However, please I would like to have a detailed answer of this. I totally agree that the intersection has probability one. But somewhere you need a careful approximation argument as Stefan Hansen suggested. Sorry for nit-picking! But I have troubles showing these things in a detailed way. –  user20869 Jun 26 '12 at 16:01
@hulik: please, tell me - is anything unclear in the new version –  Ilya Jun 27 '12 at 8:01

This follows from the fact that the complement of the event $[\forall s\leqslant t,\,X_s\leqslant Y_t]$ is the event $$ \left[\exists s\leqslant t,\,X_s\gt Y_t\right]\ =\left[\exists n\in\mathbb N,\,\exists s\in\mathbb Q,\,\exists t\in\mathbb Q,\,s\leqslant t,\,X_s\geqslant Y_t+\frac1n\right], $$ hence $$ \left[\exists s\leqslant t,\,X_s\gt Y_t\right]\ =\bigcup\limits_{s\leqslant t,\, s\in\mathbb Q,\,t\in \mathbb Q}\ \bigcup_{n\geqslant1}\ \left[X_s\geqslant Y_t+\frac1n\right]. $$ Since $\mathrm P(X_s\leqslant Y_t)=1$ for every $s\leqslant t$, $\mathrm P(X_s\geqslant Y_t+\frac1n)=0$ for every $n\geqslant1$. The union on the RHS of the displayed identity above is countable hence $\mathrm P(\exists s\leqslant t,\,X_s\gt Y_t)=0$. Considering the complement, one gets $$ \mathrm P(\forall s\leqslant t,\,X_s\leqslant Y_t)=1. $$

share|improve this answer
I do not see how I can use this to conclude that $P(X_s\le X_t,s\le t)$ a.s. And shouldn't it be an intersection over $n$ instead of an union? –  user20869 Jun 26 '12 at 15:34
The union is correct. See Edit. –  Did Jun 26 '12 at 16:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.