I have an elementary question about stochastic processes (continuous) and the notion of two processes being indistinguishable. Let $X=(X_t)_{t\geq 0}, Y=(Y_t)_{t\geq 0}$ be two stochastic processes on the same probability space $(\Omega, \mathcal{F}, P)$. I have the following definition, from Karatzas, Shreve - Brownian Motion and Stochastic Calculus.


Now there is this another definition, which comes from Kannan, Lakshmikantham - Handbook of Stochastic Analysis with Aplications.


Basically it says that there is a measurable set $A\in\mathcal{F}$ such that $P(A)=1$ and $X_t(\omega)=Y_t(\omega)$ for all $\omega\in A$ and all $0\leq t < \infty$. I've seen both definitions in different kind of formats, for example, this second definition is equivalent to this one, from Achim Klenke- Probability Theory, A Comprehensive Course.


In this case, he is writing $\mathcal{A}$ instead $\mathcal{F}$.

The last two definition are well posed but the first one it's not, because $$\{X_t=Y_t,\forall\ 0\leq t < \infty\} = \bigcap_{0\leq t < \infty}\{X_t=Y_t\}$$ is an uncountable intersections of measurable sets, and we know that this is not necessarily a measurable set. Therefore, if someone writes $P(X_t=Y_t, \forall\ 0\leq t < \infty )$, then he/she is just hoping to have a $\sigma$-algebra rich enough to comport this set or he/she is just wrong (because they want to measure a non-measurable set).

On the other hand, the last two definition (in special, the definition from Kannan's) makes we think in a way that lead us to the first (and not necessarily well posed) definition. In fact, with that set $A\in\mathcal{F}$ defined above, we can kind of say $A = \{\omega\in\Omega: \ X_t(\omega)=Y_t(\omega),\forall\ 0\leq t <\infty \}$, maybe not correct in some points of a null measure set. Now we can kind of write $P(A) = P(X_t=Y_t,\forall\ 0\leq t < \infty)$. Then we get the first definition, which would be correct, except, maybe, in a null measure set.

I have 2 interpretations of this situation, and I hope you can help me to have a correctly interpretation (maybe all my interpretations are wrong):

1) The first definition is in fact just a notation that we use when the conditions from the second definition are satisfied;

2) The first definition is not a notation, it means exactly this, but it's implied on the definition that the set $\{X_t=Y_t,\forall\ 0\leq t < \infty\}$ is measurable.

Thank you for the help.

  • $\begingroup$ In asserting "$\bigcap_{0\leq t < \infty}$" is uncountable you are implicitly assuming the parameter space is uncountable. Is that assumption actually made by the sources you are quoting? $\endgroup$ – whuber Sep 22 '15 at 16:48
  • $\begingroup$ @whuber Could you clarify? Because we have $t\in[0, \infty)$ and an intersection in this index set looks clearly not countable. $\endgroup$ – Integral Sep 22 '15 at 16:50
  • $\begingroup$ the parameter (index) space is uncountable, it's a continuous time parameter and it is explicitly said that $t$ is a real number varying continuously in $[0,\infty)$. $\endgroup$ – Integral Sep 22 '15 at 16:54
  • $\begingroup$ OK, that's good to know. (In many applications, $t$ is a natural number or integer.) BTW, the Shreve and Karatzas definition (which I presume is the "first definition" that concerns you) is identical to the second one, because "$P[X_t=Y_t\ \forall t]$" is shorthand for $P\{\omega\,|\,X_t(\omega)=Y_t(\omega)\ \forall t\}$, which can be read as the probability of the set of $\omega$ for which the two paths are the same. $\endgroup$ – whuber Sep 22 '15 at 16:55
  • 1
    $\begingroup$ Isn't it usually assumed--or explicitly stated--that all measures have been completed? Thus, whether or not the sets in question are measurable shouldn't matter; all that matters is whether they (or their complements) are subsets of sets of measure zero. I believe this notion is implicit in the definition of "almost all." $\endgroup$ – whuber Sep 22 '15 at 17:23

In view of whuber's comments, I will post an answer. I think this settled the question and I don't feel this definition is strange anymore.

Saying that $X,Y$ will be indistinguishable if $P(X_t=Y_t,\forall\ 0\leq t < \infty) = 1$ was strange to me because we are asking for uncountable intersection to be measurable. So we may try weaken the definition saying that $X,Y$ will be indistinguishable if exists a measurable set $A\in\mathcal{F}$ such that $P(A) = 1$ and $X_t = Y_t$ in $A$ for all $0\leq t < \infty$.

It's true that $A\subset\{\omega\in\Omega: \ X_t(\omega)=Y_t(\omega), \forall\ 0\leq t < \infty \}$, but it may not be equal. Therefore there is a null set $N\subset\Omega$ such that $A\cup N = \{\omega\in\Omega: \ X_t(\omega)=Y_t(\omega), \forall\ 0\leq t < \infty \}$.

As pointed out by whuber in the comentns, the probability space is supposed to be complete. So we have that $N$ is measurable, with $P(N) = 0$. In this case, we can just include $N$ in the definition and use $A\cup N$ instead of $A$. Therefore, the weak definition implies the strong, thus the definitions are equivalent.

PS: by a null set I mean a set contained in a measurable set with zero measure. Maybe the latter is called null set, I'm not sure. This is irrelevant when the space is complete, but I wanted to highlight the fact that this is relevant when the space is not complete.

| cite | improve this answer | |

The "for all" suggests an intersection as opposed to a union. $$ \{X_t = Y_t; \forall 0 \leq t < \infty\} = \bigcap_{t = 0}^{\infty} \{X_t = Y_t \}. $$

| cite | improve this answer | |
  • $\begingroup$ Yeah, you are right. It was a mistake. Still, this does not answer the question, for uncountable intersections are too not necessarily measurable. $\endgroup$ – Integral Sep 22 '15 at 16:35
  • 1
    $\begingroup$ Sigma algebras are closed under countable unions and intersections. $\endgroup$ – jlimahaverford Sep 22 '15 at 16:42
  • 2
    $\begingroup$ Yeah, countable, which is not the case here. $\endgroup$ – Integral Sep 22 '15 at 16:48
  • $\begingroup$ Ah, indeed. Then I agree with you that definitions (2) and (3) are better defined. That being said, if the processes are continuous then agreement on $\mathbb{Q}$ would mean agreement everywhere. $\endgroup$ – jlimahaverford Sep 22 '15 at 17:28
  • $\begingroup$ Intersection of disjoint sets? $\endgroup$ – Alecos Papadopoulos Sep 22 '15 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.