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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.

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

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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.

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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.

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  • $\begingroup$ 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, 2015 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, 2015 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, 2015 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, 2015 at 16:55
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    $\begingroup$ 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, 2015 at 17:23

2 Answers 2

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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.

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

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  • $\begingroup$ 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, 2015 at 16:35
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    $\begingroup$ Sigma algebras are closed under countable unions and intersections. $\endgroup$ Sep 22, 2015 at 16:42
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    $\begingroup$ Yeah, countable, which is not the case here. $\endgroup$
    – Integral
    Sep 22, 2015 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. $\endgroup$ Sep 22, 2015 at 17:28
  • $\begingroup$ Intersection of disjoint sets? $\endgroup$ Sep 22, 2015 at 18:37

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