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Under what circumstances would one say that:

The stochastic process $X$ is a version of the stochastic process $Y$?

Background: See here for a related but slightly different question on Math.SE.

Usually the word version is used most often in connection with conditional expectations, or general random variables, to mean that:

The random variable $X$ is a version of the random variable $Y$ iff: $$\mathbb{P}[X=Y]=1,$$ i.e $X=Y$ almost surely.

I have also heard the term used in reference to stochastic processes, but in this case I am not sure how it should be used, and how it relates to the terms modification and indistinguishable.

Let $X,Y$ be random functions (i.e. stochastic processes) mapping from the index set $T$ to a measurable space $\Omega$.

$X$ is a modification of $Y$ iff $$\forall\ t\in T,\ \mathbb{P}[X(t)=Y(t)]=1$$ and $X$ is indistinguishable from $Y$ iff $$\mathbb{P}[X(t)=Y(t),\ \forall\ t \in T]=1.$$

Note here the different placements of the logical quantifier $$\forall\ t\in T$$ outside vs. inside the definition of the set whose probability is in question between the two definitions.

However, under what circumstances would we say that:

The stochastic process $X$ is a version of the stochastic process $Y$?

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2 Answers 2

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The difference is really subtle. Citing Jeanblanc,Yor,Chesney (2009), they give the two following definitions:

The process X is a modification of Y if $\forall t$ $\mathbb{P}(X_t=Y_t)=1$.

The process X is indistinguishable from (or a version) of Y if {$\omega: X_t(\omega)=Y_t(\omega),\forall t$} is a measurable set and $\mathbb{P}(X_t=Y_t,\forall t)=1$

They moreover add the following relation: if $X$ and $Y$ are modifications of each other and are a.s. continuous, they are indistinguishable.

EDIT: There is an even more clear definition and explanation of the relation between the different definitions looking in Karatzas&Shreve (1998), p.2. Check it at this link.

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  • $\begingroup$ So is indistinguishability the true "analog" of one-dimensional functions being versions of each other (i.e. almost everywhere equal)? $\endgroup$
    – Chill2Macht
    May 14, 2016 at 4:35
  • $\begingroup$ Also do you know how this relates to "versions" of conditional distributions? (For example, "regular versions"?) I should probably ask this as a separate question shouldn't I? $\endgroup$
    – Chill2Macht
    May 14, 2016 at 4:37
  • $\begingroup$ Also then some people define "version" as two stochastic processes with the same distribution, where distribution refers to the finite-dimensional distributions which is equivalent to being a modification -- wait so is being a modification like "being equal in distribution" and being indistinguishable is "being equal almost surely"? Because then this definition makes more sense. $\endgroup$
    – Chill2Macht
    May 14, 2016 at 4:42
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    $\begingroup$ Regarding the first question, I think so, because there is the additional requirement of measurability of the path for indistinguishability. And in that case what you wrote in this last comment makes complete sense. $\endgroup$
    – Pasriv
    May 14, 2016 at 4:45
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    $\begingroup$ So in practice if $X$ and $Y$ are indistinguishable, it means almost all their sample paths agree. This implies that $X$ and $Y$ are modifications of each other, which in turn implies they have the same finite-dimensional distributions. $\endgroup$
    – Pasriv
    May 14, 2016 at 5:03
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Let $A_t = \{X(t)=Y(t)\}$ to see how we can generalise this.

$X$ is a modification of $Y$ iff $$\forall\ t\in T,\ \mathbb{P}[A_t]=1$$ and $X$ is indistinguishable from $Y$ iff $$\mathbb{P}[\bigcap_{t \in T} A_t]=1.$$

Remark: There's actually no difference if $T$ is countable.

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    $\begingroup$ I really like the way you wrote this -- it makes the difference very visible/visually clear. So it's a difference in the position of the quantifiers it seems. I guess similar to the difference between uniformly continuous and equicontinuous or something like that math.stackexchange.com/questions/1195848/… $\endgroup$
    – Chill2Macht
    May 3, 2018 at 2:13
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    $\begingroup$ @Chill2Macht I had the same idea of using $\bigcap$ to understand uniform continuity the other day $\endgroup$
    – BCLC
    May 3, 2018 at 2:14
  • $\begingroup$ @Chill2Macht WHO ARE YOU XD $\endgroup$
    – BCLC
    May 29, 2018 at 14:27
  • $\begingroup$ That's a somewhat random question isn't it $\endgroup$
    – Chill2Macht
    Jun 3, 2018 at 17:49
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    $\begingroup$ @Chill2Macht it's a joke. Like it's as if you're me from the future or another dimension (crickets chirp) $\endgroup$
    – BCLC
    Jun 4, 2018 at 0:32

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