# What is the Difference Between a Version and a Modification of a Stochastic Process?

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 $Y$?

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.

• So is indistinguishability the true "analog" of one-dimensional functions being versions of each other (i.e. almost everywhere equal)? May 14, 2016 at 4:35
• 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? May 14, 2016 at 4:37
• 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. May 14, 2016 at 4:42
• 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. May 14, 2016 at 4:45
• 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. May 14, 2016 at 5:03

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.

• 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/… May 3, 2018 at 2:13
• @Chill2Macht I had the same idea of using $\bigcap$ to understand uniform continuity the other day
– BCLC
May 3, 2018 at 2:14
• @Chill2Macht WHO ARE YOU XD
– BCLC
May 29, 2018 at 14:27
• That's a somewhat random question isn't it Jun 3, 2018 at 17:49
• @Chill2Macht it's a joke. Like it's as if you're me from the future or another dimension (crickets chirp)
– BCLC
Jun 4, 2018 at 0:32