Would someone be able to offer a layman's explanation of what is means when two stochastic processes are a Modification of each other and when they are Indistinguishable?

My Stochastic Analysis notes define the following:

a) $\textit{The stochastic processes X and Y are called a modification of each other if}$ \begin{align} P(X(t) = Y(t)) = 1 \quad \textit{for all t $\in$ I} \end{align} b) $\textit{The stochastic processes X and Y are called indistinguishable}$ \begin{align} P(X(t) = Y(t) \,\,\,\textit{for all $t \in I$}) = 1 \end{align}

I interpret this to mean that:

(a) for each $t$ in $I$ the probability that the processes are equal is equal to 1.

(b) the probability that the entire path of the processes X is equal to the entire path of the process Y, is equal to 1.

I don't understand the difference between the two statements. The first seems to examine the processes at each point. The second examines the paths of the processes (which are made up of the points!).

Is anyone able to provide a motivational explanation as to why (a) does not mean that the paths are the same? I find it counter intuitive since if the processes are equal at each point then surely their paths are the same?

The classic example I've been given which (apparently) shows the two processes to be a modification and not indistinguishable is:

$\textbf{Example:}$ Let $\Omega = [0,\infty), \mathcal{A} = \mathcal{B}([0,\infty))$ and $P$ be a probability measure on $\mathcal{A}$ which has a density. Define two stochastic processes $(X(t): t \ge 0)$ and $(Y(t): t \ge 0)$ by \begin{align} X(t)(\omega) = \begin{cases} 1, \text{ if $t = \omega$},\\ 0, \text{ otherwise} \end{cases} \quad Y(t)(\omega) = 0 \quad \text{for all $t \ge 0$ and all $\omega \in \Omega$.} \end{align} Then $X$ and $Y$ are modifications of each other but $X$ and $Y$ are not indistinguishable.

$\textbf{Question:}$ With regard to the example above if $t = 0$ and $\omega = 0$ then $X(0)(0)= 1$ but $Y(0)(0) = 0$, then how can their probability be equal to 1 at this point? Hence how can they be a modification of each other?

Many thanks,


  • $\begingroup$ Do you understand that b) implies a) while a) (if $I$ is uncountably infinite) does not imply b)? In that sense the statements are different. $\endgroup$
    – drhab
    Feb 16 '15 at 9:40
  • 1
    $\begingroup$ @drhab. Thanks. Unfortunately not, this is why I've specifically asked for a "layman's explanation". I'm hoping someone will be able to convey a technical construct in a clear intuitive manner. $\endgroup$
    – John Smith
    Feb 16 '15 at 9:47
  • $\begingroup$ They are modification of each other because I think $X(t)(w)$ is always equal to zero. So it always equal to $Y(t)(w)$ at each t. Because the probability that w equal to t is zero I think because the probability measure at a countable set (or a single point) is zero. $\endgroup$
    – john_w
    Dec 23 '19 at 2:47

For consistency of terminology, let me say that $X_t$ and $Y_t$ are $M$-equivalent if one is a modification of the other, and $D$-equivalent if they are indistinguishable from one another. (This is not standard terminology, but I find the difference in syntax between the two terms makes it slightly difficult to write.)

Here's a way of looking at it based on sampling.

Suppose $X_t,Y_t$ are $M$-equivalent. Now choose $t_1 \in I$ arbitrarily. Repeatedly run $X_t$ and $Y_t$ "independently", but only sample them at time $t_1$. Then as you sample more and more times, the fraction of the samples such that $X_{t_1}=Y_{t_1}$ will converge to $1$, regardless of which $t_1$ you chose.

Suppose now that they are $D$-equivalent. Repeatedly run $X_t$ and $Y_t$ again, but this time record the entire trajectory. Then as you sample more and more times, the fraction of the samples such that $X_t$ and $Y_t$ are equal at every time will converge to $1$.

So $D$-equivalence automatically implies $M$-equivalence, since almost all samples have equality at every time and hence at any particular time.

When $I$ is countable, $M$-equivalence also implies $D$-equivalence, because the event "$X_t=Y_t$ for every $t \in I$" is the intersection of the events "$X_{t_k}=Y_{t_k}$" over $t_k \in I$, and each of these has probability $1$.

But when $I$ is uncountable (as in problems in continuous time), we may have $M$-equivalence but not $D$-equivalence. To see this, suppose $X_t,Y_t$ are $M$-equivalent and let $N(t)$ be the event that $X_t \neq Y_t$. Then $N(t)$ has probability zero. (If you like, this is just saying that $\int_t^t dx = 0$.) Then we are in $N=\bigcup_{t \in I} N(t)$ if there is some $t$ such that $X_t \neq Y_t$. Now $N$ is an uncountable union of sets of probability zero. So it might have positive probability, or it might not even be measurable.

Let's imagine sampling from the example from Oksendal. Pick a $t_1 \in [0,1]$, now the random time will be $t_1$ only with probability zero. So the fraction of samples with $X_{t_1} \neq Y_{t_1}$ will get smaller as we take more and more samples. But if we look at the whole trajectory instead, then at some random time we will always have $X_t \neq Y_t$, in every single sample. We will never see the exact same trajectory from both.

  • 1
    $\begingroup$ How do you know $N=\cup_{t\in I}N(t)$ is an event? An uncountable union of events is not necessarily an event. $\endgroup$
    – Integral
    Sep 22 '15 at 2:20
  • $\begingroup$ @Integral OK that's fair, in principle you could have such a small $\sigma$-algebra that only the finite dimensional distributions of your stochastic processes can be properly defined. But even in that case, you have $M$-equivalence without having $D$-equivalence. This doesn't tend to happen, because we tend to work with sufficiently rich $\sigma$-algebras to be able to say these kinds of things. But I'll edit to clarify anyway. $\endgroup$
    – Ian
    Sep 22 '15 at 13:06
  • $\begingroup$ for $t\in \mathbb R$, one can approximate the continuous sigma algebras with the events for time in $\mathbb Q$, the latter is countable. $\endgroup$ Jan 6 '21 at 11:32
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    $\begingroup$ @MarineGalantin Whether that is sufficient depends quite strongly on the surrounding context; for example it is a useful technique when the paths are assumed continuous a priori. But in general it does not save the day. $\endgroup$
    – Ian
    Jan 6 '21 at 15:03

Let's have a closer look at the given example:

Fix $t \geq 0$. By definition, $$X(t)(\omega) = 0 = Y(t)(\omega)$$ for all $\omega \neq t$. Consequently,

$$\{\omega; X(t)(\omega) \neq Y(t)(\omega)\} \subseteq \{t\}.$$

As $X(t)(t) = 1 \neq 0 = Y(t)(t)$, we get

$$\{\omega; X(t)(\omega) \neq Y(t)(\omega)\} = \{t\}.$$

Since $P$ is a probability measure with density, this shows

$$\mathbb{P}(X(t) \neq Y(t)) = \mathbb{P}(\{t\})=0,$$

i.e. $(X_t)_{t \geq 0}$ is a modification of $(Y_t)_{t \geq 0}$.

It remains to show that $(Y_t)_{t \geq 0}$ and $(X_t)_{t \geq 0}$ are not indistinguishable. To this end, fix $\omega \in \Omega$ and note that

$$X(\omega)(\omega) =1 \neq 0= Y(\omega)(\omega),$$

i.e. for each $\omega \in \Omega$ there exists $t \in I$ such that $$X(t)(\omega) \neq Y(t)(\omega).$$ Consequently,

$$\mathbb{P}(X(t) = Y(t) \quad \text{for all} \, t \in I) = 1-\mathbb{P}(\exists t \in I: X(t) \neq Y(t)) = 1-1 = 0.$$

  • $\begingroup$ Thanks. Just to clarify your use of notation. Am I to understand that $\{\omega; X(t)(\omega) \ne Y(t)(\omega) \} \subseteq \{t\}$ means the set generated by $X(t)(\omega) \ne Y(t)(\omega)$ is a subset of all of the sets that are generated by $t$? $\endgroup$
    – John Smith
    Feb 16 '15 at 10:44
  • $\begingroup$ @JohnSmith No. Both $A:=\{\omega \geq 0; X(t)(\omega) \neq Y(t)(\omega)\}$ and $[0,\infty)$ and $B:=\{t\}$ are subsets of $[0,\infty)$. So, $A \subseteq B$ means $\omega \in A \implies \omega \in B$. $\endgroup$
    – saz
    Feb 16 '15 at 11:30
  • $\begingroup$ math.stackexchange.com/questions/3093990/… This question is about a proof in your book ;) $\endgroup$ Jan 30 '19 at 23:48
  • $\begingroup$ @Anoldmaninthesea. I can take a look at the question but it's not "my" book... $\endgroup$
    – saz
    Jan 31 '19 at 7:26

How I differentiate between modification and indistinguishable is that the former is a pointwise property whereas the latter is a uniform property (recall pointwise converegence and uniform convergence).

Let me give it a try by using definition.

For a fixed $t\geq 0,$ note that \begin{align*} \mathbb{P}\left( X(t) = Y(t) \right) & = \mathbb{P}\left( X(t) = 0 \right) \\ & = \mathbb{P}\left( \left\{ \omega\in\Omega: X(t)(\omega) = 0 \right\} \right) \\ & = \mathbb{P}\left( \left\{ \omega\in\Omega: \omega \neq t \right\} \right) \\ & = \mathbb{P}\left( \Omega \setminus \{t\} \right) \\ & = 1. %should be 1 here instead of 0 \end{align*} Therefore, $X$ and $Y$ are modifications of each other.

On the other hand, note that \begin{align*} \mathbb{P} \left( \bigcap_{t\geq 0} \left\{ \omega\in\Omega: X(t)(\omega) = Y(t)(\omega) \right\} \right) & = \mathbb{P} \left( \bigcap_{t\geq 0} \left\{ \omega\in\Omega: X(t)(\omega) = 0 \right\} \right) \\ & = \mathbb{P} \left( \bigcap_{t\geq 0} \left\{ \omega\in\Omega: \omega\neq t \right\} \right) \\ & = \mathbb{P} \left( \bigcap_{t\geq 0} \Omega\setminus \{t\}\right) \\ & = \mathbb{P}(\emptyset) \\ & = 0 \end{align*} where the second last equality is because of $\Omega = [0,\infty).$ Therefore, $X$ and $Y$ are distinguishable.


Unfortunately I did not understand the other answers given thus far to this question.

After searching I found that the example I posted seems to come from Oksendal - Stochastic Differential Equations - p18 - q2.9.

Further searching found these solutions to Oksendal: https://www.quantnet.com/threads/is-there-solution-to-sde-by-%C3%98ksendal.8066/

The solution given there to the example problem is:


With \begin{align} X_t(\omega) = \begin{cases}1 \text{ if $t=\omega$}\\0 \text{ otherwise} \end{cases} \end{align} and $Y_t(\omega) = 0$ for all $(t,\omega) \in [0,\infty) \times [0,\infty)$ we have \begin{align} P[X_t = Y_t] = P[X_t = 0] = P(\{\omega; \omega \ne t\}) = 1. \end{align} Hence $X_t$ is a version of $Y_t$.

Unfortunately I still do not undertstand why $P(\{\omega; \omega \ne t\}) = 1$ - and more importantly my original question of obtaining a layman's explanation of what "modification" or "indistinguishable" really means remains unanswered.

If there are any teachers out there or academic staff whom have had experience explaining difficult concepts then I would really appreciate any contribution you could offer.

Many thanks,


  • 1
    $\begingroup$ George Lowther has written the following on 'Stochastic Processes, Indistinguishability and Modifications'. almostsure.wordpress.com/2009/11/03/… $\endgroup$
    – John Smith
    Feb 16 '15 at 16:00
  • $\begingroup$ So why do you not ask (specific) questions if you don't understand the provided answers? $\endgroup$
    – saz
    Feb 21 '15 at 7:20
  • $\begingroup$ @saz If you read my original post, I stated "would someone be able to offer a layman's explanation..." - unfortunately you were not able to provide a layman's explanation. You ignored that part on the post and instead focused on answering the example. It didn't help. Ian's response above does offer more of a layman's explanation. Regards. $\endgroup$
    – John Smith
    Feb 27 '15 at 10:09
  • $\begingroup$ I didn't claim that I fully answered your question (although I wouldn't say that I'm not able to provide a layman's explanation). However, you say in the above answer that you still do not understand the (counter)example. So, more precisely, my question is: Why do you not ask (specific) question (concerning the example) if you do not understand it? (And why do you not upvote Ian's response if you find it a helpful?) $\endgroup$
    – saz
    Feb 27 '15 at 10:26
  • $\begingroup$ @saz Thanks for the reply. I did not press you further on your example because after my first comment I realised it simply did not help myself understand the topic, i.e. my question. I did not ask "how do I solve this?", I asked for help trying to understanding the concepts in 'very basic terms'. Your reply - which I'm sure is 100% correct, doesn't help me understand. I'm sure you're a great mathematician, I appreciate the effort. I will upvote Ian's reply once I've read it more thoroughly and asked him any questions. $\endgroup$
    – John Smith
    Feb 27 '15 at 13:25

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