Difference between Modification and Indistinguishable

Question

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,

John

Answer 1

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. 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.

Answer 2

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,

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

Answer 3

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).

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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.

Answer 4

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:

$\textbf{Solution:}$

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,

John