Difference between Modification and Indistinguishable

Solution 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. (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.

Solution 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,

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

Solution 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).

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.