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Let $(\Omega ,\mathcal F,\mathbb P)=([0,1],\mathcal B([0,1],m)$ where $m$ is the Lebesgue measure. Let $X_t=0$ for all $t$ and $Y_t(\omega )=\mathbb 1_{t=\omega }$.

Even if $X_t(\omega )=Y_t(\omega )$ for all $t$ and all $\omega \neq t$ (i.e. for all $t$, except at one point, they are the same processes), we have that $$\mathbb P(\forall t\geq 0, X_t=Y_t)=0.$$ So it looks that to be indistinguishable, we should have $X_t(\omega )=Y_t(\omega )$ for all $t$ and all $\omega $. But, if this would be the case, I guess that we would define indstinguishable processes as $X_t(\omega )=Y_t(\omega )$ for all $t$ and all $\omega $, and not as $\mathbb P(\forall t, X_t=Y_t)=1$. So, are there indistinguishable processes $(X_t)$ and $(Y_t)$ s.t. $X_t(\omega )= Y_t(\omega )$ for all $t$ and all $\omega $ doesn't hold ?

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2 Answers 2

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Two processes $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are a modification of each other if

$$\mathbb{P}(X_t=Y_t)=1 \quad \text{for all $t \geq 0$},$$

i.e. for each $t \geq 0$ there exists an exceptional null set $N=N(t)$ such that $$X_t(\omega) = Y_t(\omega) \quad \text{for all $\omega \in \Omega \backslash N(t)$}.$$

If the processes $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are indistinguishable, then the exceptional null set can be chosen independently from $t$, i.e. there exists a null set $N$ such that

$$X_t(\omega) = Y_t(\omega) \quad \text{for all $t \geq 0$, $\omega \in \Omega \setminus N$}.$$

Equivalently, $\mathbb{P}(\forall t \geq 0: X_t=Y_t)=1$. This means that the processes can differ on a null set. Note that null sets might be quite "large", depending on the measure which we consider.

Take, for instance, $\Omega = [0,1]$ with the Dirac measure $\mathbb{P}:=\delta_0$, and define

$$X_t(\omega) := \sin(\omega t) \qquad Y_t(\omega)=0.$$

We have $X_t(0)=0=Y_t(0)$ and so

$$\mathbb{P}(\forall t \geq 0: X_t=Y_t) \geq \delta_0(\{0\})=1.$$

Consequently, $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are indistinguishable although they are (from the "natural" point of view) quite different processes. In particular, $X_t(\omega) \neq Y_t(\omega)$ for "many" $t$ and $\omega$.

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In your example take $Y_t(\omega)=1_{t}(\omega)$ for $\omega\in N$ and $0$, otherwise, where $N\ne\emptyset$ but $m(N)=0$.

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