# Proving $X$ and $Y$ are indistinguishable, then $X$ is a modification of $Y,$ implying that $X$ and $Y$ have the same finite-dimensional distribution.

Can someone please help augment my understanding in modification and indistinguishability. I am not particularly confident in my proof for the following problems. I am also unsure how to tackle the third component. Below is some of my work. Does it look OK? Thank you for your time and consideration.

Let $$X$$ and $$Y$$ be two processes defined on $$(\Omega, \mathcal{F}, \mathbb{P}).$$

a. Prove that if $$X$$ and $$Y$$ are indistinguishable, then $$X$$ is a modification of $$Y,$$ implying that $$X$$ and $$Y$$ have the same finite-dimensional distribution.

$$\textit{Proof.}$$ Let $$\{X_t \colon t \in T\}$$ and $$\{Y_t \colon t \in T\}$$ be two stochastic processes defined on the same probability space $$(\Omega, \mathcal{F}, \mathbb{P})$$ and have index set $$T.$$ Suppose there exists a measurable set $$A \in \mathcal{F}$$ so $$\mathbb{P}[A] = 1$$ and so for every $$\omega \in A, t\in T$$ it follows $$X_t(\omega) = Y_t(w).$$ Hence, they are indistinguishable. If they are indistinguishable then it follows that they are modifications of one another. Therefore, $$X$$ and $$Y$$ possess the same finite dimensional distribution.

b. Prove or disprove that if $$X$$ is a modification of $$Y,$$ then $$X$$ and $$Y$$ are indistinguishable.

$$\textit{Proof.}$$ We will disprove the following statement. Let $$U$$ be a random variable uniformly distributed on $$[0,1].$$ Define our two stochastic processes $$\{X_t\colon t \in [0,1]\}$$ and $$\{Y_t \colon t \in [0,1]\}$$ by $$i. X_t(\omega) = 0 \text{ for all } t \in [0,1], \text{ and } \omega\in\Omega$$ $$ii. \text{ for all } t\in [0,1] \text{ and } \omega \in \Omega,$$ $$Y_t(\omega) = \{ \begin{array}{ll} 1 & \mbox{if t = U(\omega)};\\ 0 & \mbox{otherwise}.\end{array}$$

Then $$X$$ is a modification of $$Y$$ since for all $$t \in [0,1]$$ it follows that $$\mathbb{P}[X_t = Y_t] = \mathbb{P}[Y_t = 0] = \mathbb{P}[U\ne t] = 1.$$ Further, $$X$$ and $$Y$$ are not indistinguishable as for every $$\omega \in \Omega$$ the sample paths $$t\mapsto X_t(\omega)$$ and $$t\mapsto Y_t(\omega)$$ are not equal as functions on $$T.$$ Particularly, $$Y_{U(\omega)}(\omega) = 1,$$ while $$X_{U(\omega)}(\omega) = 0.$$

c. If the statement b. was disproven, identify a set of additional assumptions $$A$$ so if $$X$$ is a modification of $$Y$$ and assumptions $$A$$ hold, then $$X$$ and $$Y$$ are indistinguishable.

What you have done so far is fine. For c) one standard additional assumption is right (or left) continuity of paths. Suppose there is a set $$E$$ such that $$P(E)=1$$ and $$\omega \notin E$$ implies that the functions $$t \to X_t(\omega)$$ and $$t \to Y_t(\omega)$$ are both right continuous. Suppose $$(X_t)$$ is a modification of $$(Y_t)$$. Then the processes are indistinguishable. I will let you try this and add a proof if necessary. [Hint: If $$X_t(\omega)=Y_t(\omega)$$ all rational $$t$$ then $$X_t(\omega)=Y_t(\omega)$$ for all $$t$$].