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

probability distributionsrandom variablesstochastic-calculusstochastic-processesuniform 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.

Best Answer

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 $t$].