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I'm trying to understand the proof that if $X,Y$ are modifications $(P\{X_t = Y_t\}=1\,\, \forall t \in T$) of each other and are right continuous, that they are then indistinguishable ($P\{X_t = Y_t\,\, \forall t \in T\}=1)$. What I've seen so far is the following:

Right continuity of $X,Y$ allows us to write \begin{align*} P\{X_t = Y_t\,\, \forall t \in T\} &= P\{X_t = Y_t\,\, \forall t \in T \cap \mathbb{Q}\} \\ &=\bigcap_{q \in T \cap \mathbb{Q}}P\{X_q = Y_q\} \end{align*}

and the set $P\{X_q = Y_q\}=1$ for all $q \in T$ so since we take the countable union of such sets, $P\{X_t = Y_t\,\, \forall t \in T\}$ = 1.

What I'm struggling with is why right-continuity is so important here? How does that allow us to approximate the process with rational time points?

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What exactly are $X_t$? I guess, a random variable for all $t\in T$ and that $T\subseteq\Bbb R$. Are there other coniditions for $X_t$ in general? And, how exactly this right continuity is defined for these? – Berci Feb 27 '13 at 11:46
Sorry, I figured that would be clear from the context. X and Y are continuous time stochastic processes. The right continuity means that all realisations of these two processes are right continuous in the calculus sense. So for any $\omega \in \Omega$ we have $X_t(\omega)$ is right continuous. – BallzofFury Feb 27 '13 at 12:41
I think you meant intersection, instead of union. – André Caldas Feb 27 '13 at 13:30
Right you are, changed it. – BallzofFury Feb 27 '13 at 16:19
up vote 7 down vote accepted

Note: I edited this answer to be much more detailed after BallzofFury's comment.

Since $X$ and $Y$ are modifications of each other we have $P(X_t \neq Y_t) = 0$ for all $t \in T$. Then we have \begin{align*} A' :=& \{\omega : X_t(\omega) = Y_t(\omega) \,\,\forall t\in T\cap\mathbb{Q}\} \\ =&\bigcap_{t\in T\cap\mathbb{Q}} \{\omega : X_t(\omega) = Y_t(\omega)\} \\ =& \bigg(\bigcup_{t\in T\cap\mathbb{Q}} \{\omega : X_t(\omega) \neq Y_t(\omega)\}\bigg)^C. \end{align*} Since a countable union of negligible sets is negligible we have $P(A') = 1$. In addition, we can set $$ B := \{\omega : X(\omega) \text{ is right-continuous}\} $$ and $$ C := \{\omega : Y(\omega) \text{ is right-continuous}\}. $$ By assumption $P(B) = P(C) = 1$. Then their intersection $A:=A' \cap B \cap C$ is exactly the collection of those paths which are right-continuous and where the two processes agree on the index set $T\cap\mathbb{Q}$ and we have $P(A) = 1$. Since $T\cap\mathbb{Q}$ is dense in $T$ we have for every $t \in T$ a sequence $(q_n) \subset T\cap\mathbb{Q}$ that tends to $t$ from the right. Then by the construction of the set $A$ and right-continuity we have $$ \lim_{n \to \infty} X_{q_n}(\omega) = X_t(\omega) \quad \text{and} \quad \lim_{n \to \infty} Y_{q_n}(\omega) = Y_t(\omega) $$ for all $\omega \in A$ and since $X_{q_n}(\omega) = Y_{q_n}(\omega)$ we must have $X_t(\omega) = Y_t(\omega)$ for all $\omega \in A$.

For stochastic processes their properties are always understood to hold for almost all paths. So saying that a process $X$ is right-continuous means that almost all paths of $X$ are right-continuous, i.e. $$ P\{\omega : X(\omega) \text{ is right-continuous}\} = 1. $$ So to show that a process has some desirable property we have to show that almost all its paths have that property. In your question the property we want is that the paths of the process are identical to those of the other process. We do this by first showing that they agree for almost all paths on a dense, countable index set and then we use right-continuity to show that for those paths they actually agree on the whole index set T. Thus the desired property holds for almost all paths.

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Yes, I get that, but what I'm wondering is why continuous (usually right) functions are approximable by rationals in such a fashion. How can we show that this works, and preserves what we want? – BallzofFury Feb 27 '13 at 14:34
Awsome, the sequence $q_n$ in $\mathbb{Q}$ was the missing link for me between right-continuity and the equality. Many thanks for the taking time to write this up! – BallzofFury Feb 27 '13 at 16:05
No problem, it was my pleasure. – Gibarian Feb 27 '13 at 16:09

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