My question concerns the definition 1.3 page 2 of Brownian motion and stochastic calculus of Karatzas & Shreve.

Recall that a stochastic process is a collection of random variables (r.v.) $$ X = \{ X_t, t \in [0,\infty) \}. $$ Each of these r.v. is a measurable map from $(\Omega,\mathcal{F})$ (the sample space) to $(S,\mathcal{S})$ the state space.

Then the definition of indistinguishability between two stochastic processes $X$ and $Y$ defined on the same probability space $(\Omega,\mathcal{F}, \mathbb{P})$ is as follows: $X$ and $Y$ are indistinguishable if $$ \mathbb{P} (X_t = Y_t, \forall t \in [0,\infty) ) = 1. $$

I am confused by a measurability issue: Why $\cap_{t \geq 0} \{ X_t = Y_t \} \in \mathcal{F}$? (not countable intersection). I think it is implicitly assumed.


I think it is an excellent question and remark!

In practice, in the sequel, they will consider at least processes mesurables (Definition 1.6), that is $(t,\omega)\mapsto X_t(\omega)$ being mesurable : in this case the definition of indistinguishability has no such measurability issues.

But in the general case, to avoid that problem, we could easily add a condition such as :

There exists $A\in \mathcal{F}$ such that $A\subset \{X_t=Y_t, \ \forall t\geq 0\}$ and $P(A)=1$.

(Maybe that's what they had in mind?)

  • $\begingroup$ Thanks for the clarification. $\endgroup$ – megaproba Feb 4 '18 at 3:53

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