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I'm trying to get my head around the notion of "Law of a stochastic process" intuitively. This is what I got for a Brownian motion:

Denoting the law of a Brownian motion $\mathcal{L}_B:\mathcal{B}(C_\mathbb{R}{[0,1]})\rightarrow[0,1]$, given a set $A\in\mathcal{B}(C_\mathbb{R}{[0,1]})$, is the probability of the Brownian motion to take any of the paths in $A$.

Am I right? (I am just learning that stuff for the first time, so be gentle.)

If that is the case, and I'm trying to show that the laws of two different processes are mutually absolutely continuous, then I need to show that for every path that one of them wouldn't (i.e. with probability $0$) take, the other wouldn't take as well?

Thanks.

EZLearner
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  • Correct. 2) Not only every path, but every set of paths that has probability 0 under the one law has probability 0 under the other law too
    • – Bananach Apr 23 '16 at 19:26
  • For the second part - yeah, that's the definition. I'm trying to find a way to prove it. Suppose I show that for every finite substitution of times $0=t_1<...<t_n$, using the finite dimensional distribution, $\mathbb{P}((Y(t_1), ... , Y(t_n))\in A) = 0$ iff $\mathbb{P}((X(t_1), ... , X(t_n))\in A) = 0$, would that do it? – EZLearner Apr 23 '16 at 20:28
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    I have to admit I don't know. Now that you have a more succinct question I suggest you go and ask anew – Bananach Apr 24 '16 at 12:07