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If I have some variable that depends on Brownian motion, how do I see clearly that replacing that Brownian motion with a different Brownian motion won't affect the law of my variable?

To make this more concrete, consider the stochastic differential equation $X_t=X_0+\int_{0}^{t}{b(s,X_s)ds}+\int_{0}^{t}{\sigma(s,X_s)dB_s}$. Suppose we have sufficient conditions to guarantee existence and uniqueness. (If we assume $b$ and $\sigma$ are Lipschitz and $X_0\in L^p$ for $p\geq 2$, then we have existence and uniqueness of the solution $X_t$. Edit: By uniqueness in this case, I mean that, if we fix B and $X_0$ beforehand, then it is possible to show that any two processes that satisfy the same equation are equal almost everywhere. We did this in class using something like Picard iterates. But it's not obvious to me that you get uniqueness in law given the law of X_0 and B yet.)

If I switch the Brownian motion $B_s$ with some other Brownian motion $B'_s$, it feels like the law of the solution $X_t$ does not change, but how do we see this?

Edit: Similarly, I'm not even sure how to show that the law of a stochastic integral $\int_{0}^{t}{Y_sdX_s}$ depends only on the laws of $X_s$ and $Y_s$ without going back to the construction. (If we are given two different processes $Y$ and $Y'$ with the same law, we can think about how to approximate them both with elementary processes that try to make it clear that the integral converges to processes with the same law in $H^2$.)

Edit 2: Okay, I think I was asking why pathwise uniqueness implies uniqueness in law, which is standard but not obvious to prove by myself.

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You answered your own question in the statement! When you say "uniqueness" of a solution, in the context of SDE this means that the law of $X_t$ is uniquely determined by the laws of $X_0$ and and $B$. See, for example, Theorem 10.5 in the 2nd Ed. of Introduction to Stochastic Integration by Chung and Williams.

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  • $\begingroup$ I will take a look at that book when I get a chance to go to the library. However, the precise statement of uniqueness in that case that I know is that, if you have two processes $X$ and $X'$ that have the same initial conditions and satisfy the equation, then they are equal almost surely. This is done in our class using something like Picard iterates. Maybe it is obvious for most people, but I don't see how I get that the solution only depends on the law of X_0 and B for free yet. $\endgroup$
    – DCT
    Nov 20, 2013 at 1:03

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