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Let $(X,B)$, where $B$ is a standard BM and $X$ is process, satisfy the SDE $dX_t = b(t,X_t) + \sigma(t,X_t)dB_t$. Suppose that for some other standard BM $W$ and process $Y$ we have that the joint law of the processes $(X,B)$ and $(Y,W)$ are equal. How do I show that the second pair also satisfies the SDE? This was claimed in a talk, but no justification was given, and I am not sure how to prove it. Thanks!

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  • $\begingroup$ Can you clarify what you mean by "satisfy the SDE"? Are $(X,B)$ and $(Y,W)$ supported on the same probability space and you mean to show that on that space, $Y$ satisfies the SDE? $\endgroup$
    – user159517
    Apr 27 at 12:20
  • $\begingroup$ Check the concept of "weak solution" $\endgroup$
    – Chaos
    Apr 27 at 20:44

1 Answer 1

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Suppose $\Big(X,B,\Omega,\mathcal F, \big(\mathcal F_t\big)_{t\geq0}, \mathbb P\Big) \stackrel{\mathcal{Law}}{=} \Big(Y,W,\Theta,\mathcal G, \big(\mathcal G_t\big)_{t\geq0}, \mathbb Q\Big)$

Since $X$ is a solution to the SDE in the former space, we know that everything is correctly integrable in the latter one (this is simply because $\mathbb P_X = \mathbb Q_Y$).

Now, define $$\Phi(X,B) := \Big(\int_0^t b(s,X_s)ds + \int_0^t \sigma(s,X_s)dB_s\Big)_{t\geq0}$$ (and the other one analogously).

We know that $\mathbb P(X_t = \Phi(X,B)_t \forall t\geq0) = 1$. As a consequence, defining $\Phi(Y,W)$ analogously, $$\mathbb Q(Y_t = \Phi(Y,W)_t \forall t\geq0) = 1,$$ which translates into $$Y_t = \int_0^t b(s,Y_s)ds + \int_0^t \sigma(s,Y_s)dW_s$$ as processes, i.e. $(Y,W)$ is a solution to the SDE in $\Big(\Theta,\mathcal G, \big(\mathcal G_t\big)_{t\geq0}, \mathbb Q\Big)$.

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  • $\begingroup$ It's not that easy. The stochastic integral is not defined pathwise, but also depends on the filtration. It is therefore not at all clear that you can use the same map $\Psi$ to represent both $Y$ and $X$ - an argument is needed here. $\endgroup$
    – user159517
    May 13 at 12:54

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