# How to show that $(X,B)$ and $(Y,W)$ satisfy the same SDE if their joint law is equal?

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!

• 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? Apr 27 at 12:20
• Check the concept of "weak solution" Apr 27 at 20:44

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 $$\Big(\Theta,\mathcal G, \big(\mathcal G_t\big)_{t\geq0}, \mathbb Q\Big)$$.

• 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. May 13 at 12:54