How to prove two stochastic processes have the same distribution

Let $C([0,\infty), R)$ be the canonical space of continuous functions. Assume $(\Omega, \mathcal{F}, \{\mathcal{F}_{t}\}_{t\geq 0})$ is a measurable space with a filtration. Let $P, Q$ be two probability measures on $(\Omega, \mathcal{F})$. Assume $X_{t}$ and $Y_{t}$ are two stochastic processes adapted to $\{\mathcal{F}_{t}\}_{t\geq 0}$. If for any borel set $A$ and $t$, $$P(X_{t}\in A)= Q(Y_{t}\in A)$$ Can we conclude that the law on $C([0,\infty), R)$ induced by $(X_{t}, P)$ is the same as that of $(Y_{t}, Q)$? Any references are very appreciated.

• If it is not true. How about one of them is a Levy process. – Hengyu Mar 19 '14 at 5:12
• Just for Discussion. To strengthen the condition. If at the beginning, we require that\\$$X_{t} = Y_{t}$$ and $$P(X_{T} \in A ) = Q(X_{t}\in A )$$ Is $$law(X_{t}|P) = law(X_{t}|Q)$$. Here $P$ are $Q$ are mutually absolutely continuous with each other. – Hengyu Mar 19 '14 at 17:11

No. Try $(X_t)$ standard Brownian motion and $Y_t=\sqrt{t}\cdot Y_1$ for every $t$, where $Y_1$ is standard normal.
• That's ok. Thank you very much. I think when $(N_{t}, P)$ is a Levy process, the conclusion is true. The proof is as follows. If $(N_{t}, P)$ is a Levy process. Then its distribution of $(N_{t}, P)$ is infinitesimal divisible. We observe that the distribution of $(N_{t}, Q)$ is also infinitesimal divisible. Since there is a one-to-one correspondence of Levy process and infinitesimal divisible distribution( Sato's book: Levy process and infinitesimal distribution). Then we conclude that $law(N_{t}, P)=law(N_{t}, Q)$. – Hengyu Mar 19 '14 at 17:09