Another way to see that a Brownian Bridge is a Strong Markov Process

535 Views Asked by At

From this Is Brownian bridge a Markov process , I can see that $X_{t}=B_t-tB_1$, Brownian Bridge, is a Markov Process.

Does exist another way to see it without using Ito processes but just the definition? What about the strong Markov property?

EDIT: I proved without using Ito processes that the Brownian Bridge is a Markov Process. I used the follow property:

$$X_t \mbox{ satisfies the Markov Property} \quad \mbox{ iff } \quad X_t \mbox{ is indipendent of } X_z \mbox{, if we know } X_s \quad \mbox{with } z<s<t$$

Now, I want to prove (or disprove) if $X_t$ satisfies the strong Markov Property. Could you help me?