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

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?

• "What about the strong Markov property?" Yes, what about it? Sorry but what do you mean with this sentence? – Did Apr 7 '18 at 8:43
• @Did. I proved the first my problem. Could you know how I can go on? – Skills Apr 7 '18 at 9:19
• First, it is not true that the process $(X_t)$ satisfies the Markov property iff $X_t$ is independent of $X_z$ conditionally on $X_s$ for every $z<s<t$. – Did Apr 7 '18 at 9:43
• @Did $X_t$ is a BM. Maybe is not a "iff" but only a sufficient condition? – Skills Apr 7 '18 at 12:31
• Certainly not sufficient. And $X$ is not a Brownian motion. – Did Apr 7 '18 at 13:27