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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?

Did
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Skills
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    "What about the strong Markov property?" Yes, what about it? Sorry but what do you mean with this sentence? – Did Apr 07 '18 at 08:43
  • @Did. I proved the first my problem. Could you know how I can go on? – Skills Apr 07 '18 at 09:19
  • 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 07 '18 at 09:43
  • @Did $X_t$ is a BM. Maybe is not a "iff" but only a sufficient condition? – Skills Apr 07 '18 at 12:31
  • Certainly not sufficient. And $X$ is not a Brownian motion. – Did Apr 07 '18 at 13:27
  • Ok. It's just a gaussian process. So is it a necessary condition? – Skills Apr 07 '18 at 13:50
  • Yes, if the process $(X_t)$ satisfies the Markov property then $X_t$ is independent of $X_z$ conditionally on $X_s$ for every $z<s<t$. But you will not prove that $(X_t)$ is Markov using only that. – Did Apr 07 '18 at 14:03
  • @Did In my notes I wrote that there are 2 other different ways to prove the Markov property over the classic way: $1$: $\mathbb{E}[f(B_{t+h}) \mid \mathcal{F}t]=\mathbb{E}[f(B{t+h}) \mid B_t]$; $2$: $B_z$ is indipendent of $B_t$ if we know $B_s$ with $z<s<t$ – Skills Apr 07 '18 at 14:11
  • But you see that 2. is much too weak, since one only conditions on a single $B_z$ with $z<t$ instead of on the whole past $\mathcal F_t$. – Did Apr 07 '18 at 14:53

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