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I am currently studying the Brownian motion and I am stuck with a problem related to the reflection principle.

What I am trying to calculate is the probability that a standard Brownian Motion $X_t$ returns to zero given that it starts in $X_{t_a} = a$ and ends in $X_{t_b} = b$. ($t_a,t_b,a,b > 0$)

That is:

$$P [X_t = 0 \hspace{1 mm}for \hspace{1 mm}some \hspace{1 mm} t∈[t_a,t_b]|X_{t_a} = a, X_{t_b} = b], \quad t_a,t_b,a,b > 0.$$

Any answer or comment is greatly appreciated, thanks!

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    Your question is quite unclear, in particular its formulation in English and the formula do not match. Are you assuming that $X_0=a$ and asking for $P(T_0\lt T_b)$ where $T_x$ is the first hitting time of $x\ne a$?2012-08-30
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    As did said, the question is unclear: incorrectly formulated I suppose. Is what you want to ask for $P[X_t=0\text{ for some } t\in[\alpha,\beta]|X_\alpha=a, X_\beta=b]$ for some $0<\alpha<\beta$ and $a,b>0$? You should also outline what you have tried and where you got stuck: e.g. how well do you understand the reflection principle.2012-08-30
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    The reflection principle lets you calculate $\mathbb P (\tau_0 < t_b, X_{t_b} > x)$ and related quantities from which the conditional distribution can be derived. There is a more direct argument based on change of measure for brownian bridge which can be found in D Siegmund's Sequential Analysis2012-08-30
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    Thank you all and apologize for my mistake. Einar, you are right. What I’m trying to calculate is P[Xt=0 for some t∈[α,β]|Xα=a,Xβ=b] for some 0<α<β and a,b>0. I understand the basics of the reflection principle (ie: the distribution of Mt and the distribution of the stopping time given the Xt distribution). However I’m having problems with the starting point at Xta = a and ending point at Xtb = b. I also know that the solution is exp(-2ab/t), but I’m stuck with the proof. Thanks a lot!!2012-08-30
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    question edited2012-08-30
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    In fact, with the notations in the comment, the solution is exp(-2ab/(β-α)) and is indeed a consequence of Désiré André's reflection principle (which is not quite what you describe, I am afraid).2012-08-30
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    did, could you please elaborate your comment? The solution is indeed exp(-2ab/(β-α)), where t = β-α, but I am having problems with the proof. How do you get that result? Any hints will be appreciated!2012-08-31

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