Brownian bridge and first hitting times

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Discussion Overview

The discussion revolves around the probability of first hitting times in the context of a Brownian bridge, specifically examining the expression P{T_{a}

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant defines the first hitting times T_{a} and T_{b} for a standard Brownian motion and seeks to understand the probability of one hitting time occurring before the other in the context of a Brownian bridge.
  • Another participant questions the clarity of the original post, specifically regarding the notation and the condition x
  • The original poster clarifies that W(0)=0 and explains the meaning of T_{a} and T_{b} in terms of the first hitting times of the Brownian motion.
  • A participant proposes a formula involving probability density functions to express P{T_{a}
  • One participant expresses uncertainty about their ability to contribute further to the problem due to a lack of detailed study on the topic.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the clarity of the original question or the approach to solving it. There are differing interpretations of the notation and conditions presented.

Contextual Notes

There are unresolved aspects regarding the probability density functions and the relationship between the hitting times and the conditions of the Brownian bridge. The implications of continuity of Brownian motion are mentioned but not fully explored.

Tetef
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Hi,

Letting W be a standard brownian motion, we define the first hitting times
T_{a}=inf\{t:W(t)=a\} with a<0
and
T_{b}=inf\{t:W(t)=b\} with b>0

The probability of one hitting time being before an other is :
P\{T_{a}<T_{b}\}=\frac{b}{b-a}

I'm looking for this probability in the case of a brownian bridge :
P\{T_{a}<T_{b} | W(t)=x\} with x<a

Could some one help me please?

Thx !
 
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Your question looks confusing. Your last statement has x < a. Also did you mean W(0) = x? Your T definitions use t also.
 
I did mean what I wrote,

A standard brownian motion has W(0)=0

I use t in the definition T_{a}=inf\{t:W(t)=a\}to say the first hitting times are defined as : T_{a} is the smallest time t where the brownian motion does hit the level a i.e. W(t)=a

In the expression of the probability P\{T_{a}&lt;T_{b}|W(t)=x\}, t is the 'time of one observation'.
If you look for P\{T_{a}&lt;t\}, you're looking for the probability for W to hit a level a before t (t being the end of the observation).

Here, in the expression P\{T_{a}&lt;T_{b}|W(t)=x\}, I'm looking for the probability for one hitting time to be before the other, knowing that at the time t, the brownian motion will be equal to x, i.e. W(t)=x; and thus form a brownian bridge between 0 at time 0 (W(0)=0) and x at time t (W(t)=x)

P\{T_{a}&lt;T_{b}|W(t)=x\} could be asked for x\inℝ.
In my case I have x&lt;a. This implies, by continuity of the brownian motion, that T_{a} does exist before t. It might make it simpler, it might not...
 
Last edited:
I don't know if it's any use but, I've thought of this :

P\{T_{a}&lt;T_{b}|W(t)=x\}=\frac{p\{T_{a}&lt;T_{b}\cap W(t)=x\}}{p\{W(t)=x\}}

where capital P are probabilities and small p are probability density functions.

we know :
p\{W(t)=x\}=\frac{1}{\sqrt{2\pi t}}exp( -\frac{x^2}{2t})

we don't know :
p\{T_{a}&lt;T_{b}\cap W(t)=x\}

but we know that :
\int^{+∞}_{-∞} p\{T_{a}&lt;T_{b}\cap W(t)=x\}dx=\frac{b}{b-a}
Is that right?
 
I have not studied this problem in any detail, so I don't know if I can be of any further help.
 

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