Brownian bridge and first hitting times

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The discussion focuses on calculating the probability P{T_{a}<T_{b}|W(t)=x} for a Brownian bridge, where T_{a} and T_{b} are first hitting times for levels a and b, respectively. The user clarifies that a standard Brownian motion starts at W(0)=0 and defines T_{a} as the first time the motion hits level a. The probability expression is analyzed, emphasizing that the condition W(t)=x (with x<a) creates a Brownian bridge scenario. The user proposes a formula involving probability density functions to express the desired probability but acknowledges uncertainty about the joint distribution of T_{a} and T_{b}. The conversation highlights the complexities of Brownian motion and the need for further exploration of the joint probabilities involved.
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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&lt;0
and
T_{b}=inf\{t:W(t)=b\} with b&gt;0

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

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