Brownian bridge and first hitting times

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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 Recognitions: Science Advisor 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 $P\{T_{a} Brownian bridge and first hitting times I don't know if it's any use but, I've thought of this : [itex]P\{T_{a}<T_{b}|W(t)=x\}=\frac{p\{T_{a}<T_{b}\cap W(t)=x\}}{p\{W(t)=x\}}$

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 :
$p\{T_{a}<T_{b}\cap W(t)=x\}$

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

 Tags bridge, brownian, hitting time, probability
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