# Brownian bridge and first hitting times

by Tetef
Tags: bridge, brownian, hitting time, probability
 P: 3 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 : $P\{T_{a}  Sci Advisor P: 5,891 Your question looks confusing. Your last statement has x < a. Also did you mean W(0) = x? Your T definitions use t also.  P: 3 I did mean what I wrote, A standard brownian motion has [itex]W(0)=0$ 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 $P\{T_{a} P: 3 ## Brownian bridge and first hitting times I don't know if it's any use but, I've thought of this : [itex]P\{T_{a}<T_{b}|W(t)=x\}=\frac{p\{T_{a}<T_{b}\cap W(t)=x\}}{p\{W(t)=x\}}$

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?
 Sci Advisor P: 5,891 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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