# Brownian bridge and first hitting times

1. Feb 6, 2013

### Tetef

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 !

2. Feb 6, 2013

### mathman

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

3. Feb 6, 2013

### Tetef

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 $P\{T_{a}<T_{b}|W(t)=x\}$, $t$ is the 'time of one observation'.
If you look for $P\{T_{a}<t\}$, 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 $P\{T_{a}<T_{b}|W(t)=x\}$, 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 $t$ ($W(t)=x$)

$P\{T_{a}<T_{b}|W(t)=x\}$ could be asked for $x\inℝ$.
In my case I have $x<a$. 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: Feb 6, 2013
4. Feb 6, 2013

### Tetef

I don't know if it's any use but, I've thought of this :

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

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?

5. Feb 7, 2013

### mathman

I have not studied this problem in any detail, so I don't know if I can be of any further help.