Brownian Bridge's first passage time distribution

Question

Let's say we have a Brownian Bridge $Y_{b,T}(t)$ such that $Y_{b,T}(0)=0$, $Y_{b,T}(T)=b$.

Let's say we are interested in the first passage time of $Y_{b,T}(t)$ at level $b$: $\tau_b = \{\min \tau; Y_{b,T}(\tau)=b\}$.

How could I calculate the distribution of $\tau_b$?

Answer 1

What about this sketch of an answer: Let's put $T=1$ in your formula to simplify the notation. Then $Y_b(t)$ is a Brownian bridge where $Y_b(0)=0$ and $Y_b(1)=b$.

This can be written as $Y_b(t) = b\ t + Y_0(t)$, that is to say the standard Brownian bridge (from zero to zero) with an added drift $b\ t$.

The standard Brownian bridge can be written in terms of a time changed Wiener process $W$, namely $$ Y_0(t) = (1-t)\ W\left(\frac{t}{1-t}\right)$$

The hitting time $\tau$ that you are interested in can be expressed as $$\tau_{Y_b}(b) = \inf \{t : Y_b(t) = b\} = \inf\{t : b\ t + (1-t)\ W\left(\frac{t}{1-t}\right) = b \} = \inf\{t : W\left(\frac{t}{1-t}\right) = b \} $$

Hence, the hitting time of the Brownian bridge is the hitting time of a time changed Wiener process. That is to say, if $$\tau_W(b) = \inf\{s : W(s) = b \}$$ then $$\frac{\tau_{Y_b}(b)}{1-\tau_{Y_b}(b)} = \tau_W(b) \Rightarrow \tau_{Y_b}(b) = \frac{\tau_W(b)}{1+\tau_W(b)} $$

For a standard Wiener process the hitting time $\tau_W(b)$ follows a Levy distribution with density $$ f_W(\tau; b) = \frac{b}{\sqrt{2\pi\tau^3}} \exp \left\{- \frac{b^2}{2\tau} \right\}$$ hence the density of the hitting time of the Brownian bridge will be $$ f_{Y_b}(\tau; b) = \frac{b}{\sqrt{2\pi\tau^3(1-\tau)}} \exp \left\{- \frac{b^2(1-\tau)}{2\tau} \right\} $$

Hope this is right.

Edit: The density (if correct) for $b=\{0.25, 0.5, 1, 2\}$ looks quite funky actually!

Answer 2

It may be overkill, but you may find the following PhD thesis by Peter Hieber of some use.