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 $\tau_b$?

  • $\begingroup$ Can you specify "Brownian Bridge" and "$Y$"? $\endgroup$
    – emcor
    Commented Jul 28, 2014 at 12:51
  • $\begingroup$ It's just a Brownian motion with boundary condition that $Y(T)=b$ $\endgroup$
    – athos
    Commented Jul 28, 2014 at 16:03

2 Answers 2


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!


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.