The probability that an arithmetic Brownian motion process $dt = \mu dt + \sigma dW$ hits an upper Barrier $U$ before it hits a lower barrier $L$ is given by

$$ \mathbb{P}(\tau_U\leq \tau_L) = \frac{\text{Y}(x_0)-\text{Y}(L)}{\text{Y}(U)-\text{Y}(L)} $$ where $$ \text{Y}(x) = exp(\frac{-2\mu x}{\sigma^2}) $$

But what is $\mathbb{P}(\tau_U\leq T \,\cap\, \tau_U\leq\tau_L)$ if both $x_0$ and $x_T$ are known?

i.e. the probability the process hits $U$ before $L$ whilst in between the end points of a Brownian bridge.


1 Answer 1



Let $B$ be a standard brownian motion starting from $x_0=0$, $m_T = \inf_{u\leq T}B_u$ and $M_T =\sup_{u\leq T}B_u$.

Let's define if it exists for $A\in\sigma(B_u,u\leq T)$, $\mathbb{P}(A | B_T=x_T)\stackrel{\rm def}{=}\lim_{\varepsilon\to 0}\mathbb{P}(A|B_T\in(x_T-\varepsilon,x_T+\varepsilon))$

$$\begin{split} \mathbb{P}(\tau_U\leq T \cap \tau_U\leq \tau_L)= & \mathbb{P}(m_T>L;M_T\geq U |B_T =x_T)\\ = &\mathbb{P}(m_T>L|B_T=x_T)-\mathbb{P}(L<m_T<M_T<U|B_T=x_T) \end{split}$$

Computations attempt

Then, this is side computations based from results about hitting times from Chapter 3 of Mathematical Methods for Financial Markets of Monique Jeanblanc, Marc Yor and Marc Chesney.

so by denoting $p_T(y)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{y^2}{2t}}$

I have: $$\begin{split} P\stackrel{\rm def}{=} & \mathbb{P}(\tau_U\leq T \cap \tau_U\leq \tau_L)\\ =& \frac{p_T(-2L+x_T)}{p_T(-x_T)} - \frac{\sum_{n=-\infty}^{\infty}p_T(x_T+2n(U-L))-p_T(2U-x_T+2n(U-L))}{p_T(x_T)} \end{split}$$

  • $\begingroup$ Thanks MJ, I've been dragged onto other things and have not had a chance to look at this in much detail. Will come back when I've had a think and a play with this. $\endgroup$
    – OldSchool
    Aug 2, 2016 at 8:28

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