MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am wondering where to start with questions like:

Given a BM $dX_t=\mu t+\sigma dB_t$, having started at $X_0=0$. What is the probability that $X_t$ does not hit 0 in the time interval $[a,T]$ where $0\le a\le T$?

Here the hit level can be changed from 0 to any constant $b\gt 0$, or even to a space-time line $x=kt+b$. This is related to kind of "Global" distribution of $X_t$. I do not find the discussion in the references I have here, for example, Karatzas&Shreve. Would appreciate your suggestion and recommendation.

share|cite|improve this question
up vote 5 down vote accepted

A straightforward approach is to simply integrate the density of $X_t$ at time $a$ (which will be normally distributed with mean $\mu$ and variance $\sigma^2 a$) against the probability of hitting 0 conditional on the value at time $a$ (which is also known in closed-form). This will give you a messy integral (with an exponential multiplied by a cumulative-normal) but it should be reducible to a (messy sum of) bivariate cumulative normal(s).

The value we want to compute is $$\int_0^\infty \mathbb{P}[X_\xi>0 \text{ for } a\leq\xi\leq T\ |\ X_a=z] e^{-z^2/2\sigma^2T}\frac{dz}{\sigma\sqrt{T}\sqrt{2\pi}}$$ where I'm integrating the density at time $a$ for positive values against the non-hitting time.

The next step is to observe that the probability $\mathbb{P}[X_\xi>0 \text{ for } a\leq\xi\leq T\ |\ X_a=z]$ is equal to the probability $\mathbb{P}[X_\xi>-z \text{ for } 0\leq\xi\leq T-a]$ but this probability is equal to a difference of (basically) cumulative normals (it's just a hitting time computation for a (scaled) Brownian motion with drift). Then plug that formula into the above integral.

A quick calculation (might be wrong, so beware) gives me $$\mathbb{P}[X_\xi>-z \text{ for } 0\leq\xi\leq T-a] = \Phi\left[\frac{-z+\alpha (T-a)}{\sigma\sqrt{T-a}}\right] - e^{2\alpha z/\sigma^2}\Phi\left[\frac{z+\alpha (T-a)}{\sigma\sqrt{T-a}}\right] $$ where $\Phi[z]=\int_{-\infty}^z e^{-\xi^2/2}\frac{d\xi}{\sqrt{2\pi}}$ is the standard cumulative normal distribution function. (This follows from application of Girsanov to a reflection argument, a well-known result.)

share|cite|improve this answer
@Apollo: could you please write down the integral? I am not sure I understand what you meant here. Does this way guarantee for all time in the interval $[a, T]$, no 0 is hit? – Qiang Li Feb 10 '11 at 22:08
Added a little more detail. – Apollo Feb 10 '11 at 22:18
@Apollo, looks great! Can I ask you: how to understand the integral? – Qiang Li Feb 10 '11 at 22:31
I mean: how to understand it is the required probability. – Qiang Li Feb 10 '11 at 22:31
We're integrating along the distribution of $X$ at time $a$ and multiplying by the conditional probability (given our location at time $a$) that we make it further to time $T$ without hitting zero. If we're already below zero then this probability is $0$ (so the lower bound of the integral starts at $0$) if we're above zero then we just need to keep the minimum of the remaining path above zero. – Apollo Feb 10 '11 at 22:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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