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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.

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1 Answer 1

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.)

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@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

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