## probability question regarding brownian motion

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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.

-

## 1 Answer

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.)

-

@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 2011 at 22:08
Added a little more detail. – Apollo Feb 10 2011 at 22:18
@Apollo, looks great! Can I ask you: how to understand the integral?  – Qiang Li Feb 10 2011 at 22:31
I mean: how to understand it is the required probability. – Qiang Li Feb 10 2011 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 2011 at 22:34
show 1 more comment
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9040796756744385, "perplexity_flag": "head"}