# The probability that a linear Brownian motion will hit a curve

## Summary

I am trying to estimate the probability that a standard linear Brownian motion will hit some curve. To make things a bit simple, I can assume that the curve is a graph of a function, that is is positive at $t=0$, that it is bounded from left by $0$ and by right by some $T>0$, that it is continuous, or even differentiable, and many other nice curvish features that may help making this question more feasible.

## Formalizing

Let $\{B(t)\mid t\ge 0\}$ be a standard linear Brownian motion, and let $f:[0,T]\to\mathbb{R}$ be infinitely-differentiable (in $(0,T)$) real function with $T>0$ and $f(0)>0$. Let $A_f$ be the event "$\exists t\in(0,T]):\ B(t)=f(t)$", that is, the Brownian motion "hits" the graph of the function $f$.

The question is as follows: given $f$, what is $\mathbb{P}\left(A_f\right)$?

## Attempt

All I could do is solve this for $f\equiv c>0$; in that case, if we define $M(t)=\max\{B(s)\mid 0\le s\le t\}$ we have that $$\mathbb{P}\left(A_f\right)=\mathbb{P}\left(M(T)\ge c\right)$$ and by reflection principle, the last probability equals $$2\mathbb{P}\left(B(T)\ge c\right)$$ and that can be solved using straight-forward normal cdf.

However, even for non-constant linear $f$'s that trick won't do; and $f(x)=1/x$ (with something at $0$, bounded from the right by some $T$) seems much harder. This is where I stop and post a question.

• Since you can solve this for constant $f$, would approximating general $f$ with simple functions work? Dec 23, 2016 at 3:12
• Probably not, as I might need some terrible inclusion-exclusion here.
– Bach
Jan 17, 2017 at 12:51
• What are you searching for, an upper or a lower bound? Feb 2, 2018 at 21:29
• A best as possible approximation.
– Bach
Feb 4, 2018 at 8:16
• @JanStuller Thanks, let me see if I can write something up. I will try for an example, basically, of the "change of measure" phenomena. Nov 19, 2020 at 8:15

Suppose the curve is piecewise linear, i.e. there exist times $$0 = t_0 < t_1 < ...$$ and values $$f_k \in \mathbb{R}$$ with $$f_0 > 0$$ such that $$f(t) = \sum_{k=0}^\infty \chi_{[t_k, t_{k+1}]}(t) \cdot \underset{:= g_k}{\underbrace{(f_k + (t - t_k) \frac{f_{k+1} - f_k}{t_{k+1} - t_k})}}.$$ Then $$\mathbb{P}(\exists t > 0: B_t - f(t) = 0) = \mathbb{P}\left(\bigcup_{k = 0}^\infty \{\exists t \in [t_k, t_{k+1}]: B_t - g_k(t) = 0\} \right).$$ Define $$w^k_t = B_{t + t_k} - B_{t_k}$$ We know that $$W^k$$ is a BM.
Define furthermore $$c_k = \frac{f_{k+1} - f_k}{t_{k+1} - t_k}.$$ Then for $$t \in [0, t_{k+1} - t_k]$$ $$B_{t + t_k} - g_k(t + t_k) = B_{t_k} - f_k + W^k_t - t c_k$$ Then $$\mathbb{P}(\exists t \in [t_k, t_{k+1}]: B_k - g_k = 0 ) = \mathbb{P} (\exists t \in [0,t_{k+1} - t_k]: B_{t_k} - f_k + W^k_t - t c_k)$$ $$= \mathbb{E} \left[ \chi_{\exists t \in [0,t_{k+1} - t_k]: B_{t_k} - f_k + W^k_t - t c_k} \right] = \mathbb{E} \left[ \mathbb{E} \left[ \chi_{\exists t \in [0,t_{k+1} - t_k]: B_{t_k} - f_k + W^k_t - t c_k} \vert \sigma(B_{t_k})\right] \right]$$ $$= \mathbb{E}_{- f_k} \left[ \mathbb{E} \left[ \chi_{\exists t \in [0,t_{k+1} - t_k]: B_{t_k} + W^k_t - t c_k} \vert \sigma(B_{t_k})\right] \right]$$ Now we use the Strong Markov property (or maybe the weak one suffices here?) $$= \mathbb{E}_{- f_k} \left[ \mathbb{E}_{B_{t_k}} \left[ \chi_{\exists t \in [0,t_{k+1} - t_k]:W^k_t - t c_k} \right] \right]$$ $$= \mathbb{E}_{- f_k} \left[ \mathbb{P}_{B_{t_k}} \left( T_0^{c_k} \leq t_{k+1} - t_k \right) \right],$$ where $$T^k_0$$ is the hitting time of $$0$$ for a Brownian motion with drift $$-c_k$$. This can be easily computed as the density is known, you should be able to find it somewhere online.
For clarity let me rewrite which process starts where $$= \mathbb{E}_{B_0 = - f_k} \left[ \mathbb{P}_{W^k_0 = B_{t_k}} \left( T_0^{c_k} \leq t_{k+1} - t_k \right) \right].$$ So this quantity should be computable. But the problem is that the term $$\mathbb{P}\left(\bigcup_{k = 0}^\infty \{\exists t \in [t_k, t_{k+1}]: B_t - g_k(t) = 0\} \right)$$ is not easily reduced to these probabilities. However from here you can maybe start making estimates.