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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.

Bach
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  • Since you can solve this for constant $f$, would approximating general $f$ with simple functions work? – Theoretical Economist Dec 23 '16 at 03:12
  • Probably not, as I might need some terrible inclusion-exclusion here. – Bach Jan 17 '17 at 12:51
  • What are you searching for, an upper or a lower bound? – Ecthelion Feb 02 '18 at 21:29
  • A best as possible approximation. – Bach Feb 04 '18 at 08:16
  • If there was some way of sending the graph of $f$ to a straight line graph, then could do a change-of-variable transformation , by changing the measure of the Brownian motion to obtain a new BM, and reformulating the question by asking when does the new BM hit the straight line graph, for which the analysis has been done above. I will try to come up with an example, but I think this works in very simple cases. – Sarvesh Ravichandran Iyer Nov 18 '20 at 05:19
  • @TeresaLisbon Just putting something out here: if $f(t)=c\sqrt t+\epsilon$ for some real constants $c$ and $\epsilon\lll1$ then $\operatorname P(A_f)$ becomes $\operatorname P(Z\ge c+\epsilon/\sqrt t)$ where $Z=B(t)/f(t)\approx\operatorname N(0,1)$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Nov 18 '20 at 07:37
  • @TheSimpliFire Thanks for that , I would want to see what OP has to say about that. – Sarvesh Ravichandran Iyer Nov 18 '20 at 08:24
  • @TeresaLisbon: Any attempt at an answer would be nice, even if it's just for a subset of curves. I'll happily award the bounty to an attempt at an answer (otherwise the bounty will go to waste...). – Jan Stuller Nov 19 '20 at 07:04
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    @JanStuller Thanks, let me see if I can write something up. I will try for an example, basically, of the "change of measure" phenomena. – Sarvesh Ravichandran Iyer Nov 19 '20 at 08:15
  • @JanStuller No, I apologize. I can answer this, but not before the Bounty expiration. I don't want the bounty for a bad answer, so I will get some things in shape before answering this. – Sarvesh Ravichandran Iyer Nov 19 '20 at 13:42
  • @TeresaLisbon: OK :). – Jan Stuller Nov 19 '20 at 13:55
  • @JanStuller I am struggling with my own lack of rigour, it is not helping me. What I want is this : consider the function $f(t) = \mu t + c$, for $\mu , c$ some non-zero constants. Now, the question $B(t) = f(t)$ for some $t$ translates to $B(t) - \mu t = c$, thus the question is translated into one involving the hitting time of this process. It is possible, by the Girsanov theorem, to show that $B_t - \mu t$ is a Brownian motion under a different measure. Now, this allows us to transfer questions about this process to questions about a BM, which are easily answered. – Sarvesh Ravichandran Iyer Nov 19 '20 at 17:39
  • A demonstration of this, which is very similar, is there in Karatzas and Shreve, section 3.5 on Girsanov's theorem. – Sarvesh Ravichandran Iyer Nov 19 '20 at 17:39
  • Ok, thank you @TeresaLisbon – Jan Stuller Nov 20 '20 at 06:56
  • Girsanov / change of measure will tell you that the probability of $A_f$ is the same as $$\Bbb E \bigg[ 1_{{W \text{ hits } f(0) \text{ somewhere on }[0,T]}}\cdot \exp\bigg(\int_0^T f'(t)dW(t) - \frac12\int_0^T f'(t)^2 dt\bigg)\bigg],$$ where $W$ is again a standard Brownian motion. Is that simplifiable? I have no damn clue, lol. But the formula on the right somehow sheds some light on how exactly $f(0)$ and $f'$ (both of which together recover $f$) each individually contribute to the probability of hitting the barrier. – shalop Mar 01 '21 at 06:32
  • Say the curve is piecewise linear. Then on each piece we have a Brownian motion with drift. The probability that this Brownian motion hits 0 (i.e. the original BM hits the curve) on this piece is the probability that a BM with drift starting at the left end of the linear segment hits 0 before the time which is the length of the segment. Repeat this somehow for all segments.... – Robert Wegner May 14 '21 at 08:04

1 Answers1

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This is very much only a partial answer but the problem intrigued me and it wouldn't fit in a comment, so just see it as inspiration for others who have more time maybe.

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.

Then to get to smooth functions you approximate with a piecewise linear function and also make some estimates.

Robert Wegner
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