8
$\begingroup$

We know that the law of $$\tau_b := \inf\{ t >0 : B_t \geq b\} ,$$ where $b>0$ and $B$ is a standard Brownian motion, is given by a certain computable density, which is uniquely determined by $b$.

If we replace $b$ by a continuous function $b: [0,\infty) \to [0,\infty]$ and redefine $\tau_b$ as $$\tau_b := \inf\{ t >0 : B_t \geq b(t) \}$$ I would guess that $b$ also should uniquely determine the law of $\tau_b$.

Edit: What I mean with uniquely determination is that I ask whether the implication $b_1 \neq b_2 \Rightarrow$ Law$(\tau_{b_1})$ $\neq$ Law$(\tau_{b_2})$ is true. So that ($b$ $\mapsto$ law of $\tau_b$) is injective

I am not asking for a solution, I want to think about it by myself, but I don't know how to approach. Any hint, comment, idea or reference is appreciated.

$\endgroup$
9
  • 1
    $\begingroup$ Girsanov's Theorem might be relevant $\endgroup$
    – mbartczak
    Sep 27, 2019 at 11:57
  • 2
    $\begingroup$ If $b(t) \to \infty$ very fast then $\tau_b =\infty$ a.s.. You can use Law of Iterated logarithm for BM to see this. $\endgroup$ Sep 27, 2019 at 11:58
  • $\begingroup$ Thanks for the comment. I added a "bounded" to the assumptions on $b$. $\endgroup$
    – Falrach
    Sep 27, 2019 at 12:05
  • 1
    $\begingroup$ What are you asking exactly? Are you asking if $b_1 = b_2$ implies that $\tau_{b_1}$ is equal in law to $\tau_{b_2}$? $\endgroup$
    – Tom
    Sep 27, 2019 at 14:05
  • 1
    $\begingroup$ Why? $b(0) = 0$ won't imply $\tau_b \equiv 0$ I think. $\endgroup$
    – Falrach
    Oct 14, 2019 at 15:58

1 Answer 1

1
$\begingroup$

Let $\tau^X_b$ denote hitting time of $b$ for process $X$, that is $\tau^X_b = \inf\{t: X_t = b(t)\}$ and $b$ be absolutely continuous, that is $b(t) = b_0 + \int_0^t f(s)ds$.

$ \newcommand{\ang}[1]{\left\langle #1 \right\rangle} $

Process $$ X_t = B_t - \ang{B, L}_t = B_t - \ang{B, \int f(s)dB_s}_t = B_t - \int_0^tf(s)ds $$ is a brownian motion w.r.t. measure $\mathbb{Q}_T$ defined of $\mathcal{F}_T$ such as $$ d\mathbb{Q}_T = \mathcal{E}(L)_T\ d\mathbb{P} = \exp({L_T-\frac12\langle L\rangle_T})d\mathbb{P} = \exp\left(\int_0^T f(t)dB_t-\frac12 \int_0^Tf(t)^2dt \right) d\mathbb{P} $$ due to Girsanov theorem. Then it follows $$ \mathbb{P}(\tau^B_b \leq t) = \mathbb{Q}_t(\tau^X_b \leq t) = \mathbb{Q}_t(\tau^B_{b_0} \leq t) = \mathbb{E}\left[1_{\tau^B_{b_0} \leq t}\mathcal{E}(L)_t\right]. $$

Some rules about how $b$ influences $\tau^B_b$ might be extracted from this.


Correct me if I am wrong as this result seems suspicious, e.g. consider $f = 1_{(0,1)} - 1_{(1,2)}$ and $-f$.

$\endgroup$
7
  • 1
    $\begingroup$ Sorry.. I dont understand your point.. What would be your counterexample of two different function $b_1$ and $b_2$ such that $\tau^B_{b_1}$ and $\tau^B_{b_2}$ have the same law? Dont forget that we re restricting ourselves to the case of positive bounded and continuous functions. $\endgroup$
    – Tom
    Oct 3, 2019 at 7:42
  • $\begingroup$ I calculated it by myself now what you did. I always used Girsanov backwards I noticed. However I think there is a little mistake. $X$ should be $X_t = B_t + \int_0^t f (s) ds$. But unfortunately I dont know how to learn from this result. $\endgroup$
    – Falrach
    Oct 3, 2019 at 10:52
  • $\begingroup$ Btw, it does not seem suspicious to me. If you change the sign as you did, the sign of the randomness in the exponential martingale changes which corresponds to the reweighting that i necessary due to mirrorin the boundary function. $\endgroup$
    – Falrach
    Oct 3, 2019 at 10:55
  • $\begingroup$ @Falrach, check that $B_t = b(t) \Leftrightarrow X_t = b_0$. I thought that the last expected value would be invariant to exchanging $f \leftrightarrow -f$ but it's not a case since the indicator and the exponential martingale are dependent, right? $\endgroup$
    – mbartczak
    Oct 3, 2019 at 11:49
  • $\begingroup$ Your equivalence in the comment above is true, but in your calculation you do not use it. You change the measure first and make use of the fact that the hitting time has an easier form if one defines $X$ as in my comment. The way that makes use of your equivalence also works (this is what I meant by backwards) and leads to the same result if one puts the change of measure by adding $\mathcal E _t / \mathcal E_t$. $\endgroup$
    – Falrach
    Oct 3, 2019 at 12:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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