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.
Edit2: The question from the first edit is called the inverse first-passage time problem for Brownian motion. These papers [1] [2] show indirectly that, if $b_1$ and $b_2$ are lower semicontinuous functions with $$\exists 0< t \leq \max (T^{b_1},T^{b_2}): b_1(t) \neq b_2 (t), $$ where $$T^{b_i} := \inf\{t> 0 : b_i(t) = -\infty \},$$ then Law$(\tau_{b_1})$ $\neq$ Law$(\tau_{b_2})$.
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 $-f$.