# Reversal of Brownian motion from first hitting time

Let $(B_t)_{t \ge 0}$ be a (standard) Brownian motion, and for fixed $C>0$ define the first hitting time $$T_C := \inf \{t \ge 0: B_t = C\}.$$

I am interested in the reversal of the Brownian motion from the first hitting time $T_C$, i.e. a description of the distribution of the path $$(B_{T_C - s} - C)_{s \le T_C}.$$

My question is whether this has the same law as $$(\widetilde{B}_s)_{s \le S_{-C}}$$ where $(\widetilde{B}_s)_{s \ge 0}$ is a Brownian motion conditioned to stay non-positive, and $S_{-C}$ is the last time that the process $(B_s)$ hits $-C$.

My guess above comes from the analogous problem for Brownian motion with negative drift. I was told that the following is true (I would appreciate a reference for that):

Fix $m > 0$ and $C > 0$ as before. Let $T_C$ (resp. $S_{-C}$) be the first hitting time of $C$ (resp. last hittimg time of $-C$) of the Brownian motion with negative drift $(B_s -ms)_{s \ge 0}$. Then $$(B_{T_C - s} + m(T_C - s) - C)_{s \le T_C} \overset{d}{=} (\widetilde{B}_s - ms)_{s \le S_{-C}}$$ where $(\widetilde{B}_s - ms)_{s \ge 0}$ is a Brownian motion with drift $-m$ conditioned to stay non-positive.

As a heuristic I can send $m \to 0$ and that would recover my claim above, but this does not constitute a mathematical proof. It would be great if someone could tell me that my claim is correct and point me to references where a proof can be easily found.

The answer is positive, in the sense that $$(\widetilde{B}_s)_{s \ge 0}$$ is a $$BES(3)$$-process starting from $$0$$, and $$S_{-C} := \sup \{t > 0: \widetilde{B}_t = -C \}$$ is the last hitting time of $$-C$$ by the Bessel process. This is a classical result due to Williams, probably proved in his paper on path decomposition on diffusion, and can also be found in the book by Revuz-Yor.