How to show that $X_t = \frac{1}{\left| B_t -x\right|}\mathbb{1}_{\left\{ B_t \neq x\right\}}$ ("inverse brownian") is a martingale?

Question

Consider $$X_t = \frac{1}{\left| B_t -x\right|}\mathbb{1}_{\left\{ B_t \neq x\right\}}$$

where $ \left(B_{t }\right)_{t \geq 0}$ is a $ \mathcal F_t$- brownian motion in $\mathbb R ^3$, null at the origin and $x \in \mathbb R ^3 \setminus \{0\}$

Consider also the sequence of stopping times $ \left( \sigma_n \right)_{n \in \mathbb N}$ defined by $$\sigma_n = \inf \{ t\geq 0 : \left| B_t -x\right| \leq \frac {1} {n}\}$$

wich converges to the stopping time $$\tau_x = \inf \{ t\geq 0 : \left| B_t -x\right| \leq 0\}$$

The question : How to show that $ \left( X_{t \wedge \sigma_n }\right)_{t \geq 0}$ is a $ \mathcal F_t$- continuous bounded martingale?

Answer 1

Boundedness should be clear. To show the martingale property, the shortest way might be to introduce the function $u:z\mapsto1/|z-x|$ defined on $\mathbb R^3\setminus\{x\}$ and to note that, for every $n\geqslant1$, $u$ is harmonic for the Laplacian on the domain $D_n=\{z\in\mathbb R^3\mid |z-x|\gt1/n\}$. Hence, for any fixed $n\geqslant1$, $(u(B_{t\wedge\sigma_n}))_{t\geqslant0}$ is a martingale, where $\sigma_n$ denotes the first exist time of $D_n$ by the process $(B_t)_{t\geqslant0}$, that is, $\sigma_n=\inf\{t\geqslant0\mid|B_t-x|\leqslant1/n\}$.