Why $\mathbb P(X_{\tau-t}<0\mid \mathcal F_{\tau})=\frac{1}{2}$?

Question

Let $(B_t)$ a BM (Brownian Motion) and $\tau=\inf\{t>0\mid B_t=a\}$. Set $X_t=B_{t+\tau}-B_\tau$. I know that $(X_t)$ is a BM independent of $\mathcal F_\tau$. In the proof of the reflexion principle on wikipedia, they say that $\mathbb P(X_{t-\tau}<0\mid \mathcal F_\tau)=1/2$, but I don't really understand why. The fact that $\mathbb P(X_{t-s}<0\mid \mathcal F_{\tau})=1/2$ for all $s<t$ is clear to me, but in somehow, if we have $\mathbb P(X_{t-\tau} <0\mid \mathcal F_\tau)$ is rather unclear.

Answer 1

The proof of wikipedia seems to be not completely correct (even completely wrong). And is indeed not clear (as written) why $X_{t-\tau}$ is independent of $\mathcal F_{\tau}$ (but I can mistaken). Let me propose the following correction : \begin{align} \mathbb P\left\{\sup_{0\leq s\leq t}W_s\geq a\right\}&=\mathbb P\left\{\sup_{0\leq s\leq t}W_s>a\right\}\\ &=\mathbb P\left\{\sup_{0\leq s\leq t}W_s>a,W_t>a\right\}+\mathbb P\left\{\sup_{0\leq s\leq t}W_s>a, W_t\leq a\right\}\\ &=\mathbb P\left\{W_t>a\right\}+\mathbb P\left\{\sup_{0\leq s\leq t}W_s>a, W_t\leq a\right\}\\ &=\mathbb P\left\{W_t\geq a\right\}+\mathbb P\left\{\sup_{0\leq s\leq t}W_s>a, W_t\leq a\right\}. \end{align} Then, \begin{align*} \mathbb P\left\{\sup_{0\leq s\leq t}W_s>a,X_{t-\tau}<0\right\}&=\mathbb P\left\{\sup_{0\leq s\leq t}W_s>a,X_{t-\tau}\leq 0,\tau<t\right\}\\ &=\mathbb E\left[\boldsymbol 1_{\{\sup_{0\leq s\leq t}W_s> a\}}\mathbb P\{X_{t-\tau}\leq 0,\tau<t\mid \mathcal F_\tau\}\right]\\ &=\mathbb E\left[\boldsymbol 1_{\{\sup_{0\leq s\leq t}W_s>a\}}\boldsymbol 1_{\{u<t\}}\left.\mathbb P\{X_{t-u}\leq 0\}\right|_{u=\tau}\right]\\ &=\frac{1}{2}\mathbb P\left\{\sup_{0\leq s\leq t}W_s>a,\tau<t\right\}\\ &=\frac{1}{2}\mathbb P\left\{\sup_{0\leq s\leq t}W_s>a\right\}\\ &=\frac{1}{2}\mathbb P\left\{\sup_{0\leq s\leq t}W_s\geq a\right\}. \end{align*}

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Now, $X_{t-\tau}$ is independent of $\mathcal F_{\tau}$ on $\{\tau<t\}$. The idea is to take a sequence of stopping time $(\tau_n)$ that take finitely many value and s.t. $\tau_n\searrow \tau$ (for example $\tau_n=\frac{\lfloor 2^n\tau\rfloor }{2^n}\boldsymbol 1_{\{\tau\leq n\}}+\infty \boldsymbol 1_{\{\tau>n\}}$). Take $N$ large enough to have $\tau_n<t$ a.s. on $\{\tau<t\}$ for all $n\geq N$, and prove that $$\mathbb E\left[\boldsymbol 1_{A\cap \{\tau<t\}}e^{iX_{t-\tau_n}\xi}\right]=\mathbb E\left[e^{iX_{t-\tau_n}\xi}\right]\mathbb P(A\cap\{\tau<t\}),$$ for all $n\geq N$ and all $\xi\in \mathbb R$. Taking $n\to \infty $ gives the wished result.