Prove that $(\tilde W_t)$ is a Brownian motion where $\tilde W_t=2\alpha -W_t$ if $t>\tau_\alpha $ Let $(W_t)$ a Brownian motion and $\tau_\alpha =\inf\{t\geq 0\mid W_t=\alpha \}$. I would like te prove that $$\tilde W_t= W_t\boldsymbol 1_{t\leq \tau_\alpha }+(2\alpha -W_t)\boldsymbol 1_{\tau_\alpha >\alpha }$$
is a Brownian motion. The fact that there are the $\boldsymbol 1_{t\leq \tau_a}$ and $\boldsymbol 1_{\tau_a>t}$ is quite disturbing. For example, I tried to prove that $\tilde W_t$ is Gaussian, but I don't see how to manage those indicator function. Any idea ?
 A: Here is a sketch of the proof; for details see for instance Theorem 6.12 in the book by Schilling & Partzsch on Brownian motion.
For fixed $\alpha$ write $\tau:=\tau_{\alpha}$. Denote by $C_{\{0\}}$ the set of continuous functions $f:[0,\infty) \to \mathbb{R}$ with $f(0)=0$, and define a continuous mapping $\Phi: C_{\{0\}} \times [0,\infty) \times C_{\{0\}} \to C_{\{0\}}$ by
$$\Phi(f,t,g)(s) := \begin{cases} f(s), & s \in [0,t), \\ f(t)+g(s-t), & s \geq t. \end{cases}$$
($\Phi$ is the concatenation of the two functions $f,g$; draw a picture!) Now notice that


*

*$\omega \mapsto (W(\cdot \wedge \tau(\omega),\omega),\tau(\omega))$ is $\mathcal{F}_{\tau+}$-measurable,

*$\omega \mapsto W(\cdot+\tau(\omega),\omega)-W(\tau(\omega),\omega)$ is independent of $\mathcal{F}_{\tau+}$ (by the strong Markov property)

*$\omega \mapsto W(\cdot+\tau(\omega),\omega)-W(\tau(\omega),\omega)$ has the same distribution as $\omega \mapsto W(\cdot,\omega)$ (because the restarted Brownian motion is also a Brownian motion). By the symmetry of Brownian motion, this implies that  $W(\cdot+\tau)-W(\tau)$ has the same distribution as $W(\tau)-W(\cdot+\tau)$.


Combining these facts we find that the processes $$(W(\cdot \wedge \tau),\tau,W(\tau)-W(\cdot+\tau)) \quad \text{and} \quad (W(\cdot \wedge \tau),\tau,W(\cdot+\tau)-W(\tau))$$ have the same distribution. Hence,
$$\tilde{W} = \Phi(W(\cdot \wedge \tau),\tau,W(\tau)-W(\cdot+\tau)) \quad \text{and} \quad W=\Phi(W(\cdot \wedge \tau),\tau,W(\cdot+\tau)-W(\tau))$$
have the same distribution. Since $W$ is by assumption a Brownian motion and $\tilde{W}$ has continuous sample paths, we conclude that $\tilde{W}$ is also a Brownian motion.
