In this proof they use the strong markov property but i don't understand why we need it. Could anyone explain it to me? Thank you
Reflection principle brownian motion
brownian motionprobabilitystatisticsstochastic-processes
Related Solutions
We define a sequence of (discrete) stopping times
$$\tau_j := \frac{\lfloor 2^j \tau \rfloor+1}{2^j}, \qquad j \in \mathbb{N}.$$
It is not difficult to see that $\tau_j$ is indeed a stopping time and $\tau_j \downarrow \tau$ as $j \to \infty$. Since the Brownian motion has continuous paths, this implies $B(\tau) = \lim_{j \to \infty} B(\tau_j)$.
Let $\xi,\eta \in \mathbb{R}$. Then, by the dominated convergence theorem,
$$\begin{align*} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg) &= \lim_{j \to \infty} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau_j+t)-B(\tau_j))} \cdot e^{\imath \, \eta B(\tau_j)} \bigg) \\ &= \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E} \bigg( e^{\imath \, \xi (B(k \cdot 2^{-j} +t)-B(k \cdot 2^{-j}))} \cdot e^{\imath \, \eta B(k \cdot 2^{-j})} \cdot 1_{\{\tau_j = k \cdot 2^{-j}\}} \bigg) \end{align*}$$
where we used in the last step that $\tau_j$ is a discrete stopping time. By assumption, $B(k \cdot 2^{-j}+t)-B(k \cdot 2^{-j})$ and $B(k \cdot 2^{-j}) \cdot 1_{\{\tau_j=k2^{-j}\}}$ are independent. Therefore, we obtain
$$\begin{align*} \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg) &= \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg) \lim_{j \to \infty} \sum_{k=1}^{\infty} \mathbb{E} \bigg( e^{\imath \, \eta B(k 2^{-j})} \cdot 1_{\{\tau_j=k \cdot 2^{-j}\}} \bigg) \\ &= \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg) \cdot \mathbb{E}\bigg(e^{\imath \, \eta B(\tau)} \bigg). \end{align*}$$
(In the second step we used again dominated convergence, similar to the above calculation.) If we choose $\eta = 0$, then we get
$$ \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \bigg) = \mathbb{E}\bigg(e^{\imath \, \xi B(t)} \bigg);$$
hence,
$$ \mathbb{E}\bigg( e^{\imath \, \xi \cdot (B(\tau+t)-B(\tau))} \cdot e^{\imath \, \eta B(\tau)} \bigg)= \mathbb{E}\bigg(e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) \cdot \mathbb{E}\bigg(e^{\imath \, \eta B(\tau)} \bigg)$$
i.e. $B(\tau+t)-B(\tau)$ and $B(\tau)$ are independent. Therefore, the strong Markov property gives
$$\begin{align*} \mathbb{E}\bigg(1_F e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) &= \mathbb{E}\bigg(1_F \mathbb{E} \bigg[ e^{\imath \, \xi (B(\tau+t)-B(\tau))} \mid \mathcal{F}_{\tau} \bigg] \bigg) \\ &= \mathbb{E}\bigg(1_F \mathbb{E} \bigg[ e^{\imath \, \xi (B(\tau+t)-B(\tau))} \mid B_{\tau} \bigg] \bigg)\\ &= \mathbb{P}(F) \cdot \mathbb{E}\bigg(e^{\imath \, \xi (B(\tau+t)-B(\tau))} \bigg) \end{align*}$$
for any $F \in \mathcal{F}_{\tau}$. Consequently, $B(\tau+t)-B(\tau)$ is independent of $\mathcal{F}_{\tau}$.
A very similar calculation shows that
$$\mathbb{E} \left( \exp \left( \imath \sum_{j=1}^n \xi_j \cdot (B(\tau+t_j)-B(\tau+t_{j-1})) \right) \right) = \prod_{j=1}^n \mathbb{E}e^{\imath \, \xi_j B(t_j-t_{j-1})}$$
for any $\xi_j \in \mathbb{R}$, $0 \leq t_0 < \ldots \leq t_n$. This means that $(B(\tau+t_j)-B(\tau+t_{j-1}))_{j=1,\ldots,n}$ are independent normal distributed random variables.
Best Answer
We use the strong Markov property to remove the condition on $\mathcal{F}^W_{\tau_a}$ and say that $X(t-\tau_a)$ is just another Brownian motion, hence $$ \mathbb{P}(X(t-\tau_a)<0\mid\mathcal{F}^W_{\tau_a})=\frac12 $$ and we conclude $$ \mathbb{E}\left[\chi_{\sup_{s\in[0,t]} W(s)\geq a}\mathbb{P}(X(t-\tau_a)<0\mid\mathcal{F}^W_{\tau_a})\right]= \mathbb{E}\left[\chi_{\sup_{s\in[0,t]} W(s)\geq a}\cdot\frac12\right] =\frac12\mathbb{P}\left(\sup_{s\in[0,t]} W(s)\geq a\right). $$