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I was able to understand Brownian Motion $\{B(t):t\geq0\}$ has Strong Markov Property i.e.

For any stopping time $\tau$, $P(B(t+\tau)\leq y | \mathcal{F}_{\tau})=P(B(t+\tau)\leq y|B(\tau))$ a.s. , $y \in \mathbb{R}$.

I want to prove the following assertions

$\cdot$ $\hat{B}(t):=B(\tau+t)-B(\tau)$, $t\geq0$ is independent of $\mathcal{F}_{\tau}$

$\cdot$ $0\leq \forall s < \forall t$, $\hat{B}(t)-\hat{B}(s)$ has the normal distribution of $N(0, t-s)$.

$\cdot$ $0\leq \forall s < \forall t$, $\hat{B}(t)-\hat{B}(s)$ is independent of $\hat{B}(u),\,u\in[0,s]$

How do I use Strong Markov property? Please teach me.

My Brownian motion definition is as follows:

$\cdot$ $\forall \omega \in \Omega$, $t \mapsto B(t,\omega)$ is continuous.

$\cdot$ $0\leq \forall s < \forall t$, $B(t)-B(s)$ has the normal distribution of $N(0, t-s)$.

$\cdot$ $0\leq \forall s < \forall t$, $B(t)-B(s)$ is independ of $B(u),\,u\in[0,s]$.

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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.

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  • $\begingroup$ Thank you for your advice. I will try to calculate as you told. $\endgroup$ – ko4 Feb 20 '14 at 7:22
  • $\begingroup$ @kouhei Don't hesitate to ask if you don't get along with it. The calculations are lengthy, but not that complicated... $\endgroup$ – saz Feb 20 '14 at 16:20

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