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Let $T$ be a finite stopping time and $B_t$ be a Brownian motion (with continuous paths).

The strong Markov Property states that $$ B^T_t = B_{T+t} - B_{T} $$ is a Brownian motion and independent of $\mathscr{F}_T$.

In the proof, it claims if the following condition holds, then $B^T$ is independent of $\mathscr{F}_T$:

$$\mathbb{E}[1_{A}F(B^T_{t_1},\cdots,B^T_{t_p})] = \mathbb{P}(A) \mathbb{E}[F(B_{t_1},\cdots,B_{t_p})] $$

for every $A \in\mathscr{F}_T$ and $0 \le t_1 < \cdots < t_p$, and every bounded nonnegative continuous function $F$.

However, I think the above condition means $$\mathbb{E}[F(B^T_{t_1},\cdots,B^T_{t_p}) \mid \mathscr{F}_T]=\mathbb{E}[F(B_{t_1},\cdots,B_{t_p})]$$?

(I am reading Measure Theory, Probability, and Stochastic Processes by Jean-François Le Gall and the strong Markov property shows up in page 367.)

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  • $\begingroup$ I just found out that I missed something. First, it is shown that $B^T_{t}$ is a Brownian motion. Since the law of the Brownian motion is uniquely defined, $\mathbb{E}[F(B_{t_1},\cdots,B_{t_p})] = \mathbb{E}[F(B^T_{t_1},\cdots,B^T_{t_p})]$ $\endgroup$ Feb 2, 2023 at 7:40

1 Answer 1

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A rv $X$ is independent from a $\sigma$-algebra $\mathscr{G}$ (i.e. $\sigma(X)$ is independent of $\mathscr{G}$) iff $E[\mathbf{1}_Gf(X)]=P(G)E[f(X)],\,\forall G\in \mathscr{G},\,\forall f$ nonnegative continuous and bounded.

Indeed: $(\Rightarrow)$ since $\mathbf{1}_G,f(X)$ are nonnegative and respectively are $\mathscr{G}$- and $\sigma(X)$-measurable, the statement follows from the result $E[Zf(X)]=E[Z]E[f(X)]$ for all $Z$ nonnegative s.t. $\sigma(Z)\subseteq \mathscr{G}$ and $f:\mathbb{R}\to [0,\infty)$ Borel (recall continuous implies Borel measurable). $(\Leftarrow)$: let $y$ be arbitrary, and choose a sequence $f_n$ continuous nonnegative bounded by $1$ s.t. pointwise $f_n\downarrow \mathbf{1}_{(-\infty,y]}$. We get by DCT $$E[\mathbf{1}_G\mathbf{1}_{(-\infty,y]}(X)]=\lim_{n \to \infty}E[\mathbf{1}_Gf_n(X)]=\lim_{n \to \infty}E[\mathbf{1}_G]E[f_n(X)]=P(G)E[\mathbf{1}_{(-\infty,y]}(X)]$$ This means $P(G\cap\{X\leq y\})=P(G)P(X\leq y)$ for all $y \in \mathbb{R}$. Since the family $\{(-\infty,y],y \in \mathbb{R}\}$ is $\cap$-stable and generates the Borel sets, it follows that $\sigma(X)$ and $\mathscr{G}$ are independent.

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