I'm reading about the strong markov property in Billingsley. I'm confused about some aspects of the proof:

Theorem (Strong Markov Property): Let $\tau$ be a stopping time and $$B^*_t(\omega) = B_{\tau(\omega)+t}(\omega) - B_{\tau(\omega)}(\omega)$$ Then $(B^*_t)_{t \geq 0}$ is a Brownian motion and independent of $\mathcal{F}_{\tau}$ which is to say: $$\textbf{P}(\{B^*_{t_1},...,B^{*}_{t_k}\} \in A \cap M) = \textbf{P}(\{B^{*}_{t_1},...,B^{*}_{t_k}\} \in A) \textbf{P}(M) = \textbf{P}(\{B_{t_1},...,B_{t_k}\} \in A) \textbf{P}(M)$$ for any $A\subset \mathbb{R}^n$ and for any $M \in \mathcal{F}_{\tau}$

Proof: Supposed first that $\tau$ has countable range $V$ and let $t_0$ be some point in $V$. Since $$\{\omega:B^*_t(\omega)\in A\} = \bigcup_{t_o\in V}\{\omega:B_{t_0+t}(\omega)-B_{t_0}(\omega),\tau(\omega)=t_o\}$$ $B^*_t$ is a random variable. Also if $M \in \mathcal{F}_{\tau}$ and $i$ is a finite index: $$ \bigg( \bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\} \bigg) \cap M = \bigcup_{t_o\in V}\{\bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\},\tau(\omega)=t_o,M\}$$ so that $$\textbf{P}(\bigg( \bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\} \bigg) \cap M) = \sum_{t_o\in V}\textbf{P}(\{\bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\},\tau(\omega)=t_o,M\})$$ If $M \in \mathcal{F}_{\tau}$, then $M \cap \{\tau = t_0\} \in \mathcal{F}_{t_0}$. Further, if $\tau = t_0$, then $B^*_t$ is just a translated Brownian motion so that we can use the usual markov property: $$\sum_{t_o\in V}\textbf{P}(\{\bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\},\tau(\omega)=t_o,M\}) \\ = \sum_{t_o\in V}\textbf{P}(\{\bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\})\textbf{P}(\tau(\omega)=t_o,M\}) \quad (*) \\ = \textbf{P}(\{\bigcup_{i}\{\omega:B^*_{t_i}(\omega)\in A_i\}) \textbf{P}(M) \quad (**)$$ This proves the first and third equality, to get the second, consider the case $M = \Omega$.

Then there's some work on moving to the continuous case which I haven't gotten to yet. My questions:

  1. Why is the first bolded statement showing that $B^*_t$ is a random variable? Is it showing that $B^*_t$ is measurable for any $\tau$?

  2. In equation $(*)$, I thought the $t_i$ were fixed, do they vary so they're always "ahead" of $t_0$? If the $t_i$ are fixed, how does one invoke markov property for $t_0$ sufficiently large?

  3. This is related to the last question -- for equation $(**)$, for the sum to push through to $\textbf{P}(\tau(\omega)=t_o,M\})$, the $t_i$'s would have to be constant right?

  4. I don't understand the final bolded statement. The three equalities that they wish to show must be equivalent for any $M\ \in \mathcal{F}_{\tau}$. How is it sufficient to choose one case?

  1. $B_t^*$ is a random variable because for any Borel set $A$, $\{B^*_t\in A\}$ is a countable union of measurable sets:

$$ \bigcup_{t_0\in V}\{(B_{t_0+t}-B_{t_0})\in A,\tau=t_0\}. $$

  1. The indices $\{t_i\}$ are fixed but their "independent" of the index set of the original process and you care about $B_{t_i}'=B_{t_i+t_0}-B_{t_0}$.

  2. The second line (*) is wrong. The whole argument is as follows: \begin{align} &\mathsf{P}(\{(B_{t_1}^{*},\ldots,B_{t_k}^{*})\in A\}\cap M) \\ &\qquad=\sum_{t_0\in V}\mathsf{P}(\{(B_{t_1}^{*},\ldots,B_{t_k}^{*})\in A\}\cap M\cap\{\tau=t_0\}) \\ &\qquad=\sum_{t_0\in V}\mathsf{P}(\{(B_{t_1}',\ldots,B_{t_k}')\in A\}\cap M\cap\{\tau=t_0\}) \\ &\qquad=\sum_{t_0\in V}\mathsf{P}((B_{t_1}',\ldots,B_{t_k}')\in A)\mathsf{P}(M\cap\{\tau=t_0\}) \\ &\qquad=\mathsf{P}((B_{t_1},\ldots,B_{t_k})\in A)\sum_{t_0\in V}\mathsf{P}(M\cap\{\tau=t_0\}) \\ &\qquad=\mathsf{P}((B_{t_1},\ldots,B_{t_k})\in A)\mathsf{P}(M). \end{align}

  3. Taking $M=\Omega$ and using the equality between the first and the third terms, \begin{align} \mathsf{P}((B_{t_1}^{*},\ldots,B_{t_k}^{*})\in A)&=\mathsf{P}(\{(B_{t_1}^{*},\ldots,B_{t_k}^{*})\in A\}\cap \Omega) \\ &=\mathsf{P}((B_{t_1},\ldots,B_{t_k})\in A). \end{align}

  • $\begingroup$ 1. Okay so each member of the union is a measurable set of $\Omega \times T$? 2. I'm still a little confused about this. Lets pick two elements $a$ and $b$ in $V$. Then we will have $B^*_{t_i}=B_{t_i+a} - B_{a}$. Also there's another $b$: $B'^*_{t_i}=B_{t_i+b} - B_{b}$. Are the $t_i$'s chosen in $B^*$ are different from $B'^*$? 3. This calculation is much clearer to me except for second to last line - I have a question but it depends on the answer 2. 4. Ah - okay I see what's happening. $\endgroup$
    – yoshi
    Jul 10 '18 at 3:40
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    $\begingroup$ 1. If $(\Omega,\mathcal{F},\mathsf{P})$ is the underlying prob. space, then $\{\tau=t_0\}\in\mathcal{F}$ and $\{B_t'\in A\}\in \mathcal{F}$ and so $\{\tau=t_0\}\cap\{B_t'\in A\}\in \mathcal{F}$. $\endgroup$
    – d.k.o.
    Jul 10 '18 at 3:49
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    $\begingroup$ 2. Look at these notes $\endgroup$
    – d.k.o.
    Jul 10 '18 at 4:21
  • $\begingroup$ okay, BM is translation invariant. 3. So for the calculation, in the second to last line the term $\mathsf{P}((B_{t_1},\ldots,B_{t_k})\in A)$ pulls out because of translation invariance then right? Technically, for each $t_0 \in V$, $(B_{t_1},\ldots,B_{t_k})$ represent different time points $\endgroup$
    – yoshi
    Jul 10 '18 at 12:37
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    $\begingroup$ $(B_{t_1}',\ldots,B_{t_k}')$ and $(B_{t_1},\ldots,B_{t_k})$ have the same distribution and the latter does not depend on $t_0$. $\endgroup$
    – d.k.o.
    Jul 10 '18 at 15:53

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