Strong Markov Property and Product of Expectations Let $(B_{t})_{t\geq0}$
  be a Brownian motion and let $\tau=\inf\left\{ t\geq0:B_{t}\leq-4\right\} $ be a stopping time.
Then the strong Markov property ensures that e.g. $A:=\left\{ (B_{t})_{t\in[0,\tau]}\;\mbox{goes above 5}\right\}$
  is independent of $B:=\left\{ (B_{t})_{t\in(\tau,\tau+10)}\;\mbox{goes below -8}\right\}$. In other words, $B$ only depends on $\tau$ and not what went on before that. Is that not correct? 
Hence
$\mathbb{E}\left(1_{A}1_{B}\right)=\mathbb{E}\left(1_{A}\right)\mathbb{E}\left(1_{B}\right).$
But is it also correct that 
$\mathbb{E}\left(1_{A}\mathbb{E}^{\tau}\left(1_{\left\{ (B_{t})_{t\leq10}\mbox{goes below}-8\right\} }\right)\right)=\mathbb{E}\left(1_{A}\right)\mathbb{E}\left(\mathbb{E}^{\tau}\left(1_{(B_{t})_{t\leq10}\mbox{goes below}-8}\right)\right)$?
Or more generally,
${\mathbb{E}\left(1_{\left\{ (B_{t})_{t<\tau}\in E_{1}\right\} }\mathbb{E}^{\tau}\left(1_{\left\{ (B_{t})_{t\leq T}\in E_{2}\right\} }\right)\right)=\mathbb{E}\left(1_{\left\{ (B_{t})_{t<\tau}\in E_{1}\right\} }\right)\mathbb{E}\left(\mathbb{E}^{\tau}\left(1_{\left\{ (B_{t})_{t\leq T}\in E_{2}\right\} }\right)\right)}$?
where $E_{i}$ are Borel sets s.t. the events are independent.
The examples are just arbitrary. I'm interested in more general examples of independence, so if anyone can formulate a more general statement you are very welcome to do so.
 A: First of all, note that the strong Markov property of Brownian motion does not state that $(B_t)_{t \in [0,\tau]}$ and $(B_{t+\tau})_{t \geq 0}$ are independent, but that $(B_t)_{t \in [0,\tau]}$ and $(B_{t+\tau}-B_{\tau})_{t \geq 0}$ are independent. This independence yields in fact
$$\mathbb{E}(1_A \cdot 1_B) = \mathbb{E}(1_A) \mathbb{E}(1_B)$$
for any two events $A,B$ of the form
$$A := \{B_{t \in [0,\tau]} \in E\} \qquad B := \{(B_{t+\tau}-B_{\tau})_{t \geq 0} \in F\}. \tag{1}$$
This means that $A$ may depend on the (whole) path up to time $\tau$ and the set $B$ on the (whole) path of the restarted and shifted Brownian motion $(B_{\tau+t}-B_{\tau})_{t \geq 0}$. In particular, we can choose e.g.
$$\begin{align*} A &:= \{B_{t \in [0,\tau]} \, \text{goes above $5$}\} \\ B &:= \{B_{t+\tau}-B_{\tau} \, \text{goes below $-8$}\} \\ &= \{B_{t+\tau} \, \text{goes below $-12$}\}. \end{align*}$$

Concerning the conditional expectation: Note that we cannot condition on $\tau$ (the notion $\mathbb{E}^{\tau}$ doesn't make sense in this context); instead we have to take the conditional expectation with respect to $\mathcal{F}_{\tau}$ (or $B_{\tau}$). 
Suppose that $A$ and $B$ are as in $(1)$. Then $A \in \mathcal{F}_{\tau}$ and therefore
$$\begin{align*} \mathbb{E}(1_A 1_B) &= \mathbb{E}( \mathbb{E}(1_A 1_B \mid \mathcal{F}_{\tau})) \\ &= \mathbb{E}(1_A \mathbb{E}(1_B \mid \mathcal{F}_{\tau})). \tag{2} \end{align*}$$
Now, since $(B_t)_{t \geq 0}$ has the strong Markov property, we have
$$\mathbb{E}(1_B \mid \mathcal{F}_{\tau}) = \mathbb{E}^x( 1_{\{B_t-B_0 \in F\}}) \big|_{x=B_{\tau}}.$$
Hence,
$$\mathbb{E}(1_A 1_B) = \mathbb{E} \left( 1_A \mathbb{E}^x( 1_{\{B_t-B_0 \in F\}}) \big|_{x=B_{\tau}} \right). \tag{3}$$
So far the calculation holds for any stopping time $\tau$. Now


*

*if $\tau = \inf\{t \geq 0; B_t = -4\}$, then $B_{\tau}=-4$ and $(3)$ yields $$\begin{align*} \mathbb{E}(1_A 1_B) &= \mathbb{E}(1_A \mathbb{E}^{-4}( 1_{\{B_t-B_0 \in F\}})) \\ &= \mathbb{E}(1_A) \mathbb{E}(1_{\{B_t-(-4) \in F\}}) \\ &= \mathbb{E}(1_A 1_B). \end{align*}$$ (The important point is that $B_{\tau}$ is deterministic!)

*if $\tau$ is not of such an easy form (i.e. $B_{\tau}$ is not deterministic), then the right-hand side of $(3)$ does in general not give the product of the expectations. This is, roughly, due to the fact that we have only used the strong Markov property and not the independence of the increments (which means that a similar argumentation applies for strong Markov processes and in this more general setting we cannot expect that the expectation equals the product of the expectations). In this case, we really have to use that $(B_{\tau+t}-B_{\tau})_{t \geq 0}$ is independent from $\mathcal{F}_{\tau}$ and therefore in $(2)$ $$\mathbb{E}(1_B \mid \mathcal{F}_{\tau}) = \mathbb{E}(1_B)$$ which gives again $\mathbb{E}(1_A 1_B) = \mathbb{E}(1_A 1_B)$.

