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In the proof of the reflection principle in Durrett's textbook (Probability: Theory and Examples (4e), Theorem 8.4.1, page 317), there's a step which I'm a little shaky on. Basically, this proof invokes the strong Markov property so to set this up in Durrett's notation, let $B_s$ denote Brownian motion and define (I think for fixed $t$ here, but correct me if I'm wrong): $$ Y_s(\omega) = \begin{cases} 1, & \textrm{if } s<t,\, \omega(t-s)>a\\ 0, & \textrm{otherwise} \end{cases} $$ Then defining the stopping time $S = \inf\{s<t: B_s=a\}$ (and $\inf \emptyset = \infty$), we get: $$ Y_S(\theta_S \omega) = \begin{cases} 1, & \textrm{if } S<t,\, B_t>a\\ 0, & \textrm{otherwise} \end{cases} $$ where $\theta_S$ is the usual random shift operator for elements of $\mathcal{C}[0,\infty)$. So far, so good. This took a while to wrap my head around but it makes sense. Applying the strong Markov property we get $$ E_0(Y_S \circ \theta_S | \mathcal{F}_S) = E_{B_S} Y_S \textrm{ on } \{S < \infty\}$$ Now taking expectations, Durrett gets: $$ P_0(T_a < t, B_t \geq a) = E_0(Y_S \circ \theta_S; S < \infty) $$ I'm not seeing the right hand side. I'm trying to see how $$ E_0 (E_{B_S} Y_S; S < \infty) = E_0(Y_S \circ \theta_S; S<\infty) $$ but I'm getting confused. Does this somehow involve translation invarance of $B_s$ or is it straight from the definition of $Y_S$ and I'm just missing something obvious?

EDIT: Just putting @saz's solution here to clear up exactly where the strong Markov property is used. Essentially integrating $Y_S(\theta_S \omega)$ on $\{S < \infty\}$ and writing it as: $$E_0(Y_S \circ \theta_S; S < \infty) = E_0(E_0(Y_S \circ \theta_S | \mathcal{F}_S); S < \infty)$$ It becomes obvious where to apply the Markov property.

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  • $\begingroup$ How is $T_a$ defined? $\endgroup$
    – saz
    Jul 30, 2015 at 14:23
  • $\begingroup$ Sorry, $T_a = \inf\{t: B_t = a\}$, so essentially the same as $S$. $\endgroup$
    – gogurt
    Jul 30, 2015 at 16:25
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    $\begingroup$ $ E_0 (E_{B_S} Y_S; S < \infty) = E_0(Y_S \circ \theta_S; S<\infty) $ follows directly by taking expectations of the equation (two lines above) that defines the strong Markov property. $\endgroup$
    – user940
    Jul 30, 2015 at 17:34
  • $\begingroup$ @ByronSchmuland Ouch. That's obvious. Thanks for straightening that out. $\endgroup$
    – gogurt
    Jul 30, 2015 at 17:51
  • $\begingroup$ @ByronSchmuland ... but that's exactly what I wrote in the first line of my answer; I simply added some more details. $\endgroup$
    – saz
    Jul 30, 2015 at 18:34

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Actually, Durrett takes the expectation in the previous identity:

$$Y_S(\theta_S \omega) = \begin{cases} 1, & \text{if} \, S(\omega)<t, B_t(\omega)>a \\ 0, & \text{otherwise} \end{cases}. \tag{1}$$

Indeed: By $(1)$, we have $$Y_S(\theta_S \omega)= 1_{\{S<t,B_t>a\}}(\omega). $$

Since $T_a(\omega) = S(\omega)$ for all $\omega \in \{S<t\}$, this shows

$$Y_S(\theta_S \omega) = 1_{\{T_a<t, B_t>a\}}(\omega). \tag{2}$$

As $\{T_a < t\} = \{S<\infty\}$, this implies

$$Y_S(\theta_S \omega) 1_{\{S<\infty\}} \stackrel{(2)}{=} 1_{\{T_a<t, B_t>a\}}(\omega) 1_{\{S<\infty\}}(\omega) = 1_{\{T_a<t, B_t>a\}}(\omega). \tag{3}$$

Integrating both sides yields

$$\begin{align*} \mathbb{P}_0(T_a < t, B_t>a) &\stackrel{(3)}{=} \mathbb{E}_0(Y_S \circ \theta_S 1_{\{S<\infty\}}) \\ &= \mathbb{E}_0(Y_S \circ \theta_S; S< \infty). \end{align*}$$

Finally, note that $\mathbb{P}_0(B_t=a)=0$ and therefore the claim follows.

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  • $\begingroup$ ARGH. I see. I just got confused by the way it's worded in the book. Thanks. $\endgroup$
    – gogurt
    Jul 30, 2015 at 17:48

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