Strong Markov property, Brownian motion I have a question about the strong Markov property of Brownian motions.
Let $(\{X_t\}_{t \ge 0}, P_x)$ be a $d$-dimensional Brownian motion starting from $x \in \mathbb{R}^d$.
Let $\tau=\inf\{t>0 \mid |X_t-x|>r\}$, $r>0$. Here, $|\cdot|$ denotes the $d$-dimensional Euclidean norm.
In a paper, the authors claim that
\begin{align*}
P_{x}(\tau<t, |X_t-X_{\tau}| \ge r/2) 
&= \int_{0}^{t}E_{x}[P_{X_s}(|X_{t-s}-X_0| \ge r/2);\tau \in ds]. \tag{1}
\end{align*}
However, I do not know the proof.
By the strong Markov property, 
\begin{align}
P_{x}(\tau<t, |X_t-X_{\tau}| \ge r/2) 
&=P_{x}(\tau<t, |X_{(t-\tau)+\tau}-X_{\tau}| \ge r/2) \\
&=E_{x}[P_{X_\tau}(|X_{t-\tau}-X_0| \ge r/2);\tau<t] \tag{2}.
\end{align}
Can we obtain the equality (1) from the identity (2)?
 A: You didn't apply the strong  Markov property correctly.  This is indicated by the term
$$\mathbb{P}_{X_{\tau}}(|X_{t-\tau}-X_0| \geq r/2)$$
which appears on the right-hand side of $(2)$, note that this term is not well-defined since $t-\tau$ is not a non-negative random variable (it is only non-negative on $\{\tau \leq t\}$), and therefore $X_{t-\tau}$ is not well-defined.
There is the following stronger version of the strong Markov property which is needed here, you can find it for instance in the book by Schilling & Partzsch on Brownian motion (Theorem 6.11)

Theorem Let $(X_t)_{t \geq 0}$ be a $d$-dimensional Brownian motion with canonical filtration $(\mathcal{F}_t)_{t \geq 0}$. Let $\tau$ be a stopping time, and let $\eta \geq \tau$ be an $\mathcal{F}_{\tau+}$-measurable random variable. Then it holds for any bounded Borel measurable function $u$ that $$\mathbb{E}^x (u(X_{\eta}) \mid \mathcal{F}_{\tau+})(\omega) = (T_{\eta(\omega)-\tau(\omega)}u)(X_{\tau}(\omega)) \quad \text{a.s.} \tag{3}$$ where $$(T_s u)(x):= \mathbb{E}^x(u(X_s))$$

The above theorem gives the following corollary:

Corollary Under the assumptions of the theorem it holds that $$\mathbb{E}^x(f(X_{\eta},X_{\tau}) \mid \mathcal{F}_{\tau+})(\omega) = (G_{\eta(\omega)-\tau(\omega)} f)(X_{\tau}(\omega)) \quad \text{a.s.} \tag{4}$$ for any bounded Borel measurable function $f$ where $$(G_s f)(x) := \mathbb{E^x}(f(X_s,x)).$$

The idea is to prove $(4)$ first for functions of the form $f(x,y) = u(x) v(y)$ (... for such functions the assertion is a direct consequence of the above theorem and the pull out property), and then to use a density (or monotone class) argument to extend the identity to any bounded Borel measurable function $f$.

Now for the stopping time $\tau$ from your question define
$$\eta := \max\{t,\tau\}= \begin{cases} \tau, & \{\tau \geq t\},  \\ t, & \{\tau \leq t\}.\end{cases}$$
Then clearly $\eta \geq \tau$ and moreover $\eta$ is $\mathcal{F}_{\tau+}$-measurable. Using the tower property and  the pull out property we find that
\begin{align*} \mathbb{P}^x(\tau<t, |X_{t}-X_{\tau}| \geq r/2) &= \mathbb{E}^x \big[ \mathbb{E}^x( 1_{\{\tau <t\}} 1_{\{|X_{t}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}) \big] \\ &= \mathbb{E}^x \big[ 1_{\{\tau <t\}} \mathbb{E}^x(  1_{\{|X_{t}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}) \big] \end{align*}
Since $$1_{\{|X_{t}-X_{\tau}| \geq r/2\}} = 1_{\{|X_{\eta}-X_{\tau}| \geq r/2\}} \tag{5}$$ we have $$\mathbb{E}^x(  1_{\{|X_{t}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}) = \mathbb{E}^x(  1_{\{|X_{\eta}-X_{\tau}| \geq r/2\}} \mid \mathcal{F}_{\tau+}),$$ and therefore it follows from the above corollary that
\begin{align*} \mathbb{P}^x(\tau<t, |X_{t}-X_{\tau}| \geq r/2) &= \int_{\Omega} 1_{\{\tau <t\}}(\omega) \mathbb{P}^{X_{\tau}(\omega)}(|X_{t-\tau(\omega)}-X_0| \geq r/2) \, d\mathbb{P}^x(\omega) \\ &= \int_{(0,t)} \mathbb{E}^x(\mathbb{P}^{X_s}(|X_{t-s}-X_0| \geq r/2); \tau \in ds)  \end{align*}
