Stopping Times and the Brownian Restart

Proof.

Item (i).
For every \(t \geq 0\),

\[ \begin{align*} \{ \tau \wedge \rho \leq t \} &= \{ \tau \leq t \} \cup \{ \rho \leq t \} \in \mathcal{M}_t , \\\\ \{ \tau \vee \rho \leq t \} &= \{ \tau \leq t \} \cap \{ \rho \leq t \} \in \mathcal{M}_t . \end{align*} \]

For the shift, \(\{ \tau + v \leq t \}\) is empty when \(t \lt v\), and for \(t \geq v\) it equals \(\{ \tau \leq t - v \} \in \mathcal{M}_{t-v} \subseteq \mathcal{M}_t\).

Item (ii).
\(\Omega \cap \{\tau \leq t\} = \{\tau \leq t\} \in \mathcal{M}_t\), so \(\Omega \in \mathcal{M}_\tau\). If \(A \in \mathcal{M}_\tau\), then

\[ (\Omega \setminus A) \cap \{ \tau \leq t \} = \{ \tau \leq t \} \setminus \bigl( A \cap \{ \tau \leq t \} \bigr) \in \mathcal{M}_t , \]

and complements in \(\mathcal{M}_\infty\) are inherited, so \(\Omega \setminus A \in \mathcal{M}_\tau\). Countable unions pass through the intersection with \(\{\tau \leq t\}\) directly. For the measurability of \(\tau\), fix \(s \geq 0\); the set \(\{\tau \leq s\}\) satisfies

\[ \{ \tau \leq s \} \cap \{ \tau \leq t \} = \{ \tau \leq s \wedge t \} \in \mathcal{M}_{s \wedge t} \subseteq \mathcal{M}_t , \]

so \(\{\tau \leq s\} \in \mathcal{M}_\tau\), and the sets \(\{\tau \leq s\}\) for \(s \geq 0\) together with \(\{\tau = \infty\} = \Omega \setminus \bigcup_m \{\tau \leq m\}\) generate the Borel structure of \([0, \infty]\).

Item (iii).
Let \(A \in \mathcal{M}_\tau\). Since \(\tau \leq \rho\), \(\{\rho \leq t\} \subseteq \{\tau \leq t\}\), so

\[ A \cap \{ \rho \leq t \} = \bigl( A \cap \{ \tau \leq t \} \bigr) \cap \{ \rho \leq t \} \in \mathcal{M}_t , \]

the second factor lying in \(\mathcal{M}_t\) because \(\rho\) is a stopping time.

Item (iv).
First, \(\{\tau \lt s\} = \bigcup_m \{\tau \leq s - 1/m\} \in \mathcal{M}_s\) for every \(s \gt 0\), the union running over \(m \in \mathbb{N}\) with \(1/m \lt s\). For a finite grid point \(g = (k+1) 2^{-n}\),

\[ \{ \tau_n = g \} = \{ \tau \lt g \} \setminus \{ \tau \lt g - 2^{-n} \} \in \mathcal{M}_g , \]

reading \(\{\tau \lt 0\} = \emptyset\) for \(k = 0\). Hence for \(t \geq 0\),

\[ \{ \tau_n \leq t \} = \bigcup_{g \in G_n ,\; g \leq t} \{ \tau_n = g \} \in \mathcal{M}_t , \]

a finite union of sets lying in \(\mathcal{M}_g \subseteq \mathcal{M}_t\), so \(\tau_n\) is a stopping time. On \(\{\tau \lt \infty\}\) the defining half-open window gives \(\tau \lt \tau_n \leq \tau + 2^{-n}\) directly. For the monotonicity, fix \(\omega\) with \(\tau(\omega) \in [k 2^{-n-1}, (k+1) 2^{-n-1})\), so that \(\tau_{n+1}(\omega) = (k+1) 2^{-n-1}\); the window \([k 2^{-n-1}, (k+1) 2^{-n-1})\) is contained in a single window of the coarser grid whose right endpoint is at least \((k+1) 2^{-n-1}\), whence \(\tau_{n+1}(\omega) \leq \tau_n(\omega)\). Finally, let \(A \in \mathcal{M}_\tau\) and \(g = (k+1)2^{-n}\). Then

\[ A \cap \{ \tau_n = g \} = \bigl( A \cap \{ \tau \lt g \} \bigr) \setminus \bigl( A \cap \{ \tau \lt g - 2^{-n} \} \bigr) , \]

and for every \(s \gt 0\),

\[ A \cap \{ \tau \lt s \} = \bigcup_{m} \bigl( A \cap \{ \tau \leq s - 1/m \} \bigr) \in \mathcal{M}_s , \]

each term lying in \(\mathcal{M}_{s - 1/m} \subseteq \mathcal{M}_s\) by the definition of \(\mathcal{M}_\tau\). Taking \(s = g\) and \(s = g - 2^{-n}\) closes item (iv).

Item (v).
Write \(Y_\tau\) for the map in question and fix a Borel set \(B \subseteq \mathbb{R}\) and \(t \geq 0\). On \(\{\tau \leq t\}\) we have \(Y_\tau = Y_{\tau \wedge t}\), so

\[ \{ Y_\tau \in B \} \cap \{ \tau \leq t \} = \{ Y_{\tau \wedge t} \in B \} \cap \{ \tau \leq t \} . \]

The map \(\omega \mapsto (\tau(\omega) \wedge t, \omega)\) from \(\Omega\) to \([0, t] \times \Omega\) is measurable from \(\mathcal{M}_t\) to \(\mathcal{B}([0,t]) \otimes \mathcal{M}_t\): its second component is the identity, and its first component is \(\mathcal{M}_t\)-measurable because \(\{\tau \wedge t \leq s\}\) equals \(\{\tau \leq s\} \in \mathcal{M}_s \subseteq \mathcal{M}_t\) for \(s \lt t\) and equals \(\Omega\) for \(s \geq t\). Progressive measurability says precisely that the restriction of \(Y\) to \([0,t] \times \Omega\) is \(\mathcal{B}([0,t]) \otimes \mathcal{M}_t\)-measurable, so the composition \(Y_{\tau \wedge t}\) is \(\mathcal{M}_t\)-measurable, and the displayed intersection lies in \(\mathcal{M}_t\). Letting \(t\) run over \(\mathbb{N}\) shows \(\{Y_\tau \in B\} \cap \{\tau \lt \infty\} \in \mathcal{M}_\infty\), and the display then verifies the defining condition of \(\mathcal{M}_\tau\). Here measurability is meant on the event \(\{\tau \lt \infty\}\), which itself lies in \(\mathcal{M}_\tau\) because its intersection with \(\{\tau \leq t\}\) is \(\{\tau \leq t\}\) again.

Proof.

If \(U = \mathbb{R}^m\), then \(\tau_U \equiv \infty\) and \(\{\tau_U \leq t\} = \emptyset \in \mathcal{M}_t\), so we may assume the complement \(C = \mathbb{R}^m \setminus U\) is nonempty and closed. Below \(| \cdot |\) is the Euclidean length on \(\mathbb{R}^m\).

The exit position.
Suppose \(\tau_U(\omega) \lt \infty\) and abbreviate \(\tau = \tau_U(\omega)\). By the definition of the infimum there are times \(s_j \in [\tau, \tau + 1/j)\) with \(Y(s_j, \omega) \in C\). Continuity of the path gives \(Y(s_j, \omega) \to Y(\tau, \omega)\), and \(C\) is closed, so \(Y(\tau, \omega) \in C\). This proves the second claim and is the pivot of the first inclusion below.

A countable test.
Write \(d(y) = \inf_{c \in C} |y - c|\) for the distance to \(C\), a continuous function of \(y\), and note \(y \in C\) exactly when \(d(y) = 0\), because \(C\) is closed. We claim that for every \(t \gt 0\),

\[ \{ \tau_U \leq t \} = \bigcap_{m \in \mathbb{N}} \bigcup_{r \in \mathbb{Q} \cap [0, t)} \Bigl\{ \omega : d\bigl( Y(r, \omega) \bigr) \lt \tfrac{1}{m} \Bigr\} . \]

Grant the claim for a moment. Each set on the right lies in \(\mathcal{M}_r \subseteq \mathcal{M}_t\), because \(Y(r, \cdot)\) is \(\mathcal{M}_r\)-measurable by adaptedness and \(d\) is continuous; the countable intersection of countable unions therefore lies in \(\mathcal{M}_t\). The remaining case \(t = 0\) is direct: if \(Y(0, \omega) \in U\), then continuity of the path and openness of \(U\) keep the path inside \(U\) on an initial interval, so \(\tau_U(\omega) \gt 0\), while \(Y(0, \omega) \in C\) forces \(\tau_U(\omega) = 0\) outright; hence \(\{\tau_U \leq 0\} = \{\omega : Y(0, \omega) \in C\} \in \mathcal{M}_0\).

The inclusion from left to right.
Suppose \(\tau = \tau_U(\omega) \leq t\) and fix \(m\). The first step gives \(Y(\tau, \omega) \in C\), so \(d(Y(\tau, \omega)) = 0\). By continuity of \(s \mapsto d(Y(s, \omega))\) there is a neighborhood of \(\tau\) on which \(d(Y(\cdot, \omega)) \lt 1/m\). If \(\tau \gt 0\), this neighborhood contains rationals \(r \lt \tau \leq t\); if \(\tau = 0\), it contains \(r = 0\) itself. Either way a witness \(r \in \mathbb{Q} \cap [0, t)\) exists.

The inclusion from right to left.
Suppose that for every \(m\) there is a rational \(r_m \in [0, t)\) with \(d(Y(r_m, \omega)) \lt 1/m\). The sequence \((r_m)\) lies in the compact interval \([0, t]\), so a subsequence \(r_{m_j}\) converges to some \(s \in [0, t]\). Continuity of the path and of \(d\) give

\[ d\bigl( Y(s, \omega) \bigr) = \lim_{j \to \infty} d\bigl( Y(r_{m_j}, \omega) \bigr) = 0 , \]

so \(Y(s, \omega) \in C\) and hence \(\tau_U(\omega) \leq s \leq t\).

Proof.

The shifted filtration.
Each \(\tau + v\) is a stopping time by item (i) of the calculus, and \(\tau + a \leq \tau + b\) for \(a \leq b\), so item (iii) makes \(\{\mathcal{G}_v\}\) an increasing family. If \(N\) is a \(\mathbb{P}\)-null set, then \(N \cap \{\tau + v \leq t\}\) is again a null set, hence lies in \(\mathcal{F}_t\), which contains every null set; so \(N \in \mathcal{G}_v\). For the adaptedness, recall that the track works with the continuous version of the motion; modifying \(w\) on a null event if necessary, every path is continuous, and the modification is harmless because null events lie in every \(\mathcal{F}_t\). A continuous adapted process is progressively measurable, so item (v) of the calculus, applied at the stopping time \(\tau + v\), makes \(w_{\tau+v}\) measurable with respect to \(\mathcal{G}_v\). The variable \(w_\tau\) is \(\mathcal{F}_\tau\)-measurable for the same reason, and \(\mathcal{F}_\tau \subseteq \mathcal{G}_v\) by monotonicity, so the difference \(\tilde{B}_v\) is \(\mathcal{G}_v\)-measurable on \(\{\tau \lt \infty\}\). On the remaining event the process was set to zero, and both that event and its subsets are null events lying in \(\mathcal{G}_v\), so the patch preserves measurability. This proves (iii).

The factorization at a discrete stopping time.
The core of the theorem is the following claim: for every \(A \in \mathcal{F}_\tau\), every finite grid \(0 \leq v_1 \lt \cdots \lt v_k\), and every bounded Borel \(\varphi : \mathbb{R}^k \to \mathbb{R}\),

\[ \mathbb{E} \bigl[ \mathbf{1}_A\, \varphi( \tilde{B}_{v_1}, \ldots, \tilde{B}_{v_k} ) \bigr] = \mathbb{P}(A)\, \mathbb{E} \bigl[ \varphi( w_{v_1}, \ldots, w_{v_k} ) \bigr] , \]

which we first prove with the dyadic discretization \(\tau_n\) of item (iv) of the calculus in place of \(\tau\), writing \(\tilde{B}^n_v = w_{\tau_n + v} - w_{\tau_n}\), and for bounded Borel \(\varphi\). Since \(\tau \lt \infty\) almost surely, so is \(\tau_n\), and the events \(\{\tau_n = g\}\) for finite grid values \(g \in G_n\) partition \(\{\tau_n \lt \infty\}\). On \(\{\tau_n = g\}\) the shifted process is \(\tilde{w}^g_v = w_{g + v} - w_g\), the shift of the motion by the deterministic amount \(g\), so countable additivity of the integral over the partition gives

\[ \mathbb{E} \bigl[ \mathbf{1}_A\, \varphi( \tilde{B}^n_{v_1}, \ldots, \tilde{B}^n_{v_k} ) \bigr] = \sum_{g} \mathbb{E} \bigl[ \mathbf{1}_{A \cap \{\tau_n = g\}}\, \varphi( \tilde{w}^g_{v_1}, \ldots, \tilde{w}^g_{v_k} ) \bigr] , \]

the series converging absolutely because \(\varphi\) is bounded and the events are disjoint. Fix a value \(g\). The slicing half of item (iv) places \(A \cap \{\tau_n = g\}\) in \(\mathcal{F}_g\), and the \(\sigma\)-algebra \(\sigma( \tilde{w}^g_v : v \geq 0 )\) is independent of \(\mathcal{F}_g\). The indicator of \(A \cap \{\tau_n = g\}\) and the variable \(\varphi( \tilde{w}^g_{v_1}, \ldots, \tilde{w}^g_{v_k} )\) are measurable with respect to these two independent \(\sigma\)-algebras, hence are independent random variables, every event determined by the one lying in the first \(\sigma\)-algebra and every event determined by the other in the second. Both are bounded, so the expectation of their product factors:

\[ \mathbb{E} \bigl[ \mathbf{1}_{A \cap \{\tau_n = g\}}\, \varphi( \tilde{w}^g_{v_1}, \ldots, \tilde{w}^g_{v_k} ) \bigr] = \mathbb{P} \bigl( A \cap \{\tau_n = g\} \bigr)\, \mathbb{E} \bigl[ \varphi( \tilde{w}^g_{v_1}, \ldots, \tilde{w}^g_{v_k} ) \bigr] . \]

The shifted process \(\tilde{w}^g\) is itself a standard Brownian motion, as the previous page verified when shifting the clock by a deterministic amount, and the joint distribution of a Brownian motion at the times \(v_1, \ldots, v_k\) is the Gaussian distribution that (W1) prescribes, the same for \(\tilde{w}^g\) as for \(w\). Expectations of bounded Borel functions depend on the vector only through its distribution, so \(\mathbb{E}[ \varphi( \tilde{w}^g_{v_1}, \ldots, \tilde{w}^g_{v_k} ) ] = \mathbb{E}[ \varphi( w_{v_1}, \ldots, w_{v_k} ) ]\), a factor no longer depending on \(g\). Summing over \(g\) and using countable additivity once more, we obtain

\[ \mathbb{E} \bigl[ \mathbf{1}_A\, \varphi( \tilde{B}^n_{v_1}, \ldots, \tilde{B}^n_{v_k} ) \bigr] = \mathbb{P}(A)\, \mathbb{E} \bigl[ \varphi( w_{v_1}, \ldots, w_{v_k} ) \bigr] . \]

Passing to the stopping time.
Now let \(\varphi\) be bounded and continuous. Off a null event, \(\tau_n \downarrow \tau\) with every path of \(w\) continuous, so \(\tilde{B}^n_{v_i} = w_{\tau_n + v_i} - w_{\tau_n} \to w_{\tau + v_i} - w_\tau = \tilde{B}_{v_i}\) for each \(i\), and \(\varphi( \tilde{B}^n_{v_1}, \ldots, \tilde{B}^n_{v_k} ) \to \varphi( \tilde{B}_{v_1}, \ldots, \tilde{B}_{v_k} )\) almost surely. The integrands are bounded by \(\sup |\varphi|\), a constant, so dominated convergence passes the display to the limit: the factorization holds for \(\tau\) itself and every bounded continuous \(\varphi\).

From continuous tests to the \(\sigma\)-algebra.
Let \(R = F_1 \times \cdots \times F_k\) be a product of closed sets \(F_i \subseteq \mathbb{R}\), and set

\[ \varphi_j( z_1, \ldots, z_k ) = \prod_{i=1}^{k} \max \bigl( 0,\; 1 - j\, d_i( z_i ) \bigr) , \]

where \(d_i(z) = \inf_{c \in F_i} |z - c|\). Each \(\varphi_j\) is continuous with values in \([0, 1]\), and \(\varphi_j \downarrow \mathbf{1}_R\) pointwise as \(j \to \infty\), because \(d_i(z) = 0\) exactly on the closed set \(F_i\). Dominated convergence applied to both sides of the factorization gives, for every \(A \in \mathcal{F}_\tau\),

\[ \mathbb{P} \bigl( A \cap \{ ( \tilde{B}_{v_1}, \ldots, \tilde{B}_{v_k} ) \in R \} \bigr) = \mathbb{P}(A)\, \mathbb{P} \bigl( ( w_{v_1}, \ldots, w_{v_k} ) \in R \bigr) . \]

Events of the displayed kind, taken over all finite grids and all products of closed sets, are closed under intersection: two grids merge into one, a coordinate left unconstrained carries the closed factor \(\mathbb{R}\), and the intersection of two products of closed sets is again one. They generate \(\sigma( \tilde{B}_v : v \geq 0 )\). The extension of the factorization from this family to the generated \(\sigma\)-algebra is the monotone-class step this track has taken on faith since the construction of the integral, and we extend the same trust to this application. This proves (ii).

The restarted motion is standard.
Take \(A = \Omega\) in the last display: the vectors \(( \tilde{B}_{v_1}, \ldots, \tilde{B}_{v_k} )\) and \(( w_{v_1}, \ldots, w_{v_k} )\) assign the same probability to every product of closed sets. Products of closed sets are closed under intersection and generate the Borel \(\sigma\)-algebra of \(\mathbb{R}^k\), so the two distributions coincide, by the same trusted extension. The finite-dimensional distributions of \(\tilde{B}\) are therefore those of the standard motion: (W1) and (W2) hold for \(\tilde{B}\) because both are statements about finite-dimensional distributions, and \(\tilde{B}_0 = 0\). For (W3), every path of \(\tilde{B}\) is continuous: on \(\{\tau \lt \infty\}\) it is a difference of values of the continuous path of \(w\), and on the remaining event it is identically zero. This proves (i).

Restarting at a later stopping time.
Fix \(0 \leq a \lt b\). The time \(\rho = \tau + a\) is a stopping time with \(\rho \lt \infty\) almost surely, and the three preceding steps used no property of \(\tau\) beyond these two. Applied at \(\rho\), they show that \(\sigma( w_{\rho + u} - w_\rho : u \geq 0 )\) is independent of \(\mathcal{F}_\rho = \mathcal{G}_a\). The increment \(\tilde{B}_b - \tilde{B}_a = w_{\rho + (b - a)} - w_\rho\) is measurable with respect to that \(\sigma\)-algebra, so it is independent of \(\mathcal{G}_a\). This proves (iv).