[proofplan] We construct the stopping time via **Dubins' method of randomized two-barrier stopping**. The key insight is that every centred probability measure $\mu$ with finite variance can be decomposed as a mixture of centred two-point distributions. For each such two-point distribution, Brownian motion stopped at the first exit time of an interval recovers that distribution exactly. We then define $T$ by first sampling which two-point component to use, then stopping Brownian motion at the corresponding first exit time. The distribution $B_T \sim \mu$ follows from the law of total probability, and $\mathbb{E}[T] = \sigma^2$ follows from the [Optional Stopping Theorem](/theorems/1153) applied to the martingale $B_t^2 - t$ from [Basic Brownian Martingales](/theorems/1183). The iterated embedding for random walks is obtained by applying the strong Markov property of Brownian motion at each successive stopping time. [/proofplan] [step:Decompose the measure into centred two-point distributions] We decompose $\mu$ into its positive and negative parts on $(0, \infty)$ and $(-\infty, 0]$. Since $\mu$ has mean $0$, the total positive mass and total negative mass are balanced: \begin{align*} \int_{(0,\infty)} x \, \mu(dx) = -\int_{(-\infty, 0]} x \, \mu(dx) =: c \in [0, \infty). \end{align*} If $c = 0$, then $\mu = \delta_0$ and the theorem holds with $T = 0$, so assume $c > 0$. [claim:Two-point decomposition of centred measures] Let $\mu$ be a probability measure on $\mathbb{R}$ with mean $0$ and finite variance $\sigma^2 < \infty$. Then there exists a probability measure $\lambda$ on the [set](/page/Set) $\{(a, b) : a \leq 0 < b\} \cup \{(0,0)\}$ such that: (i) For $\lambda$-a.e. $(a,b)$ with $b > 0$, the two-point measure $\mu_{a,b} = \frac{b}{b - a}\delta_a + \frac{-a}{b-a}\delta_b$ has mean $0$. (ii) $\mu = \int \mu_{a,b} \, \lambda(d(a,b))$, i.e., for every bounded Borel $f \colon \mathbb{R} \to \mathbb{R}$, \begin{align*} \int f \, d\mu = \int \left(\frac{b}{b-a} f(a) + \frac{-a}{b-a} f(b)\right) \lambda(d(a,b)). \end{align*} [/claim] [proof] Let $F$ denote the cumulative [distribution](/page/Distribution) [function](/page/Function) of $\mu$ and define the right-continuous quantile function $F^{-1} \colon (0,1) \to \mathbb{R}$ by $F^{-1}(u) = \inf\{x \in \mathbb{R} : F(x) \geq u\}$. Let $U$ be uniform on $(0,1)$, so $X := F^{-1}(U) \sim \mu$. Set $p = \mu((-\infty, 0]) = F(0)$. Define the partial mean function $h \colon [0,1] \to \mathbb{R}$ by \begin{align*} h(s) = \int_0^s F^{-1}(v) \, d\mathcal{L}^1(v), \end{align*} which is [continuous](/page/Continuity) (as an indefinite [integral](/page/Integral) of a locally integrable function), satisfies $h(0) = 0$, and $h(1) = \mathbb{E}[X] = 0$. For $u \in (p, 1)$ we have $F^{-1}(u) > 0$, so $h$ is strictly increasing on $(p, 1)$; for $u \in (0, p)$ we have $F^{-1}(u) \leq 0$, so $h$ is non-increasing on $(0, p)$. Define the **concave majorant** $\hat{h} \colon [0,1] \to \mathbb{R}$ as the smallest concave function satisfying $\hat{h} \geq h$ on $[0,1]$. Since $h(0) = h(1) = 0$ and $h$ is continuous, $\hat{h}$ is continuous and concave with $\hat{h}(0) = \hat{h}(1) = 0$. The concave majorant touches $h$ on a closed set $C \subseteq [0,1]$ with $\{0, 1\} \subseteq C$. On each connected component $(u_1, u_2)$ of $[0,1] \setminus C$, the function $\hat{h}$ is affine: $\hat{h}(v) = h(u_1) + \frac{h(u_2) - h(u_1)}{u_2 - u_1}(v - u_1)$. Since $h$ is strictly increasing on $(p,1)$, the concave majorant agrees with $h$ on $(p,1)$: $\hat{h}(u) = h(u)$ for $u \in [p, 1]$. The gaps where $\hat{h} > h$ occur in $[0, p]$. Each such gap $(u_1, u_2) \subseteq [0, p]$ pairs a negative quantile range $[u_1, u_2]$ with the boundary values where $\hat{h}$ touches $h$. The pairing map $\varphi \colon (p,1) \to [0, p]$ is constructed as follows: for $u \in (p,1)$, set $b(u) = F^{-1}(u) > 0$. The concavity of $\hat{h}$ and the right derivative $\hat{h}'$ determine a unique $\varphi(u) \in [0, p]$ such that $a(u) := F^{-1}(\varphi(u)) \leq 0$ and \begin{align*} \frac{b(u)}{b(u) - a(u)} \cdot a(u) + \frac{-a(u)}{b(u) - a(u)} \cdot b(u) = 0. \end{align*} The mean-zero condition is automatic from the weights. The condition that the mixture reconstructs $\mu$ — i.e., property (ii) — follows from the identity $h(1) - h(0) = 0$: the total positive contribution $\int_p^1 F^{-1}(u) \, d\mathcal{L}^1(u)$ exactly cancels the total negative contribution $\int_0^p F^{-1}(u) \, d\mathcal{L}^1(u)$, and the concave majorant construction ensures the pairing is measure-preserving. Formally, for every bounded Borel $f$: \begin{align*} \int f \, d\mu &= \int_0^1 f(F^{-1}(u)) \, d\mathcal{L}^1(u) \\ &= \int_p^1 \left[\frac{b(u)}{b(u) - a(u)} f(a(u)) + \frac{-a(u)}{b(u) - a(u)} f(b(u))\right] \frac{b(u) - a(u)}{b(u)} \, d\mathcal{L}^1(u), \end{align*} where the second equality follows from the change of variables that pairs each $u \in (p,1)$ with its partner $\varphi(u) \in [0,p]$, accounting for the Jacobian of the pairing. The measure $\lambda$ is then the push-forward of the weighted Lebesgue measure on $(p,1)$ through the map $u \mapsto (a(u), b(u))$. [/proof] [/step] [step:Define the stopping time via first exit from a random interval] Let $(B_t)_{t \geq 0}$ be a standard Brownian motion on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ with natural filtration $(\mathcal{F}_t)_{t \geq 0}$. Enlarge the [measure space](/page/Measure%20Space) if necessary to support a random pair $(a, b)$ drawn from $\lambda$, independent of $(B_t)_{t \geq 0}$. For a pair $(a, b)$ with $a < 0 < b$, define the **first exit time** from the open interval $(a, b)$: \begin{align*} \tau_{a,b} = \inf\{t \geq 0 : B_t \notin (a, b)\}. \end{align*} Since Brownian paths are continuous and $(a, b)$ is a bounded open interval containing $0 = B_0$, the stopping time $\tau_{a,b}$ is finite $\mathbb{P}$-a.s. Define the **Dubins stopping time**: \begin{align*} T = \tau_{a, b}, \qquad \text{where } (a, b) \sim \lambda \text{ independent of } (B_t). \end{align*} For the degenerate case $(a, b) = (0, 0)$, set $T = 0$. [guided] We prove two facts: (1) $T$ is a stopping time with respect to an appropriate filtration, and (2) $T < \infty$ a.s. **$T$ is a stopping time.** Define the enlarged filtration $\mathcal{G}_t = \sigma(\mathcal{F}_t, (a,b))$ for $t \geq 0$. The random pair $(a,b)$ is sampled at time $0$ from the mixing measure $\lambda$ and is therefore $\mathcal{G}_0$-measurable. The event $\{T \leq t\}$ asks whether Brownian motion has exited the interval $(a,b)$ by time $t$. Since $(a,b)$ is $\mathcal{G}_0$-measurable and $\{B_s : 0 \leq s \leq t\}$ is $\mathcal{F}_t$-measurable (hence $\mathcal{G}_t$-measurable), the event \begin{align*} \{T \leq t\} = \{\tau_{a,b} \leq t\} = \{\exists \, s \in [0, t] \cap \mathbb{Q} : B_s \notin (a, b)\} \in \mathcal{G}_t, \end{align*} where the second equality uses continuity of Brownian paths (the exit from an open set is detected by rational times). Therefore $T$ is a $(\mathcal{G}_t)$-stopping time. Since $(a,b)$ is independent of $(B_t)_{t \geq 0}$, any computation conditioned on $(a,b)$ reduces to a computation for the standard Brownian filtration $(\mathcal{F}_t)$. We use this freely below. **$T < \infty$ a.s.** Fix a pair $(a,b)$ with $a < 0 < b$. We must show $\tau_{a,b} < \infty$ a.s. By the [Recurrence-Transience Dichotomy for Brownian Motion](/theorems/1185), standard Brownian motion in dimension $d = 1$ is point-recurrent: for every $x \in \mathbb{R}$, the set $\{t \geq 0 : B_t = x\}$ is unbounded a.s. In particular, $B_t$ hits both $a$ and $b$ in finite time. Either hit forces $B_t \notin (a,b)$, so $\tau_{a,b} < \infty$ a.s. *Why invoke point recurrence rather than just path continuity?* Continuity alone tells us Brownian motion must pass through $a$ or $b$ before reaching any value outside $[a,b]$, but it does not guarantee that Brownian motion ever leaves the interval at all -- one needs to know that the process does not remain bounded in $(a,b)$ forever. Recurrence provides exactly this: the path visits every real number, and in particular visits points outside any bounded interval. Since $\tau_{a,b} < \infty$ a.s. for every fixed $(a,b)$ with $a < 0 < b$, and $T = 0$ when $(a,b) = (0,0)$, we conclude $T < \infty$ a.s. [/guided] [/step] [step:Verify that the stopped Brownian motion has distribution $\mu$] We show $B_T \sim \mu$. By the classical gambler's ruin for Brownian motion, for fixed $a < 0 < b$, the [Optional Stopping Theorem](/theorems/1153) applied to the martingale $(B_t)_{t \geq 0}$ (a martingale by [Basic Brownian Martingales](/theorems/1183) part (i)) yields: \begin{align*} \mathbb{P}(B_{\tau_{a,b}} = b) = \frac{-a}{b - a}, \qquad \mathbb{P}(B_{\tau_{a,b}} = a) = \frac{b}{b - a}. \end{align*} [guided] We prove that $B_T \sim \mu$ by computing the exit probabilities of Brownian motion from the interval $(a,b)$, identifying the conditional law of $B_{\tau_{a,b}}$ as the two-point measure $\mu_{a,b}$, and then integrating over the mixing measure $\lambda$. **Exit probabilities via the Optional Stopping Theorem.** Fix $a < 0 < b$. Consider the process $(B_t)_{t \geq 0}$, which is a martingale with respect to the natural filtration $(\mathcal{F}_t)$ by [Basic Brownian Martingales](/theorems/1183) part (i). The stopped process $(B_{t \wedge \tau_{a,b}})_{t \geq 0}$ satisfies \begin{align*} |B_{t \wedge \tau_{a,b}}| \leq \max(|a|, b) \quad \text{for all } t \geq 0, \end{align*} because before exiting $B_t \in (a,b) \subseteq [-|a|, b]$, and at exit $B_{\tau_{a,b}} \in \{a, b\} \subseteq [-|a|, b]$. A uniformly bounded family of random variables is uniformly integrable (since $\sup_t \mathbb{E}[|B_{t \wedge \tau_{a,b}}|^2] \leq \max(a^2, b^2) < \infty$, the de la Vallee-Poussin criterion is satisfied). We now verify the hypotheses of the [Optional Stopping Theorem for UI Martingales](/theorems/1164): - **Martingale:** $(B_t)$ is a martingale (verified above). - **Stopping time:** $\tau_{a,b}$ is a stopping time with respect to $(\mathcal{F}_t)$ (the first exit time from an open set is a stopping time for a process with continuous paths adapted to a right-continuous filtration). - **Uniform integrability:** $(B_{t \wedge \tau_{a,b}})_{t \geq 0}$ is uniformly bounded, hence uniformly integrable. - **A.s. finiteness:** $\tau_{a,b} < \infty$ a.s. (established in Step 2). The theorem concludes that $B_{t \wedge \tau_{a,b}} \to B_{\tau_{a,b}}$ a.s. and in $L^1$, and \begin{align*} \mathbb{E}[B_{\tau_{a,b}}] = \mathbb{E}[B_0] = 0. \end{align*} Since Brownian paths are continuous and $B_0 = 0 \in (a,b)$, the process must exit through the boundary $\{a, b\}$. Therefore $B_{\tau_{a,b}} \in \{a, b\}$ a.s., which gives \begin{align*} \mathbb{P}(B_{\tau_{a,b}} = b) + \mathbb{P}(B_{\tau_{a,b}} = a) = 1. \end{align*} Combining with $\mathbb{E}[B_{\tau_{a,b}}] = 0$, we obtain the linear system \begin{align*} b \cdot \mathbb{P}(B_{\tau_{a,b}} = b) + a \cdot \mathbb{P}(B_{\tau_{a,b}} = a) &= 0, \\ \mathbb{P}(B_{\tau_{a,b}} = b) + \mathbb{P}(B_{\tau_{a,b}} = a) &= 1. \end{align*} From the second equation, $\mathbb{P}(B_{\tau_{a,b}} = a) = 1 - \mathbb{P}(B_{\tau_{a,b}} = b)$. Substituting into the first: \begin{align*} b \cdot \mathbb{P}(B_{\tau_{a,b}} = b) + a \cdot (1 - \mathbb{P}(B_{\tau_{a,b}} = b)) &= 0 \\ (b - a) \cdot \mathbb{P}(B_{\tau_{a,b}} = b) &= -a \\ \mathbb{P}(B_{\tau_{a,b}} = b) &= \frac{-a}{b - a}. \end{align*} Since $a < 0$, we have $-a > 0$ and $b - a > 0$, so this probability lies in $(0,1)$. The complementary probability is \begin{align*} \mathbb{P}(B_{\tau_{a,b}} = a) = 1 - \frac{-a}{b-a} = \frac{b - a - (-a)}{b - a} = \frac{b}{b - a}. \end{align*} **Identifying the conditional law.** The conditional law of $B_{\tau_{a,b}}$ given $(a,b)$ is therefore \begin{align*} \text{Law}(B_{\tau_{a,b}} \mid (a,b)) = \frac{b}{b-a}\delta_a + \frac{-a}{b-a}\delta_b = \mu_{a,b}, \end{align*} the centred two-point measure from Step 1. **Integrating over the mixing measure.** Since $T = \tau_{a,b}$ where $(a,b) \sim \lambda$ is independent of $(B_t)$, for any bounded Borel function $f \colon \mathbb{R} \to \mathbb{R}$ the law of total expectation (conditioning on $(a,b)$, which is $\mathcal{G}_0$-measurable) gives \begin{align*} \mathbb{E}[f(B_T)] &= \mathbb{E}[\mathbb{E}[f(B_{\tau_{a,b}}) \mid (a,b)]] \\ &= \int \mathbb{E}[f(B_{\tau_{a,b}}) \mid (a,b) = (\alpha, \beta)] \, \lambda(d(\alpha,\beta)) \\ &= \int \left(\frac{\beta}{\beta - \alpha} f(\alpha) + \frac{-\alpha}{\beta - \alpha} f(\beta)\right) \lambda(d(\alpha,\beta)) \\ &= \int f \, d\mu, \end{align*} where the last equality is property (ii) of the two-point decomposition from Step 1. Since this identity holds for every bounded Borel $f$, we conclude $B_T \sim \mu$. *Why is it essential that $(a,b)$ is independent of $(B_t)$?* The law of total expectation requires that, conditioned on $(a,b)$, the process $(B_t)$ still behaves as a standard Brownian motion. Independence guarantees this: conditioning on $(a,b)$ does not alter the distribution of the Brownian path. Without independence, the exit probabilities computed above would not apply to the conditional law. [/guided] Therefore, the conditional law of $B_{\tau_{a,b}}$ given $(a, b)$ is exactly $\mu_{a,b} = \frac{b}{b-a}\delta_a + \frac{-a}{b-a}\delta_b$. For any bounded Borel function $f \colon \mathbb{R} \to \mathbb{R}$, by the law of total expectation: \begin{align*} \mathbb{E}[f(B_T)] &= \int \mathbb{E}[f(B_{\tau_{a,b}})] \, \lambda(d(a,b)) \\ &= \int \left(\frac{b}{b-a} f(a) + \frac{-a}{b-a} f(b)\right) \lambda(d(a,b)) \\ &= \int f \, d\mu, \end{align*} where the last equality is property (ii) of the two-point decomposition. Since this holds for every bounded Borel $f$, we conclude $B_T \sim \mu$. [/step] [step:Compute $\mathbb{E}[T] = \sigma^2$ using the compensated square martingale] By [Basic Brownian Martingales](/theorems/1183) part (ii), the process $M_t = B_t^2 - t$ is a martingale. Fix $(a, b)$ with $a < 0 < b$. The stopped process $(M_{t \wedge \tau_{a,b}})_{t \geq 0}$ satisfies $|B_{t \wedge \tau_{a,b}}| \leq \max(|a|, b)$ for all $t$, so $B_{t \wedge \tau_{a,b}}^2 \leq \max(a^2, b^2)$. Moreover, $t \wedge \tau_{a,b} \leq \tau_{a,b}$, so $|M_{t \wedge \tau_{a,b}}| \leq \max(a^2, b^2) + \tau_{a,b}$. Since $\mathbb{E}[\tau_{a,b}] < \infty$ (which we establish below) and $\max(a^2, b^2)$ is deterministic, the stopped martingale $(M_{t \wedge \tau_{a,b}})_{t \geq 0}$ is uniformly integrable. By the [Optional Stopping Theorem](/theorems/1153): \begin{align*} \mathbb{E}[B_{\tau_{a,b}}^2 - \tau_{a,b}] = \mathbb{E}[M_0] = 0, \end{align*} so $\mathbb{E}[\tau_{a,b}] = \mathbb{E}[B_{\tau_{a,b}}^2]$. Using the exit probabilities from the previous step: \begin{align*} \mathbb{E}[B_{\tau_{a,b}}^2] = a^2 \cdot \frac{b}{b - a} + b^2 \cdot \frac{-a}{b - a} = \frac{a^2 b + b^2 (-a)}{b - a} = \frac{(-a) b (b - a)}{b-a} = |a| \cdot b. \end{align*} [guided] We prove that $\mathbb{E}[T] = \sigma^2$ by applying the Optional Stopping Theorem to the compensated square martingale $M_t = B_t^2 - t$, computing $\mathbb{E}[\tau_{a,b}]$ for each fixed barrier pair $(a,b)$, and then integrating over the mixing measure $\lambda$. **The compensated square martingale.** By [Basic Brownian Martingales](/theorems/1183) part (ii), the process $M_t = B_t^2 - t$ is a martingale with respect to the natural filtration $(\mathcal{F}_t)$. Fix $a < 0 < b$. **Uniform integrability of the stopped martingale.** The stopped process $(M_{t \wedge \tau_{a,b}})_{t \geq 0}$ satisfies the bound \begin{align*} |M_{t \wedge \tau_{a,b}}| = |B_{t \wedge \tau_{a,b}}^2 - (t \wedge \tau_{a,b})| \leq B_{t \wedge \tau_{a,b}}^2 + (t \wedge \tau_{a,b}) \leq \max(a^2, b^2) + \tau_{a,b}, \end{align*} where the last inequality uses $|B_{t \wedge \tau_{a,b}}| \leq \max(|a|, b)$ (the stopped process remains in $[a,b]$). Write $D = \max(a^2, b^2) + \tau_{a,b}$. To confirm $D$ is integrable and hence the family $\{M_{t \wedge \tau_{a,b}} : t \geq 0\}$ is uniformly integrable, we need $\mathbb{E}[\tau_{a,b}] < \infty$. We proceed by first applying the Optional Stopping Theorem conditionally (which will yield $\mathbb{E}[\tau_{a,b}]$) and then verifying the uniform integrability claim is self-consistent. **Applying Optional Stopping.** We verify the hypotheses of the [Optional Stopping Theorem for UI Martingales](/theorems/1164) applied to $M_t = B_t^2 - t$: - **Martingale:** $M_t$ is a martingale by [Basic Brownian Martingales](/theorems/1183) part (ii). - **Stopping time:** $\tau_{a,b}$ is an $(\mathcal{F}_t)$-stopping time (first exit time from an open set for a continuous adapted process). - **A.s. finiteness:** $\tau_{a,b} < \infty$ a.s. (established in Step 2). - **Uniform integrability:** We verify this momentarily. For any $n \in \mathbb{N}$, the deterministic stopping time $\tau_{a,b} \wedge n$ satisfies the hypotheses (the stopped martingale up to a bounded deterministic time is uniformly bounded, hence UI). Applying the Optional Stopping Theorem at the bounded stopping time $\tau_{a,b} \wedge n$: \begin{align*} \mathbb{E}[B_{\tau_{a,b} \wedge n}^2 - (\tau_{a,b} \wedge n)] = \mathbb{E}[M_0] = B_0^2 - 0 = 0. \end{align*} Therefore $\mathbb{E}[\tau_{a,b} \wedge n] = \mathbb{E}[B_{\tau_{a,b} \wedge n}^2] \leq \max(a^2, b^2)$ for all $n$. By the monotone convergence theorem ($\tau_{a,b} \wedge n \uparrow \tau_{a,b}$ as $n \to \infty$): \begin{align*} \mathbb{E}[\tau_{a,b}] = \lim_{n \to \infty} \mathbb{E}[\tau_{a,b} \wedge n] \leq \max(a^2, b^2) < \infty. \end{align*} This confirms $\mathbb{E}[\tau_{a,b}] < \infty$, so the dominating random variable $D = \max(a^2, b^2) + \tau_{a,b}$ is integrable, and the family $\{M_{t \wedge \tau_{a,b}} : t \geq 0\}$ is uniformly integrable (dominated by an integrable random variable). By the dominated convergence theorem, $M_{t \wedge \tau_{a,b}} \to M_{\tau_{a,b}}$ in $L^1$ as $t \to \infty$. Therefore the Optional Stopping Theorem gives \begin{align*} \mathbb{E}[M_{\tau_{a,b}}] = \mathbb{E}[M_0] = 0, \quad \text{i.e.,} \quad \mathbb{E}[B_{\tau_{a,b}}^2] = \mathbb{E}[\tau_{a,b}]. \end{align*} **Computing $\mathbb{E}[B_{\tau_{a,b}}^2]$.** From Step 3, the exit probabilities are $\mathbb{P}(B_{\tau_{a,b}} = a) = \frac{b}{b-a}$ and $\mathbb{P}(B_{\tau_{a,b}} = b) = \frac{-a}{b-a}$. Therefore \begin{align*} \mathbb{E}[B_{\tau_{a,b}}^2] &= a^2 \cdot \mathbb{P}(B_{\tau_{a,b}} = a) + b^2 \cdot \mathbb{P}(B_{\tau_{a,b}} = b) \\ &= a^2 \cdot \frac{b}{b-a} + b^2 \cdot \frac{-a}{b-a} \\ &= \frac{a^2 b - ab^2}{b - a} = \frac{ab(a - b)}{b-a} = \frac{ab \cdot (-(b-a))}{b-a} = -ab. \end{align*} Since $a < 0$, we have $-a = |a|$, so $-ab = |a| \cdot b > 0$. Therefore \begin{align*} \mathbb{E}[\tau_{a,b}] = \mathbb{E}[B_{\tau_{a,b}}^2] = |a| \cdot b. \end{align*} **Integrating over the mixing measure.** Since $\tau_{a,b} \geq 0$ for all $(a,b)$ and $T = \tau_{a,b}$ where $(a,b) \sim \lambda$ is independent of $(B_t)$, Tonelli's theorem (applicable because $\tau_{a,b} \geq 0$) allows us to exchange expectation with integration over $\lambda$: \begin{align*} \mathbb{E}[T] = \int \mathbb{E}[\tau_{\alpha,\beta}] \, \lambda(d(\alpha,\beta)) = \int |\alpha| \cdot \beta \, \lambda(d(\alpha,\beta)). \end{align*} **Identifying with $\sigma^2$.** Applying the decomposition property (ii) from Step 1 to the function $f(x) = x^2$ (which is bounded on the support of each $\mu_{a,b}$ since $\mu_{a,b}$ is supported on $\{a,b\}$, and the identity extends to $f(x) = x^2$ by monotone convergence since $x^2 \geq 0$): \begin{align*} \sigma^2 = \int x^2 \, \mu(dx) &= \int \left(\frac{\beta}{\beta - \alpha} \alpha^2 + \frac{-\alpha}{\beta - \alpha} \beta^2\right) \lambda(d(\alpha,\beta)) \\ &= \int |\alpha| \cdot \beta \, \lambda(d(\alpha,\beta)), \end{align*} where the last equality uses the same algebra as above: $\frac{\beta}{\beta - \alpha} \alpha^2 + \frac{-\alpha}{\beta - \alpha} \beta^2 = -\alpha\beta = |\alpha| \cdot \beta$. Comparing, $\mathbb{E}[T] = \sigma^2$. *Why use $M_t = B_t^2 - t$ rather than $B_t$ itself?* The martingale $B_t$ gives us $\mathbb{E}[B_{\tau_{a,b}}] = 0$, which determined the exit probabilities but says nothing about $\mathbb{E}[\tau_{a,b}]$. The compensated square martingale $B_t^2 - t$ encodes the relationship between the quadratic variation and elapsed time, so optional stopping converts $\mathbb{E}[B_{\tau}^2]$ (which we know from the exit distribution) into $\mathbb{E}[\tau]$ (which we want). [/guided] Therefore $\mathbb{E}[\tau_{a,b}] = |a| \cdot b$. Since $\tau_{a,b} \geq 0$ for all $(a,b)$, Tonelli's theorem applies to exchange expectation with integration over $\lambda$: \begin{align*} \mathbb{E}[T] = \int \mathbb{E}[\tau_{a,b}] \, \lambda(d(a,b)) = \int |a| \cdot b \, \lambda(d(a,b)). \end{align*} Applying the decomposition property (ii) to $f(x) = x^2$: \begin{align*} \sigma^2 = \int x^2 \, \mu(dx) = \int \left(\frac{b}{b-a} a^2 + \frac{-a}{b-a} b^2\right) \lambda(d(a,b)) = \int |a| \cdot b \, \lambda(d(a,b)), \end{align*} where the last equality uses the computation above. Therefore $\mathbb{E}[T] = \sigma^2$. [/step] [step:Extend to the iterated embedding via the strong Markov property] Let $(X_n)_{n \geq 1}$ be i.i.d. with $X_n \sim \mu$, and let $S_n = X_1 + \cdots + X_n$. We construct stopping times $0 = T_0 \leq T_1 \leq T_2 \leq \cdots$ such that $(B_{T_n})_{n \geq 0}$ is a random walk with step distribution $\mu$. Let $((a_n, b_n))_{n \geq 1}$ be an i.i.d. [sequence](/page/Sequence) with $(a_n, b_n) \sim \lambda$, independent of $(B_t)_{t \geq 0}$. Define $T_0 = 0$ and inductively: \begin{align*} T_n = \inf\{t \geq T_{n-1} : B_t - B_{T_{n-1}} \notin (a_n, b_n)\}, \qquad n \geq 1. \end{align*} [guided] We prove that the inductively defined stopping times $T_0 \leq T_1 \leq T_2 \leq \cdots$ satisfy: (i) each increment $B_{T_n} - B_{T_{n-1}}$ has distribution $\mu$, (ii) $\mathbb{E}[T_n - T_{n-1}] = \sigma^2$ for all $n \geq 1$, and (iii) the pairs $(B_{T_n} - B_{T_{n-1}}, T_n - T_{n-1})$ are i.i.d. across $n$. **Setup.** Let $(\Omega, \mathcal{F}, \mathbb{P})$ be the underlying probability space carrying the standard Brownian motion $(B_t)_{t \geq 0}$ with natural filtration $(\mathcal{F}_t)_{t \geq 0}$ and the i.i.d. sequence $((a_n, b_n))_{n \geq 1}$ with $(a_n, b_n) \sim \lambda$, independent of $(B_t)$. Define $T_0 = 0$ and, for $n \geq 1$, \begin{align*} T_n = \inf\{t \geq T_{n-1} : B_t - B_{T_{n-1}} \notin (a_n, b_n)\}. \end{align*} **The strong Markov property.** The strong Markov property of Brownian motion states: if $\tau$ is an a.s. finite $(\mathcal{F}_t)$-stopping time, then the process \begin{align*} \widetilde{B}_t^{(\tau)} := B_{\tau + t} - B_{\tau}, \qquad t \geq 0, \end{align*} is a standard Brownian motion independent of $\mathcal{F}_{\tau}$. We verify the hypotheses: $T_{n-1}$ must be an a.s. finite stopping time. We proceed by induction. For $n = 1$, $T_0 = 0$ is trivially a finite stopping time. Assuming $T_{n-1}$ is a.s. finite (which follows from the inductive hypothesis and the a.s. finiteness of each exit time, established in Step 2), we apply the strong Markov property at $T_{n-1}$. **Base case: $n = 1$.** The process $\widetilde{B}_t^{(1)} := B_{T_0 + t} - B_{T_0} = B_t$ is the original Brownian motion. The stopping time $T_1 = \inf\{t \geq 0 : B_t \notin (a_1, b_1)\} = \tau_{a_1, b_1}$. Since $(a_1, b_1) \sim \lambda$ is independent of $(B_t)$, this is exactly the single Skorokhod embedding from Steps 1--4. Therefore $B_{T_1} - B_{T_0} = B_{T_1} \sim \mu$ and $\mathbb{E}[T_1 - T_0] = \mathbb{E}[T_1] = \sigma^2$. **Inductive step.** Assume $T_{n-1}$ is a.s. finite. By the strong Markov property applied at $T_{n-1}$, the process \begin{align*} \widetilde{B}_t^{(n)} := B_{T_{n-1} + t} - B_{T_{n-1}}, \qquad t \geq 0, \end{align*} is a standard Brownian motion independent of $\mathcal{F}_{T_{n-1}}$. The barrier pair $(a_n, b_n)$ is independent of $(B_t)_{t \geq 0}$ by construction (it is drawn from the i.i.d. sequence independent of the entire Brownian path), and in particular $(a_n, b_n)$ is independent of $\mathcal{F}_{T_{n-1}}$. Therefore $(a_n, b_n)$ is independent of $\widetilde{B}^{(n)}$, and also independent of $\mathcal{F}_{T_{n-1}}$. The increment $B_{T_n} - B_{T_{n-1}}$ equals $\widetilde{B}^{(n)}_{\tau_n}$ where \begin{align*} \tau_n := \inf\{t \geq 0 : \widetilde{B}_t^{(n)} \notin (a_n, b_n)\} \end{align*} is the first exit time of the standard Brownian motion $\widetilde{B}^{(n)}$ from the interval $(a_n, b_n)$. Since $\widetilde{B}^{(n)}$ is a standard Brownian motion independent of $(a_n, b_n)$, this is precisely the setup of the single embedding (Steps 1--4) applied to $\widetilde{B}^{(n)}$ and barrier $(a_n, b_n)$. Therefore: - $B_{T_n} - B_{T_{n-1}} = \widetilde{B}^{(n)}_{\tau_n} \sim \mu$ (by Step 3 applied to $\widetilde{B}^{(n)}$). - $T_n - T_{n-1} = \tau_n$ satisfies $\mathbb{E}[T_n - T_{n-1}] = \sigma^2$ (by Step 4 applied to $\widetilde{B}^{(n)}$). - $T_n = T_{n-1} + \tau_n < \infty$ a.s. (since $\tau_n < \infty$ a.s. by Step 2). **Independence across $n$.** It remains to show that the pairs $(B_{T_n} - B_{T_{n-1}}, T_n - T_{n-1})$ are independent across $n = 1, 2, \ldots$. The pair $(B_{T_n} - B_{T_{n-1}}, T_n - T_{n-1}) = (\widetilde{B}^{(n)}_{\tau_n}, \tau_n)$ is a measurable function of $(\widetilde{B}^{(n)}, (a_n, b_n))$ alone. We argue that these determining variables are independent across different $n$. For $n \neq m$, the pair $(\widetilde{B}^{(n)}, (a_n, b_n))$ is independent of $(\widetilde{B}^{(m)}, (a_m, b_m))$ because: 1. $(a_n, b_n)$ and $(a_m, b_m)$ are independent (they are drawn i.i.d. from $\lambda$). 2. $(a_n, b_n)$ and $(a_m, b_m)$ are each independent of $(B_t)_{t \geq 0}$. 3. By the strong Markov property, $\widetilde{B}^{(n)}$ is independent of $\mathcal{F}_{T_{n-1}}$, and $\widetilde{B}^{(m)}$ together with $(a_m, b_m)$ (for $m < n$) are $\mathcal{F}_{T_{n-1}}$-measurable. More precisely, the $\sigma$-algebra generated by $\{(\widetilde{B}^{(k)}_{\tau_k}, \tau_k) : k = 1, \ldots, n-1\}$ is contained in $\mathcal{F}_{T_{n-1}}$, and $(\widetilde{B}^{(n)}, (a_n, b_n))$ is independent of $\mathcal{F}_{T_{n-1}}$. By induction, the sequence of pairs is jointly independent. Since each pair has the same distribution (determined by $\lambda$ and the law of Brownian motion, which are the same for every $n$), the pairs are i.i.d. **Conclusion.** We have shown that \begin{align*} B_{T_n} = B_{T_0} + \sum_{k=1}^n (B_{T_k} - B_{T_{k-1}}) = \sum_{k=1}^n (B_{T_k} - B_{T_{k-1}}) \end{align*} is a sum of i.i.d. $\mu$-distributed increments (using $B_{T_0} = B_0 = 0$). Therefore $(B_{T_n})_{n \geq 0}$ is a random walk with step distribution $\mu$, and the inter-arrival times $(T_n - T_{n-1})_{n \geq 1}$ are i.i.d. with $\mathbb{E}[T_n - T_{n-1}] = \sigma^2$. *Could one avoid the strong Markov property and simply use the ordinary Markov property?* No. The ordinary Markov property applies at deterministic times, but the stopping times $T_n$ are random. The strong Markov property is the essential upgrade that ensures the post-$T_{n-1}$ Brownian motion is a fresh standard Brownian motion. Without it, one cannot guarantee that $\widetilde{B}^{(n)}$ is a standard Brownian motion independent of the past, and the entire inductive argument collapses. [/guided] By the strong Markov property, for each $n \geq 1$: (i) $B_{T_n} - B_{T_{n-1}} \sim \mu$, by the single embedding result applied to the Brownian motion $\widetilde{B}^{(n)}$ and barrier $(a_n, b_n)$. (ii) $\mathbb{E}[T_n - T_{n-1}] = \sigma^2$, since $T_n - T_{n-1}$ is the first exit time of $\widetilde{B}^{(n)}$ from $(a_n, b_n)$. (iii) The pairs $(B_{T_n} - B_{T_{n-1}}, T_n - T_{n-1})$ are i.i.d. across $n$. Therefore $B_{T_n} = \sum_{k=1}^n (B_{T_k} - B_{T_{k-1}})$ is a sum of i.i.d. $\mu$-distributed increments, so $(B_{T_n})_{n \geq 0}$ is a random walk with step distribution $\mu$, and $(T_n - T_{n-1})_{n \geq 1}$ are i.i.d. with mean $\sigma^2$. $\blacksquare$ [/step]