[proofplan] We use the [Reflection Principle](/theorems/1181) at the first hitting time $T_b$ of level $b$ to convert the event $\{S_t \geq b, B_t \leq a\}$ into a tail probability of $B_t$. The condition $a \leq b$ ensures that the reflected event $\{\widetilde{B}_t \geq 2b - a\}$ is contained in $\{T_b \leq t\}$, allowing the constraint $T_b \leq t$ to be dropped. The distributional identity for $S_t$ follows by setting $a = b$. [/proofplan] [step:Introduce the first hitting time $T_b$ and express $\{S_t \geq b\}$ in terms of $T_b$] Define the first hitting time $T_b = \inf\{s \geq 0 : B_s = b\}$. Since $S_\infty = \sup_{s \geq 0} B_s = +\infty$ almost surely (which follows from the [Recurrence-Transience Dichotomy](/theorems/1185)(i) in $d=1$), $T_b < \infty$ almost surely. Path [continuity](/page/Continuity) guarantees $B_{T_b} = b$. The event $\{S_t \geq b\}$ is equivalent to $\{T_b \leq t\}$, since the supremum of $B$ over $[0,t]$ reaches $b$ if and only if the first hitting time of $b$ occurs by time $t$. [/step] [step:Apply the reflection principle to relate $\{S_t \geq b, B_t \leq a\}$ to a tail probability] By the [Reflection Principle](/theorems/1181) applied at the stopping time $T_b$, the reflected process $\widetilde{B}_t = B_t \mathbb{1}_{t \leq T_b} + (2B_{T_b} - B_t) \mathbb{1}_{t > T_b} = B_t \mathbb{1}_{t \leq T_b} + (2b - B_t) \mathbb{1}_{t > T_b}$ is a standard Brownian motion. On the event $\{T_b \leq t\}$, we have $\widetilde{B}_t = 2b - B_t$, so \begin{align*} \{S_t \geq b, \, B_t \leq a\} = \{T_b \leq t, \, B_t \leq a\} = \{T_b \leq t, \, \widetilde{B}_t \geq 2b - a\}. \end{align*} Since $a \leq b$, we have $2b - a \geq b$, so $\widetilde{B}_t \geq 2b - a \geq b$ implies $\sup_{0 \leq s \leq t} \widetilde{B}_s \geq b$, which gives $T_b \leq t$. Therefore $\{\widetilde{B}_t \geq 2b - a\} \subset \{T_b \leq t\}$, and the constraint $T_b \leq t$ is redundant: \begin{align*} \mathbb{P}(S_t \geq b, \, B_t \leq a) = \mathbb{P}(\widetilde{B}_t \geq 2b - a) = \mathbb{P}(B_t \geq 2b - a), \end{align*} where the final equality uses $\widetilde{B} \overset{d}{=} B$. [/step] [step:Derive the distributional identity $S_t \overset{d}{=} |B_t|$] Setting $a = b$ in the identity $\mathbb{P}(S_t \geq b, B_t \leq a) = \mathbb{P}(B_t \geq 2b - a)$: \begin{align*} \mathbb{P}(S_t \geq b, \, B_t \leq b) = \mathbb{P}(B_t \geq b). \end{align*} Since $S_t \geq B_t$ always, $\{S_t \geq b\} = \{S_t \geq b, B_t \leq b\} \cup \{B_t > b\}$ (a disjoint union). Therefore \begin{align*} \mathbb{P}(S_t \geq b) = \mathbb{P}(S_t \geq b, B_t \leq b) + \mathbb{P}(B_t > b) = 2\mathbb{P}(B_t \geq b) = \mathbb{P}(|B_t| \geq b), \end{align*} where the last equality uses symmetry of the distribution $B_t \sim \mathcal{N}(0, t)$: $\mathbb{P}(|B_t| \geq b) = \mathbb{P}(B_t \geq b) + \mathbb{P}(B_t \leq -b) = 2\mathbb{P}(B_t \geq b)$. Since this holds for all $b > 0$, $S_t$ and $|B_t|$ have the same distribution. [/step]