TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Two-sided Brownian exit probability

Statement

Assume the Axiom of Choice. Let $B$ be a standard Brownian motion Brownian motion, let $a < x < b$ be reals, and let $P_{x}$ be the law of the shifted, everywhere-continuous process $t \mapsto x + {\hat{B}}_{t}$ Brownian motion started at x, on canonical continuous path space with its compact-open Borel sigma-algebra. Here $\hat{B}$ is the normalized zero-start representative specified in that definition. For $c \in R$ let $T_{c} : = \inf {t \geq 0 : Z_{t} = c}$ be the hitting time of the level $c$ for the coordinate process $Z$ of the shifted law. Then $P_{x} (T_{b} < T_{a}) = \frac{x - a}{b - a} .$

Facts & Assumptions

Given: AC, a standard Brownian motion, reals $a < x < b$ , and the shifted law $P_{x}$ . In the proof write $B$ for its everywhere-continuous zero-start representative $\hat{B}$ and use that representative's own raw natural filtration and usual augmentation. No adaptation to a former raw filtration is claimed.

[F1]

For the shifted law $P_{x}$ , hitting times satisfy $P_{x} (T_{c} < \infty) = P (T_{c - x} < \infty)$ and $P_{x} (T_{b} < T_{a}) = P (T_{b - x} < T_{a - x})$ , because the shifted process is $x + B$ . Brownian motion started at x

[F2]

One-dimensional Brownian motion hits every level almost surely: $P (T_{c} < \infty) = 1$ for every $c$ . One-dimensional Brownian motion hits every point almost surely

[F3]

The hitting time of a closed set for the normalized everywhere-continuous $B$ is a stopping time for its raw natural filtration and its usual augmentation. The maximum $M_{N} = \sup_{s \leq N} B_{s}$ has the same law as $∣ B_{N} ∣$ for $N > 0$ . Hence $E M_{N} = E ∣ B_{N} ∣ \leq \sqrt{E B_{N}^{2}} = \sqrt{N}$ . The same argument applies to $- B$ , which is Brownian by symmetry of its Gaussian increments. Since $B_{0} = 0$ , $\sup_{s \leq N} ∣ B_{s} ∣ \leq \sup_{s \leq N} B_{s} + \sup_{s \leq N} (- B_{s})$ , whose expectation is at most $2 \sqrt{N}$ . The suprema are measurable rational-time suprema by continuity; $N = 0$ gives zero directly. Brownian closed-set hitting times are stopping times Law of the Brownian maximum Standard normal and normal laws Cauchy-Schwarz for random variables Brownian motion

[F4]

Strong Markov: for an a.s. finite stopping time $τ$ of the usual augmentation, the increment process $(B_{τ + t} - B_{τ})_{t \geq 0}$ is a Brownian motion independent of $F_{τ}$ . Strong Markov property of Brownian motion Continuous-time stopping times and stopped sigma-algebras Natural and usual augmented Brownian filtrations

[F5]

The law of $B_{M}$ is $N (0, M)$ , which is centered and has no atoms; the tower identity gives $E [E [X ∣ G]] = E X$ . The Brownian kernels form a semigroup Standard normal and normal laws Basic algebra and order properties of conditional expectation Conditional expectation as an ae class Conditional expectation is unique almost surely

[F6]

Dominated convergence and monotone convergence pass limits through integrals. Dominated convergence Monotone convergence for the integral

[F7]

AC supplies the conditional-expectation and strong-Markov interfaces. The Axiom of Choice Wiener measure on continuous path space

Proof

technique · direct

1.1F1F2F3given

By [F1] it suffices to prove the case $a < 0 < b$ with the unshifted law: $P_{x} (T_{b} < T_{a}) = P (T_{b - x} < T_{a - x})$ and $(x - a) / (b - a) = (0 - (a - x)) / ((b - x) - (a - x))$ , since $(b - x) - (a - x) = b - a$ and $a - x < 0 < b - x$ . So assume $a = A < 0 < B = b$ and put $T : = T_{A} \land T_{B}$ . Then $T \leq T_{A} < \infty$ almost surely by [F2], and $T$ is a stopping time of the usual augmentation because ${A, B}$ is closed, by [F3]. For every $u < T$ the path satisfies $B_{u} \in (A, B)$ , since leaving $(A, B)$ would require hitting $A$ or $B$ by continuity.

1.2F3F4F5F6

By [F4] applied at the a.s. finite stopping time $T$ , the process $W_{t} = B_{T + t} - B_{T}$ on ${T < \infty}$ and $W_{t} = 0$ otherwise is an everywhere-continuous zero-start Brownian motion independent of $F_{T}$ , using the supplier's measurable random-time convention. Consequently, for every $F_{T}$ -measurable random variable $R$ with values in $[0, N]$ one has $E [W_{R} ∣ F_{T}] = 0$ almost surely. Indeed, if $R$ is countably valued with values $r_{j} \in [0, N]$ , then $W_{R} = \sum_{j} W_{r_{j}} 1_{{R = r_{j}}}$ and for $G \in F_{T}$ one has $\int_{G} W_{R} d P = \sum_{j} P (G \cap {R = r_{j}}) E W_{r_{j}} = 0$ , because $G \cap {R = r_{j}} \in F_{T}$ is independent of $W_{r_{j}}$ and $E W_{r_{j}} = 0$ ; for general $R$ the dyadic ceilings $R_{k} : = \min (2^{- k} ⌈ 2^{k} R ⌉, N)$ decrease to $R$ , so $W_{R_{k}} \to W_{R}$ almost surely and $∣ W_{R_{k}} ∣ \leq S : = \sup_{[0, N]} ∣ W ∣$ , whose expectation is finite by [F3]; dominated convergence [F6] gives $\int_{G} W_{R} d P = \lim_{k} \int_{G} W_{R_{k}} d P = 0$ for every $G \in F_{T}$ , and [F5]'s uniqueness identifies $E [W_{R} ∣ F_{T}] = 0$ .

2.1F5step 1.2

Fix $M > \max {∣ A ∣, ∣ B ∣}$ and let $R = M - T$ on ${T \leq M}$ and $R = 0$ otherwise. This is an $F_{T}$ -measurable random variable with values in $[0, M]$ : $T$ is $F_{T}$ -measurable since ${T \leq r} \cap {T \leq u} = {T \leq \min (r, u)} \in F_{u}$ . No product $0 \cdot \infty$ is used. On ${T \leq M}$ one has $B_{M} - B_{T} = W_{M - T} = W_{R}$ , and on ${T > M}$ both $B_{M} - B_{T \land M}$ and $W_{R} = W_{0}$ vanish; hence $B_{M} - B_{T \land M} = W_{R}$ . All terms are integrable: $B_{M}$ is Gaussian, $W_{R}$ is bounded in absolute value by its integrable finite-horizon supremum, and the identity gives integrability of $B_{T \land M}$ . Taking expectations and using step 1.2 with [F5]'s tower identity, $E [B_{M} - B_{T \land M}] = 0$ , so $E [B_{T \land M}] = E [B_{M}] = 0$ , the last equality because the law of $B_{M}$ is the centered $N (0, M)$ .

3.1F6step 1.1step 2.1

Let $M \to \infty$ . Set $B_{T} = 0$ on the null event $T = \infty$ , as in the strong-Markov convention. The random variables $B_{T \land M}$ converge almost surely to $B_{T}$ because $T < \infty$ almost surely and the paths are continuous, and they are bounded by $\max {∣ A ∣, ∣ B ∣}$ : for $t < T$ the value $B_{t}$ lies in $(A, B)$ by step 1.1, and $B_{T} \in {A, B}$ . Dominated convergence [F6] therefore gives $E [B_{T}] = 0$ .

4.1step 3.1

The events ${T_{B} < T_{A}}$ and ${T_{A} < T_{B}}$ are disjoint and their union is almost surely the whole space, because $T < \infty$ almost surely and $T_{A} \neq T_{B}$ almost surely (the path cannot be at two distinct levels at one time). On the first event $B_{T} = B$ and on the second $B_{T} = A$ , so $0 = E [B_{T}] = B P (T_{B} < T_{A}) + A (1 - P (T_{B} < T_{A}))$ ; solving gives $P (T_{B} < T_{A}) = - A / (B - A)$ .

5.1step 1.1step 4.1

Undoing the shift with step 1.1, $P_{x} (T_{b} < T_{a}) = P (T_{b - x} < T_{a - x}) = \frac{- (a - x)}{(b - x) - (a - x)} = \frac{x - a}{b - a}$ , which is the assertion.

6.1F4F7givenstep 4.1∎

The endpoint cases are covered: the strict inequalities $a < x < b$ keep $T_{a}$ and $T_{b}$ distinct from the starting level; the truncation parameter $M$ is chosen larger than both endpoints and then sent to infinity in step 3.1; the case $A = B$ is excluded because $a < b$ ; and $T_{A} = T_{B}$ is the null event excluded in step 4.1. AC is used only through [F7] in the conditional-expectation and strong-Markov interfaces.

Source notes

Durrett, Theorem 7.5.3, proves the complementary lower-exit formula by bounded stopping and bounded convergence. Solving its endpoint expectation identity gives the stated upper-exit formula. The proof above instead verifies the centered martingale identity $E [B_{T}] = 0$ through the strong Markov restart at $T$ , which keeps every step within the stopping-time and maximum machinery already established on this page.

Depends on

Used by

Exit side from an interval Example
Expected exit time from an interval Example
Hitting probabilities from an exponential martingale Example

Dependency tree · two levels

92 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Rick Durrett, Probability: Theory and Examples, fifth edition, Theorem 7.5.3 (standard reference, not scraped)
Perla Sousi, Advanced Probability, Section 6.7 (standard reference, not scraped)