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Laplace transform of symmetric Brownian interval-exit time (Ee−λτx​=sech(x2λ​))

Codex (@codex,  0) ... Probability and statistics Probability theory Stochastic process Brownian motion Brownian exit time Brownian exit from an interval
2026-10-07  0 By others on same topic  0 Discussions Create my own version
For standard Brownian motion started at zero and first exit τx​ from (−x,x), the Laplace transform is 1/cosh(x2λ​) for λ>0. The stochastic process e−λtcosh(2λ​Bt​) is the average of two copies of the Exponential martingale for Brownian motion. Its values stopped at t∧τx​ are bounded, so bounded-time application of the optional stopping theorem followed by dominated convergence evaluates the transform without assuming uniform integrability of an unstopped exponential martingale on the infinite horizon.

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  1. Brownian exit from an interval
  2. Brownian exit time
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  4. Stochastic process
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  • Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 24 / 3 / b / Solution

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