First-passage Laplace transform for Brownian motion with drift

Solution

Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 25 / 6 / d / Solution

{title2=$u(x)=\mathbb E_x[e^{-\lambda\tau}g(X_\tau);\tau<\infty]$}

For a continuous <Itô diffusion> with <diffusion generator> $\mathcal L$, let $u$ solve $\mathcal Lu=\lambda u$ with $\lambda>0$, be bounded on an open domain and its boundary, and equal $g$ on the boundary. Stop at the first boundary hit $\tau$. The <Itô formula> makes $e^{-\lambda(t\wedge\tau)}u(X_{t\wedge\tau})$ a bounded martingale. Its limit is zero on $\{\tau=\infty\}$ and equals $e^{-\lambda\tau}g(X_\tau)$ on $\{\tau<\infty\}$. Thus
$$
u(x)=\mathbb E_x[e^{-\lambda\tau}g(X_\tau)\mathbf1_{\{\tau<\infty\}}].
$$
Boundedness justifies passage to the terminal expectation even if hitting is not almost surely finite.

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 Incoming links: Discounted boundary-hitting representation