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I'm reading about Dirichlet problem and Brownian motion in these notes, i.e.,

Fact. Let $D$ be an open and bounded domain in $\mathbb{R}^n$ and $\partial D$ be its (smooth) boundary. Let $h \in \mathcal{C}(\partial D)$. Then there exists a unique function $u \in \mathcal{C}^2(D)$ such that $$ \begin{cases} \Delta u(\underline{x})=0, & \forall \underline{x} \in D, \\ u(\underline{x})=h(\underline{x}), & \forall \underline{x} \in \partial D . \end{cases} \quad (\star) $$

Remark. The function $u$ takes values in $\mathbb{R}\left(\right.$ not in $\left.\mathbb{R}^n\right)$.

Stochastic representation of the solution. Let $\underline{B}^{\underline{x}}$ be an $n$-dimensional Brownian motion starting at point $\underline{x} \in D$ at time $t=0$ (i.e., $\underline{B}_t^{\underline{x}}=\underline{x}^x+\underline{B}_t$, where $\underline{B}$ is a standard $n$-dimensional Brownian motion). Let also $$ \tau=\inf \left\{t>0: \underline{B}_t^{\underline{x}} \notin D\right\} $$ be the first exit time of $\underline{B}^{\underline{x}}$ from the domain $D ; \tau$ is a stopping time. Notice also that $\underline{B}_\tau^{\underline{x}} \in \partial D$.

Proposition 6.5. The solution of $(\star)$ reads $$ u(\underline{x})=\mathbb{E}(h(\underline{B}_\tau^\bar{x})), \quad \forall \underline{x} \in D . $$

Proof.

  • Let us first show that the process $(u(\underline{B}_t^{\underline{x}}), 0 \leq t \leq \tau)$ is a martingale. By the multidimensional Ito-Doeblin formula, we have $$ \begin{aligned} u\left(\underline{B}_t^{\underline{x}}\right)-u\left(\underline{B}_0^{\underline{x}}\right) & =\sum_{i=1}^n \int_0^t u_{x_i}^{\prime}\left(\underline{B}_s^{\underline{x}}\right) d B_s^{(i)}+\frac{1}{2} \int_0^t \Delta u\left(\underline{B}_s^{\underline{x}}\right) d s \\ & =\sum_{i=1}^n \int_0^t u_{x_i}^{\prime}\left(\underline{B}_s^{\underline{x}}\right) d B_s^{(i)}, \end{aligned} $$ since $\Delta u(\underline{x})=0, \forall \underline{x} \in D$ and $\underline{B}_s^{\underline{x}} \in D, \forall s \leq \tau$.

  • Applying therefore the optional stopping theorem, we obtain $$ u(\underline{x})=\mathbb{E}\left(u\left(\underline{B}_0^x\right)\right)=\mathbb{E}\left(u\left(\underline{B}_\tau^{\underline{x}}\right)\right)=\mathbb{E}\left(h\left(\underline{B}_\tau^{\underline{x}}\right)\right), $$ where the third equality holds since $u(\underline{x})=h(\underline{x})$ on $\partial D$ and $\underline{B} \underline{\tau} \in \partial D$.

Because $\tau$ is random, I could not make sense of the sentence "$(u(\underline{B}_t^{\underline{x}}), 0 \leq t \leq \tau)$ is a martingale". Could you elaborate on this point?

Akira
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1 Answers1

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The idea is that, from the way the problem is defined, the unique solution $u$ is defined on $D\cup \partial D$. Hence, given $\underline x\in D$ and $t\ge0$, the random variable $``u(\underline{B}_t^{\underline{x}})"$ is only well-defined (up to a modification) if $\underline{B}_t^{\underline{x}} \in D\cup\partial D $ almost surely. By the definition of $\tau$, we thus see that the process $ u(\underline{B}_t^{\underline{x}}) $ is only defined for $0\le t\le \tau$.

More formally, $(u(\underline{B}_t^{\underline{x}}), 0 \leq t \leq \tau)$ should be understood as the stochastic process $Y_t := u(\underline{B}_t^{\underline{x}})\mathbf 1_{\tau\le t} + u(\underline{B}_\tau^{\underline{x}})\mathbf 1_{\tau> t} $, or equivalently as the stopped process $Y_t := u(\underline{B}_{\tau\wedge t}^{\underline{x}})$. As explained above, this can be thought of as "conditioning $\underline{B}_t^{\underline{x}}$ to remain within $D\cup\partial D$" and ensures that the stochastic process $u(\underline{B}_t^{\underline{x}})$ is always well defined.

You can now regard the proof as an application of Itô-Doeblin formula to $\underline{B}_{\tau\wedge t}^{\underline{x}}$ (which is a martingale, since $\underline{B}_{ t}^{\underline{x}}$ is) , implying that $Y_t$ is a martingale. The optional stopping theorem then yields the desired conclusion.

Stratos supports the strike
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