Estimate occupation time of Brownian Motion

Question

Let $V$ be a simple domain (e.g. a sphere) and $\tau_V= \inf\{t\geq0:B_t\notin V\}$, where $B_t$ is Brownian Motion. I try to investigate the following term $$E^x[\int_{0}^{\tau_V} g(B_s) ds],$$ with $x \in V$ in a stochastical way. For example, I would like to have a lower and/or upper bound. I know that this can be interpreted as a solution of the Poisson problem but I don't want to use pure Analysis methods. Does anybody have an idea for an approach that can be used here or can somebody recommend any literature?

Answer 1

Note - I wrote this answer and then realised that I can't come up with anything more creative than solution of Dirichlet problem, which you don't want. I'm posting this anyway, maybe there's someone who doesn't know this "define integral out of simple function" trick.

Let us define $G(x) = \int_0^x g(t)dt$ and $F(x) = \int_0^x G(t) dt = \int_0^x (\int_0^t g(s) ds ) dt$. Assume $g(0)=0$ for clarity of notation - we're not losing much of generality, but general cleanness of writing improves. Then $G'(x) = g(x), F'(x) = G(x), F''(x) = g(x)$.

From Ito-Doublin lemma we have, for every twice differentiable function f and any semimartingale $X_t$: $$f(X_t) = f(0) + \int f'(X_s) dX_s + \frac{1}{2} \int f''(X_s) ds$$

Brownian motion is, obviously, a semimartingale. Thefore: $$ F(B_t) = F(B_0) + \int G(B_s) dB_s + \frac{1}{2} \int g(B_s) ds$$

First integral is called continuous local martingale, second one is finite variation part.

Note that we know that $\limsup B_t = \infty$, so it is going to escape any given interval, sooner or later. Therefore any writing of sort $t \wedge \tau_V \rightarrow \tau_V$ if $t \rightarrow \infty$ is true, if we assume that $B_0 \in V$

From this argument we know that $g(B_{t \wedge B_t})$ is bounded. We will change time $t$ in Ito integral above into $t \wedge \tau_V$. If $g$ is bounded, than $G$ is bounded as well. We know that bounded continuous local martingale is martingale, therefore we can write $$\mathbb{E} \int^{t \wedge \tau_V}_0 G(B_s) dB_s = 0$$.

Taking expectations on both side of Ito expansion and using dominated convergence argument (from boundedness we can do it) we therefore get: $$\mathbb{E} F(B_{\tau_V}) = F(B_0) + \frac{1}{2} \mathbb{E} \int_0^{\tau_V} g(B_s) ds$$

In dimensions $d$, $d>1$, Ito expansion looks like this: $$F(B_t) = F(B_0) + \sum_{i = 1}^d \int \frac{\partial}{\partial x_i} F(B_s) dB_s^i + \sum_{i,j}^d \frac{1}{2} \int \frac{\partial^2}{\partial x_i \partial x_j} F(B_s) d [B_i, B_j]$$

Since $[B_i, B_j] = t \delta_{ij}$ last part reduces to diagonal Dirichlet operator.

Left hand-side can be reexpressed as Dirichlet problem, but I presume this is something you don't want.