I have a question about a reflecting Brownian motion and its boundary local time.

Bass and Hsu studied the existence of Reflecting Brownian motion and boundary local time on a bounded Lipschitz domain in 1991. Although I don't state the definition of boundary local time here, I will briefly explain what it is. Roughly speaking, boundary local time $\{L_{t}\}$ is an additive functional on a probability space and satisfies the following the equation: \begin{align*} L_{t}=\int_{0}^{t}1_{\left\{X_{s} \in \partial D \right\}}\,dL_{s}, \end{align*} where $\{X_{t}\}$ is a Reflecting Brownian motion on $\bar{D}$ (closure of a bounded Lipschitz domain $D$). That is, $L_t$ increases only when $X_t \in \partial D$.


Let $D$ be a rectangle like as the follwong picture. Even in this case, we can define reflecting Brownian motion $(X_t,P_x)$ starting from $x \in \bar{D}$ and boundary local time $\{L_{t}\}$.

I am interested in the quantity $ P_{x}(L_t>M), $ where $M$ is a positive constant.

I think $P_{a}(L_t >M) \ge P_{b}(L_t >M)$, where $a,b$ are boundary points in the following picture. But I don't know how to prove this inequality. If you know related studies, please let me know.

enter image description here


In this question, $\{X_{t}\}$ is the Markov process generated by the following Dirichlet form on $L^{2}(\bar{D})$: \begin{align*} \mathcal{E}(f,g)=\frac{1}{2}\int_{D}(\nabla f, \nabla g)\,dx,\quad f,g \in H^1(D), \end{align*} where $H^{1}(D)$ is the Sobolev space on $D$ with Neumann boundary condition. $\{L_{t}\}$ is the positive continuous additive functional associated with the surface measure on $\partial D$. To be more precise, $\{L_t\}$ and $\sigma$ are in the Revuz correspondence. Futheremore, $X_{t}$ has the following Skorohod representation: \begin{align*} X_t=X_0+B_t+\frac{1}{2}\int_{0}^{t}n(X_s)\,dL_s, \end{align*} where $\{B_t\}$ is the $d$-dimensional Brownian motion and $n$ is the unit inward normal vector to $\partial D$.

  • $\begingroup$ Can you please specify the reflection (or constraint) directions? Otherwise your description of the reflecting Brownian motion is a bit incomplete. $\endgroup$ Sep 1, 2016 at 18:55
  • $\begingroup$ Thank you for pointing it out. $X_t$ is a Brownian motion in $D$ with normal reflection at the boundary. $\endgroup$
    – sharpe
    Sep 2, 2016 at 7:14
  • 1
    $\begingroup$ Is this process independent 1 d processes reflected at an upper & lower boundary? $\endgroup$
    – user83457
    Sep 2, 2016 at 10:33
  • $\begingroup$ In the Neumann case, the components of the RBM are completely decoupled. Therefore, I do not understand the point of the second component. Please clarify. $\endgroup$ Sep 2, 2016 at 12:32
  • $\begingroup$ For the moment, I tried to clarify my question. $\endgroup$
    – sharpe
    Sep 2, 2016 at 14:06

1 Answer 1


Since the components of the given planar RBM are uncoupled, it suffices to consider a scalar RBM $X$ on the spatial interval $[0,1]$. Here is a sample path of this process on the temporal interval $[0,10]$.

sample path

I will approach local time via the occupation time $$ \int_0^t 1_{[0, \epsilon]}(X(s)) ds $$ which gives the (random) amount of time the process spends in an $\epsilon$ neighborhood of zero during the interval $[0, t]$. This is a fairly complicated random variable, since it depends on the entire path of the RBM. However, its expected value is analytically available. Let $$ u^{\epsilon}(t,x) = \mathbb{E}_x \int_0^t 1_{[0, \epsilon]}(X(s)) ds \;. $$ Note that $u^{\epsilon}(t,x)$ satisfies an inhomogeneous initial boundary value problem: $$ \partial_t u^{\epsilon}(t,x) = \frac{1}{2} \partial_{xx} u^{\epsilon}(t,x) + 1_{[0, \epsilon]}(x) \quad \forall x \in [0,1] \;, \forall t \ge 0 \;, $$ with initial data $u(0,x)=0$ and pure Neumann boundary conditions $\partial_x u(t,0) = \partial_x u(t,0) = 0$. By expanding the solution and the inhomogeneity in terms of the eigenfunctions of the second derivative operator $\{ e_j(x) \}$ on $[0,1]$ with pure Neumann boundary conditions at $0$ and $1$, one obtains the following explicit solution: $$ u^{\epsilon}(t,x) = \epsilon t + 2^{3/2} \sum_{j \ge 2} \left( 1- \exp\left( \frac{t}{2} (j-1)^2 \pi^2 \right) \right) \frac{\sin( (j-1) \pi \epsilon)}{(j-1)^3 \pi^3} e_{j}(x) $$
The following figure plots the behavior of $E_x L_t := \lim_{\epsilon \downarrow 0} \frac{u^{\epsilon}(t,x)}{\epsilon}$ as a function of the initial condition of the RBM $X(0)=x$ with $t=1$. As expected, this quantity decreases with distance from zero. (Since $t=1$, I find it curious that this quantity is greater than unity.)

local time

To answer the question, just view $x$ in this figure as the vertical component of the cartoon given in the question. As we already said, what happens in the horizontal component can be treated separately.


By using Monte-Carlo simulation, it is straightforward to obtain even more detail about the random variable $L_t^{\epsilon} = \epsilon^{-1} \int_0^t 1_{[0, \epsilon]}(X(s)) ds$. Here are graphs of the cumulative distribution function $CDF(s)=\mathbb{P}_x(L_t^{\epsilon} \le s)$ with $t=1$, for the initial conditions $X(0)=x$ indicated in the figure legend, and for $\epsilon$ sufficiently small. Logarithmic scaling is adopted to make it easier to compare the CDFs.

local time cdf

  • 1
    $\begingroup$ Thank you for your kind answer. But I am interested in boundary local time. In $1$D case, boundary local time $L_t$ may be characterized by $E_{x}[L_t]=\lim_{\epsilon \to 0}u_{\epsilon}(t,x)/\epsilon$, $u_{\epsilon}(t,x)=\int_{0}^{t}1_{[0,\epsilon]}(X_s)\,ds+\int_{0}^{t}1_{[1-\epsilon,1]}(X_s)$. Futhermore, $E_{0}[L_t]=E_{1}[L_t]$ should hold and $E_{x}[L_t] \ge E_{1/2}[L_t]$ for every $x \in[0,1]$. In this case, can we show $P_{x}(L_t \ge \alpha ) \ge P_{1/2}(L_t \ge \alpha)$? $\endgroup$
    – sharpe
    Sep 3, 2016 at 8:57
  • $\begingroup$ Motivated by your question, I added a bit more detail to my answer. $\endgroup$ Sep 3, 2016 at 18:55
  • $\begingroup$ Thank you for your answer. I think it may be difficult to show $P_x(L_t \ge \alpha) \ge P_{1/2}(L_t \ge \alpha)$. $\endgroup$
    – sharpe
    Sep 3, 2016 at 20:22

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