I'll enjoy your kindness to ask this question, despite that it seems silly for you. Please show me a document or a url; emphasizing in a concise manner, the relationship between Brownian motion and Laplacian. With google, I found a lot of links; but I can not get the link to Markov chain, heat kernel and half Laplacian. friendly.
relationship between Brownian motion and $1 / 2 \Delta$
1 Answers
Brownian motion is a Markov process, since $B_{t+s}-B_t\perp B_t$ and $B_{t+s}-B_t\sim\mathcal N(0,s)$. I do not know how familiar are you with the theory of Markov processes, but here is the nice connection which does asks for this theory - only for Ito's lemma. Let us consider $f\in C^2(\mathbb R)$, then f(B_T) = \int\limits_0^Tdf(B_t) = \int\limits_0^Tf'(B_t)dB_t + \int\limits_0^T\frac12f''(B_t)dt so by taking expectation of both sides \mathsf E f(B_T) = \int\limits_0^T\frac12 \mathsf Ef''(B_t)dt. Now, if we take insted a process $X_t= x+B_t$ then \mathsf E_x f(X_T) = \int\limits_0^T\frac12 \mathsf E_xf''(X_t)dt. where $\mathsf E_x$ is an expectation under the condition that $X_0 = x$. If we denote $m(t,x) = \mathsf E_xf(X_t)$ then $ m_t = \frac12m_{xx} = \frac12\Delta m\quad (1) $
For the multidimensional case you just take $X_t = (x^1,...,x^n)+(B^1_t,...,B^n_t)$ where $B^i_t$ are mutually independent Brownian motions. $\frac12\Delta$ is called an infinitesimal generator of the stochastic process.
Edited: lets introduce a transition operator $P_t[f(x)] = \mathsf E_x[f(X_t)]$. Then an infinitesimal generator of the process $X$ is simply $ \mathcal Af(x) = \lim\limits_{t\to0}\frac{P_t[f(x)] - f(x)}{t}, $ for $X_t = x+B_t$ it follows from Ito's lemma. In that case, of course, $\mathcal A = \frac12\Delta$. Moreover, $P_t$ and $\mathcal A$ commute for all $t\geq 0$ which justifies (1).
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0@user When one writes $f''(x+B_t)$ one means the value of $f''$ at the point $x+B_t$, so your dilemna about *derivative with respect to $t$ or with respect to $x$* is moot. (But please begin your comments with this @ thing.) – 2011-07-08