I'm looking for information on how to compute the distribution of the random vector

$$Z = \int_0^t f(B_s) ds$$

where $t>0$ is fixed, $B_s$ is a 2D Brownian bridge with $B_0 = 0$, $B_t=b \in \mathbb{R}^2$, and $f : \mathbb{R}^2 \rightarrow \mathbb{R}^K$ has components $f_k : \mathbb{R}^2 \rightarrow \mathbb{R}$ that are positive, bounded, and in $C^\infty$.

Here's a reference to the local time of a 1D Brownian bridge, that gives an explicit density for the local-time $L_t(x)$ so a time-integral can be computed as $\int_\mathbb{R} f(x)L_t(x) dx$. But it doesn't appear to work for processes in the plane.

Any pointers to relevant papers or textbook chapters would be appreciated.

The motivation is a problem in experimental biophysics, which I'm happy to outline if others are interested. This is my first venture into stochastic integration.

Edit: By request, a summary of the motivating problem:

The overall problem is optical tracking of a microscopic biological process consisting of many component parts, some diffusing and others not. It requires estimating the size and rate of diffusion of an ensemble of particles in a membrane, under unfavorable conditions: particle sizes smaller than the optical resolution, other objects in the field, unknown per-particle brightness, and most significantly, a hidden process that adds new particles to the ensemble at random times.

The random vector $Z$ is one piece of our probability model of this (rather complicated) biological process. It represents the light due to a single particle, collected for a brief exposure $[0,t)$ by a digital camera attached to our microscope. Each of the scalar components $f_k$ is the parameter of a Poisson random variable modeling the number of photons detected by pixel $k$ during the exposure. A simple model of the optics has $$f_k(x) = \int_{A_k} g(x - y) dy$$ where $A_k \subset \mathbb{R}^2$ is the rectangular area "seen" by pixel $k$ (so the $A_k$ are disjoint) and $g$ is a Gaussian approximating the microscopes' point spread function.

The fixed end-point $B_t=b$ arises from the way we move from this continuous-time model to our discrete time experimental data. We use a Hidden Markov Model in which each discrete time $i \in \mathbb{N}$ corresponds to an instant $t_i$ between camera exposures; and the HMM emissions $Z$ are conditioned upon a transition $b_i \mapsto b_j$. Then we can evaluate the likelihood of an experimental observation, a complete digital movie containing $n$ exposures, in only $O(n)$ operations by using the forward algorithm to iterate over the exposures.

I welcome comments and criticism of this approach.

  • 1
    $\begingroup$ I'd love to hear about the problem. $\endgroup$ – weakstar Feb 1 '11 at 0:09

To compute the distribution of $Z$ for a general function $f$ might be difficult. To evaluate it, one could rewrite the process $(B_s)_{0\le s\le t}$ as $B_s=W_s+(b-W_t)(s/t)$ for every $0\le s\le t$, where $(W_s)_{s\ge0}$ is a standard 2D Brownian motion starting from $W_0=0$.

  • $\begingroup$ By "evaluate it" do you mean evaluate the probability density function of $Z$ at arbitrary points in $\mathbb{R}^K$? If so, could you elaborate on how to proceed after re-writing $B_s$ in terms of $W_s$? $\endgroup$ – Gabriel Feb 1 '11 at 16:33
  • $\begingroup$ Sorry if this was unclear. To evaluate the distribution of $Z$, one can simulate a large number of i.i.d. copies. To simulate one copy of $Z$, one can discretize the integral defining $Z$ and compute $f(W_s+(b-W_t)(s/t))$ at every point $s=it/N$ with $i\le N$. If $s=it/N$, $W_s-(s/t)W_t$ is $\sqrt{t/N}$ times $(1-k/N)(X_1+\cdots+X_i)+(X_{i+1}+\cdots+X_N)$, where $(X_k)_{1\le k\le N}$ is an i.i.d. sample of 2D normal random variables. For $K$ copies of $1/N$ discretizations of $Z$, one needs $2KN$ i.i.d. standard normal random variables. $\endgroup$ – Did Feb 1 '11 at 16:59
  • $\begingroup$ Ah, yes of course. I was hoping there might be some further analysis, taking advantage of the structure of $f$, that could eliminate or reduce the need for Monte Carlo simulation. But I suppose this may be the best approach. Upvoted. Will wait a few days and then mark accepted if my silver bullet hasn't arrived. $\endgroup$ – Gabriel Feb 1 '11 at 17:35
  • $\begingroup$ You are right that I did not use the structure of $f$. But even taking it into account, simply to compute $E(Z_1)$ is not that easy. Assuming that $g$ is the density of a standard Gaussian random variable, centered and with variance $I_2$, (I think that) $E(Z_1)$ is the integral from $0$ to $t$ of $g((A_1-b(s/t))/\sigma(s))$ with $\sigma^2(s)=1+(s/t)(t-s)$ (where $g(B)$ is the integral of $g$ over $B$). Sure, if the rectangle $A_1$ is parallel to the axes, each $g((A_1-b(s/t))/\sigma(s))$ factors into the product of two one-dimensional Gaussian integrals... but I do not see how to go further. $\endgroup$ – Did Feb 1 '11 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.