# Time-integral of a smooth, vector-valued function of a planar Brownian bridge

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.

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I'd love to hear about the problem. –  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 $W_0=0$.
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 $W_s$? –  Gabriel Feb 1 '11 at 16:33
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. –  Did Feb 1 '11 at 16:59
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. –  Gabriel Feb 1 '11 at 17:35
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. –  Did Feb 1 '11 at 18:04