# Probability density function of the integral of a continuous stochastic process

I am interested in whether there is a general method to calculate the pdf of the integral of a stochastic process that is continuous in time.

My specific example: I am studying a stochastic given process given by

$X(t)=\int\limits_{0}^{t}\cos(B(s))\,\text{d}s$,

where $B(t)$ is the Wiener process, which is normally distributed over an interval of length $\tau$ with zero mean and variance $\tau$:

$B(t+\tau)-B(t)\sim\mathcal{N}(0, \tau)$.

I am able to calculate the first and second moments of $X(t)$, see: Expectation value of a product of an Ito integral and a function of a Brownian motion

A couple of thoughts on the matter:

1) Integrals of Gaussian continuous stochastic processes, such as the Wiener process can be considered as the limit of a sum of Gaussians and are hence themselves Gaussian. Since $\cos(B(s))$ is not Gaussian, this doesn't seem to help here.

2) If we can derive an expression for the characteristic function of the process $X(t)$, then we can theoretically invert this to obtain the pdf. The Feynman-Kac formula enables us to describe the characteristic function in terms of a PDE. If this PDE has a unique analytic solution then we can make use of this. In my specific example, this is not the case - the PDE obtained has no analytic solution. I can provide more detail on this point if required.

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Are you still interested in this question? –  Tarasenya Mar 18 '13 at 14:26

$E[\int_a^b X(t) dh(t)] = \int_a^b E[X(t)]dh(t)$
$E[ \int_a^b X(t) dt \int_a^b Z(t) dt] = \int_{[a,b]^2}E[X(u)Z(v)]du dv$