# 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
I wasn't when you asked (I didn't have a spare moment to think about "interesting things"), but I am again now, should anyone see this! –  Gabriel Mar 11 '14 at 11:47

Well the first moment is relatively straightforward. If the integral exists than

$E[\int_a^b X(t) dh(t)] = \int_a^b E[X(t)]dh(t)$

See for example Grigoriu 3.69.

Also by for example Grigoriu 3.58

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

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I have expressions for the first two moments using a similar approach... but unless it's possible to derive expressions for an arbitrary number of moments (hence the MGF), this doesn't help with the pdf unfortunately. If such a thing is possible, then perhaps you could invert the MGF - but it seems unlikely to work nicely? –  Gabriel Mar 11 '14 at 11:51
Sorry guess I didn't read your question carefully. Thought you wanted the first two moments. –  Kai Sikorski Mar 11 '14 at 16:04
@Gabriel, I think you could generalize that idea to get arbitrary moments. Probably would be hard to do the integrals in closed form. Do you really need the pdf ? –  Kai Sikorski Mar 11 '14 at 16:11
It's more a point of academic interest! Thanks for your thoughts. –  Gabriel Mar 11 '14 at 17:53