Let $(B_t)$ be a standard Brownian motion, $f$ a continuous function and $X_t = \int_0^t f(s)B_s ds$.

I was able to prove that $(X_t)$ is a Gaussian process with zero mean and trying to find the covariance function of $(X_t)$ now. But I got so far is the following:

$\mathbb{E} \left( X_t X_s\right) = \mathbb{E}\left( \left(\int_0^t f(u)B_u du \right) \left( \int_0^s f(v)B_v dv\right)\right) = \int_0^t \int_0^s f(u)f(v)\mathbb{E}\left(B_u B_v \right)du dv$ $= \int_0^t \int_0^s f(u)f(v) \cdot \min(u,v) du dv$ (1).

(modulo Fubini's theorem justification). I am having trouble with simplification of my result in (1), not sure whether I can simplify it further, or am I doing something wrong?

I would be grateful for any ideas or suggestion. Thanks

  • 4
    $\begingroup$ Split the integral up in $u<v$ and $u>v$. Furthermore, use the symmetry $u\leftrightarrow v$. $\endgroup$ – Fabian Feb 3 '13 at 12:58

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