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I'll give it a try, but am not yet 100% sure that it's the way to go. Ansatz: Let's find the distribution of the integral of a Brownian motion with respect to time (call it $x$) and find the expectation of the product of two such integrals $x$ and $y$. Then, calculate the covariance as $Cov(x,y)=E(xy)-E(x)E(y)$. 1. Distribution of $I(t,T)\equiv\int\limits_{s=...


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I am having trouble to understand your notation $$ \int_t^Td\xi(t,s)g(s)\,ds\,. $$ What is the meaning of this when you switch from the differential form $dF_t$ to the integral form $$ F_t=F_0-\int_0^tg(s)\,\xi(s,s)\,ds\,+\quad? $$ Surely, in the deterministic case when $\sigma\equiv 0\,$ we have by ordinary calculus $$ \frac{dF}{dt}=-\xi(t,t)\,g(t)+\int_t^T\...


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