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Consider $$ G_t = \int_0 ^t \frac{B_u}{u}du$$ where $\left(B_{t} \right)_{t\geq0}$ is $\mathcal F _t $ - brownian motian in $\mathbb R$, null at the origin. It's simple to show that $\left(G_{t} \right)_{t\geq0}$ is a centred gaussien process.

But, when it comes to the covariance evaluation, it seems we can find some probles with singularity at zero.

Someone could help me ?

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Is the integral w.r.t. $du$? – Alex R. Jan 3 '13 at 23:31
Yes. Thank you Alex. – Paul Jan 4 '13 at 0:37
How did you prove that $G_t$ is well-defined (i.e. $u \mapsto \frac{B_u}{u} \in L^1[0,t]$)? – saz Jan 4 '13 at 14:31
@saz Local modulus of continuity. – Did Jan 4 '13 at 15:02
@did Thanks...! – saz Jan 4 '13 at 15:15
up vote 3 down vote accepted

For every nonnegative $t$, $$\mathbb E(G_t^2)=2\int_0^t\int_0^s\mathbb E(B_uB_s)\frac{\mathrm du}u\frac{\mathrm ds}s=2\int_0^t\int_0^su\frac{\mathrm du}u\frac{\mathrm ds}s=2\int_0^ts\frac{\mathrm ds}s=2t. $$ Likewise, for every nonnegative $s\leqslant t$, $$\mathbb E(G_s(G_t-G_s))=\int_s^t\int_0^s\mathbb E(B_uB_v)\frac{\mathrm du}u\frac{\mathrm dv}v=\int_s^t\int_0^su\frac{\mathrm du}u\frac{\mathrm dv}v=s\log\left(\frac{t}s\right). $$ Thus, for every nonnegative $s\leqslant t$, $$ \mathrm{Cov}(G_s,G_t)=2s+s\log\left(\frac{t}{s}\right). $$

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Hello did! Could you expand or just explain me the argument to derive the very first equality? Thank you! – Paul Jan 17 '13 at 13:24
Sure: $G_t^2$ is an integral on $(u,s)\in[0,t]\times[0,t]$, which is twice the integral of the same function on $0\leqslant u\leqslant s\leqslant t$, by the symmetry $(u,s)\mapsto(s,u)$ and because the diagonal $0\leqslant u=s\leqslant t$ has Lebesgue measure zero.. – Did Jan 17 '13 at 14:25

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