Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
1  
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

1 Answer 1

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 $s\leqslant t$, $$ \mathrm{Cov}(G_s,G_t)=2s+s\log\left(\frac{t}{s}\right). $$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.