Question about Ito integral I was wondering if Ito integral:
$\int_0^T B(t)dB(t) $ 
is Gaussian (in which B(t) is Brownian Motion)??
Thank you so much, I appreciate any help ^^

$d(B_t^2) = 2B_t dB_t + dt$.  Therefore your integral is $\frac12(B_T^2-T)$.  $B_T$ is Gaussian $N(0,\sqrt T)$, therefore $B_T^2$ is $T$ times a $\chi^2_1$.
