Consider the following expectation:
$E(\exp(\frac{1}{2} \int_0^T \frac{1}{W_t^2} dt))$
where $W_t$ is a Brownian motion. Is this expectation finite? And how would you go about showing whether it is finite? While $W_t = 0$ might cause a problem, the probability of that happening might also be/approach zero. Any pointers would be much appreciated.
Best
Assuming that $W_0=0$, the integral $\int_0^T W_s^{-2}\,ds$ is divergent for each $T>0$, a.s. This follows from the occupation time formula for Brownian local time $L^x_T$, $x\in\Bbb R$: $$ \int_0^T f(W_s)\,ds =\int_{\Bbb R} f(x)L^x_T\,dx, $$ valid for bounded measurable $f$. But $L^0_T>0$ a.s. because $W_0=0$; and $x\mapsto L^x_T$ is continuous a.s. Consequently, for a.e. $\omega$ there is $\epsilon(\omega)>0$ such that $L^x_T(\omega)\ge (1/2)L^0_T(\omega)>0$ for all $x\in(-\epsilon(\omega),\epsilon(\omega))$, and so $$ \int_0^T W_s^{-2}\,ds =\int_{\Bbb R} x^{-2}L^x_T\,dx\ge\int_{-\epsilon}^\epsilon x^{-2}(1/2)L^0_T\,dx =\infty. $$