the article given by wikipedia http://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula#Proof states at some point of the proof that: (line 7) ''the third term is o(dtdu) and can be dropped''
Can anyone see why the cross-variation term can be dropped, i.e is equal to zero ?
This is a non-rigorous way to think about this. On the other hand, the first time I learned about Brownian motion was using non-standard analysis, so this can be made rigorous with some effort. I wrote this using the notation $B_t$, which on the web page is written as $W_t$ ($B$ for Brown, $W$ for Weiner).
So I think of $dt$ as being an infinitesimal quantity. And $dB_t = \pm \sqrt{dt}$. In this way $$ B_t = \int_0^t dB_t = \sum_{n=0}^{t/dt} \pm \sqrt{dt} $$ is a reasonable sized value, i.e., neither infinitesimal nor infinite. This is because $$ E \left|\sum_{n=1}^N \pm 1\right| \approx \left(E \left|\sum_{n=1}^N \pm 1\right|^2\right)^{1/2} = \sqrt{N} $$ and so $$ E\left|\sum_{n=0}^{t/dt} \pm \sqrt{dt}\right| \approx \sqrt{t/dt} \times \sqrt{dt} \approx \sqrt t .$$ (Here all the $\pm$ are independent and take either value with probability $1/2$.)
Then $$ \int_0^t dt du \approx \sum_{n=0}^{t/dt} dt(U dt \pm V \sqrt{dt}) $$ for some $U$ and $V$ (which might not be constant, but depend upon $t$ and $B_t$, and this is of order $dt^2\times(t/dt) + dt\sqrt{dt}\times \sqrt{t/dt}$, and this is clearly infinitesimal, and hence in non-standard analysis converts to zero.
