I have just started studying stochastic analysis and I am stuck on trying to show that for $B$ a two-dimensional Brownian motion, $\ X_t=\log\left(\left|B_t\right|\right)_t$ is a local martingale but not a martingale.
I saw people trying to tackle such questions using Ito integrals, but being one of the first things in my course, I imagine there is an elementary way to think about this, and I would be very curious about it? In particular, what sequence of stopping times should I use? Intuitively I imagine it has to do with the fact that the stopped process is uniformly integrable, so we can apply Doob's stopping theorem.
I would be extremely grateful if someone could please explain these details to me? Thank you very much for your time and help!!
Because of the problem pointed out by @saz, let's assume that $B_0=x\not=0$. For convenience only, I'll suppose that $|x|=1$. It follows easily from Ito's formula that $X_t$ is a local martingale with $X_0=0$. If $X$ were a martingale then you would have $\Bbb E[X_t]=0$ for all $t>0$. But unless I am mistaken, a calculation using polar coordinates shows that $$ \Bbb E[X_t]={\sqrt{2\pi}\over t}\int_0^1\log(1/ r)e^{-r^2/2t}r\,dr, $$ which is strictly positive.