For a 3-dimensional Brownian motion $B = (B_t, t ≥ 0)$ and $x ∈ \mathbb{R}^3 \backslash \{0\}$ define the process $Y = (Y_t, t ≥ 0)$ via $Y_t =\frac{1}{|B_t+x|}$ how come this is a continuous local martingale but not proper martingale? and is it possible to deduce continuity from of $Y$ from the continuity of the brownian motion? I suspect this could get difficult when $B_t = -x$?
$\begingroup$
$\endgroup$
2
-
4$\begingroup$ The event $\{B_t = -x \text{ for some } t \geqslant 0\}$ has probability zero, so $Y_t$ is indeed well-defined (with probability one). By the Itô formula one easily finds that $Y_t$ is a local martingale. Finally, $\mathbb E Y_t$ converges to zero by the dominated convergence theorem, so $Y_t$ cannot be a true martingale. (This is a standard exercise, I think, so I am not sure this question fits the guidelines of this site. Thus I leave this as a comment so that others can decide.) $\endgroup$– Mateusz KwaśnickiCommented Jan 6, 2021 at 21:02
-
$\begingroup$ I saw it as an example of a local martingale but no martingale but there was no proof attached to it and i couldn't make my mind up as to why this is. Probably the proof wasn't given in the script because it is very suitable as an exercise. Still your answer has helped me understand. Thank you very much! $\endgroup$– Martin WeizengussCommented Jan 6, 2021 at 22:53
Add a comment
|