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The fundamental theorem of algebra states that is algebraically closed, that is;
For any non-constant polynomial in , there exists a such that .
Let be a Brownian motion on and suppose for a contradiction that a non-constant polynomial does not have any zero’s. Let , then is analytic and tends to 0 at infinity. Pick such that and note that and contain an open set, which can be done due to the fact that is continuous and non-constant.
Now is a continuous local martingale (by using Ito’s formula) and moreover it is bounded. Hence by the Martingale convergence we have that a.s. and in .
This last statement is contradicted by the fact that Brownian motion is recurrent on the complex plane, in particular, it visits and infinitely many times which gives that
directly contradicting the Martingale convergence.
I found this little gem in Rogers and Williams.
The optional stopping theorem (OST) gives that if is a martingale and is a bounded stopping time, then .
Now take a Brownian motion , which is a martingale, and the stopping time . Obviously and OST says that .
So what went wrong? Well quite simply, I did not check that was actually bounded. Why is it not bounded you say? No? Ok, well I’m going to tell you anyway; it is not bounded because there are many paths that trail below 0 which never hit 1. To properly show this, one must first calculate the supremum of the Brownian motion, then work out the probability that it is less than 1.
Rather nice reminder to always check that the stopping times are bounded before applying the OST. This was provided in a lecture by Nathanael Berestycki.