time changes for martingales
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From: TefJlives on 5 Jul 2010 03:41 Hello all, I am having a hard time understanding something about time changes for martingales, and would appreciate some help. I am using Revuz and Yor as my primary reference. So a time change is a family of stopping times C_s which is increasing in s. If X is a local martingale, so is X_{C_t}, according to Proposition V.1.5. What I am wondering is the following. Suppose we are given an increasing adapted process A_t and a local martingale X_t with <X>_infinity = infinity a.s. Does there then exist a time change C_t such that <X>_{C_t} = A_t? Certainly C_t can be defined but I don't see why it should be a stopping time. Can someone please help? Greg
From: Ken Pledger on 6 Jul 2010 16:51 In article <2156ce80-0602-4458-b7de-d599dcbea26d(a)i16g2000prn.googlegroups.com>, TefJlives <gmarkowsky(a)gmail.com> wrote: > .... > I am having a hard time understanding something about time changes for > martingales .... It might be worth while to send your question to <alt.sci.math.probability> or even <sci.stat.math>. Ken Pledger.
From: TefJlives on 6 Jul 2010 23:35 On Jul 6, 1:51 pm, Ken Pledger <ken.pled...(a)mcs.vuw.ac.nz> wrote: > In article > <2156ce80-0602-4458-b7de-d599dcbea...(a)i16g2000prn.googlegroups.com>, > > TefJlives <gmarkow...(a)gmail.com> wrote: > > .... > > I am having a hard time understanding something about time changes for > > martingales .... > > It might be worth while to send your question to > <alt.sci.math.probability> or even <sci.stat.math>. > > Ken Pledger. Thanks, I'll give it a shot. Greg
From: TefJlives on 11 Jul 2010 05:49
Here's a rephrasing of the question, in case it helps. Suppose f=u+iv is holomorphic and B_s is a planar Brownian motion. We can find an adapted process C_t such that \int_0^t |f'(B_s)|ds = \int_0^{C_t} |f'(B_s)|^2ds If we define V_t = u(B_{C_t}), we then have <V>_t = |f'(B_t)|dt But is V a local martingale? It is if C_t is a stopping time for each t. To show that it is a stopping time we need to show {C_t <= r} \in F_r where F is the filtration. But {C_t <= r} = {\int_0^t |f'(B_s)|ds <= \int_0^r |f'(B_s)|^2ds} \in F_{max{r,t}} So I don't see how we can conclude that C_t is a stopping time. And yet, we should be able to adjust the speed to obtain a new process V for which <V>_t = |f'(B_t)|dt, no? Can someone help me with this? Greg On Jul 6, 8:35 pm, TefJlives <gmarkow...(a)gmail.com> wrote: > On Jul 6, 1:51 pm, Ken Pledger <ken.pled...(a)mcs.vuw.ac.nz> wrote: > > > In article > > <2156ce80-0602-4458-b7de-d599dcbea...(a)i16g2000prn.googlegroups.com>, > > > TefJlives <gmarkow...(a)gmail.com> wrote: > > > .... > > > I am having a hard time understanding something about time changes for > > > martingales .... > > > It might be worth while to send your question to > > <alt.sci.math.probability> or even <sci.stat.math>. > > > Ken Pledger. > > Thanks, I'll give it a shot. > > Greg |