time change

Discussion in 'Probability' started by TefJlives, Jul 9, 2010.

  1. TefJlives

    TefJlives Guest

    Hello all,

    I wonder if anyone can help me with the question below. 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?
    Thank you.

    Greg
     
    TefJlives, Jul 9, 2010
    #1
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