1
$\begingroup$

If $X$ is a standard 1d brownian motion and $M_t$ $= \mbox{max}\{X_S: 0 \le s \le t\} $, what can we say about $M_t/t$ as $t \rightarrow \infty$?

Mainly, what can we say about the behavior of this martingale?

My attempt: P(Msubt > a) = 2P(B(t) >a), but what integral does this fit in with?

  • 0
    Hint: The reflection principle gives that $M_t$ has the same distribution as $|B_t|$.2012-05-02
  • 0
    Sorry I used $B_t$ as my notation for a Brownian motion there.2012-05-02
  • 0
    It also might help you at some point when estimating the tail of the normal distribution that there is a really stupid bound: $e^{-x^2}\leq xe^{-x^2}$ for $x$ large.2012-05-03
  • 0
    So you are saying to use Hoeffdings inequality? will it help?2012-05-03
  • 0
    Perhaps I was being too cryptic. Chris suggested that the distribution of $M_t$ is the same as $|B_t|$. You can easily calculate the probability that $|B_t|\geq at$ for $a$ a positive constant, as a Gaussian integral but you'll need to do some estimates for how this "tail" probability behaves for large $t$. I believe my prior hint will help.2012-05-03
  • 0
    mary: Any luck with the answer below?2012-05-07
  • 0
    mary: Bis repetita.2012-07-25

1 Answers 1