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$sign(x)=1$ if $x\geq0$

$sign(x)=-1$ if $x< 0$

Consider $$ X_t = \int^t_0 sign(W_u)dW_u $$ where $W_t$ is a wiener proces.

How can I determine the distribution of $X_t$ and compute $E[\exp(\lambda X_t )]$?

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  • $\begingroup$ It seems to me $X_t$ is itself a Wiener process. Isn't it? Then the distribution of $X_t$ is Normal, and the Expectation is the expecttaion of a Lognormal (for which there is a known formula). $\endgroup$
    – nbbo2
    Commented Dec 22, 2018 at 1:44
  • $\begingroup$ noob2: Can you explain why $X_t$ is still a Weiner process. Clearly it still has independent increments, a mean of zero and its stationary but is that all that one needs for a process to be a weiner process Thanks. $\endgroup$
    – mark leeds
    Commented Dec 22, 2018 at 5:52
  • $\begingroup$ I was formulating a hypothesis, to be checked or rejected by more detailed analysis. $\endgroup$
    – nbbo2
    Commented Dec 22, 2018 at 12:45
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    $\begingroup$ Hi Noob2: I don't think it's Weiner process because for that to be the case, the process has to be convtinuous in t. galton.uchicago.edu/~lalley/Courses/313/… $\endgroup$
    – mark leeds
    Commented Dec 22, 2018 at 14:32
  • $\begingroup$ I’m intuitively with @Noob2. The distribution of dW is the same as the distribution of -dW. Therefore what difference does it make if there is a Sign () function. $\endgroup$
    – dm63
    Commented Dec 22, 2018 at 23:06

1 Answer 1

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Note that $\{X_t, \, t \ge 0\}$ is continuous, square-integrable martingale with quadratic variation process \begin{align*} \langle X\rangle_t = \int_0^t {\rm sign}^2(W_s)\, ds =t. \end{align*} Then, it is a standard Brownian motion based on Levy’s Characterization of Brownian Motion. The remaining is straightforward.

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    $\begingroup$ In your link, 4) says almost sure continuity is a requirement but I don't know if he uses it in the proof and don't have time to go through it. I'm not saying you are wrong but rather just letting you know. Also, it's possible that the process described is almost surely continuous. I'm not almost sure about that. LOL. $\endgroup$
    – mark leeds
    Commented Dec 24, 2018 at 2:26
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    $\begingroup$ An Ito integral with respect to a continuous martingale is continuous. $\endgroup$
    – Gordon
    Commented Dec 24, 2018 at 13:51

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