# Compute distribution of a stochastic variable

$$sign(x)=1$$ 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 $$E[\exp(\lambda X_t )]$$?

• 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). – noob2 Dec 22 '18 at 1:44
• 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. – mark leeds Dec 22 '18 at 5:52
• I was formulating a hypothesis, to be checked or rejected by more detailed analysis. – noob2 Dec 22 '18 at 12:45
• 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/… – mark leeds Dec 22 '18 at 14:32
• 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. – dm63 Dec 22 '18 at 23:06

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.