Let $W(t)$ be a standard $\mathcal{F}_t$-Brownian Motion. Let $$g(x)=\begin{cases}1 & \text{if } x\geq 0\\ -1 & \text{if }x<0 \end{cases}$$
Consider the process $B(t)=\int_0^tg(W(s))dW(s)$. I want to show that $B(t)$ is also a Brownian Motion.
My attempt: I can justify the start at the origin and a.s. $t$-continuity as follows:
- Start at the origin: $B(0)=\int_{0}^0g(W(s))dW(s)=0$.
- a.s. $t$-continuity: as $B(t)$ is a stochastic integral, there exists a $t$-continuous version of it, so $B(t)$ is almost surely continuous.
Where I'm having a little bit of trouble is in showing the normality and independence of increments. Ideally, in my opinion, I would like to apply Ito's formula to get a relationship between $B(t)$ and $W(t)$, or between $dB(t)$ and $dW(t)$, so that by using the properties of $W(t)$, I can say something about $B(t)$, however, my issue is that $g$ is not twice-continuously differentiable (although it almost surely is). A way around this, to keep with the same strategy would be to use the definition of the Ito integral as the limit of the integral for elementary functions, although seems like an unnecessary amount of work. Any hints or help is welcomed.
Best Answer
Use the Levy's theorem (http://math.ucsd.edu/~pfitz/downloads/courses/spring05/math280c/levy.pdf).
You have that that $g(W_t)^2=1$ a.s, this argument is sufficient to prove that $B_t$ and $B_t^2-t$ are martingales.