Let $B_t$ be a Brownian motion. Define sign function as follows. $sign(0) = 0$ and $sign(x) = \frac{x}{|x|}, \forall x \neq 0$. I do not know how to show the following two questions, especially on the progressive measurability one. Could anyone provide some detailed explanation on how to go about with them, please? Thank you!
- Show that $sign(B_t)$ is a progressive process.
- Show that $sign(B_t)\cdot B_t$ is a Brownian motion.
A direct way to show that the process $X$ defined by $X_t=\mathrm{sign}(B_t)$ is progressively measurable is to consider, for every positive integer $n$, $$X^n_t=\frac{nB_t}{\sqrt{1+n^2B_t^2}}.$$ Each process $X^n$ is adapted with continuous paths hence progressively measurable. Furthermore, for every $t$, $X^n_t\to X_t$ almost surely, when $n\to\infty$, hence $X$ is progressively measurable.
(This answers item 1. Item 2. is absurd, for example because $X_tB_t=|B_t|$ and $t\mapsto E(|B_t|)$ is not constant hence $|B|$ is not a Brownian motion (much more is known but this simple argument suffices).)