Quadratic Variation of Sign Function & Brownian Motion Find the quadratic variation of a sign function  $X_t = sign(W_t)$ where $W$ is a brownian motion.
The sign function is defined as 
$$sign(y)=\begin{cases}
1 & \text{ if } y>0 \\ 
0 & \text{ if } y=0 \\ 
-1 & \text{ if } y<0 
\end{cases}.$$
The quadratic variation is defined as 
$$[sign(w)]_t = lim_{\left \| p \to 0  \right \|} \sum_{k=1}^{n} [ {sign(w_{t_{k+1}})} - {sign(w_{t_k})} ]^2$$
where the limit is taken over the partition $$max({t_{k+1}} -{t_k}) \to 0.$$
Since ${t_{k+1}} -{t_k}$ is very small, so $sign(w_{t+1})=sign(w_{t}).$
Hence the quadratic variation of a sign function is zero.
My question is:


*

*Is my explanation right? Do we need to consider that at some point, sign function is turning -1 to 0 and to 1?

*I feel like I didn't use the property that $W$ is brownian motion, do I miss anything?

*Since the quadratic variation of a Brownian motion over [0,t] is t. Can I conclude that $sign(W)$ is not brownian motion from here?
Disclaimer: This is one of the question in my exam which I have blur memory. I would like to discuss whether I am on the right track. Thanks!
 A: I want to make some conclusions myself based on comments from @zacharyselk and @saz.

Q1: Is my explanation right? Do we need to consider that at some point, sign function is turning -1 to 0 and to 1?

All the explanation seems fine until the last statement--"Hence the quadratic variation of a sign function is zero."
Since ${t_{k+1}} -{t_k}$ is very small, so $sign(w_{t+1})=sign(w_{t})$ almost surely. 
Those $sign(w_{t+1}) \neq sign(w_{t})$, then $[ {sign(w_{t_{k+1}})} - {sign(w_{t_k})} ]^2=1$. It is the circumstances when the brownian motion touch x-axis or leaving x-axis. 
Since ${t_{k+1}} -{t_k} \to 0$, we won't see $[ {sign(w_{t_{k+1}})} - {sign(w_{t_k})} ]^2=4$. This means the brownian motion cross x-axis not just touching or leaving.

Q2: I feel like I didn't use the property that $W$ is brownian motion, do I miss anything?

The main property for a brownian motion are:


*

*$W_0 = 0$

*$W_t$ is almost surely continuous

*$W_t$ has independent increments.

*$W_{t}-W_{s}\sim {\mathcal {N}}(0,t-s)$  (for $0\leq s\leq t$).
I use the second a.s. continuous assumption when I made the conclusion
${t_{k+1}} -{t_k}$ is very small, so $sign(w_{t+1})=sign(w_{t})$ almost surely.
I probably will need to use the independent and distribution assumption if I want to derive the distribution of the answer.

Q3: Since the quadratic variation of a Brownian motion over [0,t] is t. Can I conclude that $sign(W)$ is not brownian motion from here?

@Saz answers "Since $X_t$ takes only the values $−1$, $0$ and $1$, we know that $X_t$ is not Gaussian and therefore the process is not a Brownian motion."
