# 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:

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

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

3. 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!

• Well, you did use continuity. You should be more careful/explicit in using that. – user223391 Dec 24 '17 at 21:39
• No you assumed that $w_t$ was continuous, which it is. But you need to be more careful/explicit – user223391 Dec 24 '17 at 21:58
• The question is basically "how many times does $W$ cross the x axis" – user223391 Dec 24 '17 at 22:09
• 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. – saz Dec 25 '17 at 18:53
• With probability 1, the zero set of Brownian motion has no isolated points which might make the problem difficult. See e.g. stat.berkeley.edu/~pitman/s205s03/lecture18.pdf – user223391 Dec 25 '17 at 19:19

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:

1. $W_0 = 0$

2. $W_t$ is almost surely continuous

3. $W_t$ has independent increments.

4. $W_{t}-W_{s}\sim {\mathcal {N}}(0,t-s)$ (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."