Consider a bounded, continuous martingale $(X_t)_{t\ge 0}$. I was able to show that $(X^2-[X])_{t\ge 0}$ is uniformly integrable, where $[X]$ denotes the quadratic variation.
Is there an example of a bounded, continuous martingale $(X_t)_{t\ge 0}$, such that $(X^2-[X])_{t\ge 0}$ is unbounded, but still uniformly integrable?
I was thinking about a Brownian motion that is mirroring, each time being equal to $1$ or $-1$. My hope was, that the quadratic variation is still $t$, such that $(X^2-[X])_{t\ge 0}$ is unbounded. Unfortunately, I am not able to write it down properly.
I would really appreciate help on this problem. Thank you in advance!
Consider a one-dimensional Brownian motion $(B_t)_{t \geq 0}$ and the stopping time $$\tau := \inf\{t \geq 0; |B_t| \geq 1\}.$$ By the optional stopping theorem, the stopped process $X_t := B_{t \wedge \tau}$ is a continuous martingale. Moreover, $(X_t)_{t \geq 0}$ is bounded and its quadratic variation is $[X]_t = \min\{\tau,t\}$. On the other hand, the difference
$$X_t^2 - [X]_t = B_{t \wedge \tau}^2 - (t \wedge \tau)$$
fails to be bounded. Indeed, if the difference were bounded, then the boundedness of $B_{t \wedge \tau}^2$ would imply that $$\sup_{t \geq 0} (t \wedge \tau)=\tau$$ is bounded. This, however, is impossible since $\mathbb{P} \left( \sup_{s \leq t} |B_s|<1 \right)>0$ for any $t>0$.