# Unbounded quadratic variation process for bounded continuous martingale

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!

• Bounded here means that $\sup_t |X_t| < \infty$ a.s.? Or do you mean bounded in $L^2$? May 31, 2020 at 13:30
• No, bounded means, there is a constant $c$, such that for all $t\ge 0$ and $\omega\in\Omega$, it holds: $|X_t(\omega)|\le c$. May 31, 2020 at 13:36

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 $$t>0$$.