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!

  • 1
    $\begingroup$ Bounded here means that $\sup_t |X_t| < \infty$ a.s.? Or do you mean bounded in $L^2$? $\endgroup$
    – Michh
    May 31, 2020 at 13:30
  • 1
    $\begingroup$ No, bounded means, there is a constant $c$, such that for all $t\ge 0$ and $\omega\in\Omega$, it holds: $|X_t(\omega)|\le c$. $\endgroup$
    – user408858
    May 31, 2020 at 13:36

1 Answer 1


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 $t>0$.

  • $\begingroup$ Formidable! :-) $\endgroup$
    – user408858
    May 31, 2020 at 20:38

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