I've been asked with computing the $L^2$ variation of $\int_0^tW_sds$, where $W_t$ is a Brownian motion. I am not supposed to use to stochastic calculus to solve, but the definition of quadratic variation and perhaps the fact that for a Brownian motion $W_t$ the quadratic variation in $L^2$ is $t$ (the length of the interval of the index set).

Is there a slick way to use this fact to calculate the quadratic variation? So far I have gone by brute force to show that quadratic variation of the integral in $L^2$ is $0$ (which is quite lengthy so I have not yet written them).

I will appreciate any advice, or at least an indication whether $0$ is indeed the $L^2$ quadratic variation.


Fix $T>0$ and set

$$X_t := \int_0^t W_s \, ds.$$


$$|X_t-X_s| \leq |t-s| \sup_{r \in [0,T]}| W_r|$$

for any $s,t \in [0,T]$. For a given partition $\Pi = \{0=t_0< \ldots < t_n=T\}$ with mesh size $|\Pi|$ we find that

$$\sum_{j=1}^n |X_{t_j}-X_{t_{j-1}}|^2 \leq \sup_{r \in [0,T]}W_r^2 \sum_{j=1}^n (t_j-t_{j-1})^2 \leq |\Pi| \sup_{r \leq T} W_r^2 \underbrace{\sum_{j=1}^n (t_j-t_{j-1})}_{=T}.$$

Taking the expectation we obtain that

$$\mathbb{E} \left( \sum_{j=1}^n |X_{t_j}-X_{t_{j-1}}|^2 \right) \leq T |\Pi| \mathbb{E} \left( \sup_{r \leq T} W_r^2 \right),$$

and letting $|\Pi| \to 0$ we conclude that the ($L^2$-) quadratic variation of $(X_t)_{t \geq 0}$ equals $0$.

  • $\begingroup$ This may be trivial, but I don't currently see how you show that $\mathbb{E}\Big[ \underset{r\leq T}{\sup} W_r^2 \Big]<\infty$? $\endgroup$ – Keen-ameteur May 23 '18 at 19:02
  • $\begingroup$ @Keen-ameteur Use Doob's maximal inequality $\endgroup$ – saz May 23 '18 at 19:07
  • $\begingroup$ I see that now, thank you $\endgroup$ – Keen-ameteur May 23 '18 at 19:14
  • $\begingroup$ @Keen-ameteur You are welcome. $\endgroup$ – saz May 23 '18 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.