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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.

Keen-ameteur
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1 Answers1

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Fix $T>0$ and set

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

Clearly

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

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 $0$.

saz
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