Let $(B_t)_t$ be a brownian motion and: $$ X_t=\int_0^t B_u \, du. $$ What is the quadratic variation $<X>_t$ of $X$?
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It is zero. Can you see why? – Nate Eldredge Jan 14 '20 at 20:36
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See also https://math.stackexchange.com/questions/236762/quadratic-variation-of-x-t-int-0t-b-s-ds?rq=1 – Nate Eldredge Jan 14 '20 at 20:38
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I know that it's zero, but i can't see why. – Joel Jan 14 '20 at 20:38
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What is your definition of quadratic variation? What techniques do you know for computing it? – Nate Eldredge Jan 14 '20 at 20:39
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I need to compute $\lim_{\delta \to 0} \sum^{p}{i=1}(X{t_i}-X_{t_{i-1}})^2 $ with $\delta = sup_i (t_i - t_{i-1})$, i can't see why it has a finite variation. – Joel Jan 14 '20 at 20:44
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Take a look at this question or this question – saz Jan 15 '20 at 05:42