I'm asked to find the quadratic variation of the integral $\int_{0}^{t} W_s^2 ds$.


1 Answer 1


The quadratic variation of $$X_t=\int_0^t W_s^2\,ds$$ is 0. This is because it's an Ito process with no $dB_s$ term.

  • 1
    $\begingroup$ How do I prove it? Do I have to show the total variation is bounded, and then the quadratic variation is 0. $\endgroup$
    – Geoff Chen
    Oct 9, 2018 at 1:57

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