5
$\begingroup$

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

$\endgroup$

1 Answer 1

4
$\begingroup$

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.

$\endgroup$
1
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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