# Quadratic variation of an integral of a function of a Brownian motion

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

## 1 Answer

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.

• How do I prove it? Do I have to show the total variation is bounded, and then the quadratic variation is 0. – Geoff Chen Oct 9 '18 at 1:57