I am having some difficult in making sense of ${\rm{E}}\left[ {\int\limits_0^t {{W_u}du} } \right]$ where W is just your standard one dimensional brownian motion. Is interchanging the order of the integrals allowed here?

  • $\begingroup$ Sure, why not? Write the expectation $E$ as an integral with respect to the underlying probability measure, i.e. $E(Y) = \int_{\Omega} Y \, d\mathbb{P}$, and apply Fubini's theorem. Then you will see that $E(\int_0^t W_u \, du)=0$. $\endgroup$ – saz May 16 at 18:05

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