# How to find the mean of $\int_0^t W_s ds$, where $W_s$ is a Wiener process?

am trying to find the expectation of $\int_0^t W_s ds$, with $W_s$ being the Standard Wiener process. I am trying to use Ito's formula, by decomposing as:

$$\frac{W_t^3}{6} = \frac{1}{2}\int_0^t B_s^2 dB_s + \int_0^t B_s ds$$

Then, applying the expectation on both sides, I get that:

$$E\left[\int_0^t B_s ds\right]= - E\left[\frac{1}{2}\int_0^t B_s^2 dB_s\right]$$

In this case, I am wondering if I am allowed to use Fubini's theorem to move the expectation on the right into the integral. Would it be defined for the $dB_s$ integrand? Thanks.

• Fubini indeed: $$E\left(\int_0^t W_s ds\right)=\int_0^t E(W_s) ds=\int_0^t 0 ds=\ldots$$ – Did Apr 28 '16 at 6:58