I am currently working with integrated brownian motion and have to find its mean and variance. I know they are $0$ and $t^3/3$, but I am having trouble showing this.
Using the Îto formula we can rewrite $$\int_0^tW(s)\mathrm{d}s=\int_0^t(t-s)\mathrm{d}W_s.$$ Then, we can write this out as a sum to find the identity $$\lim_{n\to\infty}\sum_{i=1}^n\left(t-\frac{it}{n}\right)\left(W\left(\frac{it}{n}\right)-W\left(\frac{(i-1)t}{n}\right)\right),$$ of which we would like to take the expectation.
If we can swap said expectation and limit, we are golden and it would be easy to show both the result of the mean and the variance. My difficulty lies in showing this is allowed however. I have tried applying the DCT, but could not find a dominating function.
Is swapping limit and expectation the way this should be shown? And if so, how?
I tried to find an answer to this, but all I could find was that it was "easy to see that $\int_0^t(t-s)\mathrm{d}W_s$ is $0$", which is exactly the step I don't understand.
As d.k.o. commented, this is solved by taking $$2t\cdot\sup_{0\leq s\leq t}\left|W(s)\right|$$ as the dominating function.