I am working on the following problem which is on an introductory chapter of Brownian motion: Let B(t) be the standard Brownian motion. Define $X(t)=[1/\sqrt{t}]\int_{0 \to t}{f(B(s))}ds$ where $f$ is an $L^1$ function which integrates to 1 on the whole real line. The question is to show $EX(t)$ goes to $\sqrt{2/\pi}$ as t goes to infinity and $E[X(t)^2]$ goes to 1 as $t$ goes to infinity.
What I have done is to write $E[X(t)]=(1/\sqrt{t})\int_{0 \to t}{E[f(B(s))]}$ (by Dominated Convergence Thm and Fubini thm) and compute $E[f(B(s))]$ using the density function of $N(0,s)$.
It works fine for $E[X(t)]$. But when I do the same for the second power: $E[X(t)^2]=(1/t)\int\int_{0 \to t}{E[f(B(s))f(B(u))]}$, I have to deal with $E[f(B(s))f(B(u))]$ which makes things complicated:
$$E[f(B(s))f(B(u))] =E[f((B(u)-B(s))+B(s))f(B(s))1(u>s)+f((B(s)B(u))+B(u))f(B(u))1(u<s)]$$.
The reason I do the last step is because I need to use the independence of B(u)-B(s) and B(s) to write out the joint density function as a product of two normal density functions otherwise I do not know how to compute E[f(B(s))f(B(u))].
I don't feel I am doing this question properly but I don't see any other way to do it. The problem also asks the limit of $E[X(t)^k]$ for any integer $k$ as $t$ goes to infinity. Any help is appreciated.