Not sure how to find $\mathbb{E}[X_t]$
if $\;X_t=\sin(B_t)\;,\quad t\geqslant0\;.$
My usual assumption is: $\displaystyle\;\mathbb{E}\big(s(x)\big)=\int_{-\infty}^{+\infty}s(x)f(x)\,\mathrm{d}x\;$ where $f(x)$ is the probability distribution of $s(x)$. But then brownian motion on its own $\mathbb{E}[B_s]=0$ and $\sin(x)$ also oscillates around zero. So I'm not sure how to combine these?
Any help would be appreciated.
The brownian motion $B_t$ has a symmetric distribution arround 0 (more precisely, a centered Gaussian). Since $sin$ is an odd function, then $\mathbb{E}[\sin(B_t)] = 0$ for all $t$. To see this, since $-B_t$ has the same distribution as $B_t$, we have that $$ \mathbb{E}[\sin(B_t)] = \mathbb{E}[\sin(-B_t)] = -\mathbb{E}[\sin(B_t)] $$ which gives $\mathbb{E}[\sin(B_t)]=0$.