TITLE: Expected value of sin of brownian motion
QUESTION [3 upvotes]: I am having trouble finding $E[X_t]$, where $X_t = \sin(B_t)$ and $B_t$ is a Brownian motion. I have learned a little about Ito integrals but I don't think I really understand how to use them.
Here is my approach at the moment:
$$E[X_t] =E[\sin(B_t)] =\left(\frac{1}{\sqrt {2\pi t} } \int_{-\infty}^\infty e^\frac{-(x)^2}{2t} \sin(x) dx\right)\bigg\rvert_{x=B_t}$$
Is this the right approach? Any help would be appreciated, thank you!

REPLY [4 votes]: If you've seen Ito's formula, then you can show that 
$$
\sin(B_t) =\sin(B_0)-{1\over 2}\int_0^t\sin(B_s)\,ds+M_t,
$$
where $(M_t)$ is a martingale with $M_0=0$. Define $g(t):=E[\sin(B_t)]$; taking expectations in the above, assuming that $B_0=x$ (constant):
$$
g(t)=g(x)-{1\over 2}\int_0^t g(s)\,ds,\qquad t\ge 0.
$$
Differentiate this to obtain an ODE for $g$ that you can easily solve.
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