Related: Ito integral representation of cosine of Brownian motion and expected value
Question:
Let, $U_t = \cos(\sigma W_t)$, $W_t$ is Brownian Motion. Find $dU_t$ and hence find $\mathbb{E}[U_t]$.
I've found $dU_t$ to be:
$$
dU_t = – \sigma \sin(\sigma W_t) dW_t – \frac{1}{2} \sigma^2 \cos(\sigma W_t) dt
$$
I'm not sure how to find the expected value of $U_t$ using Ito Calculus.
In the linked question, the author makes a change of variable like so:
$X_T = e^{\frac{1}{2}t} \cos B_t$
I'm not sure where the $e^{\frac{1}{2}t}$ comes from when the original question says: $X_T = \cos(B_T)$
Best Answer
In your case, make the variable change $X_t=e^{\frac12\sigma^2t}U_t$. Then, with Ito's rule,
$$dX_t = -e^{\frac12\sigma^2t}\sigma \sin(\sigma W_t) dW_t $$
which is drift-less (or a martingale). Therefore,
$$E[e^{\frac12\sigma^2t}U_t] = E[X_t] = X_0=1$$
or
$$E[U_t] = e^{-\frac12\sigma^2t}$$