Using Ito Calculus to find $\mathbb{E}[U_t]$ if $U_t= \cos(\sigma W_t)$ where $W_t$ is Brownian Motion

Question

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: $X_T = \cos(B_T)$

Answer 1

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

$$E[U_t] = e^{-\frac12\sigma^2t}$$

Answer 2

Let $u_t = \mathbb{E}[U_t]$. Writing your results from Ito's formula in integral form and taking expectations we find that $$u_t = \mathbb{E}[U_t] = \mathbb{E}\bigg[\int_0^t - \sigma \sin(\sigma W_s)dW_s \bigg] - \mathbb{E}\bigg[\frac12 \int_0^t \sigma^2 U_s ds \bigg]$$

Now the first term on the right hand side is the expectation of a martingale started from $0$ and hence is $0$. By Fubini's theorem the second term is $-\frac12 \int_0^t \sigma^2 \mathbb{E}[U_s] ds$. Hence we find that $u_t$ satisfies the ODE $$\frac{du_t}{dt} = - \frac12 \sigma^2 u_t$$

Hence $u_t = \exp(-\frac12 \sigma^2 t)$.