$t>0$ $\sigma\neq 0$ – constant, $W_t$– Wiener process and i need to calculate $E(\cos(\sigma W_t))$.
$d(\cos(\sigma W_t))=(0+0-\frac{1}{2}\sigma^2\cos(\sigma W_t))dt-\sigma\sin(\sigma W_t)dW_t$
$\cos(\sigma W_t)=1-\frac{1}{2}\sigma^2\int_0^t\cos(\sigma W_s)ds-\sigma\int_0^t\sin(\sigma W_s)dWs$
$E(\cos(\sigma W_t))=1-\frac{1}{2}\sigma^2E(\int_0^t\cos(\sigma W_s)ds)$
($E(\sigma\int_0^t\sin(\sigma W_t)dWs)=0$ because $\sin(\sigma W_t)\in M_{[0,T]}^2$)
$M_{[0,T]}^2=\left\{f:[0,T]\times\Omega\to\mathbb{R}:\text{f is adapted}, E\left(\int_0^Tf^2(t)dt\right)<\infty\right\}$
is this the end or should it still be calculated?
Best Answer
You have
$$E(\cos(\sigma W_t))=1-\frac{1}{2}\sigma^2E(\int_0^t\cos(\sigma W_s)ds)$$
First notice that $E(\int_0^t\cos(\sigma W_s)ds)=\int_0^t E(\cos(\sigma W_s)ds$ thanks to Fubini's theorem (notice that $\cos(\sigma W_s)$ is continuous, and hence integrable in the compact $[0,t]$).
Then using the exponential representation of the cosine you have
$$\cos(\sigma W_s)=\frac{e^{i\sigma W_s}+e^{i(-\sigma)W_s}}2$$
Then $$E\cos(\sigma W_s)=E\left(\frac{e^{i\sigma W_s}+e^{i(-\sigma)W_s}}2\right)$$
Remember that $W_s$ is a gaussian r.v. hence using the characteristic function (not the moment generating function) you have
$$E\cos(\sigma W_s)=\frac 1 2 (2e^{-(1/2) \sigma^2 s})=e^{-(1/2) \sigma^2 s}$$
Then $$E(\cos(\sigma W_t))=1-\frac{1}{2}\sigma^2 \int_0^t e^{-(1/2) \sigma^2 s}$$
The result should be fairly obvious at this point.