Checking if $X(t) = \exp(t/2)\cos(W(t))$, with $W(t)$ a Wiener process, is a martingale

Question

This is what I've done:

Let $s < t$ and $F_t$ be a filtration adapted to $W(t)$ $$E[e^{t/2}\cos(W(t))|F_s] = e^{t/2} E[\cos(W(t)) - \cos(W(s)) + \cos(W(s))|F_s]$$ $$= e^{t/2} [E[\cos(W(t)) - \cos(W(s))|F_s] + \cos(W(s))]$$ Because of the independence of the increments: $$= e^{t/2} [E[\cos(W(t)) - \cos(W(s))] + \cos(W(s))]$$ This is where I'm stuck. I don't know how to calculate $E[\cos(W(t)) - \cos(W(s))]$.

Following the suggestion from Did in the comment in this question, I can do this: $$E[\cos(W(t)) - \cos(W(s))] = \frac{1}{\sqrt{2\pi (t-s)}}\int_{\mathbb{R}}(\cos(x) - \cos(y)).e^{-(x-y)^2/2(t-s)}dx$$

But I don't know if it is correct.

Edit: OMG! I made an horrible mistake. The increments $W(t) - W(s)$ are independent, not $\cos(W(t)) - \cos(W(s))$. With the hints from this answer and the one from Siron everything got clearer now.

Answer 1

To find $\mathbb{E}[\cos(W(t))]$ you could try integrating the density function. However, that might turn out to be messy. Instead, recall that $$\cos x = \frac{e^{ix}+e^{-ix}}{2},$$ and hence \begin{align} \mathbb{E}[\cos(W(t))] = \mathbb{E}\left[\frac{e^{iW(t)}+e^{-iW(t)}}{2}\right], \end{align} where $\mathbb{E}[e^{iW(t)}]$ is the characteristic function of $W(t)$ evaluated in $1$.

Answer 2

The function $u(t,x)=e^{t/2}\,\cos(x)$ satisfies the heat equation ${d\over dt}u+{1\over 2}\Delta_x u=0,$ so that, by Ito's formula, $u(t,W_t)=e^{t/2}\,\cos(W_t)$ is a martingale.