# $\mathcal{F_t}$-martingales with Itô's formula?

I need a little help with a problem. I am given some stochastic processes and supposed to show that they are $\mathcal{F_t}-$martingales. The first one is this, and they all look similar:

$$X_t=e^{t/2} \cdot \cos{W_t}.$$

Is this different from showing that it is a martingale? And how can and shall I use Itô’s formula here:

For some $f(t,x)$, $$f(t,W_t)=f(0,0)+\int_0^t\frac{\partial f}{\partial x}(s,W_s)dW_s+\int_0^{t}\frac{\partial f}{\partial t}(s,W_s)ds + \frac{1}{2}\int_0^t\frac{\partial^2f}{\partial x^2}(s,W_s)ds.$$

Marie :*

#### Solutions Collecting From Web of "$\mathcal{F_t}$-martingales with Itô's formula?"

Let’s assume that $\mathcal{F}_t$ is the natural filtration of the Brownian motion.

Notice that, for $s<t$
$$\begin{eqnarray} \mathbb{E}\left(X_t|\mathcal{F}_s\right) &=& \mathbb{E}\left(\left( \mathrm{e}^{t/2} \cos(B_s + (B_t-B_s))\right)| \mathcal{F}_s \right) \\&=& \mathrm{e}^{t/2} \mathbb{E}\left( \cos(B_s) \cos(B_{t-s}) – \sin(B_s) \sin(B_{t-s}) |\mathcal{F}_s\right) \\ &\stackrel{\color\maroon{\text{independence}}}{=}& \mathrm{e}^{t/2} \left( \cos(B_s) \mathbb{E}(\cos(B_{t-s})) – \sin(B_s) \underbrace{\mathbb{E}(\sin(B_{t-s}))}_{\text{zero}} \right) \\ &=& \mathrm{e}^{t/2} \cos(B_s) \mathbb{E}(\cos(B_{t-s})) = \mathrm{e}^{t/2} \cos(B_s) \mathrm{e}^{(s-t)/2} = X_s \end{eqnarray}$$
Expectation of the cosine function is easiest to extract from the characteristic function of the normal distribution:
$$\mathbb{E}(\cos(a B_t)) = \Re\left(\mathbb{E}\left(\mathrm{e}^{i a B_t}\right)\right) = \Re\left( \mathrm{e}^{-a^2 t/2}\right) = \mathrm{e}^{-a^2 t/2}$$

In order to establish whether $X_t$ is a square integrable martingale, we need to check that $\sup\limits_{t > 0} \left(\mathbb{E}(X_t^2)\right) < +\infty$.
$$\mathbb{E}(X_t^2) = \mathrm{e}^t \mathbb{E}\left( \cos^2 B_t\right) =\frac{\mathrm{e}^t}{2} \mathbb{E}\left( 1+ \cos(2 B_t)\right) = \frac{\mathrm{e}^t}{2} \left( 1 + \mathrm{e}^{-2 t} \right) = \cosh(t)$$
Since $\cosh(t)$ is unbounded, $X_t$ is not a square-integrable martingale for $t>0$, but it is a martingale, since $\mathbb{E}(|X_t|) < \exp(t) < \infty$ for all $0 \leqslant t < \infty$.

You could have used the Ito’s lemma to establish that $X_t$ is a semi-martingale, i.e. the drift coefficient is exactly zero.