Suppose $(B_t, \mathcal{F}_t)_{t \geq 0}$ is a classical Brownian Motion (with the canonical filtration) and consider the process $X_t=\sin(B_t)$.
What properties does our new process share with a Brownian motion? Concretely: does it follow that $(X_t, \mathcal{F}_t)_{t \geq 0}$ is also a continous martingale?
It is definetively the case that $X_t$ also has stationary, independent increments, so that: $$\mathbb{E}[X_T|X_t] = \mathbb{E}[X_t + (X_T-X_t)|X_t] = X_t + \mathbb{E}[(X_T-X_t)|X_t] = X_t + \mathbb{E}[X_T-X_t] = X_t$$
Do you agree? Continuity also seems not difficult. On the other hand, the previous quick proof does not hold for $\cos(B_t)$ since $\mathbb{E}[X_T] \neq \mathbb{E}[X_t] \neq 0$.
Does this mean $\cos(B_t)$ is really not a martingale? I am using the results on this page by the way, for the expected values of $\sin(B_t)$ and $\cos(B_t)$.
Best Answer
Continuity is immediate as the composition of continuous functions is again continuous. However, $\sin (B_t)$ and $\cos (B_t)$ are not martingales.
Let $f(x) = \sin x$, so that $f_x(x) = \cos x$ and $f_{xx}(x) = -\sin x$. Then, by Itô's lemma, $X_t$ satisfies the following stochastic differential equation:
$$dX_t = f_x(B_t) dB_t + \frac{1}{2}f_{xx}(B_t) dt = \cos (B_t)dB_t - \frac{1}{2} \sin (B_t)dt$$
In particular, $\sin B_t$ has a non-zero drift term, so that it cannot be a martingale. You may argue similarly for $\cos B_t$,