Doob decomposition of $\cos(aB_t)$

Question

Let $a$ be some real number and $B_t$ a Brownian motion. I want to compute the Doob decomposition of $\cos(aB_t)$, and the predictable variation of the local martingale part $M_t$.

The Doob decomposition is given by:

$$X_t=X_0+A_t+M_t$$

where $A_t$ is a continuous finite variation process and $M_t$ is a continuous local martingale.

By Ito-formula:

$$\text{cos}(aB_t)=-\int_0^t\text{sin}(aB_t)d(aB)_t-\int_0^t\text{cos}(aB_t)d<aB>_t$$

I believe the left term is the $A_t$ and the right $M_t$, but how to compute the predictable variation?

Answer 1

Almost there! You apply the Ito formula using the process $B$ not $aB$. It writes as follows: \begin{equation} \cos(aB_t) = \underbrace{0}_{X_0} - \underbrace{\int_0^t a\sin(aB_t)dB_t}_{M_t} - \underbrace{\frac{a^2}{2}\int_0^t\cos(aB_t)dt}_{A_t} \end{equation}

By the way, you can even show the local martingale $M$ is in fact a (square) martingale.