I have a question in my homework:
Let $X_t$ and $Y_t$ be two Brownian motions issue de $0$ and define
$$S_t=\int_0^tX_s\,dY_s-\int_0^tY_s\,dX_s$$
Show that
$$E[e^{i\lambda S_t}]=E[\cos(\lambda S_t)]$$
Does someone have an idea? I try to show that
$$E[\sin(\lambda S_t)]=0$$
By Ito's formula applied for $\sin(\lambda S_t)$, we get
$$d\sin(\lambda S_t)=\lambda\cos(\lambda S_t)dS_t-\frac{\lambda^2}{2}\sin(\lambda S_t)\,d\langle S,S\rangle_t$$
To calculate $d\langle S,S\rangle_t$, I am not sure if my calculs is correct or not:
$$d\langle S,S\rangle_t=(X_t^2+Y_t^2)\,dt$$
Since $S_t$ is a martingale we have
$$E[\sin(\lambda S_t)]=-\frac{\lambda^2}{2}E\left[\int_0^t(X_t^2+Y_t^2)\sin(\lambda S_t)\,dt\right]$$
I am not sur whether my previous calculs is correct. Could someone help me? Thanks a lot!
Since $(X,Y)$ and $(X,-Y)$ are identically distributed, the distribution of $S=(S_u)_{u\geqslant0}$ is odd. In particular, $\mathbb E(\varphi(S))=0$ for every odd integrable functional $\varphi$. For every fixed $t\geqslant0$, the functional $\varphi:(x_u)_{u\geqslant0}\mapsto\sin(\lambda x_t)$ is odd and bounded. The result follows.