I have a question in my homework:

Let $X_t$ and $Y_t$ be two Brownian motions issue de $0$ and define


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!

  • $\begingroup$ Try to eliminate the remnants of French in your post. (And, when handing back your homework, do not forget to mention the help you received from MSE.) $\endgroup$ – Did Dec 10 '12 at 9:06

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.


$S_t = -S_t$ in distribution, so $E[e^{i\lambda S_t}] = E[e^{-i\lambda S_t}]$. Notice $E[e^{i\lambda S_t}] = E[\cos(\lambda S_t)] + i E[\sin (\lambda S_t)]$. Finally, $2E[e^{i\lambda S_t}] = E[e^{i\lambda S_t}] + E[e^{-i\lambda S_t}] = 2 E[\cos(\lambda S_t)]$.


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