I am trying to understand the proof of Levy's characterization: we define the process $Y$ by $Y_t = \exp(iuM_t+\frac{1}{2}u^2t)$ where $u\in\mathbb{R}$ arbitrary and $M_t$ a continous local martingale starting at $0$ with quadratic variation $t$.
Now it should hold by applying Ito's formula, that $Y_t = 1+iu\int_0^t Y_s dM_s$, which I don't understand. To apply Ito's formula we need a twice differentiable function, in our case $f(x,t)=\exp(iux+\frac{1}{2}u^2t)$. Since $f:\mathbb{R}^2\longrightarrow \mathbb{C}$ we use the generalized Ito formula, which says:
$$f(X_t) = f(X_0) + \sum_{i=1}^d \int_0^t \frac{d}{dx_i} f(X)dX_i+\frac{1}{2}\sum_{i,j=1}^d\int_0^t\frac{d^2}{dx_idx_j}f(X)d\langle X_i,X_j\rangle,$$
where $X$ is a $d$-dimensional vector of semimartingales (I have already shown that the formula holds if $f$ maps to $\mathbb{C}$). We then get
\begin{align}
Y_t &= 1+iu\int_0^t Y_s dM_s+\frac{1}{2}u^2\int_0^tY_sds + \frac{1}{2}\left(-u^2\int_0^t Y_sds + \frac{1}{4}u^4\int_0^tY_sds + iu^3\int_0^t Y_sd\langle M,t\rangle_s \right)\\
&= 1 + iu\int_0^t Y_sdM_s + \frac{1}{8}u^4\int_0^tY_sds + iu^3\int_0^t d\langle M,t\rangle_s
\end{align}
if everything is correct. From this point I dont't know how to go on. I tried to calculate the quadratic covariation $\langle M,t\rangle$ but wasn't able to or maybe there was a mistake in the previous equation? Any help is much appreciated.
Best Answer
We suppose $\langle M,M\rangle_t=t$ so we have $$df(M_t,t)=f_t(M_t,t)dt+f_x(M_t,t)dM_t+\frac{1}{2}f_{xx}(M_t,t)\underbrace{d\langle M,M\rangle_t}_{dt}$$ and $$\begin{aligned}&f(x,t)=e^{iu x+\frac{1}{2}u^2t}\\ &f_t(x,t)=u^2f(x,t)/2\\ &f_x(x,t)=iuf(x,t)\\ &f_{xx}(x,t)=-u^2f(x,t)\end{aligned}$$ so if $Y_t=f(X_t,t)$ we get $$dY_t=\frac{1}{2}u^2Y_tdt+iu Y_tdM_t+\frac{1}{2}(-u^2)Y_tdt=iuY_tdt\implies Y_t=1+iu\int_0^tY_sdM_s$$