Hallo, does somebady know an easy proof of the following result? If $X$ and $Y$ are independent Lèvy processes then their covariation $[X,Y]$ is equal to zero. One can find such a result without proof in He, Wang and Yan, Semimartingale Theory and Stochastic Calculus, Theorem 11.43. I have a proof of it but I feel that it is too complicated. Thanks for help!! Regards, Paolo
The covariation is the unique continuous process of finite variation such that $XY [X,Y]$ is a martingale. It is therefore enough to verify that $XY$ is indeed a martingale w.r.t. the filtration $\mathscr F(X) \vee \mathscr F(Y)$ generated by $X$ and $Y$. But this follows from properties the conditional expectation: The independence of $\sigma(X_t) \vee \mathscr F(X)_s$ and $\sigma(Y_t) \vee \mathscr F(Y)_s$ yields $$ E(X_tY_t  \mathscr F(X)_s \vee \mathscr F(Y)_s) = E(X_t  \mathscr F(X)_s ) E(Y_t  \mathscr F(Y)_s )=X_s Y_s.$$ EDIT: This argument works for continuous independent martingales. 


but XY does not need to be integrable if X and Y are Lèvy processes. Moreover, if XY is a martingale the covariation is a martingale and not equal to zero. 

