Hallo, does somebady know an easy proof of the following result? If $X$ and $Y$ are independent Lèvy processes then their co-variation $[X,Y]$ 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
2 Answers
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The co-variation 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.
answered Sep 11, 2012 at 14:53
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but XY does not need to be integrable if X and Y are Lèvy processes. Moreover, if XY is a martingale the co-variation is a martingale and not equal to zero.
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$\begingroup$ By $[X,Y]$ I mean the process defined by [ [X,Y]_t:=\langle X^c,Y^c\rangle_t+\sum_{0<s\leq t}\Delta X_s\Delta Y_s. ] Sorry that I did not specify it before. $\endgroup$– PaoloCommented Sep 11, 2012 at 15:16