Martingale Decomposition

Is it possible to decompose a discrete-time martingale $(M_n)$ uniquely into two processes $$M_n=M_n^I+A_n$$ where $(M^I_n)$ is a martingale with independent increments and $(A_n)$ is a martingale? If no, under which condition(s) on $(M_n)$, do we have this kind of decomposition?

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Yes. (Hint: Let $M_n^I$ be something very trivial; or, if you want something slightly less trivial, add and subtract.) –  cardinal Oct 10 '12 at 0:19
For the first (very trivial) part, yes, $M_n^I = 0$ is valid. For the second (only slightly less trivial) part, add and subtract a martingale $(S_n)$ with independent increments that is independent of $(M_n)$. –  cardinal Oct 10 '12 at 0:39
I edited my question. Actually what I mean is an unique decomposition similar to Doob decomposition. –  Martungale Martun Oct 10 '12 at 0:44
Then, the answer is obviously "no", by considering examples in the same spirit as the ones I gave. –  cardinal Oct 10 '12 at 0:46
thanks cardinal. I again edited the question. I would like to learn under which conditions it is possible to do. Any reference would also be helpful. –  Martungale Martun Oct 10 '12 at 0:55