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?
Martingale Decomposition
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probability-theory
stochastic-processes
stochastic-analysis
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0Yes. (Hint: Let $M_n^I$ be something very trivial; or, if you want something slightly less trivial, add and subtract.) – 2012-10-10
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0For 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 $(M_n)$. – 2012-10-10
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0I edited my question. Actually what I mean is an unique decomposition similar to Doob decomposition. – 2012-10-10
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0Then, the answer is obviously "no", by considering examples in the same spirit as the ones I gave. – 2012-10-10
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0thanks 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. – 2012-10-10
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0I should be clear that in the examples of adding and subtracting another nontrivial martingale, it may require augmenting the filtration to which the process is adapted. :-) – 2012-10-10