It is clear that every (I am particularly interested in continuous) semimartingale has a well defined quadratic variation process. However, what can be said about processes that have a well defined quadratic variation process? Is there a statement that says what exactly is needed in addition to existence of the quadratic variation in order for a process to be a semimartingale?
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@ mathquestion88 : Well you say that one way is trivial but it is very dependent on the way you define a semimartingale, so you should add the definition you have in your question. Best regards – TheBridge Mar 02 '15 at 22:26
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@TheBridge Is it dependent? Doesn't every semimartingale (:= good integrator OR sum of local martingale and certain locally bounded variation term (both definitions are equivalent)) have bounded quadratic variation? – mathsquestion88 Mar 03 '15 at 17:17
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Well take a compound process (a Lévy process hence a semimartingale) let's say with independent normal laws for the jumps, this process hasn't bounded quadratic variations. The quadratic variations of this process is equal to the sum of the square of the size of the jumps, but none of those jumps is bounded. Best regards – TheBridge Mar 03 '15 at 21:42
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Ooops... I just now notice the misleading formulation of the problem in my first post: of course I mean finite quadratic variation (or the quadratic variation exists) as opposed to bounded. How silly. I am sorry. – mathsquestion88 Mar 03 '15 at 22:44