This morning there is another “simple” proof of Bichteler-Dellacherie Theorem on arxiv here :

http://arxiv.org/PS_cache/arxiv/pdf/1201/1201.1996v1.pdf

Best Regards

]]>Best regards ]]>

By the way I think there a tiny typo in the paragraph preceding theorem 2.

You wrote :

“Therefore, a process is said to be X-integrable if the exists any such decomposition with respect to which it is both M-integrable and V-integrable as just described.”

Did you mean ?

“Therefore, a process is said to be X-integrable if for any such decomposition, it is both M-integrable and V-integrable as just described.”

Best regards

]]>Then, *Y* is a semimartingale with jumps bounded by 1. So it is locally bounded (so, locally integrable) and we can apply the unique decomposition *Y* = *M* + *A* where *M* is a local martingale and *A* is a predictable FV process starting from 0. Also, *A* will have jumps bounded by 1 and both *M* and *A* are locally bounded. Writing *X* = *M* + (*A* + Σ_{s≤t}1_{{|ΔXs| ≤ 1}}Δ*X _{s}*) splits

In the case where *X* is a Lévy process then these components of *X* are all independent Lévy processes. Furthermore, *A _{t}* =

These kinds of manipulations and decompositions are a bit simpler with even more of the general theory, which I will expand on a bit in upcoming posts.

]]>I have a question about whcih makes a link between theorem 2 of this post and your post about “Failure of Pathwise Integration for FV Processes”.

Let’s take and as any of the examples defined in the “Failure of Pathwise Integration for FV Processes” post so that , but nonetheless -integrable.

As is a FV process, it is a semimartingale, and as is -Integrable then from theorem 2, this implies that there exists :

– a local(ly bounded) martingale s.t. is locally integrable

– a FV process s.t. almost surely.

Moreover we have : with the first term in the righthand equality being a locally bounded martingale and the second term a FV process.

So here is the question, do you think it is possible to explicitely express and in those counterexamples ?

Second is there a link with Lévy-Itô decomposition of Lévy processes (if they are of infinite activity type) ?

Best regards

]]>Your interpretation of boundedness in probability is correct (except that the should be after the ‘s.t.’, but I assume that’s just a typo) and, yes, this is equivalent to tightness of the measures. I was thinking about writing a post on convergence and boundedness in probability (which I might still do when I get round to it). Actually, if you understand the notion of convergence in probability, then you also get boundedness in probability for free. The space of real-valued random variables, under convergence in probability is a topological vector space (TVS). That is, convergence in probability is a vector topology, so addition and multiplication by scalars is continuous. Then, there is the notion of bounded sets in any TVS. A subset *S* of a TVS is bounded if it is *absorbed* by any neighbourhood *N* of the origin. This means that for some real . In the case of you can check that this coincides with the definition you just correctly gave.

You also get some other simple facts which are true in general TVS’s. A set *S* is bounded if and only if for any sequence and real then . Also a linear map between a normed space *V* and a TVS *W* is continuous if and only if it is bounded (i.e., the image of *T* on the unit ball is bounded). In this case you can let *V* be the set of elementary predictable processes under uniform convergence, *W* be and *T* be the stochastic integral up to a fixed time. So, in the statement of the Bichteler-Dellacherie theorem, you can equivalently say that the integral is continuous (under the respective topologies). I tend to prefer the statement in terms of boundedness in probability rather than continuity, but that’s just a matter of taste.

Also, for your second statement. Yes, it is true that you get boundedness in probability for the more general case of bounded (by 1) predictable integrands. If you look at the proof in the post I linked, it works just as well if you replace “elementary predictable” by “predictable”. It was just in the statement that I only mentioned elementary integrands, since that was the main point at the time. I think I’ll go back and edit that post with the more general statement.

Thanks for your questions, I’m sure there’ll be more when you get further through the post.

Btw, I edited the latex in your post (Using LaTeX in comments).

Regards

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