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why does one define the Ito integral as the $L^2$ limit, although it can be shown by Doob's martingale inequality and Borel-Cantelli lemma that there exists a t continuous version, which is constructed as almost sure limit. So why not define the Ito integral as the continuous almost sure limit?

Thank you in advance

Nate Eldredge
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alL
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1 Answers1

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The key advantage of Ito integrals is that they are martingales. For this, a.s. convergence would not be enough in general. You don't quite need $\mathcal{L^2}$ convergence, but something almost as strong implying some boundedness. For a reference and good discussion, maybe compare Øksendal; also this question at MathOverflow.

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gnometorule
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  • Thank you for your answer. After the proof, that there exists a continous version of the ito-integral, which is the almost sure limit, Oksendal mentioned that in the further he always considers that continous version. So why cant someone define the ito-integral from the beginning as this continous almost sure limit. As a limit of continous Martingales this almost sure limit is a martingale itself, is it? – alL Feb 26 '13 at 16:38
  • @alL: Not necessarily. If you go through the proofs, you see where he uses the stronger conditions. This, in itself, obviously doesn't show that you couldn't do as you suggest. However, it just won't work in general. To fully use the strength of Ito calculus, a.s. wouldn't be enough. – gnometorule Feb 26 '13 at 17:13
  • I am trying to understand it step by step, I think I miss the hole picture. But I am right, that for example Oksendal uses after the proof that there exists a continous Version of the ito integral, which is constructed as almost sure limit, only uses that one in the further of the book? The question is so remaining: Why not define it from the start like that. I get the feeling that I am overlooking something very fundamental. – alL Feb 27 '13 at 06:54
  • Reformulation of my question: In a manuskript I a use for learning at the moment the ito-integral is defined as the L^2 Limit and after that is stated that the process is continous in t. Is that true? I dont think so. Moreover because the unification of overcountable nullsets must not neccasseraly be a nullset, there even is no correct definition of a ito-integral process as long as we dont take the t continous version. But if we do so, I ask myself again wy we have defined the ito-integral as the L^2 limit at first. Thank you so much for any help! – alL Feb 27 '13 at 15:42
  • @alL: Arguing from Øksendal's book, as I don't have your lecture notes, go to page 35 which is in the preview linked. There is a difference between the limit being $t$-continuous, or the limit having a $t$-continuous version, as Øksendal states. If you use existence of a limit in a sense differently from how he does this (and how Ito-Integration is introduced in any source I've read), you could not guaranty that the Ito-Isometry and such holds for such limit. Hence, you go the $\mathcal{L^2}$ route, and pick an appropriate version of the limit. I hope this was of some help. – gnometorule Feb 27 '13 at 18:51
  • Thank you for your help! Would you mind to go to page 33. There is shown that there exists a t-continous version of the ito-integral process (L^2 limit). This picked version is constructed as almost sure limit. Is the following right?: Do one define the Ito-Integral as L^2 limit because it's then possible to show that the picked version (page 33, almost sure limit) is a martingale? Thank you in advance! – alL Feb 27 '13 at 19:36
  • @alL: If I understand you correctly, that is my understanding. As a disclaimer, please do keep in mind that, while something like this was actually my thesis topic, my thesis was over ten years ago, and except for casual reading, I haven't used it much since. But you're on the right track there, alL! – gnometorule Feb 27 '13 at 20:27
  • Thank you for your help. I will examine some proofs deeper to get the hole picture if its neccessary. – alL Feb 27 '13 at 20:42