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Let $G_{t}$ be a filtration and $M_{t}$ a $G_{t}$-martingale. Why do we have this decomposition: $H_{t}=\mathbb{E}[H|G_t]=\int_{0}^{t}h_{s}dM_{s}+R_{t}$ where $R_{t}$ is a martingale orthogonal with M

Thank you.

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  • $\begingroup$ Is the expectation without the conditioning on $G_t$? $\endgroup$
    – Richi Wa
    May 5, 2015 at 9:05
  • $\begingroup$ God sorry for that . Yes it is $H_{t}=\mathbb{E}[H|G_{t}]$ $\endgroup$
    – glork
    May 5, 2015 at 9:25
  • $\begingroup$ You may need some more details, such as what is $h$ etc. $\endgroup$
    – Gordon
    May 5, 2015 at 13:23
  • $\begingroup$ To be helpful, you may provide a reference where you have this decomposition. $\endgroup$
    – Gordon
    May 5, 2015 at 13:38
  • $\begingroup$ in fact it's not precised what is $h$ ...it's just written: Let $H_{t}=\mathbb{E}[H|G_{t}]=\int_{0}^{t}h_{s}dM_{s}+R_{t}$ with R martingale orthogonal to M. I don't know how to upload a pdf file where you could see the context in which it's written $\endgroup$
    – glork
    May 5, 2015 at 16:57

1 Answer 1

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As per your comments, this is the Kunita Watanabe decomposition. See the post at https://math.stackexchange.com/questions/413103/kunita-watanabe-decomposition and the presentation http://www.eurandom.nl/events/workshops/2011/ISI_MRM/Presentation/Vanmaele.pdf

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  • $\begingroup$ thank you very much for the link you provided. I can now put a name on this decomposition ^^ $\endgroup$
    – glork
    May 5, 2015 at 17:55

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