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Math Help - Martingale and uniform integrability

  1. #1
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    Martingale and uniform integrability

    Let (X_n,\mathcal{F}_n) be a uniformly integrable martingale, n\in \mathbb{N}. Let M>0 and \tau(\omega)=\min(n: |X_n(\omega)|\geq M) ( \tau(\omega)=\infty if |X_n(\omega)|<M for all n). One can check \tau is a stopping time of \mathcal{F}_n.

    How do you prove that (X_{n\wedge \tau},\mathcal{F}_{n\wedge \tau}) is also a uniformly integrable martingale?

    The martingale part of the problem comes immediately from: \mathbb{E}(X_{p\wedge \tau}-X_{q\wedge \tau}|\mathcal{F}_q)=\mathbb{E}(X_{(p\wedge \tau)\vee q}-X_{(q\wedge \tau)\vee q}|\mathcal{F}_q)=\mathbb{E}(X_{(p\wedge \tau)\vee q}|\mathcal{F}_q)-X_q= 0, when q\leq p, using the Optional Sampling Theorem.

    But how about the uniform integrability? For a > M and all n, one can write \{|X_{n\wedge \tau}|\geq a\}=\{|X_{\tau}|\geq a\}\cap\{\tau \leq n\}, but how do you get a bound for the integral remainder \mathbb{E}(|X_{n\wedge \tau}|\bold{1}_{|X_{n\wedge \tau}|\geq a}) which is independent of n?

    Thanks for your help.
    Last edited by akbar; January 10th 2010 at 03:44 PM.
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  2. #2
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    Quote Originally Posted by akbar View Post
    Let (X_n,\mathcal{F}_n) be a uniformly integrable martingale, n\in \mathbb{N}. Let M>0 and \tau(\omega)=\min(n: |X_n(\omega)|\geq M) ( \tau(\omega)=\infty if |X_n(\omega)|<M for all n). One can check \tau is a stopping time of \mathcal{F}_n.

    How do you prove that (X_{n\wedge \tau},\mathcal{F}_{n\wedge \tau}) is also a uniformly integrable martingale?
    You agree it suffices to prove E[|X_\tau|{\bf 1}_{\{\tau<\infty\}}]<\infty? (since |X_{n\wedge\tau}|\leq|X_\tau|{\bf 1}_{\{\tau<\infty\}}+M{\bf 1}_{\{\tau=\infty\}} given the definition of \tau)

    We have E[|X_\tau|{\bf 1}_{\{\tau<\infty\}}]=\sum_{n=0}^\infty E[|X_n|{\bf 1}_{\{\tau=n\}}], and X_n=E[X_{n+k}|\mathcal{F}_n] for all k\geq0, and in fact we even have X_n=E[X_\infty|\mathcal{F}_n] where X_\infty is the almost-sure (and L1) limit of (X_n)_n using the classical theorem about uniformly integrable martingales. Thus |X_n|\leq E[|X_\infty||\mathcal{F}_n], hence:

    E[|X_\tau|{\bf 1}_{\{\tau<\infty\}}]\leq\sum_{n=0}^\infty E[E[|X_\infty||\mathcal{F}_n]{\bf 1}_{\{\tau=n\}}] =\sum_{n=0}^\infty E[|X_\infty|{\bf 1}_{\{\tau=n\}}] (using the stopping time definition) and thus finally:


    E[|X_\tau|{\bf 1}_{\{\tau<\infty\}}]\leq E[|X_\infty|]<\infty.

    In fact, for any stopping time \tau, the stopped martingale is still uniformly integrable. You can deduce a proof from Theorem 12.5.4 (remark (i)) in these lecture notes.
    Last edited by Laurent; January 10th 2010 at 03:26 AM. Reason: Last paragraph
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  3. #3
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    Thanks in particular for the lectures notes and Theorem 12.5.4. This is a precious result.
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