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Thread: Martingale and uniform integrability

  1. #1
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    Martingale and uniform integrability

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

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

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

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

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

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

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

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


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

    In fact, for any stopping time $\displaystyle \tau$, 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; Jan 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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