Martingale and uniform integrability

**akbar** · January 9th 2010, 02:51 PM

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.

**Laurent** · January 10th 2010, 03:08 AM

Originally Posted by akbar

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.

**akbar** · January 11th 2010, 03:43 PM

Thanks in particular for the lectures notes and Theorem 12.5.4. This is a precious result.

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