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I've come across a statement that implies that non-negative martingales for which $\{M_{\tau}\mid \tau \ \rm{stopping} \ \rm{time}\}$ is not uniformly integrable exist. I personally can't think of an example though.

I've considered gambling strategies and Brownian Motions but none seem to work.

Is there anyone who can think of an example and help me understand the concept a bit more?

Been looking into this for a while now so help is much appreciated.

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up vote 7 down vote accepted

Let $\xi_j$ be i.i.d. with $(\forall j \in \mathbb{N}) \mbox{} \mathbb{P}(\xi_j=0)=\mathbb{P}(\xi_j=2)=1/2$, and define $M_n=\prod_{j=1}^n \xi_j$ for $n\geq 0$. Then $(M_n)$ is a non-negative martingale with $\mathbb{E}(M_n)=1$ for all $n$. But $M_n\to 0$ almost surely, so $(M_n)$ cannot be uniformly integrable.

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I've considered this before but I came to the conclusion that the stopped proces $\{M_{\tau}\mid \tau \ \rm{stopping} \ \rm{time}\}$ was UI. – user70267 Apr 6 '13 at 7:16
@user70267 Already the sequence $(M_n)_n$ is not UI, and the family $(M_\tau)_\tau$ contains it, hence... – Did Apr 6 '13 at 10:46
@CSA Please refrain from petty edits. – Did Apr 12 '15 at 16:07

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