A Closed Supermartingale might not be uniformly integrable?

Question

A supermartingale $\{X_n,\mathcal{F}_n\}$ is closed, if there exists $Z\in L^1$ such that $X_n\geq E[Z|\mathcal{F}_n]$ for all $n$.

Is there any counterexample that a closed supermartingale is not uniformly integrable?

Thanks!

Answer 1

Suppose that $\{Y_n, n\in\mathbb{N}\}$ is a sequence of i.i.d. random variables and that $Y_1$ is $N(0,1)$ distributed, then $\mathsf{E}[e^{Y_n-1/2}]=1$. Let $$X_n=\prod_{j=1}^ne^{Y_j-1/2}=\exp\Bigl[\sum_{j=1}^nY_j-\frac n2\Bigr],\qquad \mathscr{F}_n=\sigma\{Y_j,1\le j\le n\}. $$ Then $$\mathsf{E}[X_{n+1}\mid\mathscr{F}_n]=X_n\mathsf{E}[e^{Y_{n+1}-1/2}\mid \mathscr{F}_n]=X_n,\qquad \mathsf{E}[X_n]=1. $$ Hence $\{X_n,\mathscr{F}_n\}$ is a martingale and also a supermartingale too. Meanwhile, $X_n\ge 0=\mathsf{E}[0\mid \mathscr{F}_n]$, so $\{X_n,\mathscr{F}_n\}$ is a closed supermartingale(but not a closed martingale).

By the strong law of large numbers we get: \begin{gather} \lim_{n\to\infty}\biggl[\sum_{j=1}^nY_j-\frac n2\biggr]=-\infty, \qquad \text{a.s.}\\ \lim_{n\to\infty}X_n=0\qquad \text{a.s.} \end{gather} Hence $$\mathsf{E}[\lim_{n\to\infty}X_n]=0<1=\lim_{n\to\infty}\mathsf{E}[X_n]. $$ and $\{X_n\}$ is not uniformly integrable.