Let $\{\mathcal{F}_n\}_n$ be a decreasing sequence of sub-$\sigma$-fields of $\mathcal{F}$($\mathcal{F}_{n+1}\subset\mathcal{F}_n$) and let $\{X_n\}_n$ be a backward submartingale($E[|X_n|]<\infty$, $X_n$ is $\mathcal{F}_n$ measurable and $E[X_n|\mathcal{F}_{n+1}]\geq X_{n+1}$).

If we know $l=\lim_{n\rightarrow\infty}E[X_n]>-\infty$, how to prove that $\{X_n\}_n$ is uniformly integrable? Many thanks!

I have an idea, it is enough to show:

  1. $\sup_nE[|X_n|]<\infty$;

  2. $\forall\epsilon>0$, $\exists\delta>0$ s.t. $\forall A\in\mathcal{F}$ such that $\mathbb{P}(A)<\delta$, we have $\sup_nE[1_{A}|X_n|]<\epsilon$.

For $|X_n|=2X_n^+-X_n$, we have

$$E[|X_n|]=2E[X_n^+]-E[X_n]\leq 2E[X_0^+]-l\leq 2E[|X_0|]-l$$

which implies $\sup_nE[|X_n|]<\infty$.

Secondly, $\forall A\in\mathcal{F}$, we can suppose without loss of generality that $A\in\mathcal{F}_0$.


Does someone have an idea for this term? Thanks a lot.....


I think that I find a proof:

As I have showed, by Jensen's inequality we prove that $\{X_n^+\}_n$ is a backward submartingale.

For any $K>0$, we have

$$KP(|X_n|\geq K)\leq E[|X_n|]=2E[X_n^+]-E[X_n]\leq 2E[X_1^+]-l<\infty$$

It follows that

$$\sup_nP(|X_n|\geq K)\rightarrow 0,\ \ K\rightarrow\infty$$


$$E[1_{\{X_n^+\geq K\}}X_n^+]\leq E[1_{\{X_n^+\geq K\}}E[X_1^+|\mathcal{F}_n]]=E[1_{\{X_n^+\geq K\}}X_1^+]\leq E[1_{\{|X_n|\geq K\}}X_1^+]\rightarrow 0,\ \ K\rightarrow\infty$$

by the integrability of $X_1^+$, which implies the uniform intergrability of $\{X_n^+\}_n$.


$$0\geq E[1_{\{X_n^+\leq -K\}}X_n]=E[X_n]-E[1_{\{X_n^+>-K\}}X_n]\geq E[X_n]-E[1_{\{X_n^+>-K\}}X_N]=E[X_n]-E[X_N]+E[1_{\{X_n^+\leq -K\}}X_N],\ \ \forall n\geq N$$

So for any given $\epsilon>0$, we have take $N$ large enough that $-\epsilon/2\leq E[X_n]-E[X_N]\leq 0$ and for this fixed $N$ we can choose $K$ such that

$$\sup_{n\geq N}E[1_{\{X_n^+\leq -K\}}|X_N|]\leq\epsilon/2$$

which implies the uniform intergrability of $\{X_n^-\}_n$, and we show $\{X_n\}_n$ is uniformly intergrable.


in the last step one can't choose $K$ large so that $\sup_{n \geq N} \mathbb{E}[1_{\{X_n^+ \leq -K\}|X_N|}] \leq \frac{\epsilon}{2}$

and that is not just because of a typo (most likely you meant $\sup_{n \geq N} \mathbb{E}[1_{\{X_n \leq -K\}|X_N|}] \leq \frac{\epsilon}{2}$ )

It is because the negative part of the submartingale can't be bound that way

maybe you should take a look at


  • 1
    $\begingroup$ The proof from the link has a problem. I don't see why in its decomposition, $M_n$ is a backward martingale. It is not even adapted to the filtration. $\endgroup$ – Iew Apr 20 '16 at 11:51
  • $\begingroup$ I agree with @Iew. I do not understand why the processes in the quoted notes is adapted. It looks like doing Doob decomposition in the wrong direction. $\endgroup$ – Kore-N Jul 31 '16 at 14:16
  • $\begingroup$ The proof @Higgs88 presents is actually working and taken from Chung's "A course in probability theory". You have a uniform bound: $\lambda \mathbb{P}(|X_n| \ge \lambda) \le \mathbb{E}[|X_n|] = 2\mathbb{E}[X_n^+] - \mathbb{E}[X_n] \le 2\mathbb{E}[X_{-1}^+] - l$ $\endgroup$ – Kore-N Jul 31 '16 at 18:51

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