Convergence of (sub)martingales in L$^1$

Question

By the convergence theorem we know that given a L$^1$-bounded submartingale $X_t$, $t\geq1$, with $\sup_{t\geq0}\mathbb{E}[X_t^+]<\infty$, the sequence $X_t(\omega)$ converges $\mathbb{P}$-a.s. to some $X(\omega)$. This follows as an application of the upcrossing inequality. However, this theorem doesn't prove convergence in L$^1$. In fact, if we consider a martingale $M_t=(\frac{1-p}{p})^{S_t}$, where $S_t$ is an asymmetric random walk with $\mathbb{E}[X_i=1]=p, \mathbb{E}[X_i=-1]=1-p$, then $M_t\rightarrow0, \mathbb{P}$-a.s., but $\|M_t\|_1=\mathbb{E}[M_t]=\mathbb{E}[M_0]=1$, which contradicts L$^1$ convergence to $0.$

My question is, what are the assumptions required in order to have convergence in $L^1$ of $L^1$-bounded submartingales?

Answer 1

You have that the following are equivalent for a martingale (EDIT:I read the question wrong):

$1)$ Convergence in $L^1$

$2)$ Uniform integrability

$3)$ The martingale being closed, i.e. $\exists Z $ s.t. $M_t=\mathbb{E}[Z|\mathcal{F_t}]$

provided you have right continuous sample paths.

Answer 2

The most complete answer to this question in discrete time can be found in Chung, A course in Probability, 2nd Edition (2001). The following Theorem 9.4.5 gives necessary and sufficient conditions for $L^1$-convergence.

Theorem Let $(X_n)_{n \ge 1}$ be a sub-martignale. Then the following are equivalent:

1. $(X_n)_{n \ge 1}$ are uniformly integrable;
2. $(X_n)_{n \ge 1}$ converges in $L^1$ (i.e.\ is a Cauchy-sequence);
3. $X_n \to X_\infty$ almost surely with $X_\infty \in L^1$, $E[X_n] \to E[X_\infty]$ and $(X_n)_{n \in \mathbb N \cup \{ \infty\}}$ is a sub-martingale.

For a martingale, the case simplifies, 3. can be replaced by

3a. $X_n \to X_\infty$ almost surely with $X_\infty \in L^1$, and $(X_n)_{n \in \mathbb N \cup \{ \infty\}}$ is a martingale.