$L^p$-convergence of submartingale

Question

Let $p\geq1.$ Consider a $\mathcal{F}_k$-submartingale $(X_k)_k$ in $L^p.$ We can prove easily that $(X_k)_k$ converges in $L^p$ if and only if $(|X_k|^p)_k$ is uniformly integrable.

If $(X_k)_k$ was a martingale then for $p>1,L^p$-convergence is equivalent to $\sup_kE[|X_k|^p]<\infty,$ and if $X_k \geq 0,$ for all $k,$ then this is true if and only if there exists $X_{\infty} \in L^p$ such that for all $k, E[X_{\infty}|\mathcal{F}_k] \geq X_k$ a.s.

In the general case, are there conditions that are equivalent to $L^p$-convergence of any submartingale in $L^p,p>1$ ?

To be noted that $\sup_kE[|X_k|^p]<\infty$ doesn't give $L^p$-convergence: Submartingales bounded in $L^p$, $p>1$

Remark: The following is a necessary and sufficient condition for $L^1$-convergence:

$(X_k)_k$ converges in $L^1$ if and only if there exists $X \in L^1_{\mathbb{R}_+}$ such that for every $k,|X_k| \leq E[X|\mathcal{F}_k].$

We will prove $\implies$ (the converse is true since $(E[X|\mathcal{F}_k])_k$ is uniformly integrable).

If we have $L^1$-convergence then there exists $X_{\infty} \in L^1$ such that for every $k,E[X_{\infty}|\mathcal{F}_k] \geq X_k.$

Considering the Doob decomposition: $X_k=Y_k+W_k$ where $Y_k$ is a martingale and $W_k$ is positive and increasing. It follows from $L^1$-convergence that $\sup_{k}E[W_k]<\infty$ so $Y_k$ converges in $L^1$ so there exists $Y_{\infty} \in L^1$ such that $Y_k=E[Y_{\infty}|\mathcal{F}_k]$ and hence for every $k,X_k \geq E[Y_{\infty}|\mathcal{F}_k].$

So for every $k,|X_k| \leq E[X|\mathcal{F}_k],$ where $X=|X_\infty|+|Y_\infty|.$

Answer 1

Here is one alternative set of conditions (sufficient but I am not sure if they are necessary). Essentially, I move the $L^p$-uniform integrability requirement of $X_k$ to the predictable component $A_k$ given by Doob's decomposition theorem.

If $X_k$ is submartingale then, by Doob's decompositon theorem, we have $X_k = M_k + A_k$ where $M_k$ is martingale and $A_k$ is a predictable nondecreasing process.

If we assume $\sup_k E|M_k|^p < \infty$ then as you mentioned we would have $M_k \rightarrow M_{\infty}$ in $L^p$ for some limiting variable $M_{\infty}$. So now we need to make an assumption to ensure that $A_k$ converges in $L^p$. Establishing almost sure convergence of $A_k$ isn't difficult (e.g. exploit nondecreasingness). To strengthen the almost sure convergence of $A_k$ to $L^p$, we assume $A_k$ is uniformly integrable in $L^p$. This also implies that $\sup_k E|X_k|^p\lessapprox \sup_k E|M_k + A_k|^p \lessapprox \sup_k E|M_k|^p + \sup_kE|A_k| < \infty$, so as mentioned in your reference, we have $X_k \rightarrow X_{\infty}$ almost surely. But, this implies $A_k = X_k - M_k \rightarrow X_{\infty} - M_{\infty}$ almost surely, so that this condition also suffices to get almost sure convergence of $A_k$ (which we needed).

So one set of conditions, would be

1. $\sup_k E|M_k|^p < \infty$
2. $A_k$ is uniformly integrable in $L^p$

Condition 1 is analogous to what is needed when $X_k$ is martingale. Condition 2 requires the predictable part of $X_k$, $A_k$, to be uniformly integrable. As argued above, these conditions imply that $\sup_k E|X_k|^p < \infty$, but this condition itself is not sufficient. Also, noting that condition 1 is equivalent to uniform integrability (U.I.) of $M_k$ and sums of U.I. sequences are U.I., these conditions actually imply that $X_k$ is U.I.

I imagine in practice that the predictable component $A_k$ is simpler than $X_k$ or $M_k$, so verifying its uniform integrability may be significantly easier in some cases than establishing the uniform integrability of $X_k$.

Update

Actually, we can get nicer conditions. Note that $A_n$ is almost surely nondecreasing and therefore $A_n$ almost surely converges to a possibly unbounded/infinite random variable $A_{\infty}$. By monotone convergence theorem (which doesn't assume anything about $A_{\infty}$), we have $$\sup_n E|A_n|^p = lim_{n\rightarrow \infty} E|A_n|^p = E|A_\infty|^p.$$ Thus, if we assume $\sup_n E|A_n|^p < \infty$, we get $E|A_\infty|^p < \infty$, which implies $A_n$ is UI in $L^p$. Thus, the earlier conditions are equivalent to

1. $\sup_k E|M_k|^p < \infty$

2. $\sup_k E|A_k|^p < \infty$

To make it cleaner, the previous conditions are also equivalent to

1. $\sup_k \{E|M_k|^p + E|A_k|^p\} < \infty.$

Thus, the failure of the condition $\sup_k E|X_k|^p < \infty$ to give convergence of $X_k$ is due to the gap $ 0 \leq \sup_k \{\|M_k\|_{L^p} + \|A_k\|_{L^p} - \|X_k\|_{L^p} \}$ being possibly infinite.