Uniform Integrability and Conditional Expectations

Question

Let $X=\{X_n:n\in \mathbb{N_0}\}$ be a supermartingale or a submartingale on a filtered space $(\Omega,\mathcal{F},\{\mathcal{F}_{n}\},\mathbf{P}).$ Suppose the sequence $\{X_n\}_{n\in\mathbb{N_0}}$ is uniformly integrable $\left( \text{i.e.} \lim \limits_{\lambda \rightarrow \infty} \sup _{n \in \mathbb{N}_0} E\left(\left|X_{n}\right| \mathbf{1}_{\left\{\left|X_{n}\right|>\lambda\right\}}\right)=0\right)$,then we have $\sup _{n \in \mathbb{N}_0} E\left(\left|X_{n}\right|\right)<\infty$.By Doob's Martingale Convergence Theorem (discrete case),there exits a random variable $X_{\infty}$ on $(\Omega,\mathcal{F},\mathbf{P})$,such that $\lim \limits_{n \rightarrow \infty} X_{n}=X_{\infty}\text{ a.s.}$.Since uniformly integrability of $\{X_n\}_{n\in\mathbb{N_0}}$,then $\lim \limits_{n \rightarrow \infty}\left\|X_{n}-X_{\infty}\right\|_{1}=0$, that is, the sequence $\{X_n\}_{n\in\mathbb{N_0}}$ converges to $X_{\infty}$ in the $L^{1}$ norm.

For any fixed $k\in\mathbb{N_0},$can we have $$ \boxed{\mathbf{E}(X_\infty\big|\mathcal{F_{k}})=\lim \limits_{n \rightarrow \infty}\mathbf{E}(X_n\big|\mathcal{F_{k}})\text{ a.s.}\quad?} $$

I doubt the validity of this equation.Surely,$X_n \to X_{\infty}$ in $L^1$ implies $\mathbf{E}(X_n\big|\mathcal{F_{k}})\to\mathbf{E}(X_{\infty}|\mathcal{F_{k}})$ in $L^1$.But from this we only know that there exists a subsequence $\left\{\mathbf{E}(X_{n_{\ell}}\big|\mathcal{F_{k}}): \ell \in \mathbb{N}_{0}\right\}$ such that $$\mathbf{E}(X_\infty\big|\mathcal{F_{k}})=\lim \limits_{\ell \rightarrow \infty}\mathbf{E}(X_{n_{\ell}}\big|\mathcal{F_{k}})\text{ a.s.}\quad$$

I feel the condition $\lim \limits_{n \rightarrow \infty} X_{n}=X_{\infty}\text{ a.s.}$ doesn't contribute to deducing $$\mathbf{E}(X_\infty\big|\mathcal{F_{k}})=\lim \limits_{n \rightarrow \infty}\mathbf{E}(X_n\big|\mathcal{F_{k}})\text{ a.s.}$$

How to construct a counterexample to show its invalidity, or prove its correctness ?

Answer 1

The sequence $\left(\mathbb E\left[X_n\mid\mathcal F_k\right]\right)_{n\geqslant k}$ is monotonic (non-increasing in the case of a supermartingale, non-decreasing in the case of a submartingale). Moreover, $\left(\mathbb E\left[X_n\mid\mathcal F_k\right]\right)_{n\geqslant k}$ converges in probability to $\mathbb E\left[X_\infty\mid\mathcal F_k\right]$ hence the convergence takes place in the almost sure sense.

Answer 2

Hint: It is easy to check that $(E(X_n|\mathcal F_k))$ is a sub/super-martingale.

$P(E(X_n|\mathcal F_k)|>\Delta)\le \frac 1 {\Delta} E|X_n|$. Use this to show that $(E(X_n|\mathcal F_k)) $ is uniformly integrable (which gives a.s. convergence). [Recall that family $(X_i)$ is uniformly integrable if and only if

1. $\sup E|X_i| <\infty$ and

2. For any $\epsilon >0$ there exists $\delta>0$ such that $\int_{A} |X_i|dP<\epsilon$ for all $i$ whenever $P(A) <\delta$].