|
The collection of all conditional expectations of an integrable random variable forms a uniformly integrable set. More generally, we have the following result.
Theorem Let $S$ be a uniformly integrable set of random variables defined on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ Then, the set \begin{equation*} \left\{\mathbb{E}[X\mid\mathcal{G}]:\textrm{$X\in S$ and $\mathcal{G}$ is a sub-$\sigma$-algebra of $\mathcal{F}$}\right\} \end{equation*}is also uniformly integrable.
To prove the result, we first use the fact that uniform integrability implies that $S$ is $L^1$ bounded. That is, there is a constant $L>0$ such that $\mathbb{E}[|X|]\le L$ for every $X\in S$ Also, choosing any $\epsilon>0$ there is a $\delta>0$ so that \begin{equation*} \mathbb{E}[|X|1_A]<\epsilon \end{equation*}for all $X\in S$ and $A\in\mathcal{F}$ with $\mathbb{P}(A)\le\delta$
Set $K=L/\delta$ Then, if $Y=\mathbb{E}[X\mid\mathcal{G}]$ for any $X\in S$ and $\mathcal{G}\subseteq\mathcal{F}$ Jensen's inequality gives \begin{equation*} |Y|\le\mathbb{E}[|X|\mid\mathcal{G}]. \end{equation*}So, applying Markov's inequality, \begin{equation*} \mathbb{P}(|Y|>K)\le K^{-1}\mathbb{E}[|Y|]\le K^{-1}\mathbb{E}[|X|]\le L/K=\delta \end{equation*}and, therefore \begin{equation*} \mathbb{E}[|Y|1_{\{|Y|>K\}}]\le \mathbb{E}[|X|1_{\{|Y|>K\}}]<\epsilon. \end{equation*}
|