What is an upper bound for $|E(X|\mathcal{A})-E(X)|$?

Question

Let $X$ be a random variable with $|X|\le1$, and $\mathcal{A}$ be a $\sigma$-algebra. What is an upper bound for $|E(X|\mathcal{A})-E(X)|$?

Existing results:

It has been known that $E|E(X|\mathcal{A})-E(X)|\le 2\alpha(\sigma(X),\mathcal{A})$, where $\alpha(\sigma(X),\mathcal{A})$ is the $\alpha$ mixing coefficient between $\sigma(X),\mathcal{A}$, and the $\sigma(X)$ is the $\sigma$-algebra generated by $X$. This result shows that $|E(X|\mathcal{A})-E(X)|=O_P(\alpha(\sigma(X),\mathcal{A}))$.

In practice, the above upper bound is too loose, and we need an exponential type upper bound to strengthen it. Suppose we can prove the following $$ E\psi(|E(X|\mathcal{A})-E(X)|/C)\le\alpha(\sigma(X),\mathcal{A}), $$ where $\psi(x)=\exp(x^2)-1$. Then one can argue that
$$ |E(X|\mathcal{A})-E(X)|=O_P(\sqrt{\log(1+\alpha(\sigma(X),\mathcal{A})}). $$ Clearly, this largely improve the first upper bound to a sharper level.

My question is how to prove this?

Answer 1

I guess $\alpha$-mixing condition might not be sufficient. Instead, we may need $\phi$-mixing conditions.

Without loss of generality, assume $1\ge X\ge 0$ almost surely.

Define $\phi$-mixing coefficients:

$\phi(\mathcal{A},\sigma(X))=\sup_{A\in\mathcal{A},B\in\sigma(X)\textrm{ with} P(A)>0}|P(B|A)-P(B)|$.

For any $A\in\mathcal{A}$ with $P(A)>0$, $$ |E(X|A)-E(X)|=|\int_0^1 [P(X>x|A)-P(X>x)]dx|\le\phi(\mathcal{A},\sigma(X)). $$

Since the above is true for arbitrary $A\in\mathcal{A}$ with $P(A)>0$,it holds that \begin{equation}\label{eqn:0} |E(X|\mathcal{A})-E(X)|\le\phi(\mathcal{A},\sigma(X)).\tag{$***$} \end{equation} Since $\psi(x)=o(x^2)$ for $x$ within a neighborhood of $0$, $$ E\psi(|E(X|\mathcal{A})-E(X)|)=o(\phi(\mathcal{A},\sigma(X))^2). $$

The above derivation relies on argument (***), which I am not sure if is correct.