If $X_n = O_P(a_n)$ and are uniform integrability, can we get a convergence rate for $E(|X_n|)$?
By Markov inequality we get $E(|X_n|) = O (a_n) \implies X_n = O_P(a_n)$.
By Vitally Theorem we get that $E|X_n| \to 0$ if and only if $X_n \overset{P}{\to} 0$ and $\{X_n\}$ are uniform integrability.
This question is close: Rate of convergence when going from convergence in probability to convergence in the mean
But it assumes $ |X_n| \leq M$ with probability one. Which is the "best" possible uniform integrability condition I can think of.
The answer is NO.
Consider $a_n$, $b_n$ positive sequences such that $a_n \to 0$, $b_n \to 0$. Let $U \sim \mathcal{U}[0,1]$ and
$$X_n = I_{[0, b_n]}(U)$$
It follows that $E(|X_n|) = b_n$ and they are uniformly integrable because they converge to $0$ in $L_1$ sense but the are also $ O_P(a_n)$.
Given $\varepsilon > 0$ for $n_0$ such that $b_n \leq \varepsilon \, \forall n \geq n_0$ we get that $P( |X_n| > a_n ) \leq P( U \in [0, b_n]) = b_n \leq \varepsilon$ if $n \geq n_0$.