Let $\{ X_n \}$ be a sequence of random variables such that $X_n = O_p(a_n)$, with $\{ a_n \}$ a sequence of positive real numbers. That is, for any $\varepsilon > 0$ there exists a finite $C_\varepsilon$ and an integer $n_\varepsilon$ such that $P(a_n^{-1} |X_n| > C_\varepsilon) < \varepsilon$ for all $n \ge n_\varepsilon$.
Under which conditions can we conclude that $\mathbb{E} (X_n) = O(a_n)$? Does it suffice to add the condition $\mathbb{E} (|X_n|) < \infty$ for all $n$?
I am aware that $X_n = O_p(a_n)$ does not even guarantee $\mathbb{E} (X_n)$ to be finite, but from a number of examples I have encountered, it was indeed the case that $\mathbb{E} (X_n) = O(a_n)$. It just feels to me as though there should be some general relationship here.
As a simple, concrete example, suppose that $X_1,X_2,\ldots$ are iid random variables, each with mean 0 and variance 1. Let $$ \bar{X}_n := \frac{1}{n} \sum_{i=1}^n X_i. $$ By the central limit theorem we have that $\sqrt{n} \bar{X}_n$ converges (in distribution) to the standard normal distribution. Hence, $$ \sqrt{n} \bar{X}_n = O_p(1), $$ which implies that $$ \bar{X}_n^2 = O_p(n^{-1}). $$ It is well-known that $$ \mathbb{E} (\bar{X}_n^2) = \frac{1}{n} = O(n^{-1}). $$ Is it pure coincidence that the expectation is of order $O(n^{-1})$ with the same rate $n^{-1}$ as the random quantity $\bar{X}_n^2$?