Let $X_1, X_2$,... be a sequence of RV, and $a_1, a_2$,... be a sequence of non-negative and non-stochastic numbers. Show if $E(|X_n|) = O(a_n)$ then $X_n = O_p(a_n)$.
My attempt:
$|E(|X_n|)|\leq M*a_n$ for large $n$. Rewriting as $\frac{E(|X_n|)}{a_n} \leq M$
I think I should proceed using Markov's inequality: $1 \geq \frac{E(|X_n|)}{M*a_n} \geq P(\frac{|X_n|}{a_n}\geq M)$. Should I be focusing on limit M to $\infty$ such that $\frac{E(|X_n|)}{M*a_n}$ goes to 0? I think that would squeeze the rhs of the equation and show $|X_n| = O_p(a_n)$, but I'm not confident in my understanding of these concepts.
Thanks for the help.
For any $x>0$, and any $n$, Markov inequality yilds $$\mathbb P\left(\frac{|X_n|}{a_n}>x\right)\leq \frac{\frac{\mathbb E(|X_n|)}{a_n} }{x} \leq \frac{M}{x} \to 0 \ \text{ as } x\to \infty.$$ Therefore for any $\varepsilon>0$ we can take $x$ large enough such that $\frac{M}{x}<\varepsilon$. And then put $B_\varepsilon=x$.