Introduce the Big Oh notation in probability and let $X_n$ be a sequence of random variables. I was reading a paper and the author finded that $$E\lvert (X_n)\rvert\leq C a_n$$ where $C>0$ and $a_n$ is a positive real sequence converging to zero. However, from this, he concluded by Markov's inequality that $$\lvert X_n - E(X_n)\rvert=O_p(a_n).$$ I wonder why the first result directly implies the second. Can you help me?
My attempt
From the first result, Markov's inequality implies that, for any $c>0$, $$P\bigg(\bigg\lvert\frac{ X_n}{a_n}\bigg\rvert>c\bigg)\leq \frac{E\lvert X_n\rvert}{ca_n}\leq \frac{C}{c},$$ which is sufficient to conclude that $X_n=O_p(a_n)$. In addition, the first result is equivalent to say that $E(X_n)=O(a_n)$.
Using triangle inequality, $\lvert X_n - E(X_n)\rvert\leq \lvert X_n \rvert+ \lvert E(X_n)\rvert=O_p(a_n)+O(a_n)=O_p(a_n)$.
This final step uses two results (which can be easily proved):
(i) for $X_n, Y_n$ sequences of random variables with $Y_n=O_p(a_n)$, if $X_n\leq Y_n$, then $X_n=O_p(a_n)$; and
(ii) for a sequence of random variables $X_n$ and a real sequence $b_n$, if $X_n=O_p(a_n), b_n=O(a_n)$, then $X_n+b_n=O_p(a_n),$
I would like to read your suggestions or feedbacks about it. I'm not sure if it (the triangle inequality argument) is the correct way to think the above consequence (or if there is another simpler way).
Thanks in advance.