$\mathbb E [||X_n|| ^ \alpha] = O(a_n) \implies X_n = O_p (a_n)$. True? Helloo everyone, just in case, excuse my english i am a french native.
Please i'd like to have a proof, if it's possible, that
$\mathbb E [||X_n|| ^ \alpha] = O(a_n) \implies X_n = O_p (a_n)$
I saw it at
http://jeanalain.monfort.free.fr/Dicostat2005/O/Ordres_de_convergence_en_probabilite.pdf
property (5) (i)

I think there is a typo in your file. So let's say it my way.
" If $\mathbb{E}\left( \Vert X_n \Vert ^{\alpha} \right) = O( a_n^{\alpha})$ then $X_n= O_p(a_n)$ ". 
Proof is a direct consequence of Markov's.
$ \mathbb{P} \left( \frac{\Vert X_n \Vert}{a_n} \le M \right) \le \frac{1}{M^{\alpha}}.\mathbb{E}\left( \frac{\Vert X_n \Vert^{\alpha}}{a_n^{\alpha}} \right)= \frac{ O(1)}{M^{\alpha}}$
