How to prove that $\mathbb{E}[R_n]=o(\sqrt{n})$ Let $(X_n)_{n\geq 1}$ random variables i.d with non-negative integers values. We set $R_n$ the cardinal of $\{X_1,...,X_n\}$. Show that $\mathbb{E}[R_n]=o(\sqrt{n})$ if $\mathbb{E}[X]<+\infty$.
We can use
\begin{equation}
 \forall a\in\mathbb{N}, \mathbb{E}[R_n]\leq a+n\mathbb{P}(X_1\geq a)
\end{equation}
(We can see it by writing $R_n\leq Card(\{X_1,...,X_n\}\cap\{0,...,a-1\})+Card(\{X_1,...,X_n\}\cap[a,+\infty[)$ ).
I first showed that $\mathbb{E}[R_n]=o(n)$ with the result question above. I thought about Cauchy-Schwarz but it was not successful, I don't really see how to use the hypothesis. Markov inequality is not of big use.
Thanks a lot for your help.
 A: We just need to apply a Markov inequality:
\begin{equation}
\mathbb{E}[R_n]\leq a+n\dfrac{\mathbb{E}[X\mathbf{1}_{X\geq a}]}{a}
\end{equation}
So
\begin{equation}
\mathbb{E}[R_n/\sqrt{n}]\leq a/\sqrt{n}+\sqrt{n}\dfrac{\mathbb{E}[X\mathbf{1}_{X\geq a}]}{a}
\end{equation}
Taking $a=\lfloor n\varepsilon\rfloor$ for $\varepsilon>0$ gives the result since $\mathbb{E}[X\mathbf{1}_{X\geq a}]$ goes to 0 as $n$ goes to $\infty$ (as the rest of a convergent series).
A: Here I tried a thing I suppose that $\mathbb{E}[X^{\beta}]<+\infty$ for an arbitrary $\beta\geq1$ that I will try to take as small possible.
With the estimation above, I have $\mathbb{E}[R_n/\sqrt{n}]\leq a/\sqrt{n}+\sqrt{n}\mathbb{P}(X\geq a)\leq a/\sqrt{n}+\sqrt{n}\mathbb{E}(X^\beta)/a^\beta$ (by Markov inequality)
Now I have to find $a_n$ such that $a_n=o(\sqrt{n})$ and $\sqrt{n}=o(a_n^\beta)$. So we see that $\alpha$ need to be a power of $n$, $a_n=n^\alpha$ with $\alpha<1/2$ and $\beta>1$
For example if $\alpha=0.49$ and $\beta=1.03$ we obtain the 2 conditions above so the quantity goes to zero as $n$ goes to infinity and so $\mathbb{E}[R_n]=o(\sqrt{n})$ if $\exists\beta>1$ such that $X\in L^\beta$.
