Suppose random variables ${X_i}$ such that
$\sum_{i=1}^n X_i^2 = Op(n^{k})$
for $n \to \infty$. Is the sum
$\sum_{i=1}^n |X_i| $
also bounded in probability and if so, what's the rate?? Also, does a similar bound hold when $\sum_{i=1}^n X_i^2 = op(n^{k})?$
Clearly, if $|X_i| \ge 1$, we have $\sum_{i=1}^n |X_i| \le \sum_{i=1}^n X_i^2 = Op(n^{k})$
However, when $|X_i|$ can be less than 1, it's not clear to me if any bound is obtainable. How do I solve this?
From the norm inequality, we have
$\sqrt{\sum X_i^2} \le \sum | X_i| \le \sqrt{n}\sqrt{\sum X_i^2}$
If $\sum X_i^2 = Op(n^k)$, then $\sqrt{n}\sqrt{\sum X_i^2} = Op(n^{.5(k+1)})$, and so
$\sum | X_i| = Op(n^{.5(k+1)})$