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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?

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1 Answer 1

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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

$\sum | X_i| = Op(n^{.5(k+1)})$

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