Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Say we have a sequence $\{X_i\}$ independent which is mean $0$ and $E|X_i|^p < \infty$ for $p \geq 1$. Is there a bound in the form $E\left|\sum_{i=1}^n X_i\right|^p \leq C\cdot \sum_{i=1}^n E| X_i|^p$ where $C$ is not function of $n$?

share|improve this question

1 Answer 1

No. For example, if $X_i$ are standard normal, $\sum_{i=1}^n X_i$ is normal with mean 0 and variance $n$, so $E\left|\sum_{i=1}^n X_i \right|^p = K(p) n^{p/2}$ where $K(p) = E |Z|^p$ for a standard normal random variable. But the right side is $n C K(p)$. So no such bound can exist if $p > 2$.

share|improve this answer
    
Thanks for the quick reply. Could you say something if there is more structure in the sequence? If, for instance, $\{X_i\}$ are centered exponentials with $X_i \sim Exp(i)$ (with mean removed) or if $\{X_i\}$ more generally are either upper or lower bounded, could anything be said about a sum of expectation bound? –  monte Dec 8 '11 at 19:21
    
Consider e.g. $p=4$. Then $E[ |\sum_{i=1}^n X_i|^4 ] = n E[X_i^4] + 3 n(n-1) E[X_i^2]^2 \sim K n^2$ as $n \to \infty$. Similarly, for any even integer $p$ the left side of your inequality should be on the order of $n^{p/2}$ as $n \to \infty$. –  Robert Israel Dec 8 '11 at 20:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.