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.

Let $X_1, X_2, \ldots$ be independent $L_p$ random variables. I'm looking for useful conditions which imply $S_n = \sum_{i = 1} ^ n X_i$ converges in $L_p$ to some random variable $S$. If it is helpful, $X_i$ can be assumed symmetric without loss of generality, and interest is primarily in $p > 2$. For $1 \le p \le 2$ we can get upper bounds on $E|S_n|^p$ of the form $C_p\sum_{i = 1} ^ n E|X_i| ^ p$ which is useful, but nothing like that works for $p > 2$.

This may be a bit vague, mainly because I'm not entirely sure what I'm looking for. I suppose the essence is this: If one wanted to show that $S_n \to S$ in $L_p$ ($p > 2$ emphasized), where $S_n$ is a sum of independent random variables what would one try? Probably anything that gives a general method for doing this that is more substantive than "check $S_n$ is Cauchy wrt $\|\cdot\|_p$" would be useful.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Suppose that all $X_i$ are centered r.v. and $p \geq 2$. Let $S_{a,b} = \sum_{i=a}^b X_i$. We will use Rosenthal's inequality. Recall that the inequality says that

$$A_p \max\left(\sum_{i=a}^b {\mathbb E}|X_i|^p, \left(\sum_{i=a}^b {\mathbb E}X_i^2\right)^{p/2}\right) \leq {\mathbb E}[|S_{a,b}|^p] \leq B_p \max\left(\sum_{i=a}^b {\mathbb E}|X_i|^p, \left(\sum_{i=a}^b {\mathbb E}X_i^2\right)^{p/2}\right),$$ where $A_p$ and $B_p$ are some constants (for $p\geq 2$).

The sum $\sum_{i=1}^\infty X_i$ converges if and only if partial sums form a Cauchy sequence i.e. ${\mathbb E}[|S_{a,b}|^p] \to 0$ as $a \to \infty$. By Rosenthal's inequality, that happens if and only if both series $\sum_{i=1}^\infty {\mathbb E}|X_i|^p$ and $\sum_{i=1}^\infty {\mathbb E}X_i^2$ converge.

share|improve this answer
add comment

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.