3
$\begingroup$

Please help me with this question. It is simple, but I am confused right now.

For $i=1,...,n$, $\alpha_i \in C$ and $r_i$ independent and is such that $P(r_i=1)=1-P(r_i=-1)$.

I know that $E|\sum_{i=1}^n\alpha_i r_i|^p\leq C \sqrt p \| \alpha \|_2$.

I need to bound $E|\sum_{i=1}^{n-1}(\alpha_i-1) r_i|^p$.

Thank you

  • 0
    Do you mean $P(r_i=1)=1-P(r_1=-1)$? What do you assume on $r_i$? Independence?2012-03-03
  • 0
    Oh, sirry. $r_i$ are Rademaher sequence2012-03-03
  • 0
    Is supposed to be $p>1$? Or only positive?2012-03-03
  • 1
    Apply the inequality you recalled, to $n-1$ instead of $n$, and $\beta_i=\alpha_i-1$ instead of $\alpha_i$. This proves the upper bound $C\sqrt{p}\|\beta\|_2$.2012-03-03
  • 0
    @Didier Piau: yes, but I need to get the bound in terms of $\|\alpha\|$2012-03-03
  • 0
    @DavideGiraudo: we can think that $p\geq 2$2012-03-03
  • 0
    1. This should be written in your question. 2. The case $\alpha=0$ shows that there exists no upper bound linear in $\|\alpha\|_2$.2012-03-03
  • 0
    Thank you. But what about othet cases?2012-03-03
  • 0
    What kind of bound do you seek? Still very vague about this.2012-03-03
  • 0
    thank yiu . The ideal bound for me would be $C_1\sqrt p \|\alpha\|_2$2012-03-03
  • 0
    Right. Only problem is that, as already said, this is impossible.2012-03-04
  • 0
    Thank you. Alright. What if in the upper bound instead of norm$ \|\alpha\|_2$ to get simething like: $$ \min_\sigma \sqrt {\sum_{i=1}^{n-1}(\alpha_i-\sigma)^2}$$. Is ot poasible to get such a bound?2012-03-05
  • 0
    Sorry, forgot to mention in the previous comment, $\sigma$ is an average of the vector $\alpha$.2012-03-05
  • 0
    In other words, an upper bound linear in $\|\alpha\|_2$ is impossible so you are trying a **smaller** upper bound? Not sure there is much hope... :-) (Unrelated: please use the @ thing if you want the people you are talking to, to see your comments, I stumbled upon yours by chance.)2012-03-12

0 Answers 0