example of the vector (bounding expectation of a form in Rademacher functions)

Let $x\in R^{2m}$ such that $x_1+\ldots+x_{2m}=0$ and let $r_i, i=1, \ldots, 2m$ be Rademacher functions, i.e. $P(r_i=1)=P(r_i=-1)=1/2$.

I would like to find an example of the vector $x$ such that $E(\sum_{i=1}^{2m}x_ir_i)^{2q}\geq C\sqrt{2q}\|x\|_2$.

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I suspect you need to add more constraints, $x=0$ will satisfy the inequality trivially.

Alternatively, assuming that the $r_i$ are independent, then $$E(\sum_{i=1}^{2m}x_ir_i)^{2q} = E (\sum_{i_1}\cdots \sum_{i_{2q}}x_{i_1} \cdots x_{2q} r_{i_1} \cdots r_{2q}) =\sum_{i_1}\cdots \sum_{i_{2q}}x_{i_1} \cdots x_{2q} E(r_{i_1} \cdots r_{2q}) = \sum x_i^{2q},$$ where we used independence and that $r_i^2 = 1$ in the last equality.

The problem then reduces to finding an $x$ that satisfies $\sum x_i = 0$ and $\sum x_i^{2q} \geq C \sqrt{2 q} \sqrt{\sum x_i^2}$.

Let $t\geq 0$ and let $x=(t,\cdots t,-t,\cdots, -t)$ ($m$ components are $t$ followed by $m$ components of $-t$). Clearly $\sum x_i = 0$. The inequality becomes $2m t^{2q} \geq C \sqrt{2 q} \sqrt{2 m} t$. This is satisfied whenever $t=0$ or whenever the inequality $t^{2q-1} \geq C \sqrt{\frac{q}{m}}$ is satisfied, or equivalently whenever $t \geq \sqrt[2q-1]{C \sqrt{\frac{q}{m}}}$.

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Thank you. For curiosity, if one assume that $\sum_i r_i=0$ and $x\in R^{2m}$ any vector. Would be your argument true? –  Michael Jul 29 '12 at 5:55
No, because you lose independence. However, I think you could zero the components of $x$ corresponding to the dependent $r_i$ (possibly one more for evenness) and repair the argument (I haven't worked this through). –  copper.hat Jul 29 '12 at 5:56
Could you elaborate please your last statement. I did not get an idea... –  Michael Jul 29 '12 at 6:00
Some subset of the $r_i$ are independent. Set $x_j=0$ for $j$ corresponding to a dependent $r_j$, and deal with a lower dimensional problem. The only catch is to have an even number of coordinates so you can do the $\pm t$ trick. –  copper.hat Jul 29 '12 at 6:04
Thak you for your explanations. –  Michael Jul 29 '12 at 6:06