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Let $\Vert\cdot\Vert$ is a norm in $R^n$. Let $x_1,\dots,x_N$ non-independent Rademacher random variables random variables (variables which are uniform on $\{-1, 1\}$). . By $E$ we denote an expectation. There exists a constant $A_{p,q}$ depending only on $p,q$ such that for any vectors $u_1,\dots, u_N\in R^n$ $$ \left(E \left\Vert\sum_{i=1}^N x_iu_i\right\Vert^p\right)^\frac{1}{p}\leq A_{p,q} E \left(\left\Vert\sum_{i=1}^N x_iu_i\right\Vert^{q}\right)^{\frac 1q}, $$ for all $p,q \in [1, \infty)$. This is so called restricted Kahane's inequality.

What are possible applications of this inequality?

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  • $\begingroup$ Certainly you need some assumption about the $x_i$. Maybe less than independence suffices but you can't let them be arbitrary. $\endgroup$ Mar 29, 2018 at 19:09
  • $\begingroup$ The claim is false even for a single x, see my answer. $\endgroup$
    – Aryeh
    Mar 30, 2018 at 4:55
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    $\begingroup$ Kahane's inequality asks that the $x_i$'s be Rademacher r.v.'s, i.e. uniform in $\{-1,1\}$. Without this assumption, as @Aryeh points out the statement in the question is false. $\endgroup$
    – Clement C.
    Mar 30, 2018 at 17:43
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    $\begingroup$ This still doesn't make sense. For one thing the expectation sign needs to be inside the parenthesis on the right. For another the inequality still fails if you assume that $x_1, \ldots x_N$ are each Rademacher but not independent. You need to make some assumption about the dependence structure. For example the restricted Khintchine inequality considers $(x_1, \ldots, x_N)$ chosen uniformly from $\{-1,1\}^N$ such that $x_1 + \ldots + x_N = 0$. $\endgroup$ Apr 3, 2018 at 5:31
  • $\begingroup$ It's been a week, and after many edits this question is still not making sense: namely, it is asking for applications of an inequality which does not hold. $\endgroup$
    – Clement C.
    Apr 5, 2018 at 17:37

1 Answer 1

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I don't think the claim is correct. Take $n=1$ (so the norm is just absolute value), $N=1$, and $u_1=1$. Let $X$ be any random variable with a finite first moment and infinite second moment (i.e., $E|X|=m<\infty$ while $E|X|^2=\infty$; one such example is the Pareto family https://math.stackexchange.com/questions/236181/example-of-a-general-random-variable-with-finite-mean-but-infinite-variance ). The claim would imply a finite $A_2$ for which $E|X|^2\le A_2^2 m^2$, which cannot be true.

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  • $\begingroup$ You are right, thank you. I have corrected inequality, now it is true. $\endgroup$
    – user124297
    Mar 29, 2018 at 14:05
  • $\begingroup$ Presumably, you'd want $A$ to depend on both $p$ and $q$. In my example, $p=2$ and $q=1$, so it still fails. $\endgroup$
    – Aryeh
    Mar 29, 2018 at 14:12
  • $\begingroup$ Yes, thank you very much for your corrections. $\endgroup$
    – user124297
    Mar 29, 2018 at 14:14

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