Let $a_i, i=1, \ldots, n$ be real numbers. Let $\epsilon_i, \, i=1, \ldots, n$ be a random variables that take values $\pm 1$ with equal probability and $r_i, i=1, \ldots, n$ be random variables that take values $\alpha$ and $-\alpha$ with equal probability such that $\sum_{i=1}^nr_i=K\in R$. Note, $r_i$ and $\epsilon_i$ are independent from each other.

How to bound from above the following expectation for $p\geq 2$:

$$ E\left|\sum_{i=1}^na_i\epsilon_i r_i\right|^p \leq \, ? $$

  • $\begingroup$ The information given is neither clear, nor sufficient. Is there any independence assumed? $\endgroup$ – fedja Jun 4 '18 at 1:21
  • $\begingroup$ $\epsilon_i$ are independent and $r_i$ has dependence. What exactely is not clear? $\endgroup$ – user124297 Jun 4 '18 at 2:02
  • $\begingroup$ For instance whether $r_i$ are independent with $\varepsilon_j$. $\endgroup$ – fedja Jun 4 '18 at 2:07
  • $\begingroup$ yes, they are independent. I have corrected it. Thank you. $\endgroup$ – user124297 Jun 4 '18 at 2:28

$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \renewcommand{\P}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\tf}{\widetilde{f}}$

I am interpreting the problem as follows: Assume that the $\ep_1,\dots,\ep_n,r_1,\dots,r_n$ are independent, and we need an upper bound on \begin{equation} E_K:=\E\Big(\Big|\sum_{i=1}^na_i\ep_i r_i\Big|^p\Big|\sum_{i=1}^nr_i=K\Big) \end{equation} for all values of $K$ for which this conditional expectation is defined. However (because of the symmetry), given any values of the $n$-tuple $(r_1,\dots,r_n)$, the conditional distribution of $(\ep_1r_1,\dots,\ep_n r_n)$ is the same as the unconditional distribution of $\al(\ep_1,\dots,\ep_n)$. Therefore, $E_K$ does not depend on $K$ (as long as $E_K$ is defined), and so, we have the best possible upper bound on $E_K$: \begin{equation} E_K=E:=\E\Big|\sum_{i=1}^n a_i\al\ep_i\Big|^p\le \E|Z|^p|\al|^p\Big(\sum_{i=1}^n a_i^2\Big)^{p/2}, \end{equation} by Haagerup's inequality, where $Z\sim N(0,1)$.

  • $\begingroup$ Thank you for your answer. I have gotten curious now- if $r_i$ and $\varepsilon_i$ are not dependent. Say, we would like to bound $\sum_{I=1}^na_ir_i$, under condition on $r_i$. Would the same procedure work? Thank you. $\endgroup$ – user124297 Jun 5 '18 at 19:39
  • $\begingroup$ @user124297 : If the setting is different from the one described in my answer, then the argument will not hold without adjustment. $\endgroup$ – Iosif Pinelis Jun 6 '18 at 0:26
  • $\begingroup$ Haagerup inequality is for the Rademacher r.v., can you please elaborate how you get normal vector on the right hand side. Thank you. $\endgroup$ – user124297 Aug 15 '18 at 1:20
  • $\begingroup$ Haagerup mainly avoided the terminology of random variables (r.v.'s). In particular, instead of Rademacher r.v.'s, he used Rademacher functions. He apparently did not explicitly used normal r.v.'s. However, the two kinds of terminology, the analytical and probabilistic ones, are easily translatable into each other. In particular, it is not hard to see that the constant $B_p$ for $p\ge2$ in Haagerup's paper equals $(\mathsf{E}\,|Z|^p)^{1/p}$, where $Z\sim N(0,1)$. $\endgroup$ – Iosif Pinelis Aug 15 '18 at 12:01

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