<p>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., $E|X|=m&lt;\infty$ while $E|X|^2=\infty$; one such example is the Pareto family
<a href="https://math.stackexchange.com/questions/236181/example-of-a-general-random-variable-with-finite-mean-but-infinite-variance">https://math.stackexchange.com/questions/236181/example-of-a-general-random-variable-with-finite-mean-but-infinite-variance</a>
). The claim would imply a finite $A_2$ for which $E|X|^2\le A_2^2 m^2$, which cannot be true.</p>

CC BY-SA 3.0
2018-03-29T14:05:11.010Z
You are right, thank you. I have corrected inequality, now it is true.
user124297
CC BY-SA 3.0
2018-03-29T14:12:51.097Z
Presumably, you'd want $A$ to depend on both $p$ and $q$. In my example, $p=2$ and $q=1$, so it still fails.
Aryeh
CC BY-SA 3.0
2018-03-29T14:14:15.463Z
Yes, thank you very much for your corrections.
user124297
CC BY-SA 3.0
2018-03-29T13:22:30.247Z
2018-03-29T13:22:30.247Z
Aryeh
<p>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.</p>

<p>What are possible applications of this inequality?</p>

CC BY-SA 3.0
2018-03-29T19:09:05.820Z
Certainly you need some assumption about the $x_i$. Maybe less than independence suffices but you can't let them be arbitrary.
Sasho Nikolov
CC BY-SA 3.0
2018-03-30T04:55:24.633Z
The claim is false even for a single x, see my answer.
Aryeh
CC BY-SA 3.0
2018-03-30T17:43:07.933Z
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.
Clement C.
CC BY-SA 3.0
2018-04-03T05:31:58.417Z
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 $x_1 + \ldots + x_N = 0$.
Sasho Nikolov
CC BY-SA 3.0
2018-04-05T17:37:17.460Z
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.
Clement C.
<co.combinatorics><pr.probability><randomness><st.statistics><high-dimensional-geometry>
Application of the inequality with expectations
user124297
# Application of the inequality with expectations

(Asked by: username_0 on 2018-03-29)

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?

**Tags:** co.combinatorics, pr.probability, randomness, st.statistics, high-dimensional-geometry

> **Comments:**
> **username_1**: Certainly you need some assumption about the $x_i$. Maybe less than independence suffices but you can't let them be arbitrary.  
> **username_2**: The claim is false even for a single x, see my answer.  
> **username_3**: Kahane's inequality asks that the $x_i$'s be *Rademacher* r.v.'s, i.e. uniform in $\{-1,1\}$. Without this assumption, as @username_2 points out the statement in the question is false.  
> **username_1**: 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 $x_1 + \ldots + x_N = 0$.  
> **username_3**: 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.  

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## 1 answers

### Answer by username_2 on 2018-03-29. Score: 1 | Accepted answer

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.

> **Comments:**
> **username_0**: You are right, thank you. I have corrected inequality, now it is true.  
> **username_2**: Presumably, you'd want $A$ to depend on both $p$ and $q$. In my example, $p=2$ and $q=1$, so it still fails.  
> **username_0**: Yes, thank you very much for your corrections.  

---

