Comparison between Rademacher average and random average  Let $X$ be a Banach space. We let $j=e^{2i\pi/3}$.
Let $(\epsilon_i)$ be a sequence of independent Rademacher variables on a fixed probability space $\Omega$. Let $(\varphi_i)$ be a sequence of independent complex random variables such that $P(\varphi_i=1)=\frac{1}{3}$, $P(\varphi_i=j)=\frac{1}{3}$ and $P(\varphi_i=j^2)=\frac{1}{3}$ for any $i$.
Do there exist sufficient conditions on $X$ such that there exist constants $m,M>0$ such that
$$
m\Big\vert\Big\vert\sum_i \epsilon_i\otimes x_i \Big\vert\Big\vert_{L^p(\Omega,X)}\leq \Big\vert\Big\vert\sum_i \varphi_i \otimes x_i \Big\vert\Big\vert_{L^p(\Omega,X)}\leq M\Big\vert\Big\vert\sum_i \epsilon_i\otimes x_i \Big\vert\Big\vert_{L^p(\Omega,X)}
$$
for any $x_1,\ldots,x_n\in X$?
Remark: I know that if $X$ has finite cotype, the Rademacher averages and the gaussian averages are equivalent.
 A: I believe that this is true for all Banach spaces.
First, let $\psi_k = \varphi_k - \varphi'_k$, where $\varphi' = (\varphi'_k)$ is an independent copy of $\varphi = (\varphi_k)$.  Then
$$ \frac12 \|\sum \psi_k x_k\|_{L^p(X)} \le \|\sum \varphi_k x_k\|_{L^p(X)} \le \|\sum \psi_k x_k\|_{L^p(X)} $$
The first inequality follows by the triangle inequality, and the second inequality by computing the conditional expectation with respect to $\sigma(\varphi')$.
Now you need to know the comparison principle - look at Chapter 3 and Theorem 3.1 of http://ocw.tudelft.nl/courses/mathematics/stochastic-evolution-equations/lectures/ - note that the $\psi_k$ are now symmetric random variables.
We need Theorem 3.1 when the $a_i$ are complex numbers.  But this is easily accomplished by separating $a_i$ into its real and imaginary parts, as long as you don't mind sacrificing an additional factor $2^p$ on the right hand side.
Now, it seems to me that you are ready to mimic the kind of argument given for Theorem 3.2.  And I believe this should establish what you want.
