Comparison of Rademacher and Gaussian expected values under linear transformations

Question

As per suggestion, I have decided to post the following as a new question, but it is a follow-up to this one: Comparison of Rademacher and Gaussian moments under linear transformations

Let $X$ be an $n$ dimensional standard Gaussian and let $X$ be an $n \times n$ orthogonal matrix. Then, the random vector $Z= U^\top X$ is also distributed as a standard Gaussian in $R^n$. This implies that for any function $f: R \to R_+$ where $E[f(G)] < \infty$ for $G$ being a standard gaussian in $R$, we can easily compute $$ E\left[\prod_{i=1}^n f(Z_i)\right] = (E[f(G)])^n,$$ by independence.

Is there a method for bounding expected values of such functions if $X$ was an $n$ dimensional vector where each coordinate is an independent Rademacher $\pm 1$ random variable. For instance, for $Z= U^\top X$, the coordinates are not independent anymore. Is there a way of comparing $E[\prod_{i=1}^n f(Z_i)]$ with $(E[f(G)])^n$ even in this setting? For instance, is it true that $E[\prod_{i=1}^n f(Z_i)] \le \alpha(n)\cdot (E[f(G)])^n$ where $\alpha(n) \le \mathrm{poly}(n)$?

Also, any references to techniques or universality statements that are applicable in such settings will be much appreciated. Thanks!

Answer 1

At the end I'm not sure if you want $G$ to be a standard Gaussian or a Rademacher?

If $G$ were to be a Rademacher, then consider the function $f$ such that $f(-1)=f(1)=0$ and $f(x)=1$ if $x\not\in\{-1,1\}$. Then the right term is equal to $0$. However if $U$ is not too trivial then the left term is positive. (example : $n=2$ take $U$ a rotation matrix of angle $\pi/3$). The idea is that the coordinates of $U^TX$ will not be Rademacher variables (a priori), so then we can play on the support of the measures.

If $G$ were to be a standard Gaussian, then take $U=I_n$ and consider the function $f$ such that $f(1)=f(-1)=1$ and $f(x)=0$ for all $x\not\in \{1,-1\}$. The left term will be positive while the right term is zero. (I think you can adapt this phenomenon if $U$ is general: take a similar $f$ (that will depend on $U$) with an appropriate discrete support which will be all possible realizations of the coordinates of $U^TX$).

I think you will have to impose some hypothesis on $f$. You already checked that it was ok for $f(x)=x^2$ and I believe you can adapt that proof to show that your intuition is true if you have a condition like : $\alpha_1 x^2 \leq f(x)\leq \alpha_2 x^2$ (for some constants $0<\alpha_1\leq\alpha_2$). It may be true for a more general class of functions (however the $\alpha(n)$ may depend on the $f$).

Hope this helped!