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Let $X$ be an $n$ dimensional standard Gaussian and let $U$ be an $n \times n$ orthogonal matrix. Then, the random vector $Z = U^\top X$ is also distributed as a standard Gaussian in $R^n$ and we have $E[\prod_{i=1}^n Z_i^2] = 1$ by independence.

Is there a method for bounding such functions if $X$ was an $n$ dimensional vector where each coordinate is an independent Rademacher $\pm1$ random variable. For instance, for $Z = U^\top X$, the coordinates are not independent anymore. Can one show that $E[\prod_{i=1}^n Z_i^2]$ is polynomial in $n$ or $\exp(n)$ even in this setting?

More generally, is there a way of comparing such quantities to the Gaussian setting or universality statements that are applicable in such settings where the random vector may have some weak dependencies?

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By the arithmetic mean--geometric mean inequality and the condition $Z=U^\top X$ for an orthogonal matrix $U$, $$\prod_1^n Z_i^2\le\Big(\frac1n\sum_1^n Z_i^2\Big)^n =\Big(\frac1n\sum_1^n X_i^2\Big)^n=1,$$ since $X_i=\pm1$ for all $i$. So, $$E\prod_1^n Z_i^2\le1,$$ as desired.

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  • $\begingroup$ Thank you very much. I didn't realize that this had a simple proof. I need to be able to bound expectation of other functions as well where it is easy to compute them in the Gaussian case, e.g. $E[\prod_{i=1}^n f(Z_i)]$ for some function $f$, so I was wondering if there is a general method for comparing to the Gaussian setting? $\endgroup$
    – brownianmotion
    Commented Sep 16, 2022 at 22:03
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    $\begingroup$ @brownianmotion : It would be better in every respect that I can think of if the question for general $f$ be posted separately. If you think that your posted question has been fully answered, can you please mark the answer accordingly? $\endgroup$
    – Iosif Pinelis
    Commented Sep 18, 2022 at 0:58
  • $\begingroup$ Thanks for the suggestion! I have marked the questions as being answered and opened a new question: mathoverflow.net/questions/430795/… $\endgroup$
    – brownianmotion
    Commented Sep 19, 2022 at 19:35
  • $\begingroup$ @brownianmotion : That new question seems interesting but difficult. I would be (pleasantly) surprised if there is an answer to it within a few pages. $\endgroup$
    – Iosif Pinelis
    Commented Sep 19, 2022 at 19:59

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