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The Carbery–Wright inequality is a seminal result about the anti-concentration of polynomials of Gaussian random variables. See e.g. Meka, Nguyen, and Vu - Anti-concentration for polynomials of independent random variables, Theorem 1.4, for the precise statement. However, I cannot find any reference where an explicit estimate on the constant $B$ on the r.h.s. of the inequality is given. Knowing this constant is crucial for the application I have in mind. Are there known estimates on it? I should also say that the polynomial I have in mind is of the form $p(g_1,g_2,\dotsc,g_k)=\langle g_1\otimes g_2 \dotsb\otimes g_k,A g_1\otimes g_2 \dotsb\otimes g_k\rangle$, where $A\in\mathbb{R}^{d^k\times d^k}$, $\operatorname{tr}(A)=0$, and the $g_i\in \mathbb{R}^d$ are random vectors with i.i.d. gaussian entries. Maybe this special structure helps to obtain better anti-concentration estimates. Any help would be appreciated!

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Have you looked at the original paper by A. Carbery and J. Wright, Distributional and $L^q$ norm inequalities for polynomials over convex bodies in $\mathbb R^n$? Theorem 8 page 244 is the famous inequality with a sharp constant.

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    $\begingroup$ what do you mean by sharp constant? I do not see an explicit constant in that Theorem. You mean this article: pdfs.semanticscholar.org/e997/… ,right? It only says "There exists an absolute constant C". Do we know what that constant is? $\endgroup$
    – user134977
    May 26, 2020 at 12:09
  • $\begingroup$ The way it is stated suggests that C is independent of the polynomial. The constant C is numerical. $\endgroup$
    – user69642
    May 26, 2020 at 12:22
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    $\begingroup$ Oh, now I understand the issue. Sure, the constant C is numerical, but my application requires bounding this constant. Note that the statement of the inequality only becomes nontrivial for $\alpha\leq C^{-2d}$, where $d$ is the degree of the polynomial. In the application I have in mind, the degree of the polynomial grows, so the inequality is nontrivial only for exponentially small $\alpha$, with the exponent depending on $C$. That is why I actually want to know $C$ in the inequality. $\endgroup$
    – user134977
    May 27, 2020 at 13:33
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    $\begingroup$ I am not sure this is correct. The statement only tells us that there is a universal lower bound. But there is no guarantee that polynomials in dimension 1 are "representative". I.e., it could be the case that a much better constant $C$ holds for dimension $1$ compared to, say 59. Of course, this gives a lower bound on the constant, but not a universal upper bound. $\endgroup$
    – user134977
    May 28, 2020 at 22:13
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    $\begingroup$ Ok! I thought you were interested in a non-trivial lower bound. $\endgroup$
    – user69642
    May 29, 2020 at 8:10

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