The Carbery–Wright inequality is a seminal result about the anticoncentration of polynomials of Gaussian random variables. See e.g. Meka, Nguyen, and Vu  Anticoncentration 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 anticoncentration estimates. Any help would be appreciated!
1 Answer
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.

1$\begingroup$ 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? $\endgroup$ 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. $\endgroup$ May 26, 2020 at 12:22

1$\begingroup$ 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. $\endgroup$ May 27, 2020 at 13:33

1$\begingroup$ 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. $\endgroup$ May 28, 2020 at 22:13

1$\begingroup$ Ok! I thought you were interested in a nontrivial lower bound. $\endgroup$ May 29, 2020 at 8:10