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This question is regarding a special case of this question, for which it is plausible the details are known. The Carbery-Wright inequality is an "anti-concentration inequality" that states the following

There exists an absolute constant $C$ such that, if $X$ is a Banach space, $p : \mathbb{R}^n → X$ is a polynomial of degree at most $d$, $0<q<\infty$, and $\mu$ is a log-concave probability measure on $\mathbb{R}^n$, then

$$\left(\int \lVert p(x)\rVert^{q/d}\mu(x)\right)^{1/q}\alpha^{-1/d}\mu\{x\in\mathbb{R}^n\mid \lVert p(x)\rVert\leq \alpha\}\leq Cq.$$

This can be used to (among other things) obtain anti-concentration inequalities for the $\ell_2$ norm of log-concave random variables, i.e. for (a specific) degree 2 polynomial.

My application requires knowledge of the constant $C$ though (or some bound on it). It does not appear to be known in the general case. Is it known in this simpler case?

Edit: In the simpler case of Gaussian measures, there appears to be a result known here.

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  • $\begingroup$ Maybe I’m a bit late, but the constants in Lemmas 1 and 4 of the original Carbery Wright paper seem to be explicit enough to get an expression for the cosnstant. $\endgroup$
    – Cain
    Commented Jun 8, 2023 at 2:30
  • $\begingroup$ @Cain Maybe, but I am not an analyst, and the disclaimer that "The value of $C$ may change from line to line" has sufficiently intimidated me that determining $C$ may be difficult. $\endgroup$
    – Mark Schultz-Wu
    Commented Jun 9, 2023 at 17:21
  • $\begingroup$ Well, the constant C in Lemma 1 is given explicitly as something which can probably be bounded by 100. The constant C in Lemma 4 is stated to be 4. Now, in the proof of Theorem 4, Case I, they obtain the absolute constant bound C(2e)^(1+1/q), with C the constant from Lemma 4. For Case II, the constant in the first terms can be bounded by 2 times the constant from Lemma 1, so 200. The constant bounding the second term is just the constant from Lemma 4. So, up to bounding the Beta function by q, you should be able to take C = 800 in your question. It is not optimal, but not far from the truth. $\endgroup$
    – Cain
    Commented Jul 26, 2023 at 15:56

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