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Following this question: Can we apply the continuous mapping theorem for the limiting joint distribution of the Tracy-Widom law?.

We know that

$$ \lim_{N\to\infty}P(N^{2/3}(\lambda_N-2)\le s_1,\dotsc, N^{2/3}(\lambda_{N-k+1}-2)\le s_k)=F_{\beta, k}(s_1,\dotsc, s_k). $$

Can we say that for any $\epsilon>0$, there exists a constant $c>0$ such that $$ P(N^{2/3}(\lambda_N-\lambda_{N-1})\le c)\ge 1-\epsilon? $$

Or assume that $(N^{2/3}(\lambda_N-2),\dotsc, N^{2/3}(\lambda_{N-k+1}-2))\to (Y_1,\dotsc, Y_{N-k+1})$ in distribution.

The question becomes that for any $\epsilon>0$, there exists a constant $c>0$ such that $$ P(Y_N-Y_{N-1}\le c)\ge 1-\epsilon? $$

We can say $N^{2/3}(\lambda_N-\lambda_{N-1})=O_p(1)$.

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The probability distribution of the spacing $\delta_N=\lambda_N-\lambda_{N-1}$ of the eigenvalues $\lambda_N$ and $\lambda_{N-1}$ at the edge of the spectrum decays exponentially for $\delta_N\gg N^{-2/3}$, with a decay rate that is independent of $N$. So no matter how small $\epsilon$, you can always find a $c$ such that $P(N^{2/3}\delta_N\le c)\ge 1-\epsilon$.

I agree with the ChatGPT bot in the deleted answer that $c$ will depend on $\epsilon$, but I do not agree that $c$ will depend on $N$. The scaling with $N$ is fully contained in the mean level spacing $N^{-2/3}$ for large $N$ (Tracy-Widom law).

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  • $\begingroup$ So why ChatGPT delete his answer? Is that one not true? $\endgroup$
    – Hermi
    Commented Dec 9, 2022 at 18:35
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    $\begingroup$ ChatGPT is a bot, an automated machine that produces answers that sound reasonable but are nonsensical; the answer was deleted by the moderators, like several other similar fake answers; see meta.mathoverflow.net/questions/5531/… $\endgroup$
    – Carlo Beenakker
    Commented Dec 9, 2022 at 20:28
  • $\begingroup$ So for a fixed number $l$, let $\delta_{Nl}:=\lambda_N-\lambda_l$, we can similar get the same answer? $P(N^{2/3}\delta_{NL}\le c)\ge 1-\epsilon$? Here $c$ will dependent on $\epsilon$ and $l$? $\endgroup$
    – Hermi
    Commented Dec 10, 2022 at 20:01
  • $\begingroup$ no, this does not hold; the inequality does hold for $\lambda_N-\lambda_{N-l}$ with fixed $l$, but not for $\lambda_N-\lambda_l$ with fixed $l$. $\endgroup$
    – Carlo Beenakker
    Commented Dec 10, 2022 at 20:37
  • $\begingroup$ Sure, I mean like for the eigenvalues $\lambda_1\le \lambda_2\le \dots \le \lambda_l$, then we have this result for $\lambda_l-\lambda_1$, right? That is for any $\epsilon>0$, there exists $c>0$ such that $\lim_{N\to\infty} P(N^{2/3}(\lambda_l-\lambda_1)\le c)\ge 1-\epsilon$. $\endgroup$
    – Hermi
    Commented Dec 10, 2022 at 20:40

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