Following this question: [Can we apply the continuous mapping theorem for the limiting joint distribution of the Tracy-Widom law?](https://mathoverflow.net/questions/436028/can-we-apply-the-continuous-mapping-theorem-for-the-limiting-joint-distribution).


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 $N^{2/3}(\lambda\_N-\lambda\_{N-1})=O\_p(1)$.



Can we get that $ P(N^{2/3}(\lambda_N-\lambda_{N-1})\le c)\ge 1-\epsilon$?
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).


