Let $A$ be a normalized $N\times N$ GOE matrix. Let $\sigma^1<\sigma^2<\dots <\sigma^N$. We know that the largest eigenvalue converges $\sigma^N$ to 2 almost surely. Assume that $X\_1,\dots, X\_N$ are iid random variables with law $\mu$ so that $E\_{X\sim \mu}[e^{\alpha X}]<\infty$ for some $\alpha>0$.


Now, I try to prove that
$$
P(\frac{1}{N}\sum\_{i=1}^N \sigma^i X\_i^2\ge ax)\le \mbox{something like $(E[e^{\alpha X^2}])^N e^{-a\alpha xN}$}
$$
which goes to zero as $N\to \infty$.




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My reasons for having this upper bound estimate are that if we consider the following thing: By Chebyshev's inequality
$$
P(\frac{1}{N}\sum\_{i=1}^N X\_i^2\ge ax)=P(\exp(\alpha\sum\_i X\_i^2)\ge e^{a\alpha x N})\le \frac{(E[e^{\alpha X\_0^2}])^N}{e^{a\alpha x N}} \to 0 
$$
as $N\to \infty$.


The inequality I want to prove has one more random variable $\sigma^i$ that is not easy to handle.



How to bound $ P(\frac{1}{N}\sum_{i=1}^N \sigma^i X_i^2\ge ax)$ for eigenvalues of a normalized $N\times N$ GOE matrix?
Since we have the joint law [GOE eigenvalues](https://www.lpthe.jussieu.fr/%7Eleticia/TEACHING/Master2019/GOE-cuentas.pdf) density $\rho\_{N}$


$$P[\sigma\_{i}\in A\_{i}]=\iint\_{A\_{i}}\frac{1}{Z\_{N}}e^{-\frac{1}{4}\sum\_{i=1}x\_{i}^{2}}\prod\_{k>m}|x\_{k}-x\_{m}|\prod dx\_{i}$$


and so


$$P(\frac{1}{N}\sum\_{i=1}^N \sigma^i X\_i^2\ge at)=\iint\_{R^{N}} P(\frac{1}{N}\sum\_{i=1}^N x\_{i} X\_i^2\ge ax) \rho\_{N}(x)dx$$
$$\leq \iint\_{R^{N}} \frac{\prod\_{i}E[e^{\alpha x\_{i} X\_i^2}])}{e^{a\alpha t N}} \rho\_{N}(x)dx.$$


To proceed from here one needs to know explicity the MGF for $X^{2}$ so as to estimate. Do you have any estimates for it at least?


