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The Wigner semicircle law states that for a random GOE-matrix $M^N \in \mathbb{R}^{N \times N}$ in the $N \rightarrow \infty$ limit for any $f \in C^b(\mathbb{R})$

$$\lim_{N \rightarrow \infty}\frac{1}{N} \mathbb{E}^{\text{GOE}}\left(\sum_{\lambda \text{ eigenvalue of } M^N} f(\lambda)\right) = \int f(\lambda) \rho(\lambda) d\lambda$$

where $\rho$ is the semicircle distribution.

Now, I am in the situation that I would like to compute

$$\lim_{N \rightarrow \infty}\frac{1}{N} \mathbb{E}^{\text{GOE}}\left(\sum_{\lambda \text{ eigenvalue of } M^N} 1_{\lambda_0>x}f(\lambda)\right) $$

where $\lambda_0$ is the smallest eigenvalue of $M^N$.

My question is: Is there still a limiting distribution?

EDIT: It is mentioned in the comments by Carlo Beenakker that the lowest eigenvalue accumulates with high probability at $-R$ where $[-R,R]$ is the support of the semicircle distribution and thus it should follow that $$\lim_{N \rightarrow \infty}\frac{1}{N} \mathbb{E}^{\text{GOE}}\left(\sum_{\lambda \text{ eigenvalue of } M^N} 1_{\lambda_0>x}f(\lambda)\right) = 1_{\{x\le -R\}} \int f(\lambda) \rho(\lambda) d\lambda.$$

Thus, the question is whether one can show that this limit really exists (i.e. that one can combine the accumulation of the lowest eigenvalue and the convergence to the semicircle distribution as shown above).

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    $\begingroup$ in the large $N$ limit this is simply $\theta(-R-x)\int f(\lambda)\rho(\lambda)d\lambda$, with $(-R,R)$ the support of the semicircle distribution and $\theta$ the unit step function $\endgroup$ – Carlo Beenakker Dec 10 '17 at 16:33
  • $\begingroup$ @CarloBeenakker sorry, how do you see this, I suppose it is a change of variables somehow, but I currently do not see it. $\endgroup$ – user118408 Dec 10 '17 at 16:44
  • $\begingroup$ it's just that in the large-$N$ limit the probability that the smallest eigenvalue is greater than $x$ jumps from 1 to 0 when $x$ reaches the lower edge of the spectrum from below; there are fluctuations around that edge, but these vanish in the large-$N$ limit $\endgroup$ – Carlo Beenakker Dec 10 '17 at 16:46
  • $\begingroup$ @CarloBeenakker I see, thank you. But is it clear that this goes well with the convergence to the semicircle distribution. I mean is there a proof that you can combine these two properties? (convergence to semicircle distribution and jump from 0 to 1 at edge). I added that $f$ is continuous and bounded. $\endgroup$ – user118408 Dec 10 '17 at 16:49
  • $\begingroup$ @JohnDriggs You can combine any statement about a probability converging to $1$ with any statement about the expectation of a bounded random variable in this way, because the other points contribute a bounded amount * an amount converging to zero. Because your $f$ is bounded, the average of $f(\lambda)$ is also bounded. $\endgroup$ – Will Sawin Dec 11 '17 at 16:38

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