Tracy-Widom Law: Vague convergence and Ledoux Bound

This is a continuation of the previous post available here. In the previous post, we develop the ingredients required for the vague convergence proof.  Let us now return to the random matrix scenario.

Proof of Theorem 4: X_n be a sequence of n\times n GUE matrices. Let \lambda_1,\ldots, \lambda_n be the eigenvalues of X_n. Fix -\infty < t< t' < \infty. Let us quickly evaluate the limit

\displaystyle \lim_{n\to \infty} P\left[n^{2/3}\left(\frac{\lambda_i}{\sqrt{n}}-2\right) \not\in (t,t'), \ \ i=1,2, \ldots, n\right]

Observe that by using Theorem 3, we have

\displaystyle \begin{aligned} & P\left[n^{2/3}\left(\frac{\lambda_i}{\sqrt{n}}-2\right) \not\in (t,t'), \ \ i=1,2, \ldots, n\right] \\ & = P\left[\lambda_i \not\in \left(2\sqrt{n}+\frac{t}{n^{1/6}},2\sqrt{n}+\frac{t'}{n^{1/6}}\right), \ \ i=1,2, \ldots, n \right] \\ & = 1+\sum_{k=1}^{\infty} \frac{(-1)^k}{k!}\int\limits_{2\sqrt{n}+\frac{t}{n^{1/6}}}^{2\sqrt{n}+\frac{t'}{n^{1/6}}}\cdots\int\limits_{2\sqrt{n}+\frac{t}{n^{1/6}}}^{2\sqrt{n}+\frac{t'}{n^{1/6}}} \det_{i,j=1}^k K^{(n)}(x_i,x_j)\prod_{i=1}^{k}dx_i \\ & = 1+\sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \int_{t}^{t'}\cdots\int_{t}^{t'} \det_{i,j=1}^k \left[\frac1{n^{1/6}}K^{(n)}\left(2\sqrt{n}+\frac{x_i}{n^{1/6}},2\sqrt{n}+\frac{x_j}{n^{1/6}}\right)\right]\prod_{i=1}^k dx_i \end{aligned}

where in the last line we use change of variable formula. Let us define

\displaystyle A^{(n)}(x,y):=\frac1{n^{1/6}}K^{(n)}\left(2\sqrt{n}+\frac{x}{n^{1/6}},2\sqrt{n}+\frac{y}{n^{1/6}}\right)

Note that A^{(n)} are kernels since K^{(n)} is a kernel. Let us also define \displaystyle \phi_n(x) = n^{1/12}\psi_n\left(2\sqrt{n}+\frac{x}{n^{1/6}}\right). Then using proposition 1 (c) we have

\displaystyle A^{(n)}(x,y)=\frac{\phi_n(x)\phi_n'(y)-\phi_n'(x)\phi_n'(y)}{x-y}-\frac1{2n^{1/3}}\phi_n(x)\phi_n(y)

Note that Proposition 4 implies that  for every C>1, \phi_n \to Ai uniformly over a ball of radius C in the complex plane. \phi_n are entire functions. Hence \phi_n' \to Ai' and \phi_n'' \to Ai''. Hence for t\le x,y \le t' we have

\displaystyle ||A^{(n)}(x,y)-A(x,y)|| \to 0

  Hence by continuity lemma for fredholm determinants we have

\displaystyle \begin{aligned} & \lim_{n\to \infty} P\left[n^{2/3}\left(\frac{\lambda_i}{\sqrt{n}}-2\right) \not\in (t,t'), \ \ i=1,2, \ldots, n\right] \\ & = \lim_{n\to\infty} \Delta(A^{(n)}) \\ & = \lim_{n\to\infty} \Delta(A) \\ & = 1+\sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \int_{t}^{t'}\cdots\int_{t}^{t'} \det_{i,j=1}^k A(x_i,x_j) \prod_{i=1}^k dx_i \end{aligned}  

where the measure is the lebesgue measure on the bounded interval \displaystyle (t,t'). This completes the proof of Theorem 4

\square

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Tracy-Widom Law: Fredholm determinants and Airy functions

This is a continuation of the previous post available here. This section is quite different from the previous post. Here we present a series of propositions. The proofs of the propositions are not present here. I may include them later. However the proofs of the propositions (except proposition 4) are not necessary in our proof.

In the previous post, we introduced Hermite polynomials and wave functions. We noticed the Hermite polynomials are orthogonal polynomials. There are a plethora of results involving orthonormal polynomials. We merely state only few of them that we will be needing in the our proof.

Proposition 1: Let H_n(x) be the Hermite polynomial of degree n. Let \psi_n(x) be the corresponding wave function. Let \displaystyle K^{(n)}(x,y)=\sum_{k=0}^{n-1} \psi_k(x)\psi_k(y). We have the following results.

(a) H_n(x+t)=\sum_{k=0}^{n} \binom{n}{k}H_k(x)t^{n-k}

(b) \displaystyle \frac1{\sqrt{n}}K^{(n)}(x,y)= \frac{\psi_n(x)\psi_{n-1}(y)-\psi_{n-1}(x)\psi_n(y)}{x-y}

(c) \displaystyle K^{(n)}(x,y)=\frac{\psi_n(x)\psi_n'(y)-\psi_n'(x)\psi_n(y)}{x-y}-\frac12\psi_n(x)\psi_n(y)

(d) \displaystyle \frac{d}{dx}K^{(n)}(x,x)=-\sqrt{n}\psi_n(x)\psi_{n-1}(x)

Proof: Not included

Let us recall Theorem 3 once again.

Theorem 3: For any measurable subset A of \mathbb{R},

\displaystyle P\left(\bigcap_{i=1}^n \{\lambda_i \in A\}\right) =1+\sum_{k=1}^{\infty} \frac{(-1)^k}{k!}\int_{A^c}\cdots\int_{A^c} \det_{i,j=1}^k K^{(n)}(x_i,x_j)\prod_{i=1}^k dx_i

Let us replace A by A^c and rewrite the above result as

\displaystyle P\left(\bigcap_{i=1}^n \{\lambda_i \not\in A\}\right) =1+\sum_{k=1}^{\infty} \frac{(-1)^k}{k!}\int_{A}\cdots\int_{A} \det_{i,j=1}^k K^{(n)}(x_i,x_j)\prod_{i=1}^k dx_i

Now it is right time to see what exactly are we trying to show if we want to prove weak convergence. For that we need some basic idea about Airy function.

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