CLT for Eigenvalues — A Guided Tour of Modern Statistics

Notes / Random Matrix Theory

8.4 CLT for Eigenvalues

Similar to the CLT for random variables, we have the following theorem.

Theorem 8.20. Let $f$ be analytic in a small neighboring of $[-2,2]$. Let $\lambda _1\le \cdots \le \lambda _N$ be e.v. of $\sqrt {\frac {2}{\beta }}G\beta E,\beta >0$. Then \[ \sum _{i=1}^N f\left (\frac {\lambda _i}{\sqrt {N}}\right )-N\int f(x)\frac {1}{2\pi }\sqrt {4-x^2} \] converges to dist to a Gaussian r.v. $\xi _f$ jointly other several $f=f_1,\dots ,f_K$ with \[ \mathbb {E}[\xi _f]=\left (\frac {2}{\beta }-1\right )m(f),\quad \operatorname {Cov}(\xi _f,\xi _g)=\frac {2}{\beta }C(f,g) \] \[ m(f)=\frac {1}{4}(f(2)+f(-2))-\frac {1}{2\pi }\int _{-2}^2 \frac {f(x)}{\sqrt {4-x^2}}{\mathop{}\!\mathrm{d}} x \] \[ C(f,g) = \frac {1}{4\pi } \int _{-2}^2 \int _{-2}^2 \frac {(f(x)-f(y))(g(x)-g(y))}{(x-y)^2}\frac {4-xy}{\sqrt {4-x^2}\sqrt {4-y^2}}{\mathop{}\!\mathrm{d}} x{\mathop{}\!\mathrm{d}} y. \] or \[ C\left (\frac {1}{z-x},\frac {1}{w-x}\right )=-\frac {1}{2(z-w)^2}\left (1-\frac {zw-4}{\sqrt {z^2-4}\sqrt {w^2-4}}\right ) \quad \text {for $z,w$ out of }[-2,2] \] where $m(f)$ and $C(f,g)$ do not depend on $\beta $.

Proof. The main idea is fancy moments method.

Proof. By Laplacian’s transform \[ {\mathbb{E}} [e^{t_1\eta _1+\cdots +t_m\eta _m}] = \exp \left (\frac {1}{2} \sum _{i,j=1}^m t_it_j\sigma _{ij} \right ). \] Hence, we take derivative and get the expectation \[ {\mathbb{E}} [\eta _1\cdots \eta _m] = \frac {\partial ^m}{\partial t_1\cdots \partial t_m} \left [\exp \left (\frac {1}{2} \sum _{i,j=1}^m t_it_j\sigma _{ij} \right )\right ] \bigg |_{t_1 = \cdots = t_m = 0}. \] Equivalently, this is the coefficient of $t_1\cdots t_m$ in Taylor’s expansion. \[ = \sum _{ij} 2^{-\frac {m}{2}} \prod _{ij} \sigma _{ij}. \] Notice that each perfect match appears $2^{\frac {m}{2}}$ times, we prove the Lemma 8.21. □

Takeaway Moments can be reconstructed by applying a differential operator to Laplace transform.

Proof. From Lecture 6, \[ {\mathbb{E}} \exp (\operatorname {Tr(GUE\cdot Z)})={\mathbb{E}} B_{\lambda _1,\dots ,\lambda _N}(z_1,\dots ,z_N)=\exp \left (\sum _{i=1}^N\frac {\lambda _i^2}{2}\right ) \] where $Z=\operatorname {diag}(z_1,\dots ,z_N)$ and $\lambda _1,\dots ,\lambda _N$ denote the e.v. of GUE.

Then act with $\mathcal {D}$ \[ B_{\lambda _1,\dots ,\lambda _N}(z_1,\dots ,z_N)=\prod _{K=1}^N(K!)\cdot \frac {\operatorname {det}[\exp (\lambda _iz_j)]}{\prod _{i,j}(\lambda _i-\lambda _j)(z_i-z_j)}. \] \[ \mathcal {D}=\prod _{i,j}(z_i-z_j)^{-1}\left (\sum _{i=1}^NT_{a,i}\right )\prod _{i,j}(z_i-z_j). \] So $B$ is eigenfunction of $\mathcal {D}$ with eigenvalue $\sum _{i=1}^N \exp (a\lambda _i)$. Hence, \[ {\mathbb{E}} \left [ \prod _{k=1}^m \left (\sum _{i=1}^N e^{a_k\lambda _i}\right ) \right ] B_{\lambda _1 \cdots \lambda _N} (z_1,\cdots ,z_N) = \mathcal {D}_{a_m} \cdots \mathcal {D}_{a_1} \exp \left (\sum _{i=1}^N\frac {\lambda _i^2}{2}\right ). \] Plug $z_1,\dots ,z_N=0$ and notice $B(0,\cdots ,0)=1$. □

Proof. \begin{align*} \mathcal {D} f(z_1) \cdots f(z_N) &= \sum _{i=1}^N \left [\prod _{j\neq i} \frac {z_i+a-z_j}{z_i-z_j}\right ]\frac {f(z_i+a)}{f(z_i)}\cdot f(z_1)\cdots f(z_N)\\ &=\frac {a^{-1}}{2\pi i}\oint _{\{z_1,\dots ,z_N\}}\prod _{i=1}^N\frac {v+a-z_j}{v-z_j}\cdot \frac {f(v+a)}{f(v)}{\mathop{}\!\mathrm{d}} v f(z_1)\cdots f(z_N) \end{align*} □

Proof. We compute $\mathcal {D}_{a_m} \cdots \mathcal {D}_{a_1} \exp (\frac {z_i^2}{2})\cdots \exp (\frac {z_N^2}{2})$. By sequentially applying Lemma 8.23 and then setting $z_1=\cdots =z_N=0$ at the end □

Now we prove Theorem 8.20 for $\beta =2,f(\frac {\lambda }{\sqrt {N}})=\exp (a\cdot \frac {\lambda }{\sqrt {N}})$

Step 1: Expectation By Corollary 8.24 with $m=1$ \[ {\mathbb{E}} \left [\sum \exp \left (a\cdot \frac {\lambda _i}{\sqrt {N}}\right )\right ]=\frac {(\frac {a}{\sqrt {n}})^{-1}}{2\pi i}\oint _{\{0\}}\left (\frac {v+\frac {a}{\sqrt {N}}}{v}\right )^N\exp \left (\frac {a^2}{2N}+\frac {a}{\sqrt {N}}v\right ){\mathop{}\!\mathrm{d}} v \] No steepest descent needed. Set $v=u\sqrt {N}$ to get \begin{align*} &\frac {N}{a} \frac {1}{2\pi i} \oint \exp \left ( N\log \left (1+\frac {a}{Nu}\right ) + au + \frac {2a^2}{N}\right )\\ =& \frac {N}{a} \frac {1}{2\pi i} \oint \exp \left (a\left (u+\frac {1}{u}\right ) + \frac {a^2}{2N} \left ( 1-\frac {1}{u^2} \right )+O(N^{-2})\right ){\mathop{}\!\mathrm{d}} u\\ =& \frac {N}{a} \frac {1}{2\pi i} \oint \exp \left (a\left (u+\frac {1}{u}\right )\right ) {\mathop{}\!\mathrm{d}} u + \frac {a}{4\pi i} \oint \exp \left (a\left (u+\frac {1}{u}\right )\right )\left (1-\frac {1}{u^2}\right ) {\mathop{}\!\mathrm{d}} u + O(N^{-1}). \end{align*}

We only need to prove \[ \frac {N}{a}\frac {1}{2\pi i}\oint \exp \left (a\left (u+\frac {1}{u}\right )\right ){\mathop{}\!\mathrm{d}} u=N\int _{-2}^2\exp (ax)\frac {1}{2\pi }\sqrt {4-x^2}{\mathop{}\!\mathrm{d}} x. \]

We transform the contour to the unit circle and change variables $x=u+\frac {1}{u}$ ($u=\frac {1}{2}(x\pm i\sqrt {4-x^2}),{\mathop{}\!\mathrm{d}} u=\frac {1}{2}(1\pm i\frac {-x}{\sqrt {4-x^2}}){\mathop{}\!\mathrm{d}} x$) \begin{align*} \frac {a^{-1}}{2\pi i}\oint \exp \left (a\left (u+\frac {1}{u}\right )\right ){\mathop{}\!\mathrm{d}} u&=\frac {a^{-1}}{2\pi i}\left (\int _{-2} ^2e^{ax}\frac {1}{2}(1+ i\frac {-x}{\sqrt {4-x^2}}){\mathop{}\!\mathrm{d}} x+\int _{-2} ^2e^{ax}\frac {1}{2}(1- i\frac {-x}{\sqrt {4-x^2}}){\mathop{}\!\mathrm{d}} x\right )\\ &=\frac {a^{-1}}{2\pi }\int _{-2}^2e^{ax}\frac {x}{\sqrt {4-x^2}}{\mathop{}\!\mathrm{d}} x\overset {\text {by parts}}{=}\frac {a^{-1}}{2\pi }\int _{-2}^2(ae^{ax})\sqrt {4-x^2}{\mathop{}\!\mathrm{d}} x \end{align*}

Conclusion: \[ {\mathbb{E}} \left [\sum _{i=1}^N\exp \left (a\frac {\lambda _i}{\sqrt {N}}\right )\right ]=N\int _{-2}^2e^{ax}\frac {1}{2\pi }\sqrt {4-x^2}{\mathop{}\!\mathrm{d}} x+O\left (N^{-1}\right ) \]

Step 2: Variance

\begin{align*} &{\mathbb{E}} \left [ \sum _{i=1}^N \exp \left (a_1\frac {\lambda _i}{\sqrt {N}}\right ) \sum _{i=1}^N \exp \left (a_2\frac {\lambda _i}{\sqrt {N}}\right )\right ] - {\mathbb{E}} \left [ \sum _{i=1}^N \exp \left (a_1\frac {\lambda _i}{\sqrt {N}}\right )\right ] {\mathbb{E}} \left [ \sum _{i=1}^N \exp \left (a_2\frac {\lambda _i}{\sqrt {N}}\right )\right ]\\ =& N\frac {(a_1a_2)^{-1}}{(2\pi i)^2} {\mathop{\oint\!\!\oint}} \prod _{k=1}^2 \left ( \frac {v_k + \frac {a_k}{\sqrt {N}}}{v_k}\right )^N \exp \left (\frac {a_k^2}{2N} + \frac {a_kv_k}{N}\right )\cdot \left [ \frac {v_1-v_2+\frac {a_1}{\sqrt {N}}-\frac {a_2}{\sqrt {N}}}{(v_1-v_2-\frac {a_2}{\sqrt {N}})(v_1-v_2+\frac {a_1}{\sqrt {N}})} - 1\right ]{\mathop{}\!\mathrm{d}} v_1 {\mathop{}\!\mathrm{d}} v_2. \\ \approx &N^2\frac {(a_1a_2)^{-1}}{2\pi i^2}{\mathop{\oint\!\!\oint}} \exp \left (a_1\left (v_1+\frac {1}{v_1}\right )+a_2\left (v_2+\frac {1}{v_2}\right )\right ) \left [\frac {1+\frac {a_1-a_2}{N(u_1-u_2)}}{(1-\frac {a_2}{N(u_1-u_2)})(1+\frac {a_1}{N(u_1-u_2)})}\right ]{\mathop{}\!\mathrm{d}} u_1{\mathop{}\!\mathrm{d}} u_2. \\ =& N^2 \frac {(a_1a_2)^{-1}}{(2\pi i)^2} {\mathop{\oint\!\!\oint}} \exp \left (a_1\left (v_1+\frac {1}{v_1}\right )+a_2\left (v_2+\frac {1}{v_2}\right )\right ) \left [\frac {a_1 a_2}{N^2(u_1-u_2)^2}\right ]{\mathop{}\!\mathrm{d}} u_1{\mathop{}\!\mathrm{d}} u_2. \end{align*}

Conclusion: Variance $\to \frac {1}{(2\pi i)^2} {\mathop{\oint\!\!\oint}} \exp \left (a_1\left (v_1+\frac {1}{v_1}\right )+a_2\left (v_2+\frac {1}{v_2}\right )\right ) \frac {{\mathop{}\!\mathrm{d}} u_1{\mathop{}\!\mathrm{d}} u_2}{(u_1-u_2)^2}$.

Match this with the formula in Theorem 8.20. Hint: Either directly with the 1st formula for $C(f,g)$, or \[ \sum _{i=1}^Nf(\lambda _i)=\frac {1}{2\pi i}\oint _{\text {around all }\lambda _i}f(z)\sum _{i=1}^N\frac {1}{z-\lambda _i}{\mathop{}\!\mathrm{d}} z \] and use it to compute with $C(\frac {1}{z-x},\frac {1}{w-x})$ in Theorems.

Step 3: Gaussianity We need to show that \[ {\mathbb{E}} \left [\prod _{k=1}^m \left (\sum _{i=1}^N\exp \left (a_k\frac {\lambda _i}{\sqrt {N}}\right )-{\mathbb{E}} \sum _{i=1}^N\exp \left (a_k\frac {\lambda _i}{\sqrt {N}}\right )\right )\right ] \to \text {expressions of Wick's formula}. \] All of them have exactly the same $\prod _{k=1}^m$ part, but cross term part $\prod _{k<l}$ varies. As in Steps 1,2, we change variables $u_k=\sqrt {N}v_k$ and use \[ \text {crossterm}=1+\frac {a_ka_l}{N^2(u_k-u_l)^2}+O\left (N^{-3}\right ) \] The summation over $2^m$ integrals leads to cancellation at parts involving 1. The next term $\to $ perfect matching. □