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      \vbox{\vspace{2mm}
    \hbox to 6.28in { {\bf STAT7017 Big Data Statistics
	\hfill Semester 2 2018} }
       \vspace{4mm}
       \hbox to 6.28in { {\Large \hfill Lecture #1: #2  \hfill} }
       \vspace{2mm}
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\begin{document}
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%\lecture{**LECTURE-NUMBER**}{**DATE**}{**LECTURER**}{**SCRIBE**}
\lecture{3}{6 August}{Dr Dale Roberts}{Rui Qiu}
%\footnotetext{These notes are partially based on those of Nigel Mansell.}

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\textbf{\underline{Limiting Spectral Distribution}}

$\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_n$ $n$ random samples of dimension $p$ (vectors). Sample covariance matrix:

$$S_n=\frac1{n-1}\sum^n_{i=1}(\mathbf{x}_i-\mean{\mathbf{x}})(\mathbf{x}_i-\mean{\mathbf{x}})^*$$ which gives you a $p\times p$ matrix.

$\mean{\mathbf{x}}$ is sample mean given by $\mean{\mathbf{x}}:=\frac{1}{n}\sum^n_{i=1}\mathbf{x}_i$.

Many traditional \underline{multivariate statistics} are functions of the eigenvalues $(\lambda_i$) of $S_n$.

In the most basic form, $T_n=\frac{1}{p}\sum^p_{k=1}\phi(\lambda_k)$. This is just a generalized form, because we don't know what $\phi$ is. But it takes an eigenvalue and gives back a number. $\phi:\mathbb{C}\rightarrow\mathbb{R}.$\\

\underline{\textbf{Example:}} The \underline{generalized variance} (last week) can be written

$$T_n=\frac{1}{p}\log\lvert S_n\rvert=\frac{1}{p}\sum^p_{k=1}\log(\lambda_k)$$

$T_n$ is a ``\underline{linear} spectral statistic of the sample covariance matrix $S_n$ with \underline{test function} $\phi(x)=\log(x)$".\\

First order Random matrix limits are concerned with \underline{when} and \underline{how} shall $T_n\rightarrow c$ (converges to some constant $c$) as $p,n\rightarrow \infty$.

It concerns the ``joint limit" of the $p$ eigenvalues. $(\lambda_k)^p_{k=1}$.\\

\underline{\textbf{Empirical distributions and their limits}}

Let $\mathbb{M}_p(\mathbb{C})$ be $p\times p$ matrices with $\mathbb{C}$-valued entries and let $(\lambda_k)^p_{k=1}$ be the eigenvalues of $A\in\mathbb{M}_p(\mathbb{C})$.

Let $\delta_a(x)=\begin{cases}
	1\ \text{if}\ x=a\\
	0\ \text{otherwise}
\end{cases}$

The \underline{empirical spectral distribution} (ESD) of $A$ is given by

$$F^A_{(x)}:=\frac{1}{p}\sum^p_{k=1}\delta_{\lambda_k}(x)$$

Generally, $F^A$ takes $\mathbb{C}$ values. If $A\in\mathbb{H}_p$, then $F^A(x)\in\mathbb{R}.$\\

\underline{\textbf{Example:}} $A=\begin{pmatrix}
	0 & -1\\
	1 & 0\\
\end{pmatrix}$ eigenvalues are $-i, +i$.

$$F^A=\frac{1}{2}(\delta_i+\delta_{-1})$$

(empirical density ``histogram" vs limiting density)\\

Take a sequence of matrices $\left(A_n\right)_{n\geq 1}\in\mathbb{M}_p(\mathbb{C})$, if the sequence of corresponding ESD $F^{A_n}$ \underline{vaguely converges} to a (possibly defective) measure $F$, we call $F$ the \underline{limiting spectral distribution} (LSD) of $\left(A_n\right)_{n\geq 1}.$\\

\underline{\textbf{Vague convergence}} means that for \underline{any} continuous function that is compactly supported, called $\phi$,

$$F^{A_n}(\phi)\rightarrow F(\phi)\text{ as } n\rightarrow\infty.$$

Here, we use the notation

$$F(\phi):=\int_{\mathbb{R}^p}\phi(x)F(dx).$$

(\textbf{compact supported}: zero outside the range $[a,b]$.)

If the distribution $F$ is \underline{\textbf{non-defective}} (i.e. $\int F(dx)=1$.) then vague convergence becomes \underline{\textbf{weak convergence}}, that is,

$$F^{A_n}(\phi)\rightarrow F(\phi)\text{ as } n\rightarrow\infty$$

for all $\phi$ continuous and \underline{\textbf{bounded}} (below value $a$, say).

In our situation, we shall be dealing with sample covariance matrices ($S_n$). This means that:

\begin{itemize}
	\item support of $F^{S_n}$ is $\mathbb{R}_+$ since $S_n$ are Hermitian and non-negative definite.	
	\item $F^{S_n}(x)=\frac{1}{p}\sum^p_{k=1}\mathbf{1}_{(\lambda_k\leq x)}\ \ \ \ \text{ ESD}.$
	\item Eigenvalues are random variable and ESDs ($F^{S_n}$) are \underline{random} probability distributions on $\mathbb{R}_+$.
\end{itemize}

The \underline{fundamental question} is: Does the limit of $(F^{S_n})$ exist?\\

How can we show this?\\

The eigenvalues of a matrix are continuous functions of the entries of the matrix.

There is no closed-form solution for eigenvalues when dimension of a square matrix is greater than 4.

There are three main techniques used in RMT:

\begin{itemize}
	\item Method of moments.
	\item Orthogonal polynomial decomposition.
	\item Stieltjes transform. (ST)
\end{itemize}

We shall focus on the ST approach.\\

\underline{\textbf{Stieltjes transfrom (ST)}}

The ST plays nearly as useful role in RMT as the Moment generating function (MGF) or characteristic function (CF) in classic probability theory.

It is defined for a measure $\mu$ as:

$$S_\mu(z)=\int\frac{1}{x-z}\mu(dx),\ z\in\mathbb{C}^+$$

where $\mathbb{C}^+=\{x+iy: y>0\}.$

The following lemma allows us to reconstruct the distribution function from its Stieltjes transform.\\

\begin{lemma}
	(Inversion): Let $\mu$ be a probability measure on $\mathbb{R}$. If $a<b$ are points of continuity of the associated distribution, then
	
	$$\mu((a,b))=\lim_{\nu\rightarrow 0^+}\frac{1}{\pi}\int^b_a Im(S_\mu(x+i\nu))dx.$$
\end{lemma}

The following lemma gives a necessary and sufficient condition for a sequence of ST to be the ST of a probability measure.\\

\begin{lemma}
	(Geronimo and Hill, 2003): Suppose that $(\mu_n)$ is a sequence of probability measures on $\mathbb{R}$ with Stieltjes transforms $(S_{\mu_n})$. If $\lim_{n\to\infty}S_{\mu_n}(
	z)=S_{\mu}(z)$ for all $z\in\mathbb{C}^+$, then there exists a probability measure $\mu$ with ST given by $S_\mu$ if and only if
	
	$$\lim_{\nu\to\infty}i\nu S_\mu(i\nu)=-1.$$
	In which case, $\mu_n\to\mu$ in distribution.
\end{lemma}

There are some more technical results that I will now state without proof.

First, we say that a function $f$ is \underline{\textbf{holomorphic}} if it is complex differentiable at every point of its domain, i.e.

$$f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}\text{ exists}.$$

Holomorphic functions are very nice:

\begin{itemize}
	\item Infinitely differentiable.
	\item Equals to its Taylor series.
\end{itemize}

\begin{proposition}
	The Stieltjes transform has the following properties:
	\begin{itemize}
		\item $S_\mu$ is holomorphic on $\mathbb{C}\backslash\Gamma_\mu$ where $\Gamma_\mu:=Supp(\mu)$.
		\item $z\in\mathbb{C}^+\iff S_\mu(z)\in\mathbb{C}^+.$
		\item If $\Gamma_\mu\subset \mathbb{R}_+$ and $z\in\mathbb{C}^+$, then $zS_\mu(z)\in\mathbb{C}^+$.
		\item $\lvert S_\mu(z)\rvert \leq \frac{\mu(1)}{dist(z, \Gamma_\mu) \vee \lvert Im(z)\rvert}$ (Distance of $z$ to support and the maximum of imaginary part of $z$)
	\end{itemize}
\end{proposition}

\begin{proposition}
	The mass $\mu(1)$ can be recovered through the formula
	
	$$\mu(1)=\lim_{\nu\to\infty}-i\nu S_\mu(i\nu)$$
	
	Moreover, for all continuous and compactly supported $\phi:\mathbb{R}\to\mathbb{R}$
	
	$$\mu(\phi)=\int\phi(x)\mu(dx)=\lim_{\nu\to0}\frac1\pi\int\phi(x)Im[S_\mu(x+i\nu)]dx$$
\end{proposition}

\begin{proposition}
	Assume that the following conditions hold for a complex-valued $g(z)$:
	
	\begin{itemize}
		\item $g$ is holomorphic on $\mathbb{C}^+$.
		\item $g(z)\in\mathbb{C}^+$ for all $z\in\mathbb{C}^+$.
		\item $\lim_{\nu\to\infty}\sup\lvert i\nu g(i\nu)\rvert<\infty$.
	\end{itemize}
	
	The $g$ is a ST of a bounded measure on $\mathbb{R}$.
\end{proposition}

\begin{theorem}
	A sequence of measures $(\mu_n)$ converges vaguely to some positive measure $\mu\iff (S_{\mu_n})$ converges to $S_\mu$ on $\mathbb{C}^+$.
\end{theorem}

The idea is that we show $S_{\mu_n}\to S_\mu$ (vague convergence) and then show that $\mu$ is a probability measure by checking that $\mu(1)=1$.

We have $A$ positive semidefinite and symmetric. Then ESD of $A$ is $F^A=\frac{1}{p}\sum^p_{j=1}\delta_{\lambda_j}$, where $A$ is $p\times p$.

\begin{equation}
	\begin{split}
		S_A(z)&=\int\frac{1}{x-z}F^A(dx)\\
		&=\frac{1}{p}\sum^p_{k=1}\int\frac{1}{x-z}\delta_{\lambda_k}(dx)\\
		&=\frac{1}{p}\sum^p_{k=1}\frac{1}{\lambda_k-z}\\
		&=\frac{1}{p}tr[(A-zI)^{-1}]
	\end{split}
\end{equation}

Note: $tr(A+B)=tr(A)+tr(B), tr(A^k)=\sum^p_{i=1}\lambda_i^k.$\\

\underline{\textbf{Trace of an inverse matrix:}} For $n\times n$ matrix $Q$, define $Q_k$ to be the submatrix obtained by deleting $k$-th row and column.

\begin{theorem}
	(Bai and Silvestein, Thm A.U.???): If $B$ and $B_k, k=1,\dots, n,$ are nonsingular and writing $B^{-1}=[b^{kl}]$, then
	
	$$tr(B^{-1})=\sum^n_{k=1}\frac{1}{b_{kk}-B'_kB^{-1}_kB_k}$$
	
	$b_{kk}$: $k$-th diagonal entry of $B$.\\
	$B_k'$: vector obtained from $k$-th row of $B$ by deleting $k$-th entry.\\
	$B_k$: vector obtained from $k$-th column of $B$ by deleting $k$-th entry.
\end{theorem}

Applying this theorem:

\begin{equation}
	S_A(z)=\frac{1}{p}\sum^p_{k=1}\frac{1}{\alpha_{kk}-z-\mathbf{\alpha}_k'(A_k-zI)^{-1}\mathbf{\alpha}_k}
\end{equation}

We would like to show that denominator is equal to

$$g(z, S_A(z))+ o(1)$$

Then we can solve for $S_A(z)=\frac{1}{g(z, S_A(z))}$ to obtain the ST of the ESD.\\

\underline{\textbf{Marchenko-Pastur distributions}}

The \underline{Marchenko-Pastur distribution} $F_{y,\sigma^2}$ with index $y$ and scale parameter $\sigma$ has density

$$P_{y,\sigma^2}(x)=\begin{cases}
	\frac{1}{2\pi xy\sigma^2}\sqrt{(b-x)(x-a)'},&\text{ if } a\leq x\leq b\\
	0,&\text{ otherwise }
\end{cases}$$

Note $a=\sigma^2(1-\sqrt{y})^2,\ b=\sigma^2(1+\sqrt{y})^2$.

If $\sigma^2=1$: \underline{standard MP distribution.}

Special case: $y=1, \sigma^2=1$.

$$P_1(x)=\begin{cases}
	\frac{1}{2\pi x}\sqrt{x(4-x)},&\ 0<x\leq 4,\\
	0,&\ \text{otherwise}.
\end{cases}$$

$\implies$ Hence density is unbounded in region.

As $y\to 0, F_y\to \delta_1$.\\

\underline{\textbf{MP distribution for independent vectors without cross-correlation}}

\begin{equation}
	\begin{split}
		Sn=&=\frac{1}{n-1}\sum^n_{i=1}\mathbf{x}_i\mathbf{x}^*-\cdots\\
		&\approx\frac{1}{n-1}\sum\mathbf{x}_i\mathbf{x}^*
	\end{split}
\end{equation}

We shall sometimes write $n$ sample vectors as $p\times n$ random matrix

$$\mathbf{X}=(\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_n)$$

So

$$\implies S_n=\frac{1}{n}\mathbf{X}\mathbf{X}^*$$

Marchenko and Pastur found the LSD of the large sample covariance matrix $S_n$.\\

\begin{theorem}
	(MP) Suppose that the entries $[x_{ij}]$ of $\mathbf{X}$ are iid complex random variables with mean zero and variance $\sigma^2$, and $p/n\to y\in(0,\infty)$. Then, almost surely,
	
	$$F^{S_n}\to F_{y,\sigma^2}$$
\end{theorem}

This theorem was shown in a special case in 1960s but its influence in statistics was only recognized recently.\\

\underline{How does the MP dist. appear in the limit?}

$\sigma^2=1$.

\begin{equation}
	\begin{split}
		P_y(x)&=\frac{1}{2\pi xy}\sqrt{(b-x)(x-a)}, a\leq x\leq b\\
		a&=(1-\sqrt{{y}})^2, b=(1+\sqrt{y})^2
	\end{split}
\end{equation}

The Stieltjes transform is

\begin{equation}
	\begin{split}
		S(z)&=\int^b_a\frac{1}{x-z}P_y(x)dx\\
		&=\frac{(1-y)-z+\sqrt{(z-1-y)^2-4y}}{2yz}
	\end{split}
\end{equation}

rearranging notice that $s=S(z)$ satisfies the quadratic equation

$$yzs^2+(z+y-1)s+1=0$$

The ST of the ESD of $S_n$ is $S_n(z)=\frac{1}{p}tr[(S_n-zI_p)^{-1}]$.

If we can show $S_n(z)\to S(z)$ as $n\to \infty$ for every $z\in\mathbb{C}^+$, then $F^{S_n}\to F_y$.

By (3.2),

$$S_n(z)=\frac{1}{p}\sum^p_{k=1}\frac{1}{\frac{1}{n}\alpha_k'\mean{\alpha}_k-z-\frac{1}{n^2}\alpha_k'\mathbf{X}_k^*\left(\frac1n\mathbf{X}_k\mathbf{X}_k^*-zI_{p-1}\right)^{-1}\mathbf{X}_k\mean{\alpha}_k}$$

$\mathbf{X}_k=\mathbf{X}$ with $k$-th row removed.

$\alpha_k'=$ $k$-th row of $\mathbf{X}$, size $n\times 1$.

Assume $\mathbb{E}$[``denominator terms with rows removed"]$\to\mathbb{E}$[``terms with rows intact"],

i.e. random error caused by approximation is small for large $p$ and $n$.

$$\mathbb{E}\left[\frac{1}{n}\alpha_k'\mean{\alpha}_k\right]=\frac{1}{n}\sum^n_{j=1}\lvert x_{kj}\rvert ^2=1.$$ 

\begin{lemma}
	Let $u$ be a $n\times 1$ random vector with entries $u_i$ that are all independent with mean $0$ and unit variance. Let $\mathbf{Q}$ be a (non-random) $n\times n$ complex matrix. Then
	
	$$\mathbb{E}\left[u^*\mathbf{Q} u\right]=tr\mathbf{Q}.$$
\end{lemma}

Note $\mathbf{A},\mathbf{B}$ matrices, $(\mathbf{A}\mathbf{B})_{ik}=\sum^m_{j=1}a_{ij}b_{jk}.$

\begin{proof}
	As $u^*\mathbf{Q}u=\sum^n_{i=1}\sum^n_{j=1}\mean{u_i}\mathbf{Q}_{ij}u_j$.
	
	\begin{equation}
		\begin{split}
			\mathbb{E}\left[u^*\mathbf{Q}u\right] &=\sum^n_{i=1}\sum^n_{j=1}\mathbb{E}\left[\mathbf{Q}_{ij}\mean{u_i}u_j\right]\\
			&=\sum^n_{i=1}\mathbf{Q}_{ii}\mathbb{E}\left[\mean{u_i}u_i\right]\\
			&=tr\mathbf{Q}\text{ as } \mathbb{E}[\mean{u_i}u_i]=1.
		\end{split}
	\end{equation}
\end{proof}

\begin{corollary}
	$\mathbb{E}\left[u^*u\right]=n.$
\end{corollary}

\begin{proof}
	Take $\mathbf{Q}=\mathbf{I}_n$, then $tr\mathbf{Q}=tr\mathbf{I}_n=n$.
\end{proof}

\begin{equation}
	\begin{split}
		&\mathbb{E}\left[\frac1{n^2}\alpha'_k\mathbf{X}_k^*\left(\frac1n\mathbf{X}_k\mathbf{X}_k^*-z\mathbf{I}_{p-1}\right)^{-1}\mathbf{X}_k\mean{\alpha}_k\right]\\
		&=\frac{1}{n^2}\mathbb{E}\left[tr\left\{\mathbf{X}_k^*\left(\frac1n\mathbf{X}_k\mathbf{X}_k^*-z\mathbf{I}_{p-1}\right)^{-1}\mathbf{X}_k\mean{\alpha}_k\alpha_k'\right\}\right]\\
		&=\frac{1}{n^2}tr\left\{\mathbb{E}\left[\mathbf{X}_k^*\left(\frac1n\mathbf{X}_k\mathbf{X}_k^*-z\mathbf{I}_{p-1}\right)^{-1}\right]\mathbb{E}\left[\mean{\alpha}_k\alpha_k'\right]\right\}\\
		&=\frac{1}{n^2}tr\left\{\mathbb{E}\left[\mathbf{X}_k^*\left(\frac1n\mathbf{X}_k\mathbf{X}_k^*-z\mathbf{I}_{p-1}\right)^{-1}\mathbf{X}_k\right]\right\}\\
		&=\frac{1}{n^2}\mathbb{E}\left[tr\left\{\mathbf{X}_k^*\left(\frac1n\mathbf{X}_k\mathbf{X}_k^*-z\mathbf{I}_{p-1}\right)^{-1}\mathbf{X}_k\right\}\right]\\
		&=\frac{1}{n^2}\mathbb{E}\left[tr\left\{\left(\frac1n\mathbf{X}_k\mathbf{X}_k^*-z\mathbf{I}_{p-1}\right)^{-1}\mathbf{X}_k\mathbf{X}^*_k\right\}\right]
	\end{split}
\end{equation}

We note that $\frac{1}{n}\mathbf{X}_k\mathbf{X}^*_k\approx S_n$ (only 1 vector removed).

So

\begin{equation}
	\begin{split}
		&\frac{1}{n^2}\mathbb{E}\left[tr\left\{\left(\frac{1}{n}\mathbf{X}_k\mathbf{X}_k^*-z\mathbf{I}_{p-1}\right)^{-1}\mathbf{X}_k\mathbf{X}_k^*\right\}\right]\\
		&\approx\frac{1}{n^2}\mathbb{E}\left[tr\left\{\left(\frac{1}{n}\mathbf{X}\mathbf{X}^*-z\mathbf{I}_{p}\right)^{-1}\mathbf{X}\mathbf{X}^*\right\}\right]\\
		&=\frac{1}{n}\mathbb{E}\left[tr\left\{\mathbf{I}_p+z\left(\frac{1}{n}\mathbf{X}\mathbf{X}^*-z\mathbf{I}_p\right)^{-1}\right\}\right]\\
		&=\frac{p}{n}+z\frac{p}{n}\mathbb{E}\left[S_n(z)\right].
	\end{split}
\end{equation}

So denominator is roughly

$$1-z-\left\{\frac{p}{n}+z\frac{p}{n}\mathbb{E}\left[S_n(z)\right]\right\}$$

as $p\to\infty, n\to\infty$ and $p/n\to y> 0$.

$$\mathbb{E}\left[S_n(z)\right]\to S(z)$$

So denominator

$$\to 1-z-(y+zyS(z))$$

and

$$S(z)=\frac{1}{1-z-(y+yzS(z))}$$

This is ST of MP distribution $F_y$!

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\end{document}