text:
\begin{document}

\thanks{Authors supported by the Generalitat de Catalunya (grant 2014 SGR 289)  and the Spanish Ministerio de Econom\'ia y Competividad (project MTM2011-27932-C02-01)}.

\begin{abstract}
We study hyperbolic Gaussian analytic functions in the unit polydisk of $\C^n$. Following the scheme previously used in the unit ball we first study the asymptotics of fluctuations of linear statistics as the directional intensities $L_j$, $j=1,\dots,n$ tend to $\infty$. Then we estimate the probability of large deviations of such linear statistics and use the estimate to prove a hole theorem. Our proofs are inspired by the methods of M. Sodin and B. Tsirelson for the one-dimensional case, and B. Shiffman and S. Zelditch for the study of the analogous problem for compact K\"ahler manifolds.
\end{abstract}

\maketitle

\section*{Introduction}

This paper studies some properties of the zero sets of Gaussian analytic functions in the polydisk. The plan of the paper and the techniques of the proofs are the same as in \cite{BMP}, where the analogous problems in the unit ball were dealt with, so we will often just outline the proofs and refer to \cite{BMP} for the details.

Consider the unit polydisk in $\mathbb C^n$
\[
 \mathbb D^n=\bigl\{z\in\C^n : |z_j|<1, \ j=1,\dots, n\bigr\}
\]
and the normalised invariant measure
\[
 d\nu_n(z)=\frac{dm(z)}{\pi^n [1-|z|^2]^2}\ ,
\]
where $dm$ stands for the Lebesgue measure. We simply write $d\nu$ when no confusion about the dimension can arise.

Here and throughout the paper we use the standard notations
\[
 [1-|z|^2]=\prod_{j=1}^n (1-|z_j|^2)
\]
and, for $L=(L_1,\dots,L_n)$,
\[
 [1-|z|^2]^L=\prod_{j=1}^n (1-|z_j|^2)^{L_j}\ .
\]

Given a vector $L$ with $L_j>1$, $j=1,\dots, n$, consider the weighted Bergman space
\[
 B_L(\mathbb D^n)=\bigl\{f\in H(\D^n) : \|f\|_{n,L}^2:=c_{n,L} \int_{\D^n} |f(z)|^2 [1-|z|^2]^{L}  d\nu_n(z)< +\infty\bigr\} ,
\]
where $c_{n,L}=\prod_{j=1}^n(L_j-1)$ is chosen so that $\|1\|_{n,L}=1$.

Consider also the normalisation of the monomials $z^\alpha$ in the norm $\|\cdot\|_{n,L}$:
\[
 e_{\alpha}(z)=\prod_{j=1}^n \left(\frac{\Gamma(L_j+\alpha_j)}{\alpha_j!\Gamma(L_j)}\right)^{1/2} z^\alpha\ .
\]
As usual here we denote $z=(z_1,\dots,z_n)$ and use the multi-index notation $\alpha=(\alpha_1,\dots,\alpha_n)$, $\alpha !=\alpha_1!\cdots \alpha_n!$, $|\alpha|=|\alpha_1|+\cdots +|\alpha_n|$ and $z^\alpha=z_1^{\alpha_1}\cdots z_n^{\alpha_n}$.

The \emph{hyperbolic Gaussian analytic function} (GAF) of \emph{intensity} $L=(L_1,\dots, L_n)$ is defined as
\[
 f_L(z)=\sum_\alpha a_\alpha e_{\alpha}(z)=\sum_\alpha a_\alpha  \prod_{j=1}^n \left(\frac{\Gamma(L_j+\alpha_j)}{\alpha_j!\Gamma(L_j)}\right)^{1/2} z^\alpha \qquad z\in\D^n,
\]
where $a_\alpha $ are i.i.d. complex Gaussians of mean 0 and variance 1 (denoted $a_\alpha \sim N_{\C}(0,1)$).

The sum defining $f_L$ can be analytically continued to $L_j>0$, which we assume henceforth.

The probabilistic properties of the hyperbolic GAF are determined by its covariance kernel, which is given by (see \cite{ST1}*{Section 1}, \cite{Stoll}*{p.17-18}):
\begin{align*}
 K_L(z,w)&=\E[f_L(z)\overline{f_L(w)}]= \sum_\alpha  \prod_{j=1}^n \frac{\Gamma(L_j+\alpha_j)}{\alpha_j!\Gamma(L_j)} z^\alpha \bar w^\alpha=
 \prod_{j=1}^n \sum_{\alpha_j=0}^\infty   \frac{\Gamma(L_j+\alpha_j)}{\alpha_j! \Gamma(L_j)} (z_j\bar w_j)^{\alpha_j} \\
 &=
 \prod_{j=1}^n \frac 1{(1-z_j \bar w_j)^{L_j}}=
 \frac 1{[1-z \bar w]^{L}}\ .
\end{align*}

In this paper we follow the scheme of \cite{BMP} and study some statistical properties of the zero variety
\[
 Z_{f_L}=\left\{z\in \D^n;\ f_L(z)=0\right\}\ .
\]
A main feature of the hyperbolic GAF is that the distribution of $Z_{f_L}$ is invariant under a large subgroup of the holomorphic automorphisms group $\Aut(\D^n)$.
Consider the group $\mathcal A$ consisting of the automorphisms of the form
\[
 \phi_w^\theta(z)=\left(e^{i\theta_1} \frac{z_1-w_1}{1-\bar w_1 z_1}, \dots, e^{i\theta_n} \frac{z_n-w_n}{1-\bar w_n z_n}\right),\qquad w\in\D^n; \ \theta_j\in [0,2\pi)\ .
\]
We use the notation $\phi_w(z)$ in case $\theta_j=0$, $j=0,\dots,n$.

Any automorphism in $\Aut(\D^n)$ is the composition of an element of $\mathcal A$ with a permutation of the coordinates (see for instance \cite{Sha}*{Theorem 2, pag. 48}).

The transformations
\[
 T_w(f)(z)=\frac{[1-|w|^2]^{L/2}}{[1-\bar w z]^L} f(\phi_w^\theta(z))
\]
are isometries of $B_L(\D^n)$, hence the random zero sets $Z_{f_L}$ and $Z_{f_L\circ\phi_w^\theta}$ have the same distribution. More specifically, the distribution of the (random) integration current
\[
 [Z_{f_L}]=\frac {i}{2\pi}\partial\bar\partial\log |f_L|^2\ ,
\]
is invariant under the subgroup $\mathcal A$. In case $L_1=\cdots=L_n$ then $[Z_{f_L}]$ is invariant under the whole group $\Aut(\D^n)$.

The typical distribution of $Z_{f_L}$ is given by the Edelman-Kostlan formula (see \cite{HKPV}*{Section 2.4} and \cite{Sod}*{Theorem 1}): the so-called \emph{first intensity} of the GAF is
\[
\E[Z_{f_L}] =\frac{i}{2\pi}\partial \overline{\partial}\log K_L(z,z)= \omega_L (z)\ ,
\]
where $\omega_L$ is the form
\[
\omega_L(z)=\sum_{j=1}^n \frac{L_j}{(1-|z_j|^2)^2} \frac{i}{2\pi} dz_j\wedge d\bar{z_j}\ .
\]
When $L_j=1$, $j=1,\dots,n$, we simply denote $\omega$.
Notice that
\[
 \omega^n(z)= n!\bigwedge_{j=1}^n \frac{1}{(1-|z_j|^2)^2} \frac{i}{2\pi} dz_j\wedge d\bar{z_j}=\frac{n!}{\pi^n}\frac{dm(z)}{[1-|z|^2]^2}=n!\;  d\nu_n(z)\ ,
\]
which is invariant by $\Aut(\D^n)$ \cite{Stoll}*{p.19}.

In Section 1 we study the fluctuations of linear statistics as the $L_j$ tend to $\infty$. Let $\mathcal D_{(n-1,n-1)}$ denote the space of real-valued, compactly supported, $\mathcal C^2$ forms of bidegree $(n-1,n-1)$. For  $\varphi\in\mathcal D_{(n-1,n-1)}$, consider the integral of $\varphi$ over $Z_{f_L}$
\[
I_L(\varphi)=\int_{Z_{f_L}}\varphi=\int_{\D^n}  \varphi  \wedge [Z_{f_L}]
\]
and note that the Edelman-Kostlan formula yields
\begin{equation}\label{E}
\E[I_{L}(\varphi)]=\int_{\D^n}\varphi\wedge \omega_L\ .
\end{equation}

\begin{theorem}\label{LS}
 Let $\varphi\in \mathcal D_{(n-1,n-1)}$ and let $D\varphi$ be the function defined by $\frac i{2\pi}\partial\bar\partial\varphi=D\varphi d\nu$. Then
 \[
  \Var [I_L(\varphi)]= \frac{\zeta(n+2)}{\prodj}\left(\int_{\D^n} (D\varphi)^2 d\nu\right)\Bigl[1+\textrm{O}(\sum_{j=1}^n\frac{\log L_j}{L_j})\Bigr].
 \]
\end{theorem}
Since $|\varphi\wedge \omega_L|\leq C(\varphi) \sum_{j=0}^n L_j$ (see \eqref{phiomega})  this shows a strong self-averaging of the volume $I_L(\varphi)$ (which increases with the dimension), in the sense that the variance is much smaller than the square of the typical values.

The same computations involved in the proof of this theorem show the asymptotic normality of $I_L(\varphi)$, i.e., that the distribution of
\[
\frac{I_L(\varphi)-\E[I_L(\varphi)]}{\sqrt{\Var [I_L(\varphi)]}}
\]
converges weakly to the (real) standard gaussian (Corollary~\ref{normality}), for each $\varphi$.

In Section 2, we estimate the probability of large deviations for $I_L(\varphi)$.

\begin{theorem}\label{smoothlargedeviations}
For all $\varphi\in \mathcal D_{(n-1,n-1)}$ and $\delta>0$, there exist $c>0$ and $L_j^0(\varphi,\delta,n)$, $j=1,\dots,n$, such that for all $L_j\geq L_j^0$,
\[
\mathbb P\bigl[|I_L(\varphi)-\E(I_L(\varphi))|>\delta \E(I_L(\varphi))\bigr]\leq e^{-c\spj}.
\]
\end{theorem}

\begin{remarks*}
 1. In case $n=1$ the result coincides with \cite{Bu-T13}*{Theorem 5.7} (see also \cite{BMP}*{Theorem 2}). Also, fixing $L_j$, $j\neq i$, and letting $L_i\to\infty$ we see that the exponent is of order $L_i^2$, which corresponds again to the one-dimensional case (for the coordinate $z_i$).

 2. When $L_j=L$ for all $j$, the exponent $(\sum_{j=0}^n L_j)(\prod_{j=0}^n L_j)$ is of order $L^{n+1}$, as  in the ball (see \cite{BMP}*{Theorem 2}) .

\end{remarks*}

Following the scheme of \cite{SZZ}*{pag.1994} we deduce a corollary that implies the upper bound in the hole theorem (Theorem~\ref{holethm} below).

For a smooth compactly supported function $\psi$ in $\D^n$ consider the $(n-1,n-1)$-form
\[
 \varphi=\psi \frac{\omega^{n-1}}{(n-1)!}\ .
\]
In this case
\[
 \varphi (z) \wedge \omega_L (z) =\bigl(\sumj\bigr) \psi(z) d\nu(z)\ .
\]

Define
\[
 I_L(\psi)=\int_{Z_{f_L}}\psi\frac{\omega^{n-1}}{(n-1)!} =\int_{\mathbb D^n} \psi \wedge \frac{\omega^{n-1}}{(n-1)!} \wedge [Z_{f_L}]
\]
and note that \eqref{E} gives here
\[
 \E[I_L(\psi)]=\bigl(\sumj\bigr) \int_{\D^n} \psi\, d\nu\ .
\]

In particular, and for an open set $U\subset\D^n$ let $\chi_U$ denote its characteristic function and let  $I_L(U)=I_L(\chi_U)$. Then $\E[I_L(U)]=\bigl(\sum_{j=0}^n L_j\bigr) \nu(U)$.

\begin{corollary}\label{conseqSLD}
Suppose that $U$ is an open set contained in a compact subset of $\D^n$. For all $\delta>0$ there exist $c>0$ and $L_j^0$ such that for all $L_j\geq L_j^0$,
\[
\mathbb P\Bigl[\Bigl|\frac{1}{\sum_{j=0}^n L_j}I_L(U)-\nu(U)\Bigr|>\delta\Bigr]\leq e^{-c\spj}.
\]
\end{corollary}

The proof of this is as in the ball (see \cite{BMP}*{Corollary 5}), so we skip it.

In the last Section we study the probability that $Z_{f_L}$ has a pseudo-hyperbolic hole of polyradius $r$. Given $w\in \D^n$ and $r=(r_1,\dots,r_n)$, $r_j \in (0,1)$ consider the pseudo-hyperbolic polydisk
\[
 E(w,r)=\Bigl\{z\in\D^n : \bigl|\frac{z_j-w_j}{1-z_j\bar w_j}\bigr|<r_j, j=1,\dots,n\Bigr\}\ .
\]
By the invariance of the distribution of the zero variety under the automorphisms $\mathcal A$, the probability that $Z_{f_L}$ does not intersect $E(w,r)$ is the same as the probability that $Z_{f_L}\cap E(0,r)=\emptyset$.

\begin{theorem}\label{holethm}
Let $r=(r_1,\dots,r_n)$, $r_j\in (0,1)$, be fixed. There exist $C_1=C_1(n,r)>0$, $C_2=C_2(n,r)>0$ and $L_j^0$ such that for all
$L_j\geq L_j^0$,
\[
e^{-C_{1}\spj} \leq \mathbb P\bigl[Z_{f_L}\cap E(0,r)=\emptyset\bigr]\leq e^{-C_{2}\spj}.
\]
\end{theorem}

A final word about notation. By $A\lesssim B$ we mean that there exists $C>0$ independent of the relevant variables of $A$ and $B$ for which $A\leq CB$. Then $A\simeq B$ means that $A\lesssim B$ and $B\lesssim A$.

\section{Linear statistics. Proof of Theorem~\ref{LS}}

The proof is as in \cite{BMP}*{Section 1} so we keep it short. The starting point is the following bi-potential expression of the variance:
\begin{align}\label{bipotential}
 \Var[I_L(\varphi)]&=
 \int_{\D^n}\int_{\D^n}  \rho_L(z,w) \frac i{2\pi}\partial\bar\partial\varphi(z) \frac i{2\pi}\partial\bar\partial\varphi (w)\\
 &=\int_{\D^n}\int_{\D^n}  \rho_L(z,w) D\varphi(z) D\varphi(w) d\nu(z) d\nu(w) \ , \nonumber
\end{align}
where $\rho_L(z,w)=4\Cov (\log|\hat f(z)|,\log|\hat f(w)|)$. By \cite{HKPV}*{Lemma 3.5.2}
\begin{align*}
  \rho_L(z,w)=\sum_{m=1}^\infty\frac{|\theta_L(z,w)|^{2m}}{m^2}\ ,
 \end{align*}
 where
\begin{equation}\label{norm-kernel}
\theta_L(z,w):=\frac{K_L(z,w)}{\sqrt{K_L(z,z)} \sqrt{K_L(w,w)}}=\frac{[1-|z|^2]^{L/2} [1-|w|^2]^{L/2}}{[1-\bar z w]^L}
\end{equation}
is the normalised covariance kernel of $f_L$.

For $\zeta,\xi\in\D$ let
\[
 \rho(\zeta,\xi)=\Bigl|\frac{\zeta-\xi}{1-\bar \zeta\xi}\Bigr|\ .
\]
Notice  that
\[
 |\theta_L(z,w)|^2=\prod_{j=1}^n \bigl(1-\bigl|\frac{z_j-w_j}{1-z_j\bar w_j}\bigr|^2\bigr)^{L_j}
 =\prod_{j=1}^n \bigl(1-\rho^2(z_j,w_j)\bigr)^{L_j}\ ,
\]
hence
\[
 |\theta_L(z,w)|^2=|\theta_L(\phi_w(z), 0)|^2
\]

We see next that only the near diagonal part of the double integral \eqref{bipotential} is relevant.
Let $\varepsilon_{j}=1/L_j^{2}$ and define
\[
 R_L^\epsilon=\Bigl\{(z,w)\in \D^n\times \D^n : 1-\rho^2(z_j,w_j)\geq \epsilon_j^{1/L_j}, j=1,\dots, n\Bigr\}\ .
\]

Split the integral into three parts
\begin{align}
  \Var[I_L(\varphi)]&=\int\limits_{(\D^n\times \D^n)\setminus R_L^\epsilon} \rho_L(z,w) D\varphi(z) D\varphi(w) d\nu(z) d\nu(w) \tag{I1} \\
  &\quad + \int_{R_L^\epsilon} \rho_L(z,w) (D\varphi(z)-D\varphi(w)) D\varphi(w) d\nu(z) d\nu(w) \tag{I2}\\
  &\quad + \int_{R_L^\epsilon} \rho_L(z,w) ( D\varphi(w))^2 d\nu(z) d\nu(w)\tag{I3}\ .
\end{align}

The bound for the first integral is a consequence of the estimate
\begin{equation}\label{est-rho}
 |\theta_L(z,w)|^2 \leq \rho_L(z,w)\leq 2 |\theta_L(z,w)|^2\ ,
\end{equation}
which follows immediately from $x\leq\sum_{m=1}^\infty\ x^m/m^2 \leq 2x$, $ x\in[0,1]$.

Then, by the definition of $R_L^\epsilon$,
\[
 |\textrm{I1}|\leq 2 \bigl(\prod_{j=1}^n \varepsilon_{j}\bigr) \int_{(\D^n\times \D^n)\setminus R_L^\epsilon} |D\varphi(z) D\varphi(w)| d\nu(z) d\nu(w)
 \leq  \frac 2{\prod_{j=1}^n L_j^2} \left(\int_{\D^n} |D\varphi(z)|\, d\nu(z)\right)^2\ .
\]

By the uniform continuity of $i\partial\bar\partial\varphi$ there exists a regular function $\eta(x_1,\dots,x_n)$ with $ \eta(1,\dots,1)=0$ and such that for all $z,w\in\D^n$,
\[
 |D\varphi(z)-D\varphi(w)|\leq \eta\bigl(1-\rho^2(z_1, w_1),\dots, 1-\rho^2(z_n, w_n)\bigr)\ .
\]
Since $\eta(x)\lesssim \|1-x\|$ for $x=(x_1,\dots, x_n)$, $x_j$ near 1, we see that for $(z,w)\in R_L^\epsilon$
\begin{align*}
|D\varphi(z)-D\varphi(w)|&\lesssim \|(\rho^2(z_1, w_1),\dots, \rho^2(z_n, w_n))\|
 \lesssim
 \max_j\bigl|1-\varepsilon_j^{1/L_j}\bigr|\simeq\max_j \frac{\log L_j}{L_j}\ ,
\end{align*}

By the invariance by automorphisms of the measure $d\nu$, we get
\begin{align*}
 |\textrm{I2}|&\lesssim \Bigl( \max_j \frac{\log L_j}{L_j}\Bigr) \int\limits_{R_L^\epsilon\cap(supp\; \varphi\times supp\; \varphi)} [1-|\phi_z(w)|^2]^L\,  d\nu(z) d\nu(w)\\
&\lesssim \; \Bigl( \max_j \frac{\log L_j}{L_j}\Bigr)
\int_{supp\;\varphi}\Bigl(\int_{\{z: 1-|z_j|^2 \geq\varepsilon_{j}^{1/L_j}\ \forall j\}} [1-|z|^2]^L d\nu(z)\Bigr) d\nu(w)\\
&\lesssim \; \Bigl( \max_j \frac{\log L_j}{L_j}\Bigr) \int_{\{z: 1-|z_j|^2 \geq\varepsilon_{j}^{1/L_j}\ \forall j\}} [1-|z|^2]^L d\nu(z)\ .
\end{align*}

On the other hand, using again the invariance of $d\nu$, we see that
\begin{align*}
 \textrm{I3}&=

 \left(\int_{\D^n} (D\varphi(w))^2 d\nu(w)\right) \int_{\{z: 1-|z_j|^2 \geq\varepsilon_{j}^{1/L_j}\ \forall j\}} \rho_L(z,0) d\nu(z)\ .
\end{align*}
By \eqref{est-rho} we have thus $\textrm{I2}=\textrm{o}(\textrm{I3})$ and therefore
\begin{equation}\label{ls}
 \Var[I_L(\varphi)]=\textrm{I3} \bigl(1+\textrm{O}(\max_j \frac{\log L_j}{L_j})\bigr) \ .
\end{equation}

It remains to compute the second factor in I3:
\begin{align*}
 J:&=\int_{\{z: 1-|z_j|^2 \geq\varepsilon_{j}^{1/L_j}\ \forall j\}} \rho_L(z,0) d\nu(z)=\sum_{m=1}^\infty\frac 1{m^2} \int_{\{z: 1-|z_j|^2 \geq\varepsilon_{j}^{1/L_j}\ \forall j\}} [1-|z|^2]^{mL} d\nu(z)\\
 &=\sum_{m=1}^\infty\frac 1{m^2} \prod_{j=1}^n \int_{\{z_j: 1-|z_j|^2 \geq\varepsilon_{j}^{1/L_j}\}} (1-|z_j|^2)^{mL_j-2}\ \frac{dm(z_j)}{\pi}\ .
\end{align*}

Using the computations in \cite{BMP}*{Section 1} for $n=1$ we see that
\begin{align*}
 \int_{\{z_j: (1-|z_j|^2)^{L_j} \geq\varepsilon_{j}\}} (1-|z_j|^2)^{mL_j-2}\ \frac{dm(z)}{\pi}&=\int_0^{(1-\epsilon_j^{1/L_j})^{1/2}} (1-r^2)^{mL_j-2} 2r\ dr\\
 &=\int_{\epsilon_j^{1/L_j}}^1 s^{mL_j-2} ds=\frac 1{mL_j}\bigl[1+\textrm O(\frac 1{mL_j})\bigr]
\end{align*}
Therefore
\[
 J=\sum_{m=1}^\infty\frac 1{m^2} \prod_{j=1}^n \frac{1}{m L_j}\bigl[1+\textrm O(\frac 1{mL_j})\bigr]
 =\frac 1{\prodj}\; \zeta(n+2)\; \bigl[1+\textrm O(\max_j \frac 1{L_j})\bigr]
\]

This and \eqref{ls} give the stated result.

As an immediate consequence of the results of M. Sodin and B. Tsirelson and the previous computations we obtain the asymptotic normality of $I_L(\varphi)$. The proof is as in \cite{BMP}*{Corollary 5}, so we skip it.

\begin{corollary}\label{normality} As $L\to\infty$ the distribution of the normalised random variable
 \[
\frac{I_L(\varphi)-\E[I_L(\varphi)]}{\sqrt{\Var(I_L(\varphi))}}
\]
tends weakly to the standard (real) gaussian, for each $\varphi$.
\end{corollary}

\section{Large deviations. Proof of Theorem \ref{smoothlargedeviations}}

Applying Stokes' theorem, we have
\begin{align*}
I_{L}(\varphi)-\E\left[I_{L}(\varphi)\right]&
 =\int_{\D^n}\varphi\wedge \frac{i}{2\pi}\partial\overline{\partial}\log\frac{|f_L|^2}{K_L(z,z)}
=\int_{\D^n}\log\frac{| f_L|^2}{K_L(z,z)}\frac{i}{2\pi}\partial\overline{\partial}\varphi.
\end{align*}
Thus,
\[
|I_{L}(\varphi)-\E[I_{L}(\varphi)]|\leq \|D\varphi\|_\infty \int_{\text{supp}\varphi}\left|\log |\hat f_{L}(z)|^2\right|d\nu(z).
\]
Writing the form as
\[
 \varphi=\bigl(\frac i{2\pi}\bigr)^{n-1} \sum_{j,k=1}^n\varphi_{jk}\Bigl(\frac{dz_1}{1-|z_1|^2}\wedge\stackrel{\stackrel{\hat \j}{\smile}}{\dots}\wedge\frac{dz_n}{1-|z_n|^2}\wedge \frac{d\bar z_1}{1-|z_1|^2}\wedge\stackrel{\stackrel{\hat k}{\smile}}{\dots}\wedge\frac{d\bar z_n}{1-|z_n|^2} \Bigr)
\]
we see that
\[
 \varphi\wedge\omega_L=\sum_{j=1}^n\varphi_{jj} L_j\bigwedge_{k=1}^n\frac i{2\pi}\frac{dz_k\wedge d\bar z_k}{(1-|z_k|^2)^2}=\bigl(\sum_{j=1}^n L_j\varphi_{jj}\bigr) d\nu(z)\ ,
\]
and therefore
\begin{equation}\label{phiomega}
 \bigl|\varphi\wedge\omega_L\bigr|\lesssim c(\varphi) \bigl(\sum_{j=1}^n L_j \bigr)\ .
\end{equation}

This shows that the proof of Theorem \ref{smoothlargedeviations} will be completed as soon as we prove the following Lemma.

\begin{lemma}\label{mainlemmaSLD}
For any $\varphi\in\mathcal D_{(n-1,n-1)}$ and any $\delta>0$ there exists $c>0$ such that
\[
\mathbb P\left[ \int_{supp\ \varphi}\left|\log |\hat f_L(z)|^2 \right|d\nu(z)>\delta \sumj \right]\leq e^{-c\spj}.
\]
\end{lemma}
The key ingredient in the proof of this lemma is given by the following control on the average of
$\bigl|\log |\hat f_L|^2\bigr|$ over pseudo-hyperbolic polydisks.

\begin{lemma}\label{controlmean}
There exists a constant $c>0$ such that for a pseudo-hyperbolic polydisk $E=E(z_0,s)$, $z_0\in\D^n$, $s\in (0,1)$,
\[
\mathbb P\left[\frac{1}{\nu(E)}\int_{E}\left|\log |\hat f_L(\xi)|^2 \right|d\nu(\xi)>5 n(\sumj)\nu(E)^{1/n} \right]\leq e^{-c\spj}.
\]
\end{lemma}

Let us see first how this allows to complete the proof of Lemma~\ref{mainlemmaSLD}, and therefore of Theorem~\ref{smoothlargedeviations}.

\begin{proof}[Proof of Lemma~\ref{mainlemmaSLD}] Cover $K:=supp\; \varphi$ with pseudo-hyperbolic polydisks $E_k=E(\lambda_k,\eps)$, $k=1,\dots,N$ of fixed invariant volume $\nu(E_k)=\eta$ (to be determined later on). A direct estimate shows that $N\simeq \nu(K)/\eta$.

By Lemma~\ref{controlmean}, outside an exceptional event of probability $Ne^{-c\spj}\leq e^{-c'\spj}$,
\begin{align*}
\int_{K}\left|\log|\hat f_L(\xi)|^2\right|d\nu(\xi)&\leq \sum^{N}_{k=1}\int_{E_k}\left|\log |\hat f_L(\xi)|^2 \right|d\nu(\xi) \lesssim
(\sumj) N \eta^{1+1/n}\\
& \simeq (\sumj) \nu(K) \eta^{1/n}.
\end{align*}
Choosing $\eta$ such that $\nu(K)\eta^{1/n}=\delta$ we are done.
\end{proof}

Now we proceed to prove Lemma~\ref{controlmean}. A first step is the following lemma.

\begin{lemma}\label{controlmax}
Fix $r=(r_1,\dots,r_n)\in (0,1)^n$, $z_0\in\D^n$ and $\delta>0$. There exists $c>0$ and $L_j^0=L_j^0(r,\delta)$ such that for all $L_j\geq L_j^0$
\begin{itemize}
 \item[(a)] $P \bigl[\max\limits_{E(z_0,r)}\log|\hat f_L(z)|^2<-\delta \sumj \bigr]\leq e^{-c\spj }$,
 \item[(b)] $P \bigl[\max\limits_{E(z_0,r)}\log|\hat f_L(z)|^2>\delta \sumj \bigr]\leq e^{-c\spj}$.
\end{itemize}
Combining both estimates $\mathbb P\bigl[\max\limits_{E(z_0,r)}\left|\log |\hat f_L(z)|^2\right|>\delta \sumj \bigr]\leq e^{-c\spj}$.

\end{lemma}

\begin{proof} By the invariance of the distribution of $\hat f$ under $\mathcal A$, it is enough to consider the case $z_0=0$.

(a) Consider the event
 \[
  \mathcal E_1=\left\{ \max_{E(0,r)}\log|\hat f_L(z)|^2<-\delta \sumj \right\}\ .
 \]
Note that
\begin{align*}
\log |\hat f_L(z)|^2=\log\frac{|f_L(z)|^2}{K_L(z,z)}& =\log|f_L(z)|^2-\log \prod_{j=1}^n \frac{1}{(1-|z_j|^2)^{L_j}},

\end{align*}
hence, by plurisubharmonicity,
\begin{align*}
\mathcal E_1& \subset \left\{ \max_{E(0,r)} \log |f_L(z)|^2 \leq \sumj \bigl(\log\frac{1}{1-r_j^2}- \delta\bigr)  \right\}.
\end{align*}

Therefore, for a suitable $\tilde\delta=\tilde\delta(r)$,
\[
 \mathcal E_1 \subset \left\{ \max_{E(0,r)} \log |f_L(z)|^2 \leq (1-2\tilde\delta) \bigl(\sumj \log\frac{1}{1-r_j^2}\bigr)\right\}\
\]
and the estimate of $\mathbb P[\mathcal E_1]$ will be done as soon as we prove the following lemma.

\begin{lemma}\label{maxlogfL}
For $0<\delta<1/2$ and a polyradius $r=(r_1,\dots,r_n)$ there exist $c=c(\delta,r)$ and $L_j^0=L_j^0(\delta,r)$ such that for all $L_j\geq L_j^0$
\[
\mathbb P\left[\max_{E(0,r)} \log|f_L(z)| \leq \bigl(\frac{1}{2}-\delta\bigr)\sumj \log\frac{1}{1-r_j^2} \right]\leq e^{-c\spj}\ ·
\]
\end{lemma}

\begin{proof} [Proof of Lemma~\ref{maxlogfL}]
Under the event we want to estimate we have
\[
\max_{|z|=r}|f_L(z)|\leq [1-r^2]^{-L\left(\frac{1}{2}-\delta\right)}\ .
\]
We shall see that this implies that some coefficients of the series of $f_L$ are necessarily ``small", something that only happens with a probability less than $\exp(-c\spj)$.

Writing $f_L$ in Taylor series we see that
\[
a_{\alpha}=\left(\prod_{j=1}^n \frac{\alpha_j!\Gamma(L_j)}{\Gamma(L_j+\alpha_j)}\right)^{1/2}\frac{\partial^{\alpha}f_L(0)}{\alpha!}.
\]
With this and Cauchy's estimates
\[
\left| \frac{\partial^{\alpha}f_L(0)}{\alpha!}\right|\leq \frac{\max\limits_{E(0,r)} |f_L|}{r^\alpha}
\]
we have
\begin{equation*}\label{coeff}
|a_{\alpha}|\leq \left(\prod_{j=1}^n \frac{\alpha_j!\Gamma(L_j)}{\Gamma(L_j+\alpha_j)}\right)^{1/2}
\frac 1{r^\alpha [1-r^2]^{(\frac 12-\delta)L}} \ .
\end{equation*}

Stirling's formula
\begin{equation*}\label{stirling}
 \Gamma(z)=\sqrt{\frac{2\pi}z} \left(\frac ze\right)^z \bigl[1+\textrm O(\frac 1z)\bigr]
\end{equation*}
yields
\begin{align*}
\frac{\alpha_j!\Gamma(L_j)}{\Gamma(L_j+\alpha_j)}& \simeq \sqrt{2\pi}\alpha_j \sqrt{\frac{L_j+\alpha_j}{L_j \alpha_j}}
 \left(\frac{\alpha_j}{L_j+\alpha_j}\right)^{\alpha_j} \left(\frac{L_j}{L_j+\alpha_j}\right)^{L_j}\\
 &\leq \alpha_j
 \left(\frac{\alpha_j}{L_j+\alpha_j}\right)^{\alpha_j} \left(\frac{L_j}{L_j+\alpha_j}\right)^{L_j}
\end{align*}

\textit{Claim:} For $\epsilon_j$ small and for the indices in
\[
 \mathcal I=\Bigl\{\alpha : L_j\bigl(\frac 1{(1-r_j^2)^{1-2\delta}}-1\bigr)\leq \alpha_j\leq \frac {L_j}{\frac{(1+\epsilon_j)^2}{r_j^2}-1}, \ j=1,\dots, n\Bigr\}
\]
the following estimate holds
\[
 |a_{\alpha}|^2\leq \prod_{j=1}^n (1+\epsilon_j)^{-\alpha_j}\ .
\]

\textit{Proof:}
The lower bound on $\alpha_j$ implies that
\[
 \left(\frac{L_j}{L_j+\alpha_j}\right)^{L_j/2} (1-r_j^2)^{-L_j (\frac 12-\delta)}\leq 1\ ,
\]
and therefore
\[
  |a_{\alpha}| \leq \prod_{j=1}^n \alpha_j^{1/2} \left(\frac{\alpha_j}{L_j+\alpha_j}\right)^{\alpha_j/2} \frac 1{r_j^{\alpha_j}}\ .
\]
The upper bound in $\mathcal I$ yields
\[
 \alpha_j^{1/2} \left(\frac{\alpha_j}{L_j+\alpha_j}\right)^{\alpha_j/2} \frac 1{r_j^{\alpha_j}} \leq (1+\epsilon_j)^{-\alpha_j/2}\ ,
\]
which gives the claim. $\square$

\medskip

Since for $\alpha\in\mathcal I$ we have $cL_j\leq \alpha_j\leq CL_j$, we deduce that the number of indices in $\mathcal I$ is of order $\prodj$. Therefore, letting $\xi\sim N_{\C}(0,1)$, and using the Claim, we have
\begin{align*}
 \mathbb P[\mathcal E_1]&\leq \mathbb P\Bigl[|a_\alpha|^2\leq \prod_{j=1}^n (1+\epsilon_j)^{-\alpha_j},\ \alpha\in\mathcal I\Bigr]\leq
 \left( P\Bigl[|\xi|^2\leq \prod_{j=1}^n (1+\epsilon_j)^{-\alpha_j}\Bigr] \right)^{c\prodj}\ .
\end{align*}

We finish by noticing that for $\alpha\in\mathcal I$
\begin{align*}
 P\Bigl[|\xi|^2\leq \prod_{j=1}^n (1+\epsilon_j)^{-\alpha_j}\Bigr]&=1-\exp\bigl(-e^{-\sum\limits_{j=1}^n \alpha_j\log(1+\epsilon_j)}\bigr)
 \simeq e^{-\sum\limits_{j=1}^n \alpha_j\log(1+\epsilon_j)}
 \simeq e^{-c\sum\limits_{j=1}^n L_j }\ .
\end{align*}
This finishes the proof of (a) in Lemma~\ref{controlmax}.
\end{proof}

 (b) Let now
 \begin{align*}
  \mathcal E_2:&=\left\{\max_{E(0,r)}\log|\hat f_L(z)|^2>\delta \sumj\right\}\\
  &=\left\{\max_{E(0,r)}\left[\log |f_L(z)|-\sum_{j=1}^n\frac{L_j}{2}\log\bigl(\frac{1}{1-|z_j|^2}\bigr)\right]>\frac{\delta}2 \sumj\right\}.
 \end{align*}
We estimate the probability of this event by controlling the coefficients of the series of $f_L$.
Let $C_j>0$ be constants to be determined later on. Split the series defining $|f_L|$ into two families of indices:
\begin{align*}
 I_1&=\bigl\{\alpha : \alpha_j\leq C_j \delta L_j,\ j=1,\dots,n\bigr\} \\
 I_2&=\mathbb N^n\setminus I_1=\bigl\{\alpha : \exists j\in\{1,\dots,n\}\ : \alpha_j> C_j \delta L_j\bigr\}
\end{align*}

Then, using Cauchy-Schwarz for the indices $I_1$ we see that
\begin{align*}
|f_L(z)|&\leq  \left(\sum_{\alpha\in I_1}|a_{\alpha}|^2\right)^{1/2}
\left(\sum_{\alpha\in I_1} \prod_{j=1}^n \frac{\Gamma(L_j+\alpha_j)}{\alpha_j!\Gamma(L_j)} |z_j|^{2\alpha_j} \right)^{1/2}+
\sum_{\alpha\in I_2}  |a_{\alpha}|  \prod_{j=1}^n \left(\frac{\Gamma(L_j+\alpha_j)}{\alpha_j!\Gamma(L_j)}\right)^{1/2} r^{\alpha}\\
&\leq \left(\sum_{\alpha\in I_1}|a_{\alpha}|^2\right)^{1/2} \left(\prod_{j=1}^n \sum_{\alpha_j=0}^\infty \frac{\Gamma(L_j+\alpha_j)}{\alpha_j!\Gamma(L_j)} |z_j|^{2\alpha_j}\right)^{1/2} +\sum_{\alpha\in I_2}  |a_{\alpha}|  \prod_{j=1}^n \left(\frac{\Gamma(L_j+\alpha_j)}{\alpha_j!\Gamma(L_j)}\right)^{1/2} r^{\alpha}\\
&= \left(\sum_{\alpha\in I_1}|a_{\alpha}|^2\right)^{1/2} \sqrt{K_L(z,z)}+ \sum_{\alpha\in I_2}  |a_{\alpha}|  \prod_{j=1}^n \left(\frac{\Gamma(L_j+\alpha_j)}{\alpha_j!\Gamma(L_j)}\right)^{1/2} r^{\alpha}\\
&=:(I)+(II).\nonumber
\end{align*}

We shall estimate each part separately. First we shall see that, except for an event of small probability, $(II)$ is bounded (if $C_j$ are chosen appropiately).

Fix $\gamma$ (to be determined later) and consider the event
\[
 A=\bigl\{ |a_\alpha|\leq e^{\gamma |\alpha|_\infty},\ \forall\alpha\in I_2 \bigr\}.
\]
Here $|\alpha|_\infty=\max_j \alpha_j$. We also use the notation
\[
 C_*=\min_j C_j\qquad ,\qquad L_*=\min_j L_j\qquad ,\qquad  L^*=\max_j L_j\ .
\]
Notice that $L^*\simeq\sumj$ and that for $\alpha\in I_2$
\[
 |\alpha|_\infty\geq\delta\min_j C_j L_j\geq\delta L_*  C_*\ .
\]
We split the indices $I_2$ in level sets
\[
 I_2^m=\bigl\{\alpha\in I_2 : |\alpha|_\infty=m\bigr\}\ .
\]
Observe that
\begin{equation}\label{i2m}
 m^{n-1}\leq \# I_2^m\leq n m^{n-1}
\end{equation}
and that for $\alpha\in I_2^m$
\[
 \prod_{j=1}^n \frac{\Gamma(L_j+\alpha_j)}{\alpha_j! \Gamma(L_j)}\leq \left(\frac{\Gamma(L^*+m)}{m! \Gamma(L^*)}\right)^n,
\]
since $\frac{\Gamma(L+n)}{n! \Gamma(L)}$ is increasing both in $n$ and $L$.

Under the event $A$, and denoting $r_0=\max_j r_j$, we have for $z\in E(0,r)$,
\begin{align*}
(II)& \leq
\sum_{\alpha\in I_2} |a_\alpha|\left(\frac{\Gamma(L^*+|\alpha|_\infty)}{|\alpha|_\infty!\Gamma(L^*) }\right)^{n/2} r_0^{|\alpha|_\infty}
\leq \sum_{m\geq C_*\delta L_*} n m^{n-1} \left(\frac{\Gamma(L^*+m)}{m! \Gamma(L^*) }\right)^{n/2} (r_0 e^\gamma)^m\ .
\end{align*}

The asymptotics of the $\Gamma$-function
\begin{equation}\label{asymptoticsGamma}
 \lim_{m\to\infty}\frac{\Gamma(m+n)}{\Gamma(m) m^n}=1
\end{equation}
yields
\[
 (II)\leq \sum_{m\geq C_*\delta L_*}\left[\frac{m^{\frac 12-\frac 1n+\frac{L^*}2}}{\sqrt{\Gamma(L^*)}}\right]^n (r_0 e^\gamma)^m\ .
\]
By \cite{BMP}*{Lemma 10}, given $\epsilon>0$ there exists $c>0$ such that for $m\geq C_*\delta L_*$
\[
 \frac{m^{\frac 12-\frac 1n+\frac{L^*}2}}{\sqrt{\Gamma(L^*)}}\leq e^{\frac{\epsilon}m n}\
\]
and therefore
\[
 (II)\lesssim \sum_{m\geq C_*\delta L_*} (e^{\epsilon+\gamma}r_0)^m= \sum_{m\geq C_*\delta L_*} e^{-[\log\frac1{r_0}-(\epsilon+\gamma)]}\ .
\]
Choose $\epsilon=\gamma=\frac 14 \log\frac1{r_0}$, so that
\[
 (II)\lesssim \sum_{m\geq C_*\delta L_*} (r_0)^{m/2}\leq\frac 1{1-r_0^{1/2}}=:C(r)\ .
\]
With this we obtain the estimate
\[
 |f_L(z)|\leq \left(\sum_{\alpha\in I_1}|a_{\alpha}|^2\right)^{1/2} \sqrt{K_L(z,z)}+ C(r)\ .
\]
Under the event $\mathcal E_2$  we have then
\[
 e^{\frac{\delta}2 \sumj}<\frac{|f_L(z)|}{\sqrt{K_L(z,z)}}\leq \left(\sum_{\alpha\in I_1}|a_{\alpha}|^2\right)^{1/2}+
 \frac{C(r)}{\sqrt{K_L(z,z)}},
\]
and therefore, for $L_j$ big enough
\begin{equation}\label{equ}
 \sum_{\alpha\in I_1}|a_{\alpha}|^2 > e^{\frac{\delta}2 \sumj}\ .
\end{equation}

It remains to estimate the probability of this estimate, and to show that the event $A$ has ``big'' probability. The variables $|a_\alpha|^2$ are independent exponentials, hence
\[
\mathbb P[A]=\prod_{\alpha\in I_2}1-\mathbb P[|a_{\alpha}|\geq e^{\gamma|\alpha|_\infty}]
=\prod_{\alpha\in I_2}\left[1-e^{-e^{2\gamma |\alpha|_\infty}}\right].
\]

Since $x=e^{-e^{2\gamma |\alpha|_\infty}}$ is close to 0, we can use the estimate $\log(1-x)\simeq -x$. Thus,
\begin{align*}
\log \mathbb P[A]&=\sum_{\alpha\in I_2}\log\left[1-e^{-e^{2\gamma |\alpha|_\infty}}\right]\simeq -
\sum_{\alpha\in I_2} e^{-e^{2\gamma |\alpha|_\infty}}\gtrsim -\sum_{m\geq C_*\delta L_*}n m^{n-1} e^{-e^{2\gamma m}}\\
&- \sum_{m\geq C_*\delta L_*} e^{-e^{\gamma m}}\simeq - e^{-e^{\gamma C_*\delta L_*}} .
\end{align*}

Choosing $C_j$ big enough so that, in addition to the previous conditions, $\gamma C_*>\max_j \log\frac{1}{1-r_j^2}$  we have
\[
 -e^{\gamma C_*\delta L_*}< -\frac 1{(1-r_0^2)^{\delta L_*}}
\]
and therefore
\[
 \log \mathbb P[A]\gtrsim -e^{-[1-r_0^2]^{-\delta L_*}}\ .
\]
On the other hand
\[
 \mathbb P(A\cap\mathcal E_2)\leq\mathbb P\bigl[\sum_{\alpha\in I_1} |a_{\alpha}|^2>e^{\frac{\delta}2 \sumj}\bigr]\ .
\]
Since the number of indices in $I_1$ is at most $M_L:=\prod_{j=1}^n C_j\delta L_j\simeq \prodj$ we have
\begin{align*}
\mathbb P[A\cap \mathcal E_2]&\leq \mathbb P\left[ \sum_{\alpha\in I_1} |a_{\alpha}|^2 \geq \frac 1{M_L} e^{\frac{\delta}2 \sumj}  \right]
= M_L e^{-\frac 1{M_L} e^{\frac{\delta}2 \sumj}}=e^{\sum_{j=1}^n\log(C_j\delta L_j)-\frac 1{M_L} e^{\frac{\delta}2 \sumj}}\ .
\end{align*}
For $L_j$ big enough
\[
 \frac 1{M_L} e^{\frac{\delta}2 \sumj}\geq e^{\frac{\delta}4 \sumj}\
\]
and therefore
\[
 \mathbb P[A\cap\mathcal E_2]\leq e^{-e^{\frac{\delta}8 \sumj}}\ .
\]
Also
\[
 \mathbb P[A^c\cap\mathcal E_2]\leq \mathbb P(A^c)\leq 1-e^{-e^{-\min\limits_j (1-r_j^2)^{-\delta L_j}}}\simeq e^{-\min\limits_j (1-r_j^2)^{-\delta L_j}} \ .
\]
All combined
\[
\mathbb P[\mathcal E_2]=\mathbb P[A\cap \mathcal E_2]+\mathbb P[A^c\cap \mathcal E_2]\leq e^{-e^{\frac{\delta}8 \sumj}}+ e^{-\min\limits_j (1-r_j^2)^{-\delta L_j}}\leq e^{-c\spj} .
\]
\end{proof}

It remains to prove Lemma \ref{controlmean}. Before we proceed we need the following mean-value estimate of $\log |\hat f_L(\lambda)|^2$, which is obtained as \cite{BMP}*{Lemma 11}.

\begin{lemma}\label{invlemma}
Let $\lambda\in\D^n$ and $s\in (0,1)^n$ a polyradius. Then
\[
\log |\hat f_L(\lambda)|^2 \leq \frac{1}{\nu (E(\lambda,s))}\int_{E(\lambda,s)}\log |\hat f_L(\xi)|^2 d\nu(\xi)+\sumj \eps(s_j),
\]
where
\[
\eps(t)=\frac{1-t^2}{t^2}\int^{\frac{t^2}{1-t^2}}_{0} \log(1+x)dx\leq \frac{t^2}{1-t^2}.
\]
\end{lemma}

\begin{proof}[Proof of Lemma \ref{controlmean}]
According to Lemma~\ref{controlmax}(a), except for an exceptional event of probability $e^{-c\spj}$, there is $\lambda\in E:=E(z_0,s)$ such that
\[
-\sumj \frac{s_j^2}{1-s_j^2}<\log  |\hat f_L(\lambda)|^2.
\]
Therefore, using Lemma \ref{invlemma},
\[
-\sumj \frac{s_j^2}{1-s_j^2}<\frac{1}{\nu(E)}\int_{E}\log  |\hat f_L(\xi)|^2 d\nu(\xi)+\sumj \frac{s_j^2}{1-s_j^2} .
\]
Hence
\[
0 < \frac{1}{\nu(E)}\int_{E}\log |\hat f_L(\xi)|^2 d\nu(\xi) +2 \sumj \frac{s_j^2}{1-s_j^2},
\]
and separating the positive and negative parts of the logarithm we obtain:
\[
\frac{1}{\nu(E)}\int_{E}\log^{-}  |\hat f_L(\xi)|^2 d\nu(\xi)
\leq \frac{1}{\nu(E)}\int_{E}\log^{+} |\hat f_L(\xi)|^2 d\nu(\xi)+2 \sumj \frac{s_j^2}{1-s_j^2} .
\]
Finally, again by Lemma \ref{controlmax}, outside another exceptional event of probability $e^{-c\spj}$,
\begin{align*}
\frac{1}{\nu(E)}\int_{E}\left|\log  |\hat f_L(\xi)|^2 \right|d\nu(\xi)&\leq \frac{2}{\nu(E)}\int_{E}\log^{+}  |\hat f_L(\xi)|^2 d\nu(\xi)+2
\sumj \frac{s_j^2}{1-s_j^2}\\
& \leq 2\max_{E}\log^{+}  |\hat f_L|^2+2 \sumj \frac{s_j^2}{1-s_j^2}\leq 5 (\sumj) \max_j \frac{s_j^2}{1-s_j^2} \\
&\leq 5(\sumj) n\nu(E)^{1/n}.
\end{align*}
\end{proof}

\section{The hole theorem. Proof of Theorem~\ref{holethm}}

The upper bound is a direct consequence of the results in the previous section. Letting $U=E(0,r)$ and applying Corollary \ref{conseqSLD} with $\delta\mu(U)$ instead of $\delta$ we get
\[
 \mathbb P\left[Z_{f_L}\cap E(0,r)=\emptyset\right]\leq\mathbb P\left[|I_L(U)- \nu(U)\sumj|>\delta \nu(U)\sumj \right]\leq e^{-C_{2}\spj}.
\]

The method to prove the lower bound is standard (see \cite{HKPV}*{Theorem 7.2.3} and \cite{ST1}): we shall choose three events forcing $f_L$ to have a hole $E(0,r)$ and then we shall see that the probability of such events is at least $e^{-C_{1}\spj}$.

Our starting point is the estimate
\[
 |f_L(z)|\geq |a_0|-\left|\sum_{\alpha\in J_1} a_\alpha\prod_{j=1}^n \left(\frac{\Gamma(L_j+\alpha_j)}{\alpha_j! \Gamma(L_j)}\right)^{1/2} z_j^{\alpha_j}\right|-
 \left|\sum_{\alpha\in J_2} a_\alpha\prod_{j=1}^n \left(\frac{\Gamma(L_j+\alpha_j)}{\alpha_j! \Gamma(L_j)}\right)^{1/2} z_j^{\alpha_j}\right|\ ,
\]
where, for constants $C_j$ to be chosen later,
\begin{align*}
 J_1&=\bigl\{\alpha\neq 0 : \alpha_j\leq C_j L_j,\ j=1,\dots,n \bigr\}, \\
 J_2&=\bigl\{\alpha  : \exists j\in\{1,\dots,n\},\ \alpha_j> C_j L_j, \bigr\}\ .
\end{align*}

The first event is
\[
 E_1:=\left\{\ |a_0|\geq 1\right\}\ ,
\]
which has probability
\[
\mathbb P[E_1]=\mathbb P[|a_0|^2\geq 1]=e^{-1}.
\]

The second event corresponds to the tail of the power series of $f_L$. Here we use the notations of the previous sections. Let, as in the previous section,
\[
 J_2^m=\{\alpha\in J_2 ; |\alpha|_\infty=m\} .
\]
Then, for $z\in E(0,r)$,
\begin{align*}
 S_3:&=\left|\sum_{\alpha\in J_2} a_\alpha \prod_{j=1}^n \left(\frac{\Gamma(L_j+\alpha_j)}{\alpha_j! \Gamma(L_j)}\right)^{1/2} z_j^{\alpha_j}\right| \leq \sum_{\alpha\in J_2} |a_\alpha| \prod_{j=1}^n \left(\frac{\Gamma(L_j+\alpha_j)}{\alpha_j! \Gamma(L_j)}\right)^{1/2} r_j^{\alpha_j}\\
 &\leq\sum_{m\geq C_* L_*} \left(\frac{\Gamma(L_*+m)}{m!\Gamma(L_*)}\right)^{n/2} r_0^m \bigl(\sum_{\alpha\in J_2^m} |a_\alpha|\bigr) \ .
\end{align*}
By the same arguments as in \cite{BMP}*{Section 3}, for $ C_j$ big enough and for $m\geq C_*L_*$
\[
 \frac{\Gamma(L_*+m)}{m! \Gamma(L_*)}\leq \left[\frac{m^{L_*/m}}{\Gamma(L)^{1/m}}\right]^m .
\]
By Stirling, for $m\geq C_*L_*$,
\[
 \frac{m^{L_*/m}}{\Gamma(L)^{1/m}}\leq (eC_*)^{1/C_*} K^{\frac 1{2C_*}},
\]
where $K=\max\limits_{x>0} x^{1/x}= e^{-1/e}$.

Let $h(C):= (eC)^{1/C} K^{\frac 1{2C}}$ and take $C_*$ big enough so that
\[
 \bigl( h(C_*)\bigr)^{n/2} r_0\leq (1-\delta)^2\ .
\]
Then
\[
 S_3\leq \sum_{m\geq C_*L_*} \bigl(h(C_*)\bigr)^{\frac{mn}2} r_0^m \bigl(\sum_{\alpha\in J_2^m} |a_\alpha|\bigr) \leq \sum_{m\geq C_*L_*} (1-\delta)^m\bigl(\sum_{\alpha\in J_2^m} |a_\alpha|\bigr).
\]
Now impose the event
\[
 E_2=\bigl\{|a_\alpha|\leq\frac mn\ \forall \alpha\in J_2^m,\ \forall m\geq C_* L_*\bigr\}\ .
\]
Under the event $E_2$, essentially by \eqref{i2m},
\[
 S_3\leq \sum_{m\geq C_* L_*}  (1-\delta)^m m^n
\]
and there exists $C_*$ big enough so that
\[
 S_3\leq \frac 14\ .
\]

We shall see next that $\mathbb P[E_2]$ is big. We have
\begin{align*}
 \log\mathbb P(E_2)&= \sum_{m\geq cL_*} \sum_{\alpha\in J_2^m} \log\bigl(1-e^{-\frac 2n m}\bigr)\simeq
 - \sum_{m\geq C_*L_*} \sum_{\alpha\in J_2^m} e^{-\frac 2n m}\\
 &\gtrsim \sum_{m\geq C_*L_*}  n m^{n-1} e^{-\frac 2n m} ,
\end{align*}
thus for $C_*$ big enough $\mathbb P(E_3)\geq 1/2$.

The third event takes care of the middle terms in the power series of $f_L$. Let
\[
E_3:=\left\{\ |a_\alpha|^2<\frac{[1-r^2]^{L}}{4\prod_{j=1}^n C_j L_j}\quad \forall \alpha\in J_1\right\}.
\]
Using Cauchy-Schwarz's inequality we get, as in previous computations:
\begin{align*}
S_2&:= \left|\sum_{\alpha\in J_1} a_{\alpha} \prod_{j=1}^n \left(\frac{\Gamma(L_j+\alpha_j)}{\alpha_j! \Gamma(L_j)}\right)^{1/2} z_j^{\alpha_j}\right|\leq
\left(\sum_{\alpha\in J_1}|a_{\alpha}|^2\right)^{1/2}
\left(\sum_{\alpha\in J_1} \prod_{j=1}^n  \frac{\Gamma(L_j+\alpha_j)}{\alpha_j! \Gamma(L_j)}  r_j^{2\alpha_j} \right)^{1/2}\\
& \leq \left(\sum_{\alpha\in J_1} |a_{\alpha}|^2\right)^{1/2} \left(\prod_{j=1}^n\sum_{\alpha_j=0}^\infty \frac{\Gamma(L_j+\alpha_j)}{\alpha_j! \Gamma(L_j)}  r_j^{2\alpha_j}  \right)^{1/2}
\leq \left(\sum_{\alpha\in J_1}|a_{\alpha}|^2\right)^{1/2}[1-r^2]^{-L/2}.
\end{align*}
Under the event $E_3$,
\[
\sum_{\alpha\in J_1}|a_{\alpha}|^2\leq \sum_{\alpha\in J_1}\frac{1}{4} [1-r^2]^{L}
=\frac{1}{16}[1-r^2]^{L},
\]
and therefore
\[
S_3\leq \frac{1}{2}.
\]
Since $1-e^{-x}\geq x/2$ for $x\in(0,1/2)$, we get
\begin{align*}
\mathbb P[E_2]&=\prod_{\alpha\in J_1}\Bigl\{1-\exp\bigl(-\frac{[1-r^2]^{L}}{4\prod_{j=1}^n C_j L_j}\bigr)\Bigr\}\geq \prod_{\alpha\in J_1} \frac{[1-r^2]^{L}}{8\prod_{j=1}^n C_j L_j}=\left(\frac{[1-r^2]^{L}}{8\prod_{j=1}^n C_j L_j}\right)^{\prod_{j=1}^n C_j L_j}\\
&\geq \exp\Bigl[-c\prodj\bigl(\sumj\log(\frac 1{1-r_j^2})+\log 8+\sum_{j=1}^n\log (C_j L_j)\bigr)\Bigr]\geq e^{-c\spj}\ .
\end{align*}

Finally,
\[
\mathbb P[E_1\cap E_2\cap E_3]\geq \mathbb P[E_1] \mathbb P[E_2] \mathbb P[E_3] \geq e^{-c\spj},
\]
and under this event $|f_L(z)|\geq 1-1/2-1/4 >0$.

\end{document}

config: arxiv


TITLE: Frechet Differentiation and Equivalent Norm examples
QUESTION [0 upvotes]: I have stumbled upon this interesting exercise, and I cannot seem to find conditions that satisfy all the properties needed, I can find several combinations which satisfy one or two of the properties, but not one for all three, I am really interested to see the solution so any help or hints would be greatly appreciated!
So the question is as follows:
Give an example of vector spaces $\mathbb{X}$ and $\mathbb{Y}$, two norms $||\cdot||_a$ and $||\cdot||_A$ on $\mathbb{X}$, where the two norms are equivalent, two norms $||\cdot||_b$ and $||\cdot||_B$ on $\mathbb{Y}$, where the two norms are equivalent, and a function $f:\mathbb{X} \to \mathbb{Y}$ such that all of the following hold:
a)  $\:f:(\mathbb{X},||\cdot||_a ) \to (\mathbb{Y},||\cdot||_b)$ is Frechet differentiable at every $x\in\mathbb{X}$;
b)  $\:f:(\mathbb{X},||\cdot||_A ) \to (\mathbb{Y},||\cdot||_b)$ is not Frechet differentiable at any $x\in\mathbb{X}$;
c)  $\:f:(\mathbb{X},||\cdot||_a ) \to (\mathbb{Y},||\cdot||_B)$ is not Frechet differentiable at any $x\in\mathbb{X}$.
It should be noted that the first part of this exercise was to prove that if a function is freshet differentiable everywhere with respect to two certain norms then if these are changed for equivalent ones the function is still freshet differentiable with the same derivative. So this can be assumed in the answer.
Thanks!

REPLY [0 votes]: Theorem. Let $\mathrm{V}$ and $\mathrm{W}$ be two normed spaces, and let $\| \cdot \|_1$ and $\| \cdot \|_a$ be a two equialent norms in $\mathrm{V}$ and, likewise $\| \cdot \|_2$ and $\| \cdot \|_b$ is a pair of equivalent norms for $\mathrm{W}.$ Suppose $f$ is defined on an open subset $\mathrm{A}$ of $\mathrm{V}$ into $\mathrm{W}$ and let $v$ be a point of $\mathrm{A}.$ Then, $f$ is differentiable in $v$ with respect to the pair $(\| \cdot \|_1, \| \cdot \|_2)$ if and only if it is differentiable in $v$ with respect to the pair $(\| \cdot \|_a, \| \cdot \|_b).$
Proof. There are constants $c, d, p, q$ all positive such that
$$c \| \cdot \|_1 \leq \| \cdot \|_a \leq d \| \cdot \|_1$$
and
$$p \| \cdot \|_2 \leq \| \cdot \|_b \leq q \| \cdot \|_2.$$
Then
$$\dfrac{c}{q} \dfrac{\|f(v + h) - f(v) - Th\|_b}{\|h\|_a} \leq \dfrac{\|f(v + h) - f(v) - Th\|_2}{\|h\|_1} \leq \dfrac{d}{p} \dfrac{\|f(v + h) - f(v) - Th\|_b}{\|h\|_a}$$
and the proof is complete.
How this theorem imply the excercise? Easy, clearly $\| \cdot \|$ is equivalent with itself.


TITLE: $2$ relatively prime numbers and the geometric series
QUESTION [2 upvotes]: Prove that for any pair of relatively whole numbers $x$ and $n$, there is some whole number $t$ for which:
$$1+x+x^2...+x^t$$
is a multiple of n  An example: $x=168$ and $n=25$, we can find:
$$1+x+x^2+x^3=4,770,025$$
I tried to show that if no $t$ exists, then the series will go on to infinity and converge to the familiar $\frac{1}{1-x}$. I also know for some $r$ and $s$, $ns+xr=1$ (Bezout's identity). I just cant seem to put these two ideas together to prove that there will always be a $t$.

REPLY [0 votes]: $$\begin{align}  &n\mid \overbrace{(x^{\large t+1}\!-1)/(\color{#0a0}{x\!-\!1})}^{\textstyle x^{\large t}+\cdots +x+1}\\[.3em]
\smash{\overset{\!\!\large \times\ (\color{#0a0}{x-1})}\iff}\,\ \ &n(\color{#0a0}{x\!-\!1})\mid x^{\large t+1}-1\\[.4em]
\iff\ \ & x^{\color{#c00}{\large t+1}}\equiv 1\!\!\!\pmod{\!m},\,\ m = n(x\!-\!1)\\
\end{align}\ \ \ $$
$x\,$ is coprime to $\,n,x\!-\!1\,$ so such $\,t\!+\!1\,$ exists  (by Euler $\phi$ or pigeonhole), e.g. $\,t = {\rm ord}_{m}(x)-1$
E.g. in your example we have that: $\ 168^{\large\color{#c00} 4}\equiv 1\pmod{25\cdot 167},\ $ so $\,\color{#c00}{t\!+\!1=4}\!\iff\! t=3\,$ works.


TITLE: Surface groups and subgroups of fundamental groups
QUESTION [3 upvotes]: The fundamental group of any closed surface is a surface group.  Let $S_3$ be the orientable surface of genus 3.
Is $\pi_1(S_3)$ isomorphic to an index-3 subgroup of any surface group?
We have 1 2-cell, 6 1-cells, and 1 0-cell.  Thus $\chi(S_3) = -4$.
If we could show that there was necessarily of covering space $S_3 \to X$ with $\pi_1(X) = G$, $p_\ast(\pi_1(S_3))$ in index-3 subgroup of $G$, then we would know that this is not possible since $4$ is not divisible by $3$.

REPLY [0 votes]: In general (for closed orientable surfaces), we have
the short exact sequence of groups
$$1\to\pi_1(S_{k h+1})\to\pi_1(S_{h+1})\to K\to 1,$$
where $|K|=k$.
If you like, you can remember the shape from the more general formula
$$\pi_1(F)\to\pi_1(E)\to\pi_1(B)\to\pi_0(F)\to 1,$$
which is true for any fibration $F\to E\to B$
where $E$ and $B$ are connected.
This, in turn, is the last part of a long exact sequence.
The bit about $|K|=k$ can be derived from the Euler characteristic,
as you did.
So the answer to your question is yes: we're in the case $k=3$ and $h=2$
with the fibration $3\to S_7\to S_3$.
Your argument, however, can be used to show that there is no fibration $3\to S_3\to ???$.
It's very easy to get these things backwards, especially since the $\pi_1(B)\to\pi_0(F)$ is a boundary map.


TITLE: $KQ$ acylic connected finite dimension path algebra. $Cok(\oplus_{x\in Q_0}Hom_k(M_x,M'_x)\to\oplus_{a\in Q_1}Hom_k(M_{s(a)},M'_{t(a)}))=Ext^1(M,M')$
QUESTION [4 upvotes]: Let $Q=(Q_0,Q_1,s,t)$ be an acylic connected finite quiver where $Q_0$ are vertices, $Q_1$ are arrows, and $s,t:Q_1\to Q_0$ are source and target respectively. Let $K$ be a field. Let $KQ$ be the associated path algebra.
"$Cok(f:\oplus_{x\in Q_0}Hom_k(M_x,M'_x)\to\oplus_{a\in Q_1}Hom_k(M_{s(a)},M'_{t(a)}))=Ext^1(M,M')$"
The map above $f:(f_x)_{x\in Q_0}\to (f_{s(a)}\alpha-\alpha f_{t(a)})_{a\in Q_1}$ where $f:M\to M'$ is given by a family of morphisms $f_x$ and $\alpha\in Q_1$ is treated as morphism for $M_{s(\alpha)}\to M_{t(\alpha)}$ and $M'_{s(\alpha)}\to M'_{t(\alpha)}$ .
$\textbf{Q:}$ How do I see the identification $Coker(f)=Ext^1(M,M')$?
Ref. Ringel, What is a hereditary algebra? Sec 3.

REPLY [2 votes]: There is a standard resolution of $M$ of the form $$0 \rightarrow P_1 \rightarrow P_0 \rightarrow M \rightarrow 0.$$ Applying $\textrm{Hom}(\cdot,M')$, you get a long exact sequence:
$$ 0\rightarrow \textrm{Hom}(M,M') \rightarrow \textrm{Hom}(P_0,M') \rightarrow \textrm{Hom}(P_1,M') \rightarrow \textrm{Ext}^1(M,M') \rightarrow \textrm{Ext}^1(P_0,M')$$
Now $\textrm{Ext}^1(P_0,M')=0$ since $P_0$ is projective, so you actually have a four-term exact sequence. The map between the second and third terms can be identified with the one that you ask about, so its cokernel is isomorphic to $\textrm{Ext}^1(M,M')$.
Section 5.6 and 5.7 of https://www.math.uni-bielefeld.de/~sek/kau/leit5.pdf give a lot of details. This resolution of $M$ is called the "standard resolution"; you should be able to find it discussed in plenty of places.


TITLE: Expected time till absorption in specific state of a Markov chain
QUESTION [2 upvotes]: This question is a follow-up to Expected number of steps for reaching a specific absorbing state in an absorbing Markov chain  because I don't understand the answer given there.  I think I need to see a concrete example.
Suppose I play red and black at a casino.  On each play, I either win the amount I staked, with probability $p<\frac12$ or I lose my stake.  Let's say I start with a bankroll of $2$ and I decide to play until I have won $3$ or lost everything.  My strategy is to bet just enough to reach my goal, or everything I have, whichever is less.
We have a Markov chain with $6$ states, $2$ of which are absorbing.  There are well-known methods to determine the probability of winning, and of determining the average number of plays I make, but what if I want to know the average number of plays I make if I reach my goal?  The transition matrix, with $q=1-p$ is $$\begin {bmatrix}
1&0&0&0&0&0\\
q&0&p&0&0&0\\
q&0&0&0&p&0\\
0&q&0&0&0&p\\
0&0&0&q&0&p\\
0&0&0&0&0&1
\end{bmatrix}$$
Henning Malcolm suggests two approaches, neither of which I can follow.  (This is because of my limitations, and is in no way intended as a criticism of the answer.)  The first assumes that I have figured out the probability of winning starting with every possible bankroll.  Then we are to compute new transition probabilities that describe the experience of the winners, as I understand it, and compute the time to absorption in the new chain.  Let $p_k$ be the probability of winning, if my bankroll is $k$.  How should I calculate the new transition matrix?
Henning Malcolm gives an alternative method if there is only one starting state we're interested in, and I'm only interested in the actual case where I start in state $2$.  He says, "first set up a system of equations that compute for each state the expected number of times one will encounter that state before being absorbed."  If we let $e_k$ be this number for state $k=1,2,3,4,$ how do we construct the equations relating the $e_k?$  I can see how to do this if we only care about the number plays until the game ends, but how do I make it reflect only the number of plays until winning?

REPLY [1 votes]: [A partial answer illustrating the first method.]
State $1$ represents losing the game, so remove it from the system and condition the remaining transition probabilities on the event that the system does not transition from state $i$ to state $1$. In practical terms, this means that you delete from your transition matrix column $1$ and every row that had a $1$ in this column, and scale each remaining row $i$ by $1/(1-p_{1i})$. For your example system, this produces the reduced matrix $$P' = \begin{bmatrix}0&1&0&0&0 \\ 0&0&0&1&0 \\ q&0&0&0&p \\ 0&0&q&0&p \\ 0&0&0&0&1 \end{bmatrix}.$$ Applying standard techinques to this matrix, we can verify that the absorption probability is $1$ for any starting state and that the expected conditional absorption times are $$\begin{bmatrix}{3+q\over1-q^2} \\ {2+q+q^2\over1-q^2} \\ {1+3q\over1-q^2} \\ {1+q+2q^2\over1-q^2}\end{bmatrix}.$$


TITLE: Let $F$ be a field and $R$ a finitely generated $F$-algebra. Let $P$ be a maximal ideal of $R$. Then $\dim(R/P)$ as a vector space over $F$ is finite.
QUESTION [0 upvotes]: Let $F$ be a field and $R$ a finitely generated $F$-algebra. Let $P$ be a maximal ideal of $R$. Then $\dim(R/P)$ as a vector space over $F$ is finite.
$P$ is a maximal ideal of $R/P$ is a field. I know that the result directly follows from the weak nullstellensatz if we can show that $R/P$ is a finitely generated $F$-algebra, but how can we be sure that $R/P$ contains $F$?

REPLY [0 votes]: We have a homomorphism $F \to R \to R/P$. A homomorphism between fields is automatically a field extension, hence $R/P$ contains $F$.


TITLE: Convergence of sequence: $f(n+1) = f(n) + \frac{f(n)^2}{n(n+1)}$
QUESTION [9 upvotes]: I am trying to work out a problem on some old analysis qual exam, and managed to reduce it to this question, but I can't seem to figure out this final step:
Consider the sequence defined by the recurrence relation $$f(n+1) = f(n) + \frac{f(n)^2}{n(n+1)}$$ with initial condition $f(1) = 1$. Does this sequence converge or diverge?

REPLY [2 votes]: We can answer this question in the same manner as this answer.
Since $f(1)=1$ and
$$
f(n+1)=f(n)+\frac{f(n)^2}{n(n+1)}\tag{1}
$$
we have that $f(n)$ is increasing and inductively we have that
$$
1\le f(n)\le n\tag{2}
$$
Furthermore, $(1)$ implies that
$$
\begin{align}
&\left(\frac1{f(n+1)}-\frac1{n+1}\right)-\left(\frac1{f(n)}-\frac1n\right)\\
&=\left(\frac1n-\frac1{n+1}\right)-\left(\frac1{f(n)}-\frac{n(n+1)}{n(n+1)f(n)+f(n)^2}\right)\\
&=\frac1{n(n+1)}-\left(\frac{n(n+1)+f(n)}{n(n+1)f(n)+f(n)^2}-\frac{n(n+1)}{n(n+1)f(n)+f(n)^2}\right)\\
&=\frac1{n(n+1)}-\frac1{n(n+1)+f(n)}\\
&=\frac{f(n)}{n(n+1)(n(n+1)+f(n))}\\
&\ge\frac{f(n)}{n^2(n+1)(n+2)}\\
&\ge\frac1{n^2(n+1)(n+2)}\tag{3}
\end{align}
$$
Since
$$
\sum_{n=1}^\infty\frac1{n^2(n+1)(n+2)}=\frac{\pi^2}{12}-\frac58\tag{4}
$$
we have
$$
\begin{align}
\frac1{f(n)}
&\ge\frac1n+\sum_{k=1}^{n-1}\frac1{n^2(n+1)(n+2)}\\
&\ge\frac{\pi^2}{12}-\frac58\tag{5}
\end{align}
$$
Therefore,
$$
f(n)\le\frac{24}{2\pi^2-15}\tag{6}
$$
Since $f(n)$ is increasing and bounded above, it converges.


TITLE: Cesaro Means decrease slower than the sequence
QUESTION [1 upvotes]: For a strictly decreasing sequence of positive real numbers, I want to show that the Cesaro means decrease slower than the sequence itself. In particular, I need
$$\dfrac{C_n}{C_{n+1}}<\dfrac{a_n}{a_{n+1}},$$
where $C_n=\dfrac{1}{n}\displaystyle\sum_{i=1}^na_i.$
This is indeed true for $a_n=\frac{1}{n}$ or $a_n=r^n$ with $r<1$.

REPLY [1 votes]: This is false. Consider $2,1,1$ and $n=2$. Then $a_2 = a_3 = 1$ while $\frac{C_2}{C_3} = \frac{3/2}{4/3} > 1$.


TITLE: Is the given sequence a Markov chain or not? If yes, find its transition probabilities.
QUESTION [2 upvotes]: Let $\xi_n$, $n \in \mathbb{Z}_+$ be a sequence of i.i.d random variables over $\mathbb{R}$ with the density $p(x)$. Consider the sequence
$$
\eta_n := \sum\limits_{k=1}^n\left(a\xi_k + b\xi_{k + 2}\right), \quad a, b\in\mathbb{R}
$$

Is it a Markov chain?
If yes, find its transition probabilities.

While trying to solve this problem, I first noticed that
$$
\eta_n = \eta_{n - 1} + a\xi_n + b\xi_{n + 2}
$$
As far as I understand, to prove the statement, I have to show that Markov property is satisfied (or show the opposite):
$$
\mathbb{P}\left(\eta_n = x_n\mid\eta_{n - 1} = x_{n - 1}, \ldots, \eta_1 = x_1\right) = \mathbb{P}\left(\eta_n = x_n\mid\eta_{n - 1} = x_{n - 1}\right)
$$
And here I am stuck and need help - I don't know how to move further. The only idea was to rewrite the equation above as
$$
\mathbb{P}\left(a\xi_n + b\xi_{n + 2} = x_n - \eta_{n - 1}\mid \eta_{n - 1} = x_{n - 1}, \ldots, \eta_1 = x_1\right)
$$
and somehow use the independence of $\xi_i$. However, I'm confused with these large sums of random variables and I'm not sure that it is a correct way to go.
If we assume that the given sequence is a Markov chain, we have to find transitional probabilities. I guess that I can use the fact that
$$
p_{c\xi}(x) = \frac{1}{|c|}p_\xi\left(\frac{x}{|c|}\right), \quad c\in \mathbb{R}
$$
and find transitional probabilities as the value of the density of the sum $a\xi_n + b\xi_{n + 2}$ in the corresponding point:
\begin{equation}
\mathbb{P}\left(\eta_n = j\mid \eta_{n - 1} = i\right) = \mathbb{P}\left(a\xi_n + b\xi_{n+2} = j - i\right) = p_{a\xi_n + b\xi_{n + 2}}(j - i) =\\
 = \int\limits_{-\infty}^\infty\!p_{a\xi_n}(j - i - y)p_{b\xi_{n+2}}(y)\,dy = \frac{1}{|ab|}\int\limits_{-\infty}^\infty\!p\left(\frac{j - i - y}{|a|}\right)p\left(\frac{y}{|b|}\right)\,dy
\end{equation}
Am I right about this?

REPLY [2 votes]: Intuitively, the $\xi_{k+2}$ term makes $\eta_n$ related to $\eta_{n-2}$ in a way that one cannot deduce just by observing $\eta_{n-1}$.  Here is an example exploiting this fact to show that $\eta_n$ is not a Markov chain:

$a = b = 1$
Every $\xi_k$ is i.i.d. coin-flip resulting in $0$ or $1$ with equal probability $p= 1/2$.

I know the OP says there is a density $p_\xi(x)$, so just pick something highly bimodal, e.g. 50-50 mix of $U(0,0.01)$ and $U(1, 1.01)$, and the rest of the argument still goes through after minor modifications...

First we condition on $\eta_2 = 3$:

Since $\eta_2 = \xi_1 + \xi_2 + \xi_3 + \xi_4 = 3$, we know $3$ out of $4$ coin-flips give a $1$, so $P(\xi_3 = 1 \mid \eta_2 = 3) = {3 \over 4}$.
Now $\eta_3 = \eta_2 + \xi_3 + \xi_5 = 3 + \xi_3 + \xi_5$, so:

$$P(\eta_3 = 3 \mid \eta_2 = 3) = P(\xi_3 = 0 \mid \eta_2 = 3)P(\xi_5 = 0) = {1 \over 4} \cdot {1 \over 2} = {1\over 8}$$
Next, we condition on the same $\eta_2 = 3$, but also on $\eta_1 = 2$:

$\eta_1 = \xi_1 + \xi_3 = 2 \Rightarrow \xi_1 = \xi_3 = 1$.
So $\eta_3 = \eta_2 + \xi_3 + \xi_5 = 3 + 1 + \xi_5 \neq 3$, i.e.:

$$P(\eta_3 = 3 \mid \eta_2 = 3, \eta_1 = 2) = 0 \ \ \ \neq \ \ \ P(\eta_3 = 3 \mid \eta_2 = 3) = {1 \over 8}$$
This directly proves $\eta_n$ is not a Markov chain.
Aside: In fact, I think even if we allow "remembering" a finite number $m$ of $\eta$'s, i.e. $\theta_n = (\eta_n, \eta_{n+1}, \dots, \eta_{n+m-1})$, the $\theta_n$'s still do not form a Markov chain because there is still information in $\theta_{n-2}$ that is not available in $\theta_{n-1}$.  Similarly if we just take the even (or odd) indices $\phi_m = \eta_{2m}$ the subsequence $\phi$ also does not form a Markov chain.

For general $a,b,\xi_k$, it might be easier to show that the condition expectations are different.  Let me take the liberty to assume that $E[\xi_k] = \gamma$ exists (is finite).  Then $\eta_3 = \eta_2 + \xi_3 + \xi_5$ implies:

$E[\eta_3 \mid \eta_2 = x] = x + \gamma + E[\xi_3 \mid \eta_2 = x]$
$E[\eta_3 \mid \eta_2 = x, \eta_1 = y] = x + \gamma + E[\xi_3 \mid \eta_2 = x, \eta_1 = y] = x + \gamma + E[\xi_3 \mid \eta_1 = y]$

because $(\eta_2 = x, \eta_1 = y) \equiv (\xi_2 + \xi_4 = x-y, \eta_1 = y)$ and $\xi_3$ is independent of $\xi_2, \xi_4$.

So the two conditional expectations are different (proving $\eta_n$ is not a Markov chain) iff $E[\xi_3 \mid \eta_2 = x] \neq E[\xi_3 \mid \eta_1 = y]$.  I'm not sure how to show that in general but for specific $p_{\xi}(x)$ it should be easy to pick one example where they are unequal.
(In particular if $a=b$ then due to the i.i.d. property of the $\xi_k$'s we have $E[\xi_3 \mid \eta_2 = x] = {x\over 4a}$ while $E[\xi_3 \mid \eta_1 = y] = {y \over 2a}$.)


TITLE: Complete metric space and continuous function
QUESTION [0 upvotes]: Let $X$ be a complete metric space and $A_1\supseteq A_2\supseteq\cdots $ be a sequence of closed sets in $X$ with diam of $A_n$ converging to $0$. Let $f:X\rightarrow X$ be a continuous function.
Question is to show that $f\left(\bigcap A_n\right)=\bigcap f(A_n)$..
By diameter of $A_n$  we mean $\sup\{d(x,y):x,y\in A_n\}$...
Clearly, we have $f\left(\bigcap A_n\right)\subseteq\bigcap f(A_n)$..
Let $a\in f(A_n)$ for all $n\in A_n$ i.e., given $n$ there exists $x_n\in A_n$ such that $a=f(x_n)$ for all $n$...
Given $\epsilon>0$ there exists $N\in \mathbb{N}$ such that diam$(A_n)<\epsilon$ for all $n\geq N$
Let $m,n\geq N$ then $x_m,x_n\in A_N$ so $d(x_n,x_m)<\epsilon$.. Thus $(x_n)$ is a cauchy sequence..
As $X$ is complete there exists $x\in X$ such that $x_n\rightarrow x$..
As we are in metric space, $f(x_n)\rightarrow f(x)$.. As $f(x_n)=a$ for all $a$ we have $a=f(x)$..
Now, enough to prove that $x\in \bigcap A_n$...
I strongly belive this is true.. But could not come with a proof immediatley..
Help me to justify this and let me know if there are any gaps in between..

REPLY [0 votes]: We observe that, every $A_n$ is closed. Moreover, since the sequence $\{x_m\}_{m\geq n}\subset  A_n$ for every $n\in\mathbb{N}$ and $x_m\to x$, by closedness of $ A_n$ we have that $x\in A_n$ for every $n\in\mathbb{N}$, hence $x\in \bigcap A_n$.


TITLE: What do they mean by $\langle u_i |O| u_j\rangle$ in quantum mechanics
QUESTION [0 upvotes]: UCSD's online QM notes, as usual, starts stating that QM operators are Hermitian and says that operator $O$ elements can be computed by
$$O_{ij} = \langle u_j|O|u_i\rangle$$
The $u_i$ are eigenvectors, which are orthogonal to each other, thanks to being Hermitian. So, I suppose that $O$ is a Hermitian operator. But where are the eigenvalues?
I expect that when you apply the operator to its eigenvector, a $\lambda$ must appear. Yet, I see it nowhere. Neither, text says that $O$ is a diagonal, as I may expect if I apply $\langle u_i|$ to $O|u_i\rangle = \lambda_i |u_i\rangle$, I should get $\lambda_i$ on the main diagonal and 0 everyelse. Are $u_i$ eigenvectors of the $O$ or some another operator?

REPLY [3 votes]: In those notes you linked to, the $|u_i\rangle$ form an orthonormal basis. They are not necessarily the eigenkets of the operator $O$. By choosing different bases $\{u_i\}$, the "representation" of the operator $O$ (e.g.., how you write it out in a matrix) will change.
You are correct that if one chooses the set of kets that are eigenkets of the operator, you'll get a diagonal matrix. But in those notes they are trying to keep it general.


text:
\section{Technical Lemmas}\label{section:technical}

\begin{lemma}[Chi Squared Distance Between Two Gaussians] For
  Gaussian distributions $\mathcal{N}(\mu_1, \sigma^2 I)$ and
  $\mathcal{N}(\mu_2, \sigma^2 I)$, the (squared) chi-squared distance
  between $\mathcal{N}_1$ and $\mathcal{N}_2$ is:
\begin{align*}
\int \frac{(\mathcal{N}_1(z)-\mathcal{N}_2(z))^2}{\mathcal{N}_1(z)} dz
=
\exp\left( \frac{\|\mu_1-\mu_2\|^2}{2\sigma^2}\right) -1
\end{align*}
\label{lem:chi_squared_gaussian}
\end{lemma}
\begin{proof}
Observe that:
\begin{align*}
\int \frac{(\mathcal{N}_1(z)-\mathcal{N}_2(z))^2}{\mathcal{N}_1(z)} dz
  =  \int \mathcal{N}_1(z) - 2\mathcal{N}_2(z) +
  \frac{\mathcal{N}_2(z)^2}{\mathcal{N}_1(z)} dz
  = -1 + \int\frac{\mathcal{N}_2(z)^2}{\mathcal{N}_1(z) } dz.
\end{align*}
Note that for $\mathcal{N}_2^2(z) / \mathcal{N}_1(z)$, we have:
\begin{align*}
\mathcal{N}_2^2(z) / \mathcal{N}_1(z) = \frac{1}{Z} \exp\left( -\frac{1}{2\sigma^2}\left( 2\| z - \mu_2\|_2^2 - \| z - \mu_1\|_2^2 \right) \right),
\end{align*} where $Z$ is the normalization constant for
$\mathcal{N}(0,\sigma^2 I)$, i.e. $Z = \int \exp\left(
  -\frac{1}{2\sigma^2} \|z\|_2^2 \right) dz$.

For $2\| z - \mu_2\|_2^2 - \| z - \mu_1\|_2^2$, we can verify that:
\begin{align*}
2\| z - \mu_2\|_2^2 - \| z - \mu_1\|_2^2 = \| z + (\mu_1 - 2\mu_2)\|_2^2 -2\|\mu_1 - \mu_2\|_2^2.
\end{align*}This implies that:
\begin{align*}
&\int\frac{\mathcal{N}_2(z)^2}{\mathcal{N}_1(z) } dz = \frac{1}{Z}
\int \exp\left(  -\frac{1}{2\sigma^2} \left( \| z - (2\mu_2 - \mu_1)\|_2^2  - 2\|\mu_1 - \mu_2 \|\right)   \right) dz\\
& = \frac{1}{Z} \exp\left(  \frac{\|\mu_1- \mu_2\|_2^2}{\sigma^2} \right) \int\exp\left( -\frac{1}{2\sigma^2} \| z - (2\mu_2 - \mu_1)\|_2^2 \right) dz\\
& = \exp\left( \frac{\|\mu_1- \mu_2\|_2^2}{\sigma^2} \right),
\end{align*}  which concludes the proof.
\end{proof}

\begin{lemma}[Expectation Difference Under Two
  Gaussians] \label{lem:mean_difference}
For   Gaussian distribution $\mathcal{N}(\mu_1, \sigma^2 I)$ and
  $\mathcal{N}(\mu_2, \sigma^2 I)$, and for any (appropriately
  measurable) positive function $g$, it holds that:
\begin{align*}
\Exp_{z\sim \mathcal{N}_1} [g(z)] -\Exp_{z\sim \mathcal{N}_2} [g(z)]
\leq \min\left\{\frac{\|\mu_1-\mu_2\|}{\sigma} ,1\right\}
\ \sqrt{\Exp_{z\sim \mathcal{N}_1}\left[  g(z) ^2\right]}.
\end{align*}
\end{lemma}

\begin{proof}
Define $m_i = \Exp_{z\sim \mathcal{N}_i} [g(z)]$
for $i\in \{0,1\}$. We have:
\begin{align*}
m_1-m_2
&= \Exp_{z\sim \mathcal{N}_1} [g(z) (1-\frac{\mathcal{N}_2(z)}{\mathcal{N}_1(z)})]\\
&\leq \sqrt{\Exp_{z\sim \mathcal{N}_1} [g(z)^2]}
\sqrt{\int \frac{(\mathcal{N}_1(z)-\mathcal{N}_2(z))^2}{\mathcal{N}_1(z)} dz}\\
&= \sqrt{\Exp_{z\sim \mathcal{N}_1} [g(z)^2]}
\sqrt{\exp\left( \frac{\|\mu_1-\mu_2\|^2}{2\sigma^2}\right) -1}
\end{align*}
where we have used the previous chi-squared distance bound.
Also since $m_2$ is positive,
\begin{align*}
m_1-m_2\leq m_1 \leq \sqrt{\Exp_{z\sim \mathcal{N}_1} [g(z)^2]},
\end{align*}
and so
\begin{align*}
m_1-m_2
\leq \sqrt{\Exp_{z\sim \mathcal{N}_1} [g(z)^2]}
\sqrt{\min\left\{
\exp\left( \frac{\|\mu_1-\mu_2\|^2}{2\sigma^2}\right) -1
, 1\right\}}
\end{align*}
Now if the $\min$ is not achieved by $1$, then
$\frac{\|\mu_1-\mu_2\|^2}{2\sigma^2} \leq 1$. And since $\exp(x) \leq 1+2x$
for $0\leq x \leq 1$, we have:
\begin{align*}
\min\left\{
\exp\left( \frac{\|\mu_1-\mu_2\|^2}{2\sigma^2}\right) -1
, 1\right\}
\leq
\min\left\{
1+ \frac{\|\mu_1-\mu_2\|^2}{\sigma^2} -1
, 1\right\}
=
\min\left\{
\frac{\|\mu_1-\mu_2\|^2}{\sigma^2} , 1\right\} .
\end{align*}
which completes the proof.
\end{proof}

\begin{lemma}[Self-Normalized Bound for Vector-Valued Martingales; \citep{abbasi2011improved}]
Let $\{\varepsilon_i\}_{i=1}^{\infty}$ be a real-valued stochastic
process with corresponding filtration $\{\calF_{i}\}_{i=1}^{\infty}$
such that $\varepsilon_i$ is $\calF_i$ measurable, $\Exp[
\varepsilon_i | \calF_{i-1} ] = 0$, and $\varepsilon_i$ is
conditionally $\sigma$-sub-Gaussian with $\sigma\in\mathbb{R}^+$. Let
$\{X_i\}_{i=1}^{\infty}$ be a stochastic process with $X_i\in
\mathcal{H}$ (some Hilbert space) and $X_i$ being $\calF_t$ measurable. Assume that a
linear operator $V:\mathcal{H}\to\mathcal{H}$ is positive definite,
i.e., $x^{\top} V x > 0$ for any $x\in\mathcal{H}$. For any $t$,
define the linear operator $V_t = V + \sum_{i=1}^{t} X_iX_i^{\top}$ (here
$xx^{\top}$ denotes outer-product in $\mathcal{H}$). With probability
at least $1-\delta$, we have for all $t\geq 1$:
\begin{align*}
\left\|  \sum_{i=1}^{t} X_i \varepsilon_i \right\|^2_{V_t^{-1}} \leq 2\sigma^2 \log\left(\frac{ \det( V_t)^{1/2} \det(V)^{-1/2} }{ \delta  }\right).
\end{align*}
\label{lemma:self_normalized}
\end{lemma}

\iffalse
\begin{lemma}[Self-Normalized Bound for Matrix-Valued Martingales]
  Consider a stochastic process $\{\phi(x_i, u_i),
  x'_{i}\}_{i=1}^\infty$ with $x'_i = W \phi(x_i,a_i) + \epsilon_i$
  where $\epsilon_i\sim \mathcal{N}\left(0, \sigma^2 I \right)$,
  $x\in\mathbb{R}^{d}$.  Denote $V_N = \sum_{i=1}^N \phi(x_i,
  u_i)\phi(x_i,u_i)^{\top} + \lambda I$ with
  $\lambda\in\mathbb{R}^+$. Then, with probability at least
  $1-\delta$, for any $N$,  we have:
\begin{align*}
 \left\| \sum_{i=1}^N \epsilon_i \phi_i^{\top} V_N^{-1/2} \right\|_2^2 \leq 8\sigma^2 d\log\left(5\right) + 8\sigma^2\log\left( \frac{\det(V_N)^{1/2}\det(\lambda I)^{-1/2}}{\delta} \right)
\end{align*}
\label{lemma:self_norm_matrix}
\end{lemma}
\fi

We generalize this lemma as follows:
\begin{lemma}[Self-Normalized Bound for Matrix-Valued Martingales]
Let $\{\varepsilon_i\}_{i=1}^{\infty}$ be a $d$-dimensional vector-valued stochastic
process with corresponding filtration $\{\calF_{i}\}_{i=1}^{\infty}$
such that $\varepsilon_i$ is $\calF_i$ measurable, $\Exp[
\varepsilon_i | \calF_{i-1} ] = 0$, and $\varepsilon_i$ is
conditionally $\sigma$-sub-Gaussian with
$\sigma\in\mathbb{R}^+$.\footnote{We say a vector-valued, random variable $z$
  is $\sigma$-sub-Gaussian if $w\cdot z$ is $\sigma$-sub-Gaussian for
  every unit vector $w$.}
Let
$\{X_i\}_{i=1}^{\infty}$ be a stochastic process with $X_i\in
\mathcal{H}$ (some Hilbert space) and $X_i$ being $\calF_t$ measurable. Assume that a
linear operator $V:\mathcal{H}\to\mathcal{H}$ is positive definite. For any $t$,
define the linear operator $V_t = V + \sum_{i=1}^{t} X_iX_i^{\top}$
Then, with probability at least
  $1-\delta$, we have for all $t$,  we have:
\begin{align*}
 \left\| \sum_{i=1}^t \epsilon_i X_i^{\top} V_t^{-1/2} \right\|_2^2
\leq 8\sigma^2 d\log\left(5\right) + 8\sigma^2\log\left( \frac{\det(V_t)^{1/2}\det(V)^{-1/2}}{\delta} \right)
\end{align*}
\label{lemma:self_norm_matrix}
\end{lemma}

\begin{proof}
Denote $S = \sum_{i=1}^t \epsilon_i X_i^{\top} $.  Let us
form an $\epsilon$-net, in $\ell_2$ distance,  $\mathcal{C}$ over the unit ball $\{w: \|w\|_2 \leq 1,
w\in\mathbb{R}^{d} \}$. Via a standard covering argument
(e.g. \citep{shalev2014understanding}), we can choose $\mathcal{C}$
such that
$\log\left(| \mathcal{C} |\right) \leq d\log(1+ 2/\epsilon)$.

Consider a fixed $w \in \mathcal{C}$ and $w^{\top} S = \sum_{i=1}^t
w^{\top}\epsilon_i X_i^T $. Note that $w^{\top}\epsilon_i$

is a $\sigma$-sub
Gaussian due to $\|w\|_2 \leq 1$.   Hence,
Lemma~\ref{lemma:self_normalized} implies that with probability
at least $1-\delta$,   for all $t$,
\begin{align*}
\left\| V_t^{-1/2} \sum_{i=1}^t X_i \left(w^{\top}\epsilon_i \right)  \right\|_2 \leq \sqrt{2}\sigma \sqrt{ \log\left(\frac{ \det( V_t)^{1/2} \det(V )^{-1/2} }{ \delta  }\right) }.
\end{align*}
Now apply a union bound over $\mathcal{C}$, we get that with probability at least $1-\delta$:
\begin{align*}
\forall w\in\mathcal{C}: \left\| V_t^{-1/2} \sum_{i=1}^t X_i \left(w^{\top}\epsilon_i \right)  \right\|_2 \leq
\sqrt{2}\sigma \sqrt{d\log\left(1+2/\epsilon\right) + \log\left( \frac{\det(V_t)^{1/2}\det(V)^{-1/2}}{\delta} \right) }.
\end{align*}
For any $w$ with $\|w\|_2 \leq 1$, there exists a $w'\in\mathcal{C}$
such that $\|w - w'\|_2 \leq \epsilon$. Hence, for all $w$ such that $\|w\|_2\leq 1$,
\begin{align*}
 \left\| V_t^{-1/2} \sum_{i=1}^t X_i \left(w^{\top}\epsilon_i \right)  \right\|_2
&\leq \sqrt{2}\sigma \sqrt{d\log\left(1+2/\epsilon\right) + \log\left( \frac{\det(V_t)^{1/2}\det(V )^{-1/2}}{\delta} \right) } \\
& \quad + \epsilon \left\| \sum_{i=1}^t \epsilon_i X_i^{\top} V_t^{-1/2} \right\|_2.
\end{align*}
By the definition of the spectral norm, this implies that:
\begin{align*}
 \left\| \sum_{i=1}^t \epsilon_i X_i^{\top} V_t^{-1/2} \right\|_2 \leq \frac{1}{1-\epsilon} \sqrt{2}\sigma \sqrt{d\log\left(1+2/\epsilon\right) + \log\left( \frac{\det(V_t)^{1/2}\det(V)^{-1/2}}{\delta} \right) }
\end{align*}
Taking $\epsilon = 1/2$ concludes the proof.
\end{proof}

\begin{lemma}\label{lem:finite_info_gain}
For any sequence $x_0, \ldots x_{T-1}$ such that, for $t<T$, $x_t\in \mathbb{R}^d$ and $\|x_t\|_2 \leq B\in\mathbb{R}^+$, we have:
\begin{align*}
\log\det\left( I + \frac{1}{\lambda} \sum_{t=0}^{T-1}x_t x_t^{\top}\right)
\leq  d\log\left( 1 + \frac{TB^2}{ d \lambda } \right).
\end{align*}
\end{lemma}
\begin{proof}
Denote the eigenvalues of $ \sum_{t=0}^{T-1}x_t x_t^{\top}$ as
$\sigma_1, \dots \sigma_d$, and note:
\begin{align*}
\sum_{i=1}^d \sigma_i = \tr\left( \sum_{t=0}^{T-1}x_t x_t^{\top}  \right) \leq T B^2.
\end{align*}
Using the AM-GM inequality,
\begin{align*}
&\log\det\left( I + \frac{1}{\lambda} \sum_{t=0}^{T-1}x_t x_t^{\top}\right) = \log\left( \prod_{i=1}^d \left(1 + \sigma_i / \lambda \right) \right) \\
&= d\log\left( \prod_{i=1}^d \left(1 + \sigma_i / \lambda \right)
  \right)^{1/d}
\leq d\log\left( \frac{1}{d}\sum_{i=1}^d \left(1 + \sigma_i / \lambda \right) \right)
\leq d \log\left( 1 + \frac{ TB^2  }{d\lambda}  \right),
\end{align*}
which concludes the proof.
\end{proof}

\iffalse
\begin{lemma}\label{lem:finite_info_gain}
When $\phi\in \mathbb{R}^d$ and $\|\phi\|_2 \leq B\in\mathbb{R}^+$, we have:
\begin{align*}
\lambda(T) \leq  d\log\left( 1 + \frac{THB^2}{ d \lambda } \right).
\end{align*}
\end{lemma}
\begin{proof}
Consider $\log\det\left( I + \frac{1}{\lambda} \sum_{t=0}^{T-1}\sum_{h=0}^{H-1} \phi_h^t (\phi_h^t)^{\top}\right)$.

Let us denote the eigenvalue of $ \sum_{t=0}^{T-1}\sum_{h=0}^{H-1} \phi_h^t (\phi_h^t)^{\top}$ as $\sigma_1, \dots \sigma_d$. Note that:
\begin{align*}
\sum_{i=1}^d \sigma_i = \tr\left( \sum_{t=0}^{T-1}\sum_{h=0}^{H-1} \phi_h^t (\phi_h^t)^{\top}  \right) \leq TH B^2.
\end{align*} Hence,
\begin{align*}
&\log\det\left( I + \frac{1}{\lambda} \sum_{t=0}^{T-1}\sum_{h=0}^{H-1} \phi_h^t (\phi_h^t)^{\top}\right) = \log\left( \prod_{i=1}^d \left(1 + \sigma_i / \lambda \right) \right) \\
&= \sum_{i=1}^d \log\left( 1 + \frac{\sigma_i}{\lambda} \right) \leq \sum_{i=1}^d \log\left( 1 + \frac{ THB^2  }{d\lambda}  \right) = d \log\left( 1 + \frac{ THB^2  }{d\lambda}  \right).
\end{align*} Note that the right hand side of the above inequality is deterministic, which concludes the proof.
\end{proof}
\fi