text:
\section{Introduction}
\label{sec:topic1_introduction}

In this paper, we consider the problem of estimating covariance matrices under various types of contamination: we are given independent copies $Y_1, \ldots, Y_n$ of a random vector $Y \in \mathbb{R}^{d}$ which follows an unknown distribution $\mathcal D$ over $\mathbb R^{d}$
with mean $\mu$ and covariance matrix $\Sigma$, and we assume that (a) the observations $Y_1, \ldots, Y_n$ are \textit{$\varepsilon$-adversarially corrupted}, meaning that $\varepsilon$ fraction of them could have been replaced by arbitrary (possibly random) vectors, or that (b) the underlying distribution $\mathcal D$ is heavy-tailed, meaning that only the fourth moment of $Y$ is finite. Our goal is to construct a robust estimator for the covariance matrix $\Sigma$ in this framework.

As attested by some early references such as the works \citet{tukey1960survey, huber1992robust}, robust estimation has a long history. During the past two decades, increasing amount of practical applications created a high demand for the tools to recover high-dimensional parameters of interest from grossly corrupted measurements.
Robust covariance estimators in particular have been studied extensively, see e.g. \citet{huber1992robust,huber2011robust, maronna2019robust}. Although some of the proposed estimators admit theoretically optimal error bounds, they are hard to compute in general when the dimension is high because the running time is exponential in the dimension (\citet{bernholt2006robust}).

Recent work by \citet{lai2016agnostic,diakonikolas2019robust} introduced the first robust estimators for the covariance matrix $\Sigma$ that are computationally efficient in the high-dimensional case, i.e. the running time is only polynomial in the dimension, assuming that the distribution $\mathcal D$ is Gaussian or an affine transformation of a product distribution with a bounded fourth moment. Since the publication of these initial papers, a growing body of subsequent works has appeared. For instance, \citet{,cheng2019faster} developed fast algorithms that nearly match the best-known running time to compute the empirical covariance matrix, assuming that the distribution of $Y$ is Gaussian with zero mean. \citet{chen2018robust} developed efficient algorithms under significantly weaker conditions on the unknown distribution $\mathcal D$, i.e. $\mathcal D$ does not have to be an affine transformation of a product distribution. However, these algorithms either require prior knowledge on the fraction of outliers, or can only achieve a theoretically suboptimal error bound in the Frobenius norm.

The present paper continues this line of research. We design a double-penalized estimator for the covariance matrix $\Sigma$, which will be shown to admit theoretically optimal error bounds when the ``effective rank'' of $\Sigma$ (to be defined later) is small, and can be efficiently calculated using traditional numerical methods.

The rest of the paper is organized as follows.
Section \ref{sec:prelim} explains the main notations and background material. Section \ref{section:main} and  \ref{sec:heavy-tailed} displays the main results for $\epsilon$-adversarially corrupted data and heavy-tailed data, respectively.
Section \ref{sec:numerical simulation} presents analysis and algorithms for numerical experiments.
Finally, the numerical results and proofs are contained in the supplementary material.

\section{Preliminaries}
\label{sec:prelim}
\begingroup
\allowdisplaybreaks
In this section, we introduce the main notations and recall some useful facts that we rely on in the subsequent exposition.
Given two real numbers $a,b \in \mathbb R$, we define $a\vee b:=\max\{a,b\}$, $a\wedge b:= \min\{a,b\}$. Also, given $x \in \mathbb{R}$, we will denote $\lfloor x \rfloor := \max\{n\in \mathbb Z: n\leq x\}$ to be the largest integer less than or equal to $x$. We will separately introduce important results of matrix algebra and sub-Gaussian distributions in the following two subsections.

\subsection{Matrix algebra}
\label{sec:matrix algebra}
Assume that $A\in \mathbb R^{d_1\times d_2}$ is a $d_1 \times d_2$  matrix with real-valued entries. Let $A^T$ denote the transpose of $A$. A square matrix $A \in \mathbb R^{d\times d}$ is called an orthogonal matrix if $AA^T=A^TA=I_d$, where $I_d$ is the identity matrix in $\mathbb R^{d\times d}$.
	\begin{comment}

The following theorem is well known:
\begin{theorem}[{Singular Value Decomposition}]
\label{thm:svd}
Let $A\in \mathbb R^{d_1\times d_2}$ be a $d_1 \times d_2$ real matrix, then the singular value decomposition of $A$ always exists, and is of the form
\[
A=U\Sigma V^T,
\]
where
\begin{itemize}
\item U is a $d_1\times d_1$ orthogonal matrix, whose columns are the eigenvectors of $AA^T$.
\item V is a $d_2\times d_2$ orthogonal matrix, whose columns are the eigenvectors of $A^TA$.
\item $\Sigma$ is a $d_1 \times d_2$ matrix with $\Sigma_{1,1}=\sigma_1(A)$,\ldots, $\Sigma_{r,r}=\sigma_r(A)$, $\Sigma_{k,k}=0$ for all $ k>r$ and $\Sigma_{i,j}=0 \ \forall i\neq j$, where $r=\mathrm{\,rank}(A)$, and $\sigma_1(A)\geq\sigma_2(A)\geq\cdots\geq\sigma_r(A)\geq0$ are the square roots of positive eigenvalues of $A^TA$. They are called the singular values of A.
\end{itemize}
\end{theorem}

	\end{comment}
Given a square matrix $A \in \mathbb{R}^{d\times d}$, we define the trace of $A$ to be the sum of elements on the main diagonal, namely, $\mbox{tr\,}(A) := \sum_{i=1}^{d} A_{i,i}$,
where $A_{i,i}$ represents the element on the $i^{th}$ row and $i^{th}$ column of $A$.
We introduce three types of matrix norms and the Frobenius (or Hilbert-Schmidt) inner product as follows:
\begin{definition}[Matrix norms]
\label{def:matrix norms}
Given $A\in \mathbb R^{d_1\times d_2}$ with singular values $\sigma_1(A) \geq \cdots \geq \sigma_{\mathrm{\,rank}(A)}(A) \geq 0$, we define the following three types of matrix norms.
\begin{enumerate}
\item Operator norm:
\begin{equation*}
\left\|A\right\| := \sigma_1(A) = \sqrt{\lambda_{max}(A^TA)},
\end{equation*}
where $\lambda_{max}(A^TA)$ stands for the largest eigenvalue of $A^TA$.
\item Frobenius norm:
\begin{equation*}
\left\|A\right\|_F := \sqrt{\sum_{i=1}^{\mathrm{\,rank}(A)}\sigma_i^2(A)} = \sqrt{\mbox{tr\,}(A^TA)}.
\end{equation*}
\item Nuclear norm:
\begin{equation*}
\left\|A\right\|_1:=\sum_{i=1}^{\mathrm{\,rank}(A)}\sigma_i(A)=\mbox{tr\,}(\sqrt{A^TA}),
\end{equation*}
where $\sqrt{A^TA}$ is a nonnegative definite matrix such that $\Big(\sqrt{A^TA} \Big)^2=A^TA$.
\end{enumerate}
\end{definition}
\begin{definition
\label{def:Frobenius inner product}
Given $A,B\in \mathbb R^{d_1 \times d_2}$, we define the Frobenius inner product as
\begin{equation*}
\dotp{A}{B} := \dotp{A}{B}_F = \mbox{tr\,}(A^TB) =\mbox{tr\,}(AB^T).
\end{equation*}
It is clear that $\left\|A\right\|_F^2=\dotp{A}{A}$.
\end{definition}
We will now introduce matrix functions. Denote $S^d(\mathbb R):=\big\{A\in \mathbb R^{d\times d}: A^T=A\big\}$
to be the set of all symmetric matrices. The eigenvalues of $A$ will be denoted $\lambda_1,\ldots,\lambda_d$, all of which are real numbers. Next, we define functions acting on $S^d(\mathbb R)$ as follows:
\begin{definition}
\label{matrix-function}
Given a real-valued function $f$ defined on an interval $\mathbb T\subseteq \mathbb R$ and a real symmetric matrix $A\in S^d(\mathbb R)$ with the spectral decomposition
$A=U\Lambda U^T$ such that $\lambda_j(A)\in \mathbb T,\ j=1,\ldots,d$, define $f(A)$ as
$f(A)=Uf(\Lambda) U^T$, where
\[
f(\Lambda)=f\left( \begin{pmatrix}
\lambda_1 & \,  & \,\\
\, & \ddots & \, \\
\, & \, & \lambda_d
\end{pmatrix} \right)
=\begin{pmatrix}
f(\lambda_1) & \,  & \,\\
\, & \ddots & \, \\
\, & \, & f(\lambda_d)
\end{pmatrix}.
\]
\end{definition}
Finally, the effective rank of a matrix $A\in S^d(\mathbb{R})\setminus\{0\}$ is defined as
\[
\mathrm{rk}(A):=\frac{\mbox{tr\,}(A)}{\left\|A\right\|}.
\]
Note that $1\leq \mathrm{rk}(A) \leq \mathrm{\,rank}(A)$ is always true, and it is possible that $\mathrm{rk}(A) \ll \mathrm{\,rank}(A)$ for approximately low-rank matrices $A$. For instance, consider $A\in S^{d}(\mathbb R)$ with eigenvalues $\lambda_1 = 1, \lambda_2=\cdots=\lambda_d=1/d$, whence we have $\mathrm{rk}(A) = {2-1/d} \ll d = \mathrm{\,rank}(A)$.

\subsection{Sub-Gaussian distributions}
\label{sec:sub-gaussian}

Given a random variable $X$ on a probability space $(\Omega,\mathcal{A},\mathbb P)$, and a convex nondecreasing function $\psi: \mathbb R_+ \mapsto \mathbb R_+$ with $\psi(0)=0
, we define the $\psi$-norm of $X$ as (\citet[Section 2.7.1]{vershynin2018high})
\begin{equation*}
\left\|X\right\|_\psi := \inf \left\{C>0: \expect{\psi\left(\frac{|X|}{C} \right)} \leq 1 \right\} \nonumber.
\end{equation*}
In particular, in what follows we will mainly consider $\psi_1(u):=\exp{\{u\}}-1,u\geq 0$ and $\psi_2(u):=\exp{\{u^2\}}-1,u\geq0$, which correspond to the sub-exponential norm and sub-Gaussian norm respectively.
We will say that a random variable $X$ is sub-Gaussian (or sub-exponential) if $\left\|X\right\|_{\psi_2}<\infty$ (or $\left\|X\right\|_{\psi_1}<\infty$). Also, let $\left\|X\right\|_{L_2}:=\left(\expect{|X|^2}\right)^{1/2}$ be the $L_2$ norm of a random variable $X$. The sub-Gaussian (or sub-exponential) random vector is defined as follows:
\begin{definition}
\label{def:sub-gaussian}
A random vector $Z$ in $\mathbb R^d$ with mean $\mu=\expect{Z}$ is called L-sub-Gaussian if for every $v\in \mathbb R^d$, there exists an absolute constant $L>0$ such that
\begin{equation}\label{L-sub-gaussian}
\left\|\dotp{Z-\mu}{v}\right\|_{\psi_2} \leq L \left\|\dotp{Z-\mu}{v}\right\|_{L_2}.
\end{equation}
Moreover, Z is called L-sub-exponential if $\psi_2$-norm in \eqref{L-sub-gaussian} is replaced by $\psi_1$-norm.
\end{definition}
It is clear that if $Z$ is L-sub-Gaussian, then $(-Z)$ is also L-sub-Gaussian. We introduce some important results for sub-Gaussian distributions.
\begin{proposition}{(\citet[pp.24]{vershynin2018high})}
\label{prop:sub-gaussian}
A mean zero random variable Z is L-sub-Gaussian if and only if there exists an absolute constant $K(L)>0$ depending only on $L$ such that
\[
P\left(|Z|\geq t\right)\leq 2\exp\{-t^2/K(L)^2\},\quad \text{for all } t\geq 0.
\]
\end{proposition}
\begin{proposition} {(\citet[Theorem 2.6.3]{vershynin2018high})}
\label{thm:hoeffding}
Let  $Z_1,\ldots,Z_n$ be i.i.d L-sub-Gaussian random variables with mean zero, and $a=\left(a_1,\ldots,a_n\right)\in \mathbb R^{n}$. Then for any $t\geq0$, there exists a constant $K(L)>0$ depending only on $L$ such that
\[
P\left( \left| \sum_{i=1}^{n}a_iZ_i \right| \geq t \right) \leq 2 \exp\left\{ -\frac{t^2}{K(L)^2\left\|a\right\|_2^2} \right\},
\]
where $\left\|a\right\|_2^2 = a_1^2 + \cdots + a_n^2$.
\end{proposition}
\begin{corollary}\label{sum of sub-gaussians}
Let $Z_1,\ldots,Z_n$ be i.i.d L-sub-Gaussian random variables with common mean $\expect{Z_1}=\mu$ and sub-Gaussian norm $\left\|Z_1\right\|_{\psi_2}=K$. Let $a = (a_1,\ldots,a_n)$ be a vector in $\mathbb R^d$ such that $\left\|a\right\|_2\leq 1$. Then
\begin{enumerate}
\item $Y:=\sum_{i=1}^{n}a_i(Z_i-\mu)$ is still L-sub-Gaussian.
\item There exists an absolute constant $c>0$, such that $\left\|Y\right\|_{\psi_2}\leq cK$.
\end{enumerate}
\end{corollary}
\begin{proof}
This corollary immediately follows by a combination of Theorem \ref{thm:hoeffding} and Proposition \ref{prop:sub-gaussian}.
\end{proof}
\endgroup

\section{Problem formulation and main results}
\label{section:main}

Let $Z_1,\ldots,Z_n\in \mathbb R^d$ be i.i.d. copies of an L-sub-Gaussian random vector $Z$ such that $\mathbb EZ = \mu$ and $\mathbb E (Z-\mu)(Z-\mu)^T = \Sigma$. Assume that we observe a sequence
\begin{equation}
\label{outlier model}
Y_j = Z_j + V_j, \ j=1,\ldots,n ,
\end{equation}
where $V_j$'s are arbitrary (possibly random) vectors such that only a small portion of them are not equal to zero. Namely, we assume that there exists a set of indices $J\subseteq \{1,\ldots,n\}$ such that $|J| \ll n$ and $V_j = 0$ for $j\notin J$. In what follows, the sample points with $j\in J$ will be called \textit{outliers} and $\varepsilon:={|J|}/{n}$ will denote  the proportion of such points.
In this case,
\[
Y_j Y_j^T = Z_j Z_j^T + \underbrace{V_j V_j^T + V_j Z_j^T + Z_j V_j^T}_{:=\sqrt{n} U^\ast_j} := X_j + \sqrt{n} U^\ast_j,
\]
where $\mathrm{\,rank}(U_j^*)\leq 2$ and the $\sqrt{n}$ factor is added for technical convenience. Our main goal is to construct an estimator for the covariance matrix $\Sigma$ in the presence of outliers $V_j$.
In practice, we usually do not know the true mean $\mu$ of $Z$. To address this problem, we first recall the definition of U-statistics.
\begin{definition
Let $Y_1,\ldots,Y_n$ $(n\geq2)$ be a sequence of random variables taking values in a measurable space $(\mathcal S, \mathcal B)$. Assume that $H: \mathcal S^{m}\mapsto \mathbb S^{d}(\mathbb R)$ $(2\leq m\leq n)$ is an $\mathcal S^{m}$-measurable permutation-symmetric kernel, i.e. $H(y_1,\ldots, y_m) = H(y_{\pi_1},\ldots,y_{\pi_m})$ for any $(y_1,\ldots,y_m) \in \mathcal S^{m}$ and any permutation $\pi$. The U-statistic with kernel $H$ is defined as
\[
U_n := \frac{(n-m)!}{n!} \sum_{(i_1,\dots,i_m) \in I_n^{m}} H(Y_{i_1},\ldots, Y_{i_m}),
\]
where $I_n^{m} := \left\{(i_1,\ldots,i_m): 1\leq i_j\leq n, i_j\neq i_k \text{ if } j\neq k \right\}$.
\end{definition}
A particular example of U-statistics is the sample covariance matrix defined as
\begin{equation}\label{sample covariance matrix}
\widetilde{\Sigma}_s := \frac{1}{n-1}\sum_{j=1}^{n}(Y_j-\bar Y)(Y_j-\bar Y)^T ,
\end{equation}
where $\bar Y:=\frac{1}{n}\sum_{j=1}^{n}Y_j$. Indeed, it is easy to verify that
\begin{equation}\label{sample covariance matrix u-stat}
\widetilde{\Sigma}_s  = \frac{1}{n(n-1)} \sum_{(i,j)\in I_n^{2}}\frac{(Y_i-Y_j)(Y_i-Y_j)^T}{2} ,
\end{equation}
hence the sample covariance matrix is a U-statistic with kernel
\begin{equation*
H(x,y) := \frac{(x-y)(x-y)^T}{2} \text{ for any } x,y\in \mathbb R^{d}.
\end{equation*}
Note that $\expect{ {(Y_i-Y_j)}/{\sqrt{2}}}=0$ and $\expect{{(Y_i-Y_j)} {(Y_i-Y_j)}^T / 2}=\Sigma$ for all $(i,j)\in I_n^2$. Namely, by expressing the sample covariance matrix as a U-statistic in \eqref{sample covariance matrix u-stat}, the explicit estimation of the unknown mean $\mu$ can be avoided. Therefore, we consider the following settings:
\begin{multline}\label{U-stat form of Y}
\widetilde{Y}_{i,j}:=\frac{Y_i-Y_j}{\sqrt{2}},\quad\widetilde{Z}_{i,j}:=\frac{Z_i-Z_j}{\sqrt{2}},\quad\widetilde{V}_{i,j}:=\frac{V_i-V_j}{\sqrt{2}}, \quad
\forall(i,j)\in I_n^2.
\end{multline}
Then
\begin{equation*}
\widetilde{Y}_{i,j}\widetilde{Y}_{i,j}^T = \widetilde{Z}_{i,j} \widetilde{Z}_{i,j}^T + \underbrace{\widetilde{V}_{i,j} \widetilde{V}_{i,j}^T + \widetilde{V}_{i,j} \widetilde{Z}_{i,j}^T + \widetilde{Z}_{i,j} \widetilde{V}_{i,j}^T}_{:=\sqrt{n(n-1)} \widetilde{U}^\ast_{i,j}}
:= \widetilde{X}_{i,j} + \sqrt{n(n-1)} \widetilde{U}^\ast_{i,j} ,
\end{equation*}
where the $n(n-1)=|I_n^2|$ factor equals the total number of $\widetilde{Y}_{i,j}$'s, and is added for technical convenience. The followings facts can be easily verified:
\begin{enumerate}\label{facts for Y tilde}
\item $\widetilde{Y}_{i,j}=\widetilde{Z}_{i,j}+\widetilde{V}_{i,j}$, with $\expect{\widetilde{Z}_{i,j}}=0$ and $\expect{\widetilde{Z}_{i,j}\widetilde{Z}_{i,j}^T}=\Sigma$, for any $(i,j)\in I_n^2$. Moreover, $\widetilde{Z}_{i,j},(i,j)\in I_n^2$ has sub-Gaussian distribution according to Corollary \ref{sum of sub-gaussians}.
\item $\widetilde{Z}_{i,j}$'s are identically distributed, but not independent.
\item Denote $\widetilde{J}=\left\{(i,j)\in I_n^2:\widetilde{V}_{i,j}\neq0\right\}$ to be the set of indices such that $\widetilde{V}_{i,j}=0$, $\forall (i,j)\notin \widetilde{J}$. Then $|\widetilde{J}|$ represents the number of outliers in $\left\{\widetilde{Y}_{i,j}: (i,j)\in I_n^2 \right\}$, and we have that
\begin{equation}\label{bound_J}
|\widetilde{J}| = 2|J|(n-|J|) + |J|(|J|-1) = |J|(2n-|J|-1).
\end{equation}
\item $\mathrm{\,Rank}(\widetilde{U}_{i,j}^*)\leq 2$. This follows from the fact that for any vector $v\in \mathbb {R}^d$, $\widetilde{U}_{i,j}^*v \in \mathrm{span}\left\{ \widetilde{V}_{i,j}, \widetilde{Z}_{i,j} \right\}$.
\end{enumerate}
In what follows, we will use the notation $\mathbf{U_{I_n^2}}:=(U_{1,2},\ldots,U_{n,n-1})$
to represent the $n(n-1)$-dimensional sequence with subscripts taking from $I_n^2$. Similarly, the notation $(S,\mathbf{U_{I_n^2}})$ will represent the $(n^2-n+1)$-dimensional sequence $(S,U_{1,2},\ldots,\allowbreak U_{n,n-1})$.
Now we are ready to define our estimator. Given $\lambda_1,\lambda_2>0$, set
\begin{multline}\label{robust covariance estimation}
(\widehat{S}_\lambda,\mathbf{\widehat{U}_{I_n^2}}) = \argmin_{S,U_{1,2},\ldots,U_{n,n-1}} \Bigg[ \frac{1}{n(n-1)}\sum_{i\neq j} \left\| \widetilde{Y}_{i,j} \widetilde{Y}_{i,j} ^T - S - \sqrt{n(n-1)}U_{i,j} \right\|^2_\mathrm{F} \\
+ \lambda_1 \left\|S\right\|_1 + \lambda_2\sum_{i\neq j} \left\|U_{i,j} \right\|_1 \Bigg],
\end{multline}
where the minimization is over $S, U_{i,j} \in S^{d}(\mathbb{R})$, $\forall (i,j) \in I_n^2$.
\begin{remark}\label{relation to penalized huber}
The double penalized least-squares estimator in \eqref{robust covariance estimation} is indeed a penalized Huber's estimator (previously observed by \citet[Section 6]{donoho2016high} in the context of linear regression). To see this, we express \eqref{robust covariance estimation} as
\begin{multline}\label{double minimum}
(\widehat{S}_\lambda,\mathbf{\widehat{U}_{I_n^2}}) = \mathrm{arg}\min_{S} \min_{\mathbf{{U}_{I_n^2}}} \Bigg[ \frac{1}{n(n-1)}\sum_{i\neq j} \left\| \widetilde{Y}_{i,j} \widetilde{Y}_{i,j} ^T - S - \sqrt{n(n-1)}U_{i,j} \right\|^2_\mathrm{F} \\
+\lambda_1 \left\|S\right\|_1 + \lambda_2\sum_{i\neq j} \left\|U_{i,j} \right\|_1 \Bigg],
\end{multline}
and observe that the minimization with respect to $\mathbf{{U}_{I_n^2}}$ in \eqref{double minimum} can be done explicitly. It yields that
\begin{equation}\label{penalized Huber}
\widehat{S}_\lambda = \argmin_{S} \Bigg\{\frac{2}{n(n-1)}\mbox{tr\,} \bigg[ \sum_{i\neq j} \rho_{\frac{\sqrt{n(n-1)}\lambda_2}{2}}(\widetilde{Y}_{i,j}\widetilde{Y}_{i,j}^T - S) \bigg]
+ \lambda_1\left\|S\right\|_1\Bigg\},
\end{equation}
where
\begin{equation}\label{huber's function}
\rho_\lambda(u) := \left\{
			    \begin{array}{ll}
			    	\frac{u^2}{2},\quad\left|u\right|\leq\lambda\\
			    	\lambda\left|u\right|-\frac{\lambda^2}{2},\quad\left|u\right|>\lambda
			    \end{array}
			    \right. \quad \forall u\in \mathbb R, \lambda \in \mathbb R^+
\end{equation}
is the Huber's loss function. Details of the derivation are presented in section \ref{sec:derivation of penalized huber} of the supplementary material.
\end{remark}

\subsection{Main results}
\label{sec:estimation unkown mean}
\begingroup
\allowdisplaybreaks

We are ready to state the main results related to the error bounds for the estimator in \eqref{robust covariance estimation}.
We will compare performance of our estimator to that of the sample covariance matrix $\widetilde{\Sigma}_s$ defined in \eqref{sample covariance matrix}. When there are no outliers, it is well-known that $\widetilde{\Sigma}_s$ is a consistent estimator of $\Sigma$ with expected error at most $\mathcal{O}(d/\sqrt{n})$ in the Frobenius norm, namely, $\expect{\left\|\widetilde{\Sigma}_s-\Sigma \right\|_F} \leq Cd/\sqrt{n}$
for some absolute constant $C>0$ (see for example, \citet{cai2010optimal}). However, in the presence of outliers, the error for $\widetilde{\Sigma}_s$ can be large (see section \ref{sec:numerical results with graphs} in the supplementary material for some examples). On the other hand, recall that $\widetilde X_{i,j} = \widetilde{Z}_{i,j}\widetilde{Z}_{i,j}^T$, and our estimator in \eqref{robust covariance estimation} admits the following bound.
\begin{theorem}
\label{thm:lasso u-stat}
Let $\delta>0$ be an absolute constant. Assume that $n \geq 2$
and $|J| \leq c_1(\delta) n$, where $c_1(\delta)$ is a constant depending only on $\delta$. Then on the event
\begin{multline*}
\mathcal E=\Bigg\{ \lambda_1 \geq  \frac{140\left\|\Sigma\right\|}{\sqrt{n(n-1)}}\sqrt{\mathrm{rk}(\Sigma)} + 4\left\| \frac{1}{n(n-1)}\sum_{(i,j)\in I_n^2}\widetilde{X}_{i,j} - \Sigma\right\|, \\
\lambda_2 \geq \frac{140\left\|\Sigma\right\|}{n(n-1)} \sqrt{\mathrm{rk}(\Sigma)} + \frac{4}{\sqrt{n(n-1)}} \max_{(i,j)\in I_n^2} \left\| \widetilde{X}_{i,j} - \Sigma \right\| \Bigg\},
\end{multline*}
the following inequality holds:
\begin{equation*}
\left\| \widehat S_\lambda - \Sigma \right\|_\mathrm{F}^2 \leq \inf_{ S: \mathrm{\,rank}(S)\leq \frac{c_2 n^2\lambda_2^2}{\lambda_1^2} } \Big\{ (1+\delta) \left\| S-\Sigma \right\|_\mathrm{F}^2
+ c(\delta) \big( \lambda_1^2 \mathrm{\,rank}(S)
+ \lambda_2^2{|{J}|^2} \big) \Big\},
\end{equation*}
where $c_2$ is an absolute constant and $c(\delta)$ is a constant depending only on $\delta$.
\end{theorem}
The proof of Theorem \ref{thm:lasso u-stat} is presented in section \ref{proof:lasso-ustat} of the supplementary material.
\begin{remark}\label{remark for thm:u-stat}
The bound in Theorem \ref{thm:lasso u-stat} contains two terms:
\begin{enumerate}
\item
The first term,
$(1+\delta)\left\| S-\Sigma \right\|_\mathrm{F}^2 +  c(\delta)\lambda_1^2 \mathrm{\,rank}(S)$,
does not depend on the number of outliers. When there are no outliers, i.e. $|J|=0$, the bound will only contain this part. In such a scenario \citet{lounici2014high} proved that the theoretically optimal bound is
\begin{equation*}
 \left\| \widehat S_\lambda - \Sigma \right\|_\mathrm{F}^2 \leq \inf_{S} \bigg \{ \left\| \Sigma - S\right\|_F^2
 + C \left\| \Sigma\right\|^2 \frac{(\mathrm{rk}(\Sigma)+t) }{n} \mathrm{\,rank}(S)   \bigg \}
\end{equation*}
with probability at least $1-e^{-t}$.
By making the smallest choice of $\lambda_1$ as specified in \eqref{lambda_1_choice}, one sees that the first term of our bound coincides with the theoretically optimal bound.
\item The second term,
$c(\delta)\lambda_2^2 |J|^2$,
controls the worst possible effect due to the presence of outliers. When more conditions on the outliers are imposed (for example, independence), this bound can be improved. Moreover, \citet{diakonikolas2017being} proved that when $Z$ is Gaussian with zero mean, there exists an estimator $\widehat{\Sigma}$ achieving theoretically optimal bound $\left\|\widehat{\Sigma}-\Sigma\right\|_F \leq \mathcal{O}(\varepsilon)\left\|\Sigma\right\|$, which is independent of the dimension $d$.
In our case,  by making the smallest choice of $\lambda_2$ as specified in \eqref{lambda_2_choice}, we can show that the error bound scales like $\allowbreak \mathcal{O} \Big( \big(\log(n) +  \mathrm{rk}(\Sigma) \big) \varepsilon \Big)  \left\|\Sigma\right\|$.
The additional factor $\big( \log(n) + \mathrm{rk}(\Sigma) \big)$ shows that our bound is sub-optimal in general. However, if $\mathrm{rk}(\Sigma)$ is small,
our bound is essentially optimal up to a logarithmic factor.
\end{enumerate}
\end{remark}
Note that in Theorem \ref{thm:lasso u-stat} the regularization parameters $\lambda_1,\lambda_2$ should be chosen sufficiently large such that the event $\mathcal E$ happens with high probability. Under the assumption that $Z_j,j=1,\ldots,n$ are independent, identically distributed L-sub-Gaussian vectors, we can prove the following result which gives an explicit lower bound on the choice of $\lambda_1$.
\begin{theorem}\label{thm:mean_bound}
Assume that Z is L-sub-Gaussian with mean $\mu$ and covariance matrix $\Sigma$. Let $Z_1,\ldots,Z_n$ be independent copies of $Z$, and define $\widetilde{Z}_{i,j} := {\left( Z_i - Z_j \right)}/{\sqrt{2}}$ for all $(i,j)\in I_n^2$. Then $\widetilde{Z}_{i,j}, (i,j)\in I_n^2$ are mean zero L-sub-Gaussian random vectors with the same covariance matrix $\Sigma$. Moreover, for any $t\geq 1$, there exists $c(L)>0$ depending only on L such that
\begin{equation*}
\left\| \frac{1}{n(n-1)}\sum_{i\neq j}\widetilde{Z}_{i,j}\widetilde{Z}_{i,j}^T - \Sigma \right\|
\leq c(L)\left\|\Sigma\right\| \bigg( \sqrt{\frac{\mathrm{rk}(\Sigma)+t}{n}} + \frac{\mathrm{rk}(\Sigma) + t}{n}  \bigg)
\end{equation*}
with probability at least $1-2e^{-t}$.
\end{theorem}
Theorem \ref{thm:mean_bound} along with the definition of event $\mathcal E$ indicates that it suffices to choose $\lambda_1$ satisfying
\begin{equation}\label{lambda_1_choice}
\lambda_1 \geq c(L) \left\|\Sigma\right\| \sqrt{\frac{\mathrm{rk}(\Sigma)+t}{n}},
\end{equation}
given that $n \geq \mathrm{rk}(\Sigma)+t$. The next theorem provides a lower bound for the choice of $\lambda_2$:
\begin{theorem}
\label{thm:max covariance bound}
Assume that $Z$ is L-sub-Gaussian with mean zero and $Z_1,\ldots,Z_n$ are samples of $Z$ (not necessarily independent). There exists $c(L)>0$ depending only on L, such that for any $t\geq1$,
\begin{equation*}
\max_{j=1,\ldots,n} \|Z_jZ_j^T - \Sigma\| \leq c(L)\left\|\Sigma\right\|\left( \mathrm{rk}(\Sigma) + \log(n) + t\right)
\end{equation*}
with probability at least $1-e^{-t}$.
\end{theorem}
Note that Theorem \ref{thm:max covariance bound} does not require independence of samples, so it can be applied to the mean zero, L-sub-Gaussian vectors $\widetilde{Z}_{i,j}, (i,j)\in I_n^2$ to deduce that
\begin{equation*}
\max_{i\neq j}\left\|\widetilde{Z}_{i,j}\widetilde{Z}_{i,j}^T-\Sigma\right\| \leq c(L) \left\|\Sigma\right\| \left[ \mathrm{rk}(\Sigma) +\log(n(n-1)) + t \right]
\end{equation*}
with probability at least $1-e^{-t}$. Combining this bound with the definition of event $\mathcal E$, we conclude that it suffices to choose $\lambda_2$ satisfying
\begin{equation}\label{lambda_2_choice}
\lambda_2 \geq c(L) \left\|\Sigma\right\| \frac{ \left( \mathrm{rk}(\Sigma)+\log(n)+t \right)}{n}.
\end{equation}
By choosing the smallest possible $\lambda_1,\lambda_2$ as indicated in \eqref{lambda_1_choice}\eqref{lambda_2_choice}, we deduce the following corollary:
\begin{corollary}\label{cor:3-1}
Let $\delta>0$ be an absolute constant. Assume that $n \geq \mathrm{rk}(\Sigma)+\log(n)$ and $|J| \leq c_1(\delta) n$, where $c_1(\delta)$ is a constant depending only on $\delta$. Then we have that
\begin{multline}\label{cor:3-1 eq1}
\left\| \widehat S_\lambda - \Sigma \right\|_F^2 \leq \inf_{ \substack {S: \mathrm{\,rank}(S) \leq
c_2' n \big( \mathrm{rk}(\Sigma) + \log(n) \big) }} \Bigg\{ (1+{\delta}) \left\| S-\Sigma \right\|_\mathrm{F}^2 \\
 + c(L,\delta) \left\|\Sigma\right\|^2  \Big[ {\frac{\mathrm{rk}(\Sigma)+\log(n)}{n}} {\mathrm{\,rank}(S)}
 +  \frac{\left(\mathrm{rk}(\Sigma)+\log(n)\right)^2}{n^2} {|{J}|^2} \Big] \Bigg\}
\end{multline}
with probability at least $1-{3}/{n}$, where $c_2'$ is an absolute constant and $c(L,\delta)$ is a constant depending only on $L$ and $\delta$.
\end{corollary}
Note that the last term in \eqref{cor:3-1 eq1} can be equivalently written in terms of $\varepsilon$, the proportion of outliers, as
\begin{equation}\label{cor:3-1 eq2}
c(L,\delta) \left\|\Sigma\right\|^2 \left(\mathrm{rk}(\Sigma)+\log(n)\right)^2 \, \varepsilon^2.
\end{equation}

\endgroup

\section{The case of heavy-tailed data}
\label{sec:heavy-tailed}
In this section, we consider the application as well as possible improvements of the previously discussed results to heavy-tailed data. Let $Y\in \mathbb R^d$ be a random vector with mean $\expect{Y} = \mu$, covariance matrix $\Sigma = \expect{(Y-\mu)(Y-\mu)^T}$, and such that $\expect{\left\|Y-\mu\right\|_2^4}<\infty$. Assume that $Y_1,\ldots,Y_n$ are i.i.d copies of $Y$, and as before our goal is to estimate $\Sigma$. Since $\mu$ is unknown and the estimation of $\mu$ is non-trivial for the heavy tailed-data, we consider the setting $\widetilde{Y}_{i,j} = (Y_i-Y_j)/\sqrt{2}$ and denote, for brevity, $H_{i,j}:=\widetilde{Y}_{i,j}\widetilde{Y}_{i,j}^T$.
We have previously shown that $\expect{\widetilde Y_{i,j}}=0$ and $\expect{H_{i,j}}=\Sigma$, so the mean estimation is no longer needed for $\widetilde{Y}_{i,j}$.
Given $\lambda_1,\lambda_2>0$, we propose the following estimator for $\Sigma$:
\begin{equation}\label{heavy-tailed estimator}
\widehat{S}_\lambda = \argmin_{S} \Bigg\{\frac{1}{n(n-1)} \mbox{tr\,} \bigg[ \sum_{i\neq j} \rho_{\frac{\sqrt{n(n-1)}\lambda_2}{2}}(\widetilde{Y}_{i,j}\widetilde{Y}_{i,j}^T - S) \bigg]
+ \frac{\lambda_1}{2}\left\|S\right\|_1\Bigg\},
\end{equation}
which is the minimizer of the penalized Huber's loss function:
\begin{equation}\label{heavy-tailed loss}
L(S) =  \frac{1}{n(n-1)}\mbox{tr\,} \bigg[ \sum_{i\neq j} \rho_{\frac{\sqrt{n(n-1)}\lambda_2}{2}}(\widetilde{Y}_{i,j}\widetilde{Y}_{i,j}^T - S) \bigg] + \frac{\lambda_1}{2}\left\|S\right\|_1.
\end{equation}

Recall that the estimator $\widehat{S}_\lambda$ in \eqref{heavy-tailed estimator} is equivalent to the double-penalized least-squares estimator in \eqref{robust covariance estimation}. The key idea to derive an error bound for $\widehat{S}_\lambda$
is motivated by \citet{prasad2019unified}, which suggests that it is possible to decompose any heavy-tailed distribution as a mixture of a ``well-behaved'' and a contamination components. The decomposition bridges the gap between the heavy-tailed model and the outlier model \eqref{outlier model}, allowing us to follow an argument similar to that in Section \ref{section:main}. To be precise,
we consider the decomposition
\begin{equation}\label{eq:truncation}
\widetilde{Y}_{i,j} = \underbrace{\widetilde{Y}_{i,j}\mathds{1}\left\{\left\|\widetilde{Y}_{i,j}\right\|_2\leq R\right\}}_{:=\widetilde{Z}_{i,j}} + \underbrace{\widetilde{Y}_{i,j}\mathds{1}\left\{\left\|\widetilde{Y}_{i,j} \right\|_2> R\right\}}_{:=\widetilde{V}_{i,j}},
\end{equation}
where $R>0$ is the truncation level that will be specified later.
In the following two subsections, we will separately show that the estimator $\widehat{S}_\lambda$ in \eqref{heavy-tailed estimator} is close to $\Sigma$ in both the operator norm and the Frobenius norm.

\subsection{Bounds in the operator norm}
\label{sec: Bound in operator norm}
In this subsection we show that $\widehat{S}_\lambda$ is close to $\Sigma$ in the operator norm with high probability.
We will be interested in the effective rank of the matrix $\expect{(H_{1,2}-\Sigma)^2}$, and denote it as
\[
r_H := \mathrm{rk}(\expect{(H_{1,2}-\Sigma)^2}) = \frac{\mbox{tr\,}(\expect{(H_{1,2}-\Sigma)^2})}{\left\|\expect{(H_{1,2}-\Sigma)^2}\right\|}.
\]
\citet[Lemma 4.1]{minsker2020robust} suggest that under the bounded kurtosis assumption (to be specified later, see \eqref{cond:bounded kurtoses}), we can upper bound $r_H$ by the effective rank of $\Sigma$, namely, $r_H \leq C \cdot \mathrm{rk}(\Sigma)$ with some absolute constant $C$.
We first present a lemma which shows that if the tuning parameter $\lambda_1$ is too large, the estimator $\widehat{S}_\lambda$ will be a zero matrix with high probability.
\begin{lemma}
\label{lemma:large tuning}
Assume that $t\geq0$,  $\sigma\geq \left\|\expect{(H_{1,2}-\Sigma)^2}\right\|^{\frac{1}{2}}$ and
\[
n\geq \max \left\{64a^2 r_H t ,\frac{4b^2t^2\left\|\Sigma\right\|^2}{\sigma^2} \right\},
\]
where $a$,$b$ are sufficiently large constants. Then for any
$\lambda_1> (\sigma/4)\sqrt{n/t} ,$
we have that $\argmin_{S}L(S)=0$ with probability at least $1-e^{-t}$.
\end{lemma}
Lemma \ref{lemma:large tuning} immediately implies that for the choice of $\lambda_1>(\sigma/4)\sqrt{n/t}$, $\left\| \widehat S_\lambda - \Sigma\right\| \allowbreak = \allowbreak \left\|\Sigma\right\|$ with high probability, which is bounded by the largest singular value of $\Sigma$.
The following theorem provides a bound  for the choice of $\lambda_1\leq (\sigma/4)\sqrt{n/t}$.
\begin{theorem}
\label{thm:heavy-tailed operator bound}
Assume that $t\geq 1$ is such that $r_H{t} \leq c_3 n$
for some sufficiently small constant $c_3$, $\sigma\geq \left\|\expect{(H_{1,2}-\Sigma)^2}\right\|^{\frac{1}{2}}$,
and $n\geq \max \big\{64a^2  r_H t,{4b^2t^2\left\|\Sigma\right\|^2}/{\sigma^2} \big\}$ for some sufficiently large constants $a$, $b$. Then for $\lambda_1\leq (\sigma/4)\sqrt{n/t}$ and
$\lambda_2 \geq \sigma / \sqrt{(n-1)t}$, we have that
\[
\left\|\widehat{S}_\lambda-\Sigma\right\|\leq \frac{20}{39}\lambda_1 + \frac{80}{39} \sigma \sqrt{\frac{t}{n}} + \frac{40}{39} \lambda_2 t
\]
with probability at least $1-\left({8}r_H/3 + 1\right)e^{-t}$.
\end{theorem}
The proofs of Lemma \ref{lemma:large tuning} and Theorem \ref{thm:heavy-tailed operator bound} are presented in section \ref{sec:proof or heavy-tailed operator bound} of the supplementary material.
\begin{remark}
The bound in Theorem \ref{thm:heavy-tailed operator bound} is close to that in \citet[Corrolary 4.1] {minsker2020robust}, with an additional term ${20}\lambda_1 / 39$. This term comes from the penalization ${\lambda_1}\left\|S\right\|_1 / 2$, showing that by ``shrinking'' our estimator to a low rank matrix, we introduce a bias term bounded by a multiple of the tuning parameter $\lambda_1$.
\end{remark}
\begin{remark}\label{remark:bounded kurtosis}
According to \citet[Lemma 4.1] {minsker2020robust}, the ``matrix variance'' parameter $\sigma^2$ appearing in the statement of Theorem \ref{thm:heavy-tailed operator bound} can be bounded by
$\left\| \Sigma \right\| \mbox{tr\,}(\Sigma) = \mathrm{rk}(\Sigma) \left\| \Sigma \right\|^2$ under the bounded kurtosis assumption. More precisely, if we assume that the kurtoses of the linear forms $\dotp{Y}{v}$ are uniformly bounded by $K$, meaning that
\begin{equation}\label{cond:bounded kurtoses}
\sup_{v:\left\|v\right\|_2=1}\frac{\expect{\dotp{Y-\expect{Y}}{v}}^4}{\Big[\expect{\dotp{Y-\expect{Y}}{v}}^2\Big]^2} \leq K
\end{equation}
for any $v\in\mathbb{R}^d$.
Then we have that
\[
\left\|\expect{(H_{1,2}-\Sigma)^2}\right\| \leq K \mathrm{rk}(\Sigma)\left\|\Sigma\right\|^2
\]
and $\sigma$ can be chosen as $C\sqrt{K \mathrm{rk}(\Sigma)}\left\|\Sigma\right\|$ with some absolute constant $C$.
Moreover, in this case the assumptions on $n$ and $t$ in Lemma \ref{lemma:large tuning} and Theorem \ref{thm:heavy-tailed operator bound} can be reduced to a single assumption that $r_H {t} \leq c_3' n$ for some sufficiently small constant $c_3'$.
We will formally state condition \eqref{cond:bounded kurtoses} in the next subsection and derive additional results based on it.
\end{remark}

\subsection{Bounds in the Frobenius norm}
\label{sec: Bound in Frobenius norm}
In this subsection we show that $\widehat S_\lambda$ is close to the covariance matrix of $Y$ in the Frobenius norm with high probability, under a slightly stronger assumption on the fourth moment of $Y$.
\begin{definition}
A random vector $Y\in\mathbb{R}^d$ is said to satisfy an $L_4-L_2$ norm equivalence with constant $K$ (also referred to as the bounded kurtosis assumption), if there exists a constant $K\geq 1$ such that
\begin{equation}\label{L4-L2 norm equivalence}
\left(\expect{\dotp{Y-\mu}{v}^4}\right)^{\frac{1}{4}} \leq K\left(\expect{\dotp{Y-\mu}{v}^2}\right)^{\frac{1}{2}}
\end{equation}
for any $v\in\mathbb R^{d}$, where $\mu =\expect{Y}$.
\end{definition}
As previously discussed in Remark \ref{remark:bounded kurtosis}, condition \eqref{L4-L2 norm equivalence} allows us to connect the matrix variance parameter $\sigma^2$ with $\mathrm{rk}(\Sigma_Y)$, the effective rank of the covariance matrix $\Sigma_Y$. We will assume that $Y$ satisfies \eqref{L4-L2 norm equivalence} with a constant $K$ throughout this subsection.

Recall the decomposition
\begin{equation}\label{eq:truncation_2}
\widetilde{Y}_{i,j} = \underbrace{\widetilde{Y}_{i,j}\mathds{1}\left\{\left\|\widetilde{Y}_{i,j}\right\|_2\leq R\right\}}_{:=\widetilde{Z}_{i,j}} + \underbrace{\widetilde{Y}_{i,j}\mathds{1}\left\{\left\|\widetilde{Y}_{i,j} \right\|_2> R\right\}}_{:=\widetilde{V}_{i,j}},
\end{equation}
where $R>0$ is the truncation level that will be specified later. Denote $\Sigma_Y:= \expect{\widetilde{Y}_{1,2}\widetilde{Y}_{1,2}^T}$, $\Sigma_Z := \expect{\widetilde{Z}_{1,2}\widetilde{Z}_{1,2}^T} $ and recall that our goal is to estimate $\Sigma_Y$.  Note that $\left\|\widetilde{Z}_{i,j}\right\|_2\leq R$ almost surely, so equation \eqref{eq:truncation_2} represents $\widetilde{Y}_{i,j}$ as a sum of a bounded vector $\widetilde{Z}_{i,j}$ and a ``contamination'' component $\widetilde{V}_{i,j}$, which is similar to \eqref{outlier model}. On the other hand, we note that the truncation level $R$ should be chosen to be neither too large (to get a better truncated distribtuion) nor too small (to reduce the bias introduced by the truncation). \citet{mendelson2020robust}  suggest that a reasonable choice is given as follows:
\begin{equation}\label{choice of truncation level}
R = \left( \frac{\mbox{tr\,}(\Sigma_Y)\left\|\Sigma_Y\right\|n}{\log\big(\mathrm{rk}(\Sigma_Y) \big)+\log(n)}\right)^{\frac14}.
\end{equation}
Denote
$\widetilde{J} = \Big\{(i,j)\in I_n^2: \left\|\widetilde{Y}_{i,j}\right\|_2 > R \Big\}$ to be the set of indices corresponding to the nonzero outliers (i.e. $\widetilde{V}_{i,j} \neq 0$), and $\varepsilon := {|\widetilde{J}|}/{\big( n(n-1) \big)}$ to be the proportion of outliers.
Under the above setup, we can derive the following lemma which provides an upper bound on $\varepsilon$ with high probability:
\begin{lemma}\label{lemma:upper bound of outliers}
Assume that $Y$ satisfies the $L_4-L_2$ norm equivalence with constant K, and $R$ is chosen as in \eqref{choice of truncation level}
. Then
 \begin{equation}\label{heavy tailed-bound of outliers}
 \varepsilon \leq c(K) {\frac{\mathrm{rk}(\Sigma_Y) \left[ \log\big( \mathrm{rk}(\Sigma_Y) \big) + \log(n) \right]}{n}}
 \end{equation}
with probability at least $1-{1}/{n}$, where $c(K)$ is a constant only depending on $K$.
\end{lemma}
The proof of Lemma \ref{lemma:upper bound of outliers} is presented in section \ref{section:proof of upper bound of outliers} of the supplementary material. It is worth noting that the proportion of ``outliers'' (in a sense of the definition above) in the heavy-tailed model can be pretty small when the sample size $n$ is large.
Consequently, we can derive the following bound:
\begin{theorem}\label{thm:lasso u-stat improved}
Given $A \geq 1$, assume that $Y\in\mathbb{R}^d$ is a random vector with mean $\expect{Y}=\mu$, covariance matrix $\Sigma_Y = \expect{(Y-\mu)(Y-\mu)^T}$, and satisfying an $L_4-L_2$ norm equivalence with constant $K$. Let $Y_1,\ldots,Y_n$ be i.i.d samples of $Y$, and let $\widetilde{Z}_{i,j}$ be defined as in \eqref{eq:truncation_2}.
Assume that
$n\geq c_4(K)\mathrm{rk}(\Sigma_Y) \big( \log( \mathrm{rk}(\Sigma_Y) ) + \log(n) \big)$ for some constant $c_4(K)$ depending only on $K$, and
$\mathrm{\,rank}(\Sigma_Y) \leq c_2(K) {n}$ for some constant $c_2(K)$ depending only on $K$. Then for $\lambda_1 = c(K) \left\|\Sigma_Y\right\| {\Big[ \mathrm{rk}(\Sigma_Y) \big( \log( \mathrm{rk}(\Sigma_Y)) + \log(n) \big) \Big] ^{1/2}}{n^{-1/2}}$ and $\lambda_2 = c(K) \left\|\Sigma_Y\right\| \big( \mathrm{rk}(\Sigma_Y)\log(n) \big)^{1/2} (An)^{-1/2}$, we have that
\begin{multline*}
\left\|\widehat{S}_\lambda- \Sigma_Y\right\|_F^2
\leq  c(K)\left\|\Sigma_Y\right\|^2 \Bigg[ \frac{\mathrm{rk}(\Sigma_Y) \Big( \log\big( \mathrm{rk}(\Sigma_Y)\big) + \log(n) \Big) }{n} \mathrm{\,rank}(\Sigma_Y) \\
+ \frac{A \cdot \mathrm{rk}(\Sigma_Y)^2 \log(n)^3}{n} \Bigg]
\end{multline*}
with probability at least $1- {({8} r_H / 3 +1)}{n^{-A}} - {4}{n^{-1}}$, where $c(K)$ is a constant depending only on $K$.
\end{theorem}
The proof of Theorem \ref{thm:lasso u-stat improved} is given in section \ref{sec:proof of the improved frobenius bound} of the supplementary material.
\begin{remark}
Let us compare the result in Theorem \ref{thm:lasso u-stat improved} to the bound of Corollary \ref{cor:3-1}.
\begin{enumerate}
	\item The first part of the bound,
	\[
	c(K) \left\|\Sigma_Y\right\|^2 \frac{\mathrm{rk}(\Sigma_Y) \Big( \log\big( \mathrm{rk}(\Sigma_Y)\big) + \log(n) \Big) }{n}\mathrm{\,rank}(\Sigma_Y) ,
	\]
	has the same order as in Corollary \ref{cor:3-1} (up to a logarithmic factor), under the assumption that $\Sigma_Y$ has low rank. This part of the bound is theoretically optimal according to Remark \ref{remark for thm:u-stat}.
	\item The second part of the bound,
	\begin{equation}\label{improved bound for the outlier}
	c(K)\left\|\Sigma_Y\right\|^2 \frac{\mathrm{rk}(\Sigma_Y)^2  \log(n)^3 }{n},
	\end{equation}
	controls the error introduced by the outliers.
	It is much smaller than the corresponding quantity in Corollary \ref{cor:3-1} when $\varepsilon$, the proportion of the outliers, is only
	assumed to be a constant. The improvement is mainly due to the special structure of the heavy-tailed data, namely,
	the ``outliers'' $\widetilde{V}_{i,j}$ are mutually independent as long as the subscripts do not overlap, and hence there are many cancellations among them. Without this special structure, one can
	only apply Theorem \ref{thm:lasso u-stat} directly and derive a sub-optimal bound of order
	\[
		c(K)\left\|\Sigma_Y\right\|^2 \frac{\mathrm{rk}(\Sigma_Y)^3 {\log(n)^3} }{n}.
	\]

\end{enumerate}
\end{remark}

\section{Numerical experiments}
\label{sec:numerical simulation}
In this section we present analysis and algorithms for our numerical experiments. Recall that our loss function is
\begin{multline}\label{original_simulation_loss}
\widetilde{L}(S,\mathbf{U_{I_n^2}}) = \frac{1}{n(n-1)}\sum_{i\neq j}\left\|\widetilde{Y}_{i,j}\widetilde{Y}_{i,j}^T-S-\sqrt{n(n-1)}U_{i,j}\right\|_F^2 \\
+ \lambda_1\left\|S\right\|_1 + \lambda_2 \sum_{i\neq j}\left\|U_{i,j}\right\|_1.
\end{multline}
We are aiming to find $(\widehat{S}_\lambda,\boldsymbol{\widehat{U}_{I_n^2}})$, the minimizer of \eqref{original_simulation_loss}, numerically. Since we are only interested in $\widehat{S}_\lambda$, equation \eqref{penalized Huber} suggests that it suffices to minimize the following function:
\begin{equation}\label{simulation_loss}
L(S) :=  \frac{1}{n(n-1)}\mbox{tr\,}\sum_{i\neq j} \rho_{\frac{\sqrt{n(n-1)}\lambda_2}{2}}(\widetilde{Y}_{i,j}\widetilde{Y}_{i,j}^T - S) + \frac{\lambda_1}{2}\left\|S\right\|_1,
\end{equation}
where $\rho_\lambda(\cdot)$ is the Huber's loss function defined in \eqref{huber's function}.

\subsection{Numerical algorithm}
\label{sec:numerical algorithm}

In this section, we will state our algorithm for minimizing $L(S)$. We start with an introduction to the proximal gradient method (see for example, \citet{combettes2005signal}). Suppose that we want to minimize the function $f(x)=g(x)+h(x)$, where
\begin{itemize}
\item $g$ is convex, differentiable
\item $h$ is convex, not necessarily differentiable
\end{itemize}
We define the proximal mapping and the proximal gradient descent method as follows:
\begin{definition}
The proximal mapping of a convex function $h$ at the point $x$ is defined as:
\begin{equation*}
\mathrm{prox}_h(x) = \argmin_{u}\left(h(u) + \frac{1}{2}\left\|u-x\right\|_2^2\right).
\end{equation*}
\end{definition}
\begin{definition}[Proximal gradient descent (PGD) method]
The proximal gradient descent method for solving the problem $\argmin_xf(x) = \argmin_xg(x)+h(x)$ starts from an initial point $x^{(0)}$, and updates as:
\begin{equation*}
x^{(k)} = \mathrm{prox}_{\alpha_kh}\left(x^{(k-1)} - \alpha_k\nabla g(x^{(k-1)}) \right),
\end{equation*}
where $\alpha_k>0$ is the step size.
\end{definition}
We have the following convergence result.
\begin{theorem}\label{pgd convergence}
Assume that $\nabla g$ is Lipschitz continuous with constant $L>0$:
\[
\left\|\nabla g(x) - \nabla g(y) \right\| \leq L \left\|x-y\right\|
\]
and the optimal value $f^*$ is finite and achieved at the point $x^*$. Then the proximal gradient algorithm with constant step size $\alpha_k=\alpha \leq L$ will yield an $\mathcal{O}({1}/{k})$ convergence rate, i.e.
\[
f(x^{(k)}) - f^* \leq \frac{C}{k},\quad \forall k \in \{1,2,\ldots\}.
\]
\end{theorem}
Theorem \ref{pgd convergence} is well known (see for example, \citet[Chapter 10]{beck2017first}), but a detailed proof in our case is given in section \ref{sec:convergence analysis of pgd} of the supplementary material for the convenience of the reader.
Moreover, when $g(x) = \frac{1}{n}\sum_{i=1}^{n}g_i(x)$, where $g_1,\ldots,g_n$ are convex functions and $\nabla g_1,\ldots,\nabla g_n$ are Lipschitz continuous with a common constant $L>0$, the update step of PGD will require the evaluation of $n$ gradients, which is expensive for large $n$ values. A natural improvement is to consider the stochastic proximal gradient descent method (SPGD), where at each iteration $k=1,2,\ldots$, we pick an index $i_k$ randomly from $\{1,2,\ldots,n\}$, and take the following update:
\[
x^{(k)} = \mathrm{prox}_{\alpha_kh}\left(x^{(k-1)} - \alpha_k\nabla g_{i_k}(x^{(k-1)}) \right).
\]
The advantage of SPGD over PGD is that the computational cost of SPGD per iteration is $1/n$ that of the PGD. On the other hand, since the random sampling in SPGD introduces additional variance, we need to choose a diminishing step size $\alpha_k = \mathcal{O}({1}/{k})$. As a result, the SPGD only converges at a sub-linear rate (see \citet{nitanda2014stochastic}). To this end, we will consider the ``mini-batch" PGD, which has been previously explored and widely used in large-scale learning problems (see, e.g., \citet{shalev2011pegasos, gimpel2010distributed, dekel2012optimal, khirirat2017mini}). This method picks a small batch of indices rather than one at each iteration to calculate the gradient, and in such a way we are able balance the computational cost of PGD and the additional variance of SPGD. The algorithm is summarized in Algorithm \ref{alg: SPGD}.
\begingroup
\allowdisplaybreaks
\begin{algorithm}[H]
	\caption{Stochastic proximal gradient descent (SPGD)}
	\label{alg: SPGD}
\begin{flushleft}
	\quad\textbf{Input:} number of iterations $T$, step size $\eta_t$, batch size $b$, tuning parameters $\lambda_1$ and $\lambda_2$, initial estimation $S^0$, sample size $n$, dimension $d$.
\end{flushleft}
	\begin{algorithmic}[1]
		\FOR {$t=1,2,\ldots,T$}
			\STATE (1) Randomly pick $i_t,j_t\in \{1,2,\ldots,n\}$ without replacement.
			\STATE (2) Compute $G_t = -\nabla g_{i,j}(S^t) = -\rho'_{\frac{\sqrt{n(n-1)}\lambda_2}{2}}(\widetilde{Y}_{i,j}\widetilde{Y}_{i,j}^T-S^t)$.
			\STATE (3) If $b>1$, then repeat (1)(2) for b times and save the average gradient in $G_t$.
			\STATE (4) (\textbf{gradient update})
					$T^{t+1} = S^{t} - G_t.$
			\STATE (5) (\textbf{proximal update})
				\[
				S^{t+1} = \argmin_S\Big\{\frac{1}{2}\left\|S-T^{t+1}\right\|_F^2 +\frac{\lambda_1}{2}\left\|S\right\|_1 \Big\} = \gamma_{\frac{\lambda_1}{2}}(T^{t+1}),
				\]
				where $\gamma_\lambda(u) = \mathrm{sign}(u)(|u|-\lambda)_+$.

		\ENDFOR
	\end{algorithmic}
\begin{flushleft}
	\quad\textbf{Output:} $S^{T+1}$
\end{flushleft}
\end{algorithm}
\endgroup

\subsection{Rank-one update of the spectral decomposition}
\label{sec:rank-one update}

Note that at each iteration of our algorithm, we need to compute the spectral decomposition of $(\widetilde{Y}_{i,j}\widetilde{Y}_{i,j}^T-S^t)$, which is computationally expensive. However, since $\widetilde{Y}_{i,j}\widetilde{Y}_{i,j}^T$ is a rank-one matrix and $S^t$ was already saved in the spectral decomposition form after previous iteration, the problem of computing the spectral decomposition of ($\widetilde{Y}_{i,j}\widetilde{Y}_{i,j}^T-S^t$) can be viewed as a rank-one update of the spectral decomposition, which has been extensively studied (see for example, \citet{bunch1978rank}, and \citet{stange2008efficient}). In this subsection we will show how to use this idea to improve our algorithm.

Consider $\widetilde{B}=B + \rho bb^T$, where the spectral decomposition $B=QDQ^T$ is known, $\rho\in\mathbb{R}$ and $b\in\mathbb R^d$. Our target is to compute the spectral decomposition of $\widetilde{B}$. Note that
\begin{equation}
\widetilde{B} = B +\rho bb^T = Q(D+\rho zz^T)Q^T,
\end{equation}
where $b=Qz$, so it suffices to compute the spectral decomposition of $D+\rho zz^T$. We denote $z=(\zeta_1,\ldots,\zeta_d)^T$, and without loss of generality, we can assume that $\left\|z\right\|_2=1$. The following theorem is fundamental for our algorithm.
\begin{theorem}{(\citet[Theorem 1]{bunch1978rank})}
\label{rank-1 update}
Let $C=D+\rho zz^T$, where $D$ is diagonal, $\left\|z\right\|_2=1$. Let $d_1\leq d_2\leq\ldots\leq d_d$ be the eigenvalues of D, and let $\widetilde{d}_1\leq\widetilde{d}_2\leq\ldots\widetilde{d}_d$ be the eigenvalues of C. Then $\widetilde{d}_i=d_i+\rho\mu_i$, $1\leq i \leq d$ where $\sum_{i=1}^n \mu_i =1$ and $0\leq\mu_i\leq1$. Moreover, $d_1\leq\widetilde{d}_1\leq d_2\leq\widetilde{d}_2\leq\ldots\leq d_d \leq \widetilde{d}_d$ if $\rho>0$ and $\widetilde d_1\leq {d}_1\leq \widetilde d_2\leq d_2\leq\ldots\leq \widetilde d_d \leq {d}_d$ if $\rho<0$. Finally, if $d_i$'s are distinct and all the elements of z are nonzero, then the eigenvalues of $C$ strictly separate those of $D$.
\end{theorem}
There are several cases where we can deflate the problem (i.e. reduce the size of the problem):
\begin{enumerate}
	\item If $\zeta_i=0$ for some $i$, then $\widetilde{d}_i = d_i$ and the corresponding eigenvector remains unchanged. This is because $(D+\rho zz^T)e_i = d_i e_i$ as $\zeta_i=0$.
	\item If $|\zeta_i|=1$ for some $i$, then $\widetilde{d}_i = d_i +\rho$ and the corresponding eigenvector remains unchanged. Moreover, in this case $\zeta_j=0$ for all $j\neq i$, so $\widetilde{d}_j = d_j$ and their eigenvectors are the same, so the problem is done.
	\item If $d_i$ has a multiplicity $r\geq 2$, we can reduce the size of the problem via the following steps:
	\begin{enumerate}
		\item Let $Q_1 = [q_{i_1},\ldots,q_{i_r}]\in\mathbb R^{d\times r}$, where $\{q_{i_1},\ldots,q_{i_r}\}$ are the eigenvectors corresponding to $d_i$. Also, set $z_1 = Q_1^Tz$, i.e. $z_1$ contains rows corresponding to $d_i$.
		\item Construct an Householder transformation $H\in \mathbb R^{r\times r}$ such that $Hz_1 = -\left\|z_1\right\|_2 e_1$, and define $\bar{Q}_1=Q_1H^T$.
		\item Replace $q_{i_1},\ldots,q_{i_r}$ by the columns of $\bar{Q}_1$, and $z_1$ by $\bar{Q}_1^Tz_1=-\left\|z_1\right\|_2e_1$. This introduces $(r-1)$ more zero entries to $z$ and possibly an entry with absolute value equals one. An application of (1)(2) gives us $(r-1)$ (or $r$) more eigen-pairs of $D+\rho zz^T$.
	\end{enumerate}
\end{enumerate}
After the deflation step, it remains to work with a $k\times k$ problem ($k\leq d$), in which the eigenvalues $d_i$ are distinct and $\zeta_i\neq 0$ for all $i$. We will compute the eigenvalues and eigenvectors separately.

First, \citet{golub1973some} showed that the eigenvalues of $C = D+\rho zz^T$ are the zeros of $\omega(\lambda)$, where
\[
\omega(\lambda) = 1+\rho\sum_{j=1}^{k}\frac{\zeta_j^2}{d_j-\lambda}.
\]
Alternatively, since $\widetilde{d}_1 < \ldots < \widetilde{d}_k$ and $\widetilde{d}_i = d_i +\rho\mu_i$, for each $i=1,\ldots,k$ we can compute $\mu_i$ by solving $\omega_i(\mu_i) = 0$, where
\begin{equation}\label{compute_eigenvalue}
\omega_i(\mu) = 1 + \sum_{j=1}^{k}\frac{\zeta_j^2}{\delta_j-\mu}.
\end{equation}
and $\delta_j = {(d_j-d_i)}/{\rho}$. \citet{bunch1978rank} proved that we can solve $\omega_i(\mu)=0$ with a numerical method that converges quadratically. Details of the numerical method are presented in section \ref{sec:eigenvalues} of the supplementary material.

Second, after computing the eigenvalues $\widetilde{d}_1,\ldots,\widetilde{d}_k$, we can calculate the corresponding eigenvectors of $C = D+\rho zz^T$ by solving $C\widetilde{q}_i = \widetilde{d}_i\widetilde{q}_i$, $i=1,\ldots,k$. Theorem 5 in \citet{bunch1978rank} shows that $\widetilde{q}_i$ can be computed via
\begin{equation}\label{compute_eigenvector}
\widetilde{q}_i = \frac{D_i^{-1}z}{\left\|D_i^{-1}z\right\|_2},
\end{equation}
where $D_i := D - \widetilde{d}_i I$. Finally, once we obtained the spectral decomposition of $D+\rho zz^T = \bar{Q}\widetilde{D}\bar{Q}^T$, we can easily get the decomposition of $B +\rho bb^T= (Q\bar{Q})\widetilde{D}(Q\bar{Q})^T$. Note that computing $k$ eigenvectors via \eqref{compute_eigenvector} costs $\mathcal{O}(k^3)$ and the matrix multiplication $Q\bar{Q}$ in the last step costs $\mathcal{O}(d^3)$, so the overall complexity of the algorithm is still $\mathcal{O}(d^3)$. This can be further improved by exploiting the special structure of  $\bar{Q}$, which is given by the product (see for example, \citet{stange2008efficient} and \citet{gandhi2017updating}):
\begin{equation}\label{eq:special structure of Q bar}
\bar{Q} =
\underbrace{
\begin{bmatrix}
\zeta_1 &  \\
   & \ddots  \\
   & 		&\zeta_d
\end{bmatrix}
} _{:= A}
\underbrace{
\begin{bmatrix}
\frac{1}{d_1-\widetilde{d}_1} & \cdots & \frac{1}{d_1-\widetilde{d}_d} \\
\vdots & & \vdots \\
\frac{1}{d_d - \widetilde{d}_1} & \cdots & \frac{1}{d_d - \widetilde{d}_d}
\end{bmatrix}
} _{:= C}
\begin{bmatrix}
\left\|\bar{c}_{\cdot 1}\right\|_2 &  \\
   & \ddots  \\
   & 		&\left\|\bar{c}_{\cdot d}\right\|_2
\end{bmatrix}^{-1}
\end{equation}
where $\bar{C} := AC = [\bar{c}_{\cdot 1},\ldots,\bar{c}_{\cdot d}]$, $\bar{c}_{\cdot i}$ represents the $i^{th}$ column of $\bar{C}$, and $\left\|\bar{c}_{\cdot i}\right\|_2$ is the Euclidean norm of $\bar{c}_{\cdot i}$.
Using \eqref{eq:special structure of Q bar}, we can evaluate the matrix multiplication $Q\bar{Q}$ through the following steps:
\begin{enumerate}
\item Compute
$QA := U = [u_{\cdot 1}, \ldots, u_{\cdot d}]$,
where $u_{\cdot i} = \zeta_i q_{\cdot i}$ and $q_{\cdot i}$ is the $i^{th}$ column of $Q$. This step is straightforward and requires $\mathcal{O}(d^2)$ computational time.
\item Let $u_{i \cdot}$ be the $i^{th}$ row of $U$. Define
\[
\widetilde{U} = UC =
\begin{bmatrix}
u_{1\cdot} C \\
\vdots \\
u_{d\cdot} C
\end{bmatrix},
\]
which requires to evaluate the product of a vector $u_{i\cdot}$ and a Cauchy matrix $C$ $d$ times. The problem of multiplying a Cauchy matrix with a vector is called Trummer's problem, and \citet{gandhi2017updating} provide an algorithm which efficiently computes such matrix-vector product in $\mathcal{O}(d\log^2d)$ time. Consequently, the complexity of this step is  $\mathcal{O}(d^2\log^2d)$.
\item Compute the matrix product
\[
\widetilde{U}
\begin{bmatrix}
\left\|\bar{c}_{\cdot 1}\right\|_2 &  \\
   & \ddots  \\
   & 		&\left\|\bar{c}_{\cdot d}\right\|_2
\end{bmatrix}^{-1}.
\]
This step is again straightforward and can be done in $\mathcal{O}(d^2)$ time.
\end{enumerate}
The overall complexity for the computation of $Q\bar{Q}$ is now reduced to $\mathcal{O}(d^2\log^2d)$, which is much smaller than $\mathcal{O}(d^3)$ when $d$ is large.
We summarized our rank-one update algorithm in Algorithm \ref{alg: rank-one update}. Numerical experiments were performed and the results are presented in section \ref{sec:numerical results with graphs} of the supplementary material.
\begingroup
\allowdisplaybreaks
\begin{algorithm}\label{rank-one update}[H]
	\caption{Rank-one update of the spectral decomposition of $B + \rho bb^T$}
	\label{alg: rank-one update}
\begin{flushleft}
	\quad\textbf{Input:} orthogonal matrix Q and vector d such that $B=Q\text{diag}(d)Q^T$, constant $\rho$, vector $b$
\end{flushleft}
	\begin{algorithmic}[1]
		\STATE Set $Qb = z$. If $\left\|z\right\|_2 \neq 1$, then further set $\rho = \rho\left\|z\right\|_2^2$ and $z = z/\left\|z\right\|_2$.
		\STATE Handle deflation cases, and record indices that have not done as a vector $d_{sub}$.
		\STATE Compute eigenvalues of the $d_{sub}\times d_{sub}$ sub-problem by solving \eqref{compute_eigenvalue} numerically.
		\STATE Compute eigenvectors of the $d_{sub}\times d_{sub}$ sub-problem with \eqref{compute_eigenvector}.
		\STATE Combine the resulting eigenvalues in $\widetilde{d}$ and eigenvectors in $\bar{Q}$.
		\STATE Compute $\widetilde{Q}=Q\bar{Q}$.
	\end{algorithmic}
\begin{flushleft}
	\quad\textbf{Output:} orthogonal matrix $\widetilde{Q}$ and vector $\widetilde{d}$.
\end{flushleft}
\end{algorithm}
\endgroup

\begin{comment}

\section*{Supplementary Materials}

	Contain
	the brief description of the online supplementary materials.
	\par

	\section*{Acknowledgements}

	The authors.
	\par

\end{comment}

\section*{Acknowledgements}
\it{Authors acknowledge support by the National Science Foundation grants CIF-1908905 and DMS CAREER-2045068.}
\par

{
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natexlab\endcsname\relax\def\natexlab#1{#1}\fi
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  \def\url#1{\texttt{#1}}\fi
\expandafter\ifx\csname urlprefix\endcsname\relax\def\urlprefix{URL}\fi

\bibliographystyle{chicago}

redpajama_set_name: RedPajamaArXiv


text:
Special status the answer to two years of Brexit chaos - Anderson
22 June, 2018 - by Martina Anderson MEP
Sinn Féin MEP Martina Anderson has said two years on from the disastrous Brexit referendum the EU must move to protect the rights of people in the north through special status.
Martina Anderson
"Tomorrow will mark two years since Britain took the disastrous decision to leave the EU. The majority of people in the north, of course, voted to remain but since then the Tories have ignored that democratic position and are intent on dragging us out with them.
"In the last two years we have had nothing but chaos and division from the British government. Theresa May's cabinet is a battlefield and they have brought little to the ongoing Brexit negotiations.
"Instead we have had two years of putting Ireland at risk economically, politically and socially.
"It is time to say enough is enough. We need to move away from the reckless Tory Brexiteers and their allies through a special arrangement for the north of Ireland.
"The EU must ensure the protection of the Good Friday Agreement in all of its parts and protect the rights of people in the north of Ireland. The Irish government too must stand firm and ensure we are not dragged over the Brexit cliff by the Tories.
"Brexit is Britain's problem and the people of the north must not become its casualties."

redpajama_set_name: RedPajamaCommonCrawl


text:
Flujo de energía y nutrientes en un ecosistema es el nombre que recibe la totalidad de la cadena trófica, así como el flujo de cualquier fuente aprovechable de energía. El flujo de energía es el aprovechamiento de los productos primarios y secundarios por organismos que a su vez utilizaron consumidores primarios herbívoros de los cuales se alimentan los consumidores secundarios o carnívoros.

La energía es básica para el funcionamiento de cualquier ecosistema, gracias a las diferentes interacciones que se dan entre diferentes organismos, la energía fluye de especie a especie. Sin embargo, a medida de que esta va entrando al ecosistema, su cantidad disminuye. La cantidad de nutrientes y energía en la Tierra es muy pequeña, y por eso tiene diferentes ciclos. El ciclo empieza en los productores, los cuales captan la luz solar y la utilizan en un 100%. Luego al ser consumidos por un consumidor del primer orden el 10%  aproximadamente de esa energía pasa a ese ser vivo. Si seguimos con la cadena trófica, nos encontramos con los consumidores de segundo orden que al alimentarse de los del primer orden, toman también un 10% de su energía, lo que sería 1% de la original. El siguiente eslabón son los consumidores de tercer orden, que obtienen un 0,1% de la energía primeramente obtenida por el productor. Así los descomponedores nada más pueden sacar el 0,01% de la energía, lo que significa que esta se pierde a medida que se avanza en la cadena alimentaria, o sea, se libera al ambiente en otras formas de energía, tales como la térmica y la luz.

Flujo de Energía

La energía primaria proviene del sol.

Niveles tróficos

Los niveles tróficos son el tipo de clasificación según el tipo de alimentación que tiene cada especie. Son:
 Autótrofos o productores
 Heterótrofos o consumidores
 Consumidores
 Descomponedores

Los productores

Constituyen el primer nivel trófico. Toman la energía del sol y la transforman en moléculas orgánicas ricas en carbohidratos, lípidos y azúcares. Los principales productores en los diferentes ecosistemas son:
 Ecosistemas acuáticos: algas.
 Ecosistemas terrestres: plantas

Los consumidores

Son aquellos que se clasifican en:
 Consumidores Primarios (herbívoros).- Se alimentan de los organismos productores.
 Consumidores Secundarios(carnívoros).- Se alimentan de herbívoros.
 Consumidores Terciarios.- Se alimentan de los consumidores secundarios.
 Descomponedores.- Son principalmente bacterias y hongos. Se alimentan de los seres muertos, y de sus desechos; así forman una conexión entre lo orgánico y lo inorgánico y ayudan al medio ambiente.

Representación gráfica

El flujo de energía, se puede graficar mediante pirámides alimenticias, cadenas alimenticias, redes alimenticias y en las sedes alimentarias
 Las pirámides alimenticias, reflejan el número de individuos, presentes en cada nivel(menos los descomponedores), trófico. Mientras más alto se llega en la pirámide (mayor nivel), menos integrantes se tienen, y menos energía.
 Las cadenas alimenticias, reflejan la transferencia de energía, desde productores, hasta descomponedores, pasando por todos los niveles tróficos.
 Las redes alimenticias, son uniones de cadenas alimenticias. Esta muestra la relación entre diferentes cadenas alimenticias.

Flujo de Nutrientes

Como ya se explicó, la energía aquí en la tierra, y los nutrientes, se encuentran en cantidades limitadas. Por eso, deben ser reciclados y reutilizados .

Ciclos biogeoquímicos

Gracias a las interacciones entre diferentes especies y organismos, los nutrientes se acaban, se desplazan, y se reutilizan cumpliendo así un movimiento cíclico, en los ecosistemas. Hay unos nutrientes de especial importancia, que son:
 Agua
 Carbono
 Oxígeno
 Nitrógeno
 Fósforo
 Hidrógeno

Ciclo del agua (H2O)

El agua, es un compuesto fundamental para la vida, ya que actúa como solvente para las reacciones químicas que se dan dentro de los organismos. El agua le permite tomar los nutrientes del suelo a las plantas, y el oxígeno del aire a los animales. Se da en 4 pasos básicos:
 Evotranspiración: proceso, mediante el cual, el agua pasa de estado líquido a gaseoso, (vapor de agua). En este proceso, se dan dos pasos importantes: evaporación directa del agua, y transpiración de los organismos, especialmente, plantas, que liberan agua. La evotranspiración, aumenta con la temperatura, y la velocidad del viento.
 La precipitación: procedimiento, por el cual, el vapor de agua se condensa, y cae a la tierra en forma de lluvia.
 Almacenamiento: tiene lugar en océanos y ríos. También, en los casquetes polares, en forma de hielo. Tal es la cantidad de agua almacenada en el polo sur, que si este se fundiese,, el nivel del mar, subiría 40 m (el polo norte fundido no hace subir 1 mm tal nivel, por el principio de Arquímedes. Los polos cubren 17 millones de km² y tienen una profundidad de más o menos 1,5 km . El agua también se almacena en los picos con nieves perpetuas.
 Escorrentía: proceso por el cual, el agua "rueda" hasta el océano. Se da en ríos y quebradas, principalmente.
El agua forma casi tres cuartas partes del mundo. Está distribuida de esta manera:
 Salada: 97 %
 Dulce: 3 %
 Hielo polar y glaciares: 77,5 %
 Subterránea:22 %
 Continental, superficial, y atmosférica: 0,5 %
 Lagos y zonas húmedas: 92 %
 Atmósfera: 7 %
 Ríos: 1 %

Ciclo del carbono (CO2)

Este ciclo comienza cuando los organismos productores, toman dióxido de carbono, para realizar la fotosíntesis, y lo incorporan a sus tejidos, en forma de azúcares. El carbono, al igual que el fósforo, pasa de un nivel trófico a otro nivel trófico en las cadenas y redes alimenticias. Vale decir, que parte del carbono absorbido por las plantas es expulsado luego por las mismas, en el proceso de la respiración. Igual sucede con los consumidores; almacenan parte del carbono consumido, y el resto lo liberan en la respiración. Al final, los descomponedores desarman las moléculas, y liberan el dióxido de carbono a la atmósfera. El dióxido de carbono puede entrar también al agua.

Ciclo del oxígeno (O2)

Se puede dividir en las siguientes partes:
 El O2, sale de las plantas, en el proceso de fotosíntesis. En el agua también hay bastante oxígeno (la molécula de agua está compuesta por un átomo de oxígeno, y dos de hidrógeno). Esta sube en el proceso de evotranspiración a la atmósfera. En las partes muy altas, los rayos solares descomponen la molécula del agua, separando así el oxígeno del hidrógeno. Los organismos fotosintéticos también producen oxígeno.

 Los organismos aeróbicos, utilizan el oxígeno para la respiración, desechando así dióxido de carbono. Por esto se dice que el ciclo del oxígeno está muy ligado con el del carbono y el agua.
 El dióxido de carbono es usado por los organismos fotosintéticos, y sale como desecho oxígeno.
Así, pueden suceder dos cosas:
 Los organismos aeróbicos lo reutilizan, y luego los fotosintéticos, completando el ciclo, o
 El oxígeno sea incorporado al agua, junto con dos átomos de hidrógeno. Así, esta agua sube, y completa y repite el susodicho ciclo.

Ciclo del nitrógeno(N),(N2)

El nitrógeno, es un elemento muy importante en la tierra. Forma el 78% de la atmósfera. Es fundamental en la estructura de los aminoácidos, las proteínas, y los ácidos nucléicos. Sin embargo, no puede ser utilizado directamente por los organismos. Así, tiene que ser transformado para el uso de los organismos. El ciclo sigue los siguientes pasos:

1 Transformación. Diferentes bacterias transforman el nitrógeno:
 Los Clostridios, fijan el nitrógeno al suelo.
 Los Rizobios, que viven en los nódulos de algunas legumbres, y transforman el nitrógeno, para que la planta lo pueda utilizar
2 Cadena trófica. Luego, el nitrógeno toma el camino de las redes y cadenas alimenticias, de herbívoros, a carnívoros. Luego, el nitrógeno regresa al suelo, en forma de desechos, y cuerpos muertos.

3 Regreso al suelo. El nitrógeno vuelve al suelo en forma de amoníaco. El amoniaco puede ser utilizado por las plantas otra vez, o permanecer en el suelo, convirtiéndose en nitratos. Estos nitratos son regresados a la atmósfera gracias a las pseudomonas, que restituyen el nitrógeno a la atmósfera.

Ciclo del fósforo (P)

El fósforo, es indispensable para la vida en la tierra. Hace parte de los huesos, de los ácidos nucleicos, de los fosfolípidos de las membranas celulares, es el principal componente del ATP, del cual los seres vivos toman energía y nutrientes. Este ciclo, es el único que no tiene movimientos sobre la faz terrestre. El fósforo se mueve a través de sus sitios de almacenamiento. Son las rocas sedimentarias, y los organismos vivos. Se da en los siguientes pasos:
 Erosión. Las rocas ricas en fósforo, se erosionan con el tiempo. El fósforo, por lo tanto, también se disuelve, y se incorpora en la tierra en forma de fosfatos.
 Cadenas tróficas. Las plantas absorben los fosfatos de la tierra, y luego, el fósforo se mueve con los organismos, en las cadenas y redes tróficas, hasta que llega a los organismos descomponedores (como por ejemplo: hongos, y bacterias).

También puede suceder, que no todo el fósforo sea absorbido por las plantas, sino que sea arrastrado por las corrientes acuíferas. En estos casos, es transportado al mar, en donde es depositado junto con los sedimentos marinos, al fondo del océano. Luego de varios años, se incorpora a las rocas, que más tarde, pueden subir a la superficie por algún fenómeno geológico, comenzando así, nuevamente el ciclo. Al estar el fósforo en el mar, puede suceder que los animales lo beban, por esto, los organismos marinos son ricos en fósforo.

Véase también

 Energía
 Nutrientes
 Autótrofos
 Heterótrofos
 Ciclos biogeoquímicos
 Ciclo del agua
 Ciclo del nitrógeno
 Ciclo del fósforo
 Ciclo del carbono
 Ciclo del oxígeno
 Metabolismo social

Notas

Bibliografía

Enlaces externos

 CiclodelCarbono.com Información acerca del Ciclo del Carbono
 Forma más adecuada de aportar nitrógeno a las plantas: Amoníaco vs Nitrato
 Las plantas acuáticas prefieren el amonio
 El nitrógeno en la nutrición de las plantas de cultivo
 Información acerca del ciclo del Agua

Ecosistemas

redpajama_set_name: RedPajamaWikipedia


text:
Q: Unable to start the Weblogic Admin Server after DEV_OAM and DEV_OPSS DB Password Reset I have Reset the DEV_OAM and DEV_OPSS Passwords
Using the command
alter user dev_oam identified by Xxxxxxx

alter user dev_opss identified by Xxxxxx

After the Password Reset i have tried to restart the Weblogic Admin Server and Oam managed server.When i am trying to restart the Admin Server, I encountered the below error.

Internal Exception: java.sql.SQLException: ORA-01017: invalid username/password; logon denied
Error Code: 1017
   Feb 20, 2015 10:58:08 PM oracle.security.jps.internal.credstore.ldap.LdapCredentialStore init
      WARNING: Could not create credential store instance. Reason oracle.security.jps.service.policystore.PolicyStoreConnectivityException: JPS-10000: There was an internal error in the policy store.
      JPS-01055: Could not create credential store instance. Reason oracle.security.jps.service.policystore.PolicyStoreConnectivityException: JPS-10000: There was an internal error in the policy store.
      Error: Diagnostics data was not saved to the credential store.
      Error: Validate operation has failed.
      [oracle@oamps2 bin]$
      [oracle@oamps2 bin]$ tail -f AdminServer.out
      [EL Severe]: 2015-02-20 22:46:27.246--ServerSession(899981417)--Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.3.1.v20111018-r10243): org.eclipse.persistence.exceptions.DatabaseException
      Internal Exception: java.sql.SQLException: ORA-01017: invalid username/password; logon denied
Error Code: 1017
      Feb 20, 2015 10:46:27 PM oracle.security.jps.internal.credstore.ldap.LdapCredentialStore init
      WARNING: Could not create credential store instance. Reason oracle.security.jps.service.policystore.PolicyStoreConnectivityException: JPS-10000: There was an internal error in the policy store.
      JPS-01055: Could not create credential store instance. Reason oracle.security.jps.service.policystore.PolicyStoreConnectivityException: JPS-10000: There was an internal error in the policy store.
      Error: Diagnostics data was not saved to the credential store.
      Error: Validate operation has failed.
      Need to do the security configuration first!

Can anyone suggest me what should i do to resolve this.

A: Have you altered all of your connection pool configuration passwords in weblogic to match the database password?  This document may help you, particularly the referenced documents at the bottom (You need an oracle support account to view these).

redpajama_set_name: RedPajamaStackExchange


Amazon Breakthrough Novel Award: 1st round results posted. They liked my pitch letter. So Redemption for the Hypnotist will be moving into the second round. I wouldn't be surprised if the book gets eliminated in the second round because of the sexual theme. But you never know, the judges might just like my writing.


This was a product I did during my internship at Overlake Christian Church in Redmond, WA. I worked in a website management program to create an information page for the church events during December 2012.
I used their assets to create a masthead image for the page and then added text and further images below that.


text:
Oenochroa är ett släkte av fjärilar. Oenochroa ingår i familjen praktmalar, Oecophoridae.

Kladogram enligt Catalogue of Life:

Källor

Praktmalar
Oenochroa


text:
Share this Story: Haunted by Diana's death, Prince Harry talks of how he feared losing Meghan, too
Haunted by Diana's death, Prince Harry talks of how he feared losing Meghan, too
'It was like I was outside of my body'
Jill Serjeant
May 21, 2021 • May 21, 2021 • 3 minute read • 82 Comments
Prince Harry at Buckingham Palace on Jan. 16. Photo by Jeremy Selwyn - WPA Pool/Getty Images
From the sound of horses' hooves on the day of Princess Diana's funeral, to chases by paparazzi when he was a child in the back of a car, Prince Harry is still haunted by the trauma of losing his mother.
Speaking in detail of how he failed to deal with her loss for more than a decade, Harry says in a new TV documentary series that the fear of also losing his wife Meghan was one of the main reasons the couple quit their royal duties and moved to California last year.
Haunted by Diana's death, Prince Harry talks of how he feared losing Meghan, too Back to video
"My mother was chased to her death when she was in a relationship with someone who wasn't white, and now look what's happened. You're talking about history repeating itself? They're not going to stop until she dies," Harry says in the mental health documentary series "The Me You Can't See."
"It's incredibly triggering to potentially lose another woman in my life," he added, referring to Meghan. The "Me You Can't See" series, which Harry produced with U.S. talk show host Oprah Winfrey, is released on Apple TV+ on Friday.
Prince Harry: pain of Diana's death pushed me to drink and drugs
Prince Harry and Oprah open up for mental health documentary
Princess Diana died in 1997 at age 36 in a crash in Paris after the car in which she was traveling with her Egyptian-born boyfriend Dodi Fayed was chased by paparazzi. Harry was 12 years old.
Harry, whose American wife Meghan is bi-racial, spoke of racist reporting in the mainstream British press about his wife, as well as abuse on social media. Meghan has said her experience in Britain drove her to thoughts of suicide when she was pregnant with their first son Archie.
In the documentary, Harry spoke of walking behind Diana's coffin through the streets of London with his brother William, father Prince Charles and uncle Charles Spencer.
"The thing I remember most was the sound of the horses' hooves going along the Mall," he said. "It was like I was outside of my body, just walking along, doing what was expected of me, showing one tenth of the emotion that everyone was showing."
Years earlier, he recalled sitting in the back of his mother's car while she, in tears, was pursued by photographers.
No justice at all
"One of the feelings that comes up for me always is the helplessness. Being a guy and being too young to help a woman, in this case your mother, and that happened every single day," he said.
Harry said he buried his feelings, but drank heavily and suffered panic attacks and anxiety in his 20s, and still freaks out when he sees cameras.
"I was so angry with what happened to her (Diana) and the fact there was no justice at all… the same people that chased her through that tunnel photographed her dying on the back seat of that car," he said.
"The clicking of cameras, and the flash of cameras makes my blood boil. It makes me angry. It takes me back to what happened to my mum, what I experienced as a kid."
Even now, at age 36, he said that returning to London makes him feel tense and hunted.
Harry said he started serious therapy almost five years ago, when he met Meghan.
"I quickly established that if this relationship was going to work, I had to deal with my past," he said. The couple married in May 2018.


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It's normal to feel a degree of sadness and anxiety after a child is born, even if you have been excitedly waiting to become a parent — and perhaps especially if you haven't. As much as you try to prepare for your little one, you'll still need to make a lot of adjustments when you realize the responsibilities of parenting aren't always as easy as you thought they would be.
But, when are your emotions a sign of something bigger? How do you know if what you are experiencing is normal "baby blues" vs. postpartum depression?
Both baby blues and postpartum depression are common and 100 percent normal after a child is born. However, each requires a different path for treatment and healing — which is why it's so important to understand the difference between postpartum "blues" vs. depression after childbirth to get the help you need. Remember, whatever you are feeling, you are not alone.
In the article below, you can find some helpful information for determining between the different types of depression you may experience after bringing your new child home. However, the information in this article should not be used to diagnose your mental condition and should not be taken as medical advice. If you are struggling with depression after your pregnancy, please reach out to your medical professional right away. You can also reach out to the 24-hour suicide hotline at 1-800-273-8255 if you are having thoughts of harming yourself or your child.
And these feelings aren't specific only to those who raise their children — new birth mothers can also experience feelings of sadness after birth. Their emotions may even be heightened, as they are simultaneously dealing with the grief and loss of placing their child for adoption with adoptive parents.
If you are feeling these emotions after birth, they are likely just caused by the new situation you are in, your changing hormones, and the fatigue from pregnancy and childbirth. However, there is also the chance that they are symptoms of more serious postpartum depression.
So, how can you tell the difference when it comes to baby blues vs. postpartum depression?
Normal postpartum "baby blues" often result in quick mood swings and the feelings stem from the normal stressors of raising a newborn. Postpartum depression, on the other hand, involves long-term feelings of sadness, worthlessness and anxiety. You may feel like you're not doing a good job as a mom or even having trouble bonding with your baby. Perhaps your post-placement feelings in adoption are stronger than you anticipate, and you are constantly worrying whether or not you made the right choice. These feelings can interfere with your sleeping and eating schedule.
If you are unsure whether what you are feeling is baby blues vs. postpartum depression, we encourage you to reach out to your medical professional. They can evaluate your symptoms to determine what kind of treatment is necessary in your situation. Don't worry — postpartum depression is 100 treatable, should you be diagnosed with it.
Sometimes, when new mothers worry about depression after pregnancy, they worry that it's something even more extreme — postpartum psychosis. They've likely heard horror stories about women who suffer from postpartum psychosis and are terrified at the idea of harming themselves or their child should this condition develop.
The good news is that postpartum psychosis is extremely rare. It only affects about 1 or 2 in every 1,000 women and usually occurs within two to three days of delivery. So, it is highly unlikely that you will experience postpartum psychosis after giving birth.
Still, how do you determine between postpartum psychosis vs. postpartum depression? Remember: Both of these conditions require medical intervention. Whether you think you may have postpartum depression or postpartum psychosis, do not hesitate to contact your medical professional right away.
While postpartum psychosis is rare, it is possible. If you are experiencing any symptoms of depression during pregnancy, however extreme they may be, please contact your medical professional. They can decide what condition you may be facing — baby blues vs. postpartum depression vs. postpartum psychosis. Whatever it is, there is help out there to make your postpartum recovery period happier and safer.


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Q: Solve $\frac{1}{k^{2}}r^{4}-r^{2}+\frac{\varrho S}{g}(-\omega)=0$ for r and reformulate $\frac{1}{k^{2}}r^{4}-r^{2}+\frac{\varrho S}{f}(-\omega)=0$
Solve the characteristic equation for r.
Using Mathematica I get:

In the end, one should be able to show that, if we name the real roots of the characterisitc equation as "q" and the imaginary roots as "p", then:
$q=\sqrt{p^2+k^2}$
Can anyone help me ?

A: The following assumes that $\varrho, S, f, \omega \gt 0$ (which seems a sensible assumption in the context).
Let $s = r^2$ and $t = \frac{\varrho S}{f}(-\omega) \lt 0$, then the equation reduces to the quadratic:
$$\frac{1}{k^{2}}s^{2}-s+t=0$$
Since $t \lt 0$ the quadratic will always have $2$ real roots of opposite signs, $s_1 \lt 0$ and $s_2 \gt 0$. It follows by Vieta's formulas that: $$s_1 + s_2 = k^2 \tag{1}$$
The roots in $r$ will then be $r_{1,2} = \pm i \sqrt{-s1}\;$ and $r_{3,4} = \pm \sqrt{s_2}\;$. Let $p = \sqrt{-s_1}\;$ be the imaginary part of $r_1$, and $q=\sqrt{s_2}$ the positive real root $r_3$. Substituting $s_1 = -p^2$ and $s_2 = q^2$ into $(1)$:
$$q^2 - p^2 = k^2 \quad \iff \quad q = \sqrt{p^2 + k^2}$$


text:
In the 2010–11 season, USM El Harrach is competing in the Ligue 1 for the 27th season, as well as the Algerian Cup. They will be competing in Ligue 1, and the Algerian Cup.

Squad list
Players and squad numbers last updated on 18 November 2010.Note: Flags indicate national team as has been defined under FIFA eligibility rules. Players may hold more than one non-FIFA nationality.

Competitions

Overview

{| class="wikitable" style="text-align: center"
|-
!rowspan=2|Competition
!colspan=8|Record
!rowspan=2|Started round
!rowspan=2|Final position / round
!rowspan=2|First match
!rowspan=2|Last match
|-
!
!
!
!
!
!
!
!
|-
| Ligue 1

|
| 4th
| 25 September 2010
| 8 July 2011
|-
| Algerian Cup

| Round of 64
| style="background:silver;"| Runners–up
| 1 January 2011
| 1 May 2011
|-
! Total

Ligue 1

League table

Results summary

Results by round

Matches

Algerian Cup

Squad information

Playing statistics

|-
! colspan=10 style=background:#dcdcdc; text-align:center| Goalkeepers

|-
! colspan=10 style=background:#dcdcdc; text-align:center| Defenders

|-
! colspan=10 style=background:#dcdcdc; text-align:center| Midfielders

|-
! colspan=10 style=background:#dcdcdc; text-align:center| Forwards

|-
! colspan=10 style=background:#dcdcdc; text-align:center| Players transferred out during the season

Goalscorers
Includes all competitive matches. The list is sorted alphabetically by surname when total goals are equal.

Transfers

In

Out

References

USM El Harrach seasons
Algerian football clubs 2010–11 season


Description : Robert Nelson, from The Age, described Augustine Dall'Ava's work as "often floating at the extremity of architectonic elements". Personally, when I first stumbled across the sculpture, I thought it looked like something out of a children's toy tool box (brightly colored spanners etc) and oddly out of place on such a busy street. But everytime I passed it, a strange thing happened, I began to like it. A bold, brightly coloured sculpture is just what the rather drab street needed, reminding passerbys of the beautiful colours and hues of the northern land. The three mild steel/stainess steel sculptures, of varying sizes, feature unusual shapes (a cube and a flame), balancing, sometimes precariously, from atop their colorful columns. The colors are symbolic of the landscape and environment of Queensland; the metallic reds, blues and yellows reflect the irridescent colors of the tropical fish of the Great Barrier Reef, the green the rainforests, the white the beautiful beaches, blue the water and yellow the sun. As night falls and long shadows are cast on the city, the sculpture take on a more interesting form. I just can't understand why the sculpture was placed next to a hideous light pole, a rubbish bin and a street sign, killing the ambience. Oh, and I still don't know why the cube has little holes in it, all I keep thinking of is, tea sprainer.
Location : On the footpath outside Queens Plaza, Edward Street, Brisbane, Queensland, Australia.
environment of the State of Queensland.
History of the Echoes From The North sculpture : The Echoes From the North sculpture was originally commissioned by the Gandel Retail Management in 2004, but since then, the artwork has been gifted to the City of Brisbane, and now is part of the city's cultural heritage.
A little observation : Don't you think the Echoes From the North sculpture, looks a little like the Trickle, Flow and Fall sculpture in Midland, Western Australia ?


Kings Community Church in Letchworth is participating in 31 days of fasting with the aim to raise funds for the Teenage Cancer Trust in the month of August. During this time, the congregation will abstain from food for varying periods with the view to raise donations for our chosen charities.
The Teenage Cancer Trust operates 28 specialist units across the country that provide care and treatment for young people aged between 13 and 24 who have been diagnosed with cancer.
The charity is currently only able to offer its vital support to half of these young cancer patients. However, they are in the process of expanding their nursing and support services within the NHS that will enable them to reach every single young person who is affected by cancer. Therefore, Kings Community Church in Letchworth want to assist the Teenage Cancer Trust in reaching their goal to have the funds to provide this invaluable nursing and support care for every young person affected by cancer.


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Connect with other COV4 attendees at the FACEBOOK EVENT page!


The builders of American Samoa's future food processing plant have held a ground-breaking ceremony and promised to generate 700 jobs.
The Filipino company, AVM Bernardo Engineering, first expressed interest in setting up a factory in the territory seven years ago.
The factory is being built at the Industrial Park in Tafuna near the Public Works compound.
The owner of the company, Anthony Bernardo, says it will be completed in a year's time, and will turn out American Samoan-made food and juice products for the US and other overseas markets.
Governor Lolo Matalasi Moliga says it's a welcome development at a time when the tuna industry faces uncertainty.
The owner of the Philippines company, AVM Bernardo, hopes it can start building a food processing plant in American Samoa before the end of this year.


Recently, Brazilian researchers have reported about another potential – and "politically correct" – source of stem cells – discarded Fallopian tubes.
The Fallopian tubes are the tubes that "catch" the eggs from the ovaries, to be transported to the uterus. Sometimes the tubes have to be removed surgically, an intervention called salpingectomy. In many cases, it is performed in connection with oophorectomy (removal of the ovary) and/or hysterectomy (removal of the uterus) in women with cervical or breast cancer. The researchers found abundant immature cells in Fallopian tubes during surgery, cells which can potentially develop into bones, muscles, and other organs.
Stem cell research is cutting edge technology in the quest for curing many diseases, from cancer, to cardiovascular disorders, to neurodegenerative diseases. Identifying more "ethical" sources of stem cells can give this field of research a boost and hopefully produce results. Currently, a few stem cell-based therapies are undergoing clinical trials.
Posted by Science-mom on Monday, June 22nd, 2009 at 11:14 am and is filed under Current Events, News.


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Your reading in Ch 8 of SECS will introduce the concept of Meteorological Years, while the Ch 2 reading in Kalogirou offers a condensed description of how a TMY data set is formed and how the files are formatted. The Supplemental readings address the most recent version of the TMY data sets in the USA (TMY3), and point to the National Solar Radiation Database maintained by NREL.
The data that you will find in your SAM simulation software, and the data that is all over the web in resources like the NREL Dynamic Maps of the USA solar resource, come from a single database, called the NSRDB, or the National Solar Radiation Database. There are currently three generations of TMY.
What is a Typical Meteorological Year? Why would we use a synthesized year of data for solar resource simulations?
A typical meteorological year (TMY) data set provides designers and other users with a reasonably sized annual data set that holds hourly meteorological values that typify conditions at a specific location over a longer period of time, such as 30 years. TMY data sets are widely used by building designers and others for modeling renewable energy conversion systems. Although not designed to provide meteorological extremes, TMY data have natural diurnal and seasonal variations and represent a year of typical climatic conditions for a location. The TMY should not be used to predict weather for a particular period of time, nor is it an appropriate basis for evaluating real-time energy production or efficiencies for building design applications or solar conversion systems.
...The TMY data set is composed of 12 typical meteorological months (January through December) that are concatenated essentially without modification to form a single year with a serially complete data record for primary measurements. These monthly data sets contain actual time-series meteorological measurements and modeled solar values, although some hourly records may contain filled or interpolated data for periods when original observations are missing from the data archive.
Estimation can often be evaluated relative to 30 year averages of weather conditions at specific locations, termed the Typical Meteorological Year (TMY). These data are not reasonable estimates of extreme conditions (e.g., hurricanes, tornadoes) and may also be inaccurate for evaluating site or time-specific data. The most common database is TMY3, now collected from the period of 1991 to 2010.
TMY data was initially developed to aid in building simulation, for modeling the energy demands in counterpoint with the solar/meteorological gains. As in your software SAM (and the source code, TRNSYS), TMY is also used by SECS design teams for initial estimates of energy and financial returns on investment. We can use SAM's TMY data set to evaluate PV, solar hot water, and CSP systems.
The source data for the TMY set in the USA comes from the NSRDB, or the National Solar Radiation Database.
Let's go back to the SAM software and explore the solar resource data in its browser.
You can open up SAM at this point and click the button "Start a new project"
Click on "Photovoltaics (Detailed)" on the left of the pop up window ("Choose a performance model").
Click on "Residential (Distributed)" on the right of the pop up window ("then choose from the available financial models").
You will now have a default residential PV project, based in Phoenix, AZ. There will likely be a SunPower (California company) PV module, and an SMA America dc-ac power inverter listed on the left column of shortcut tabs. They're not really "tabs" like the tab at the top of the window, but I'm going to use that term for convenience in the class.
Click on the shortcut tab called "Location and Resource" (will list "Location," latitude ($\phi$), longitude ($\lambda$), and elevation above sea level in the shortcut.
Leave the location on Phoenix.tm2 and look inside the section "Tools."
Click the button "View hourly data...," and a new window will pop up.
Reminder: This information will be used when you create your final project.