text:
\section{Introduction}

In this paper we are concerned with the reduced rank variant of the
multivariate response regression model. We are given $N$ observations of the
predictors $X_{i}\in \mathbb{R}^{p}$ and responses $Y_{i}\in \mathbb{R}^{r},
$ which are assumed to be related by the linear regression model:
\begin{equation}
Y=XA+U,  \label{regression_model}
\end{equation}%
where $A$ is an unknown $p$-by-$r$ matrix and $U$ is a noise matrix. This model is ubiquitous in statistics, signal
processing, and numerical analysis.

On methodological grounds one often postulates that
the responses depend only on a small number of factors which are linear
combinations of the predictors. This postulate leads to a
model, in which $A$ is assumed to be a low-rank matrix:
\begin{equation}
A=\sum_{j=1}^{s}\theta _{j}u_{j}v_{j}^{\ast },  \label{low_rank_model}
\end{equation}%
where $\left\{ u_{j}\in \mathbb{R}^{p}\right\} $ and $\left\{ v_{i}\in
\mathbb{R}^{r}\right\} $ are two fixed orthonormal vector systems. This
model appeared already in Anderson (1951) \cite{anderson51}, and it was named \textit{reduced-rank regression%
} in Izenman
(1975) \cite{izenman75}.  In some contexts, this model is also known under the names
\textit{simultaneous linear prediction}  (Fortier (1966)
\cite{fortier66}) and \textit{redundancy analysis} (van den Wollenberg
(1977) \cite{wollenberg77}), both of which assume that $U$ has the covariance matrix equal to $\sigma^2 I$.
The reduced-rank model has been intensively studied, and many results are collected in the
monograph by Reinsel and Velu (1998) \cite{reinsel_velu98}.

In this paper, we assume that $U$ has the covariance matrix equal to $\sigma^2 I$, and we are interested in the situation in which all three
variables, $p,$ $r,$ and $N,$ grow at the same rate.

\textbf{Assumption A1.} \textit{It is assumed that  as $N\rightarrow
\infty$,  $\frac{N}{p}\rightarrow $ $%
1+\lambda \geq 1$ and $\frac{N}{r}\rightarrow \mu >0$.}

It is also useful to define $\beta \df \lim_{N\rightarrow \infty }\frac{p}{r}=\mu
/\left( 1+\lambda \right).$

The studies devoted to
the reduced-rank regression in this setup are relatively recent and include
Bunea, She, and Wegkamp (2011) \cite{bunea_she_wegcamp11} and Giraud (2011)
\cite{giraud11}.

We address the following questions. First, is it possible to detect that the
true matrix $A$ is not zero? If yes, then how do we how do we estimate the rank and singular values of $A$?

Our approach to these questions is based on the study of the statistical
properties of the standard least squares estimator
\begin{equation*}
\widehat{A}\df X\backslash Y\equiv \left( X^{\ast }X\right) ^{-1}X^{\ast }Y
\end{equation*}%
and the matrix of fitted responses:
\begin{equation*}
\widehat{Y}\df X\widehat{A}
\end{equation*}

By using this approach, we will develop a rank-selection algorithm
which
performs better than the algorithm from \cite{bunea_she_wegcamp11} in a
certain range of parameters and is simpler than the algorithm in \cite{yelm07}. In
addition, we will develop tools for consistent estimation of singular values $\theta
_{i}.$ The paper \cite{bunea_she_wegcamp11} does not address this issue,
since its focus is on minimizing the prediction error, in particular on
bounds for $\mathbb{E}\left\Vert XA-X\widetilde{A}\right\Vert ,$ where $\widetilde{A}$
is an estimator of $A$ and the expectation is over randomness in $U$.

The rest of the paper is organized as follows. Section \ref{section_major_results} describes the major results. Section \ref%
{subsection_largest_eigenvalue_nonnull} provides the details of the proofs. Section \ref{conclusion} recapitulates the results.  And Appendix \ref%
{section_limiting_distribution} provides a proof for the theorem about the limiting
distribution of singular values of $\widehat{A}$.

\section{Major results}
\label{section_major_results}

\subsection{Tests of the null hypothesis}

Let $X$ be a $p$-by-$r$ real Gaussian matrix: each row is an independent observation from $\mathcal{N}(0,\Sigma)$. Then, an $r$-by-$r$ matrix $X^{\ast}X$ is said to be a \emph{Wishart matrix} with distribution $W_r(\Sigma,p)$.

A random $m$-by-$m$ matrix $X$ is said to belong to the \emph{(real) Jacobi ensemble} with parameters $\alpha_1$ and $\alpha_2$, if its distribution is invariant with respect to orthogonal transformations and the distribution of its eigenvalues is given by

	\begin{equation}
f^{(\alpha_1,\alpha_2)}\left( \lambda_{1},\ldots ,\lambda_{m}\right) =\frac{1}{c}\prod\limits_{j=1}^{m}
\lambda_{j}^{\alpha_1}( 1-\lambda_i) ^{\alpha_2}
\prod\limits_{1\leq j<k\leq m}\left\vert
\lambda_{j}-\lambda_{k}\right\vert.  \label{pdf_Jacobi}
\end{equation}%

The following result is fundamental for the analysis of matrices $\widehat{A}
$ and $\widehat{Y}.$

  \begin{theorem}
	\label{theorem_fundamental}
	(i) Suppose that $U$ is an $N$-by-$r$ matrix with i.i.d standard real Gaussian entries, and $X$ is an  $N$-by-$p$ full-rank matrix ($N\geq p$) independent of $U$. Then the squared singular values of $\widehat{Y}\df X(X\backslash U)$ are distributed as the eigenvalues of the Wishart matrix with distribution $W_r(I,p)$. \\
	(ii) In addition, suppose that $X$ has i.i.d standard real Gaussian entries.
	Let $s^2_i$ be the squared singular values of $\widehat{A}\df X\backslash U$ and $f_i=s^2_i/(1+s^2_i)$. Then, the positive $f_i$ are distributed as eigenvalues of
			the Jacobi ensemble with parameters $m=\min\{p,r\}$, $\alpha_1=(|r-p|-1)/2$ and $\alpha_2=(N-p-1)/2$.
  \end{theorem}
	\textbf{Proof}: 	The matrix $ \widehat{Y}=X(X\backslash U)$ is the orthogonal projection of $r$ column vectors of $U$ on the
	$p$-dimensional column span of $X$. Hence, in an
	appropriate basis, $\widehat{Y}$ is a block matrix with one block given by a $p$-by-$r$ matrix with i.i.d. standard Gaussian entries and another block of $N-p$-by-$r$ matrix of zeros. This proves the first part of the theorem. 	For the second part, note that positive eigenvalues of $\widehat{A}^{\ast} \widehat{A} = U^{\ast }X\left( X^{\ast }X\right) ^{-2}X^{\ast }U$
have the same distribution as positive eigenvalues of $B^{-1}C,$ where $B$ and $C$ are independent Wishart matrices.

Indeed, the rank of matrices  $U^{\ast}X(X^{\ast}X)^{-2}X^{\ast}U$ and \\ $X(X^{\ast}X)^{-2}X^{\ast}UU^{\ast}$ is
$\min\{p,r\}$, and their positive eigenvalues are the same. Let $W$ be an orthogonal $N$-by-$p$ matrix formed by the eigenvectors of $X(X^{\ast}X)^{-2}X^{\ast}$ and such that the matrix $W^{\ast}X(X^{\ast}X)^{-2}X^{\ast}W$ is diagonal with positive eigenvalues on the diagonal.
  These eigenvalues coincide with positive eigenvalues of the inverse of a Wishart matrix, $(X^{\ast}X)^{-1}$, where the Wishart matrix has the distribution $W_p(I,N)$. The matrix $W^{\ast}UU^{\ast}W$ is Wishart with distribution $W_p(I,r)$.

In addition, matrices  $W^{\ast}X(X^{\ast}X)^{-2}X^{\ast}W$ and $W^{\ast}UU^{\ast}W$ are independent because the eigenvalues and eigenvectors of $X(X^{\ast}X)^{-2}X^{\ast}$ are independent.  Finally, since similarity transformations do not change eigenvalues, the distribution of positive eigenvalues of  $X(X^{\ast}X)^{-2}X^{\ast}UU^{\ast}$ is the same as the distribution of positive eigenvalues of $B^{-1}C$ for two independent Wishart matrices $B$ and $C$ with distributions $W_p(I,N)$ and $W_p(I,r)$, respectively.

Next, note that the eigenvalues of $B^{-1}C$ are the same as those
of $F\left( I-F\right) ^{-1},$ where $F=(B+C)^{-1}C.$ Hence the
eigenvalues of $B^{-1}C,$ denoted by $l_{i},$ are
related to the eigenvalues of $\left( B+C\right) ^{-1}C,$ denoted by $f_i$ by the transformation,%
\begin{equation}
l_{i}=\frac{f_{i}}{1-f_{i}}.  \label{relation_of_eigenvalues}
\end{equation}%
Then one can use the classical fact that the eigenvalues of $\left( B+C\right) ^{-1}C$
are distributed as in (\ref{pdf_Jacobi}) with $\alpha_1=(|r-p|-1)/2$ and $\alpha_2=(N-p-1)/2$. (See Theorem 3.3.1 in Muirhead
\cite{muirhead82}).
\hfill $\square$

There are several ways to test the null hypothesis that $A=0.$ The simplest
way is to compute the largest eigenvalue of $Y^{\ast }Y,$ called $l_{Y,1},$
and compute
\begin{equation*}
s_{Y,1}=\frac{l_{Y,1}-\mu ^{\left( 1\right) }}{\sigma ^{\left(
1\right) }},
\end{equation*}%
where
\begin{eqnarray}
\mu ^{\left( 1\right) } &=&\left( \sqrt{N-1}+\sqrt{r}\right) ^{2},
\label{test1_scaling}
\\
\sigma ^{\left( 1\right) } &=&\left( \sqrt{N-1}+\sqrt{r}\right) \left[ \frac{%
1}{\sqrt{N-1}}+\frac{1}{\sqrt{r}}\right] ^{1/3}.  \notag
\end{eqnarray}

If $A=0$ and the noise matrix $U$ has i.i.d. standard Gaussian entries,  then by
Theorem 1.1 in Johnstone (2001) \cite{johnstone01} $s_{Y,1}$ converges in distribution to the Tracy-Widom
distribution $\mathcal{F}_{1}.$\footnote{Some quantiles of the Tracy-Widom distribution are: $x_{50\%}=-1.3,$ $%
x_{10\%}=0.45,$ $x_{5\%}=0.98,$ $x_{2\%}=1.60,$ $x_{1\%}=2.02,$ where $%
\mathbb{P}\left( x\geq x_{c}\right) =c.$} Note that this test does not depend on  $X$ and that the normality assumption on the entries of $U$ can be significantly relaxed. Indeed, by results of Pillai and Yin \cite{pillai_yin14}, the convergence to the Tracy-Widom distribution holds provided that the entries are independent, have zero mean, unit variance, and subexponential decay: $\mathbb{P}\{U_{ij}>t\}\leq \kappa^{-1}\exp\{-t^\kappa\}$ for a positive $\kappa$ and all $t>1$. (The first universality result of this type is due to Soshnikov \cite{soshnikov02}.)

Two other methods to test the null hypothesis are conditional on $X$. They are based on the singular values of
matrices $\widehat{A}\df X\backslash Y$ and $\widehat{Y}\df X \widehat{A}.$ Let us define two statistics,
$s_{\widehat{Y},1}$ and $s_{\widehat{A},1}:$

\begin{equation*}
s_{\widehat{Y},1}\df\frac{l_{\widehat{Y},1}-\mu ^{\left( 2\right) }}{\sigma ^{\left(
2\right) }},
\end{equation*}%
where  $l_{\widehat{Y},1},$ is the square of the largest singular value of  $\widehat{Y}$ and
\begin{eqnarray}
\mu ^{\left( 2\right) } &=&\left( \sqrt{p-1}+\sqrt{r}\right) ^{2},
\label{test2_scaling}
\\
\sigma ^{\left( 2\right) } &=&\left( \sqrt{p-1}+\sqrt{r}\right) \left[ \frac{%
1}{\sqrt{p-1}}+\frac{1}{\sqrt{r}}\right] ^{1/3}.  \notag
\end{eqnarray}%

Similarly,
\begin{equation*}
s_{\widehat{A},1}=\frac{\log l_{\widehat{A},1}-\mu ^{\left( 3\right) }}{%
\sigma ^{\left( 3\right) }},
\end{equation*}%
where $l_{\widehat{A},1}$ is the square of the largest singular value of $\widehat{A}$ and
\begin{eqnarray}
\mu ^{\left( 3\right) } &=&2\log \tan \left( \frac{\phi +\gamma }{2}\right) ,
\label{centering_and_scaling}
\\
\sigma ^{\left( 3\right) } &=&\left[ \frac{16}{\left( N+r-1\right) ^{2}}%
\frac{1}{\sin ^{2}\left( \phi +\gamma \right) \sin \phi \sin \gamma }\right]
^{1/3},  \notag
\end{eqnarray}%
and the angle parameters $\gamma $ and $\phi $ are defined by
\begin{eqnarray*}
\sin ^{2}\left( \frac{\gamma }{2}\right) &=&\frac{\min \left( p,r\right) -1/2%
}{N+r-1}, \\
\sin ^{2}\left( \frac{\phi }{2}\right) &=&\frac{\max \left( p,r\right) -1/2}{%
N+r-1}.
\end{eqnarray*}

\bigskip

\begin{theorem}
(i) Suppose that $A=0$, Assumption A1 is satisfied, $U$ is a matrix with
independent standard Gaussian entries, and $X$ is a full-rank matrix independent of $U.$
Then the random variable $s_{\widehat{Y},1}$ converges in distribution to
the Tracy-Widom distribution. \\
(ii) In addition, suppose that $X$ has i.i.d. standard Gaussian entries. Then $s_{\widehat{A},1}$ converges in distribution to
the Tracy-Widom distribution.
\label{theorem_TW}
\end{theorem}

\textbf{Proof:} Both claims follow from Theorem \ref{theorem_fundamental} above and Johnstone's  work on the largest eigenvalues of the Wishart and Jacobi ensembles (specifically, Theorem 1.1 in \cite{johnstone01} and Theorem 1 in \cite{johnstone08}). \hfill  $\square $

\begin{table}[htbp]
\begin{tabular}{cc}
\hline
\begin{minipage}[t]{0.45\textwidth} \centering Singular values of
$\widehat{Y}$ \\ \includegraphics[width=1\textwidth]{cdf_largest_s_Y.jpg}
\end{minipage} & \begin{minipage}[t]{0.46\textwidth} \centering Singular
values of $\widehat{A}$ \\
\includegraphics[width=\textwidth]{cdf_largest_s_A.jpg} \end{minipage} \\
\hline
\end{tabular}%
\captionsetup{name=Figure}
\caption{ The blue solid line shows the cumulative distribution function for
the scaled largest singular value. The red dashed line is the Tracy-Widom
distribution $TW_{1}.$ The parameters are $N=100$, $p=66$, $r=133$.}
\label{fig:largest_null}
\end{table}

Corresponding to each of these three tests, we can devise a procedure for
the choice of the rank of the model (\ref{low_rank_model}):

1. Calculate the statistics $s_{Y,i}$, $s_{\widehat{Y},i}$, or $s_{\widehat{A},i}$ for the singular values of corresponding  matrices.

2. Check how many of these statistics exceed the $10\%$ quantile
of the Tracy-Widom distribution ($x_{10\%}=0.45$) and take this number as the
rank of the model (\ref{low_rank_model}).

This should be compared to the procedure suggested in \cite%
{bunea_she_wegcamp11}. The first part of their procedure is to compute $l_{\widehat{Y},i}$. Next, they prescribe to choose the
rank of the model equal to the number of statistics  $l_{\widehat{Y},i}$ that exceed a
threshold $t $. This is similar to our prescription. However, our
choice of the threshold is different from the choice in \cite%
{bunea_she_wegcamp11}. They suggest either choosing $t$ by cross-validation or using $t =2(p+r)$. We use $t =2(p+r)$
to replicate their method in numerical experiments and call this algorithm BSW (``Bunea-She-Wegkamp"). Our choice of the threshold is based on the $10$-th quantile
of the Tracy-Widom distribution.

It is not guaranteed that any of these algorithms will estimate the rank of the model successfully. First of all, as we will see later, there is a certain threshold, so that if a true singular value $\theta$ is below this threshold then it cannot be estimated consistently. Second, since Theorem \ref{theorem_TW} is about the null case, hence it is not applicable in the situation with nonzero true singular values and it does not guarantee that the largest of the remaining singular value estimates is distributed according to the Tracy-Widom law.

The good news is that if the true singular values exceed the threshold, then Theorem 2 in \cite{onatski06} and Proposition 5.8 in \cite{BenaychGeorges_Guionnet_Maida11} suggest that our
Theorem  \ref{theorem_TW} can be extended to the non-null case by perturbative methods, and therefore the distribution of the largest of the remaining singular values of $\widehat{Y}$ or $\widehat{A}$ is indeed Tracy-Widom. (An analogous result also holds for deformations of Wigner matrices, see  Proposition 5.3 in  \cite{BenaychGeorges_Guionnet_Maida11} and Theorem 2.7 in
\cite{knowles_yin13}.)
 However, if one of the true singular values is at the threshold (precisely, if it is in a neigborhood of the threshold that shrinks as the matrix size grows), then the limiting distribution is a deformation of the Tracy-Widom law.
For the real-valued Wishart matrices, the full description of this deformation is still an active research problem (\cite{forrester13a}).

In order to compare the performance of the rank-selection algorithms we run
several numerical experiments. Their results are summarized below.

In the first experiment, we assumed that the null hypothesis is satisfied
and $A=0$. The number of observations in this experiment is $N=100$, and the
number of predictors is $p=25.$

\setlength{\tabcolsep}{9pt}
\begin{table}[htbp]
\begin{tabular}{c c |p{2cm}p{2.2cm}p{2.2cm}p{2.2cm}}
\hline
&  & BSW algorithm & Algorithm based on singular values of $Y$ & Algorithm based on singular values of $\widehat{Y}$ & Algorithm based on singular values of $\widehat{A}$
\\ \hline
$r=25$ &  $\overline{\widehat{s}}$ & 0.12 & 0.08 & 0.10 & 0.09 \\
$r=75$ &  $\overline{\widehat{s}}$ & 0.02 & 0.10 & 0.10 & 0.09 \\ \hline
\end{tabular}%
\captionsetup[table]{name=Figure}
\caption{Comparison of rank selection algorithms. \newline
Null hypothesis case: rank $s=0$. The entries in the table show the rank estimator $\widehat{s}$ averaged over $100$ repetitions of the experiment.}
\label{table_rank_selection_1}%
\end{table}

Figure \ref{table_rank_selection_1} displays the estimated rank $\widehat{s}$ averaged
over many simulations of the model. The results show that for the null case $A=0$ the algorithms based on the Tracy-Widom
distribution falsely detect about $10\%$ of model realizations as having
rank 1 (both in the case
$p=r=25$ and in the case $p=25<r=75$). This behavior should be expected since the threshold was set at the 10-percentile of
the Tracy-Widom distribution. In other words, the significance level of the tests is $10\%$.

 In contrast, the results of the BSW
(``Bunea-She-Wegkamp") algorithm change from about $12\%$ of false detections in
the case when $p=r=25$ to $2\%$ in the case when $p=25<r=75$. Hence, in our application of the BSW algorithm, the proportion of false positives varies depending on parameters. That is, one cannot easily pinpoint the significance level of the test based on the BSW algorithm.

\setlength{\tabcolsep}{9pt}
\begin{table}[htbp]
\begin{tabular}{cc|p{2cm}p{2.2cm}p{2.2cm}p{2.2cm}}
\hline
&  & BSW algorithm & Algorithm based on singular values of $Y$ & Algorithm based on singular values of $\widehat{Y}$ & Algorithm based on singular values of $\widehat{A}$
 \\ \hline
$r=25$ & $\overline{\widehat{s}}$ & 0.56 & 0.24 & 0.53 & 0.17 \\
$r=75$ & $\overline{\widehat{s}}$ & 0.78 & 0.56 & 0.92 & 0.19 \\
$r=200$ & $\overline{\widehat{s}}$ & 0.84 & 0.93 & 1.05 & 0.20 \\ \hline
\end{tabular}%
\captionsetup[table]{name=Figure}
\caption{Comparison of rank selection algorithms. \newline
Non-Null hypothesis case: $s=1$, $\protect\theta=0.025$}
\label{table_rank_selection_2}%
\end{table}

The results for non-null case with rank $s=1$ are shown in Figure \ref%
{table_rank_selection_2}. In this experiment, the strength of the signal $%
\theta$ was chosen to make it difficult but not impossible to detect the
signal. As before, $N=100$ and $p=25.$

The values in Figure \ref{table_rank_selection_2} are simulation-based estimates of the expected values of the rank estimators. For our choice of parameters, in most repetitions the rank estimator takes the values $0$ or $1$.
Hence, the numbers in this table are approximations for the power of the test, that is, for the proportion of the
times when the test detects the signal when it is in fact present. (Note, however, that this is only an
approximation since the rank estimator $\widehat{s}$ can take values greater than $1$. In particular, this explain the value $1.05$
in the table.)

For $r=25$, the values in Figure \ref{table_rank_selection_2} show that the estimated expected values of the rank estimators based on the BSW and the fitted values algorithms are $0.56$ and $0.53$, respectively. The estimated expected values for the other two rank estimators are much smaller. Since the true rank is $1$, this finding suggests that the power of the tests  based on the BSW and the fitted values algorithms is larger than for the tests based on the other two rank estimators.

As the number of responses, $r$, increases, it becomes easier to detect the
signal. The best performer for larger $r$ is the algorithm based on the
distribution of the largest singular value of the fitted responses. In
contrast, the algorithm based on the singular values of the coefficient matrix
estimator $\widehat{A}$ detects the signal poorly. The
performance of the BSW algorithm is also not very satisfactory as it appears
to be too conservative and biased in favor of the null hypothesis. For
example, for $r=200$, the BSW algorithm is outperformed by the simple
algorithm based on the singular values of responses, which does not use any
information about the design matrix $X$.

\setlength{\tabcolsep}{9pt}
\begin{table}[htbp]
\begin{tabular}{cc|p{2cm}p{2.2cm}p{2.2cm}p{2.2cm}}
\hline
&  & BSW algorithm & Algorithm based on singular values of $Y$ & Algorithm based on singular values of $\widehat{Y}$ & Algorithm based on singular values of $\widehat{A}$
\\ \hline
$r=25$ &  $\overline{\widehat{s}}$ & 6.25 & 5.00 & 6.21 & 4.51 \\
$r=75$ &  $\overline{\widehat{s}}$ & 8.26 & 7.41 & 8.51 & 6.07 \\
$r=200$ &  $\overline{\widehat{s}}$ & 9.06 & 9.03 & 9.69 & 6.97 \\ \hline
\end{tabular}%
\captionsetup[table]{name=Figure}
\caption{Comparison of rank selection algorithms. \newline
Non-Null hypothesis case: $s=10$, $\protect\theta_1=\ldots=\protect\theta%
_{10}=0.05$;  the entries in the table show the rank estimator $\widehat{s}$ averaged over $100$ repetitions of the experiment.}
\label{table_rank_selection_3}%
\end{table}

Finally, we consider the setup, in which $A$ has $10$ nonzero singular values, each with
value $\theta = 0.05.$ The results, summarized in Figure \ref%
{table_rank_selection_3}, are similar to results for one nonzero singular value in Figure \ref{table_rank_selection_2}. The best estimates are produced by the BSW algorithm and the algorithm based on fitted
responses $\widehat{Y}$. For small $r$, the signal detection is difficult and both algorithms
perform roughly similar. For large $r$, the algorithm based on fitted
responses outperform the BSW algorithm.

\subsection{Estimation of singular values}

\subsubsection{Limit distribution of the squared singular values}

Recall that the empirical eigenvalue distribution of a square Hermitian $r$-by-$r$ matrix $M$ is the measure $\mathbb{P}=\frac{1}{r}\sum_{i=1}^{r}\delta_{  \lambda_i} $ where $\lambda _i$ are eigenvalues of $M$ and $\delta_x$ is the Dirac measure concentrated on $x$, that is, for every continuous function $f$, $\int_{\mathbb{R}} f \delta_x = f(x).$

\begin{table}[htbp]
\begin{tabular}{cc}
\hline
\begin{minipage}[t]{0.45\textwidth} \centering Singular values of
$\widehat{Y}$ \\
\includegraphics[width=1%
\textwidth]{cumulative_distribution_N100_m66_r66_Yhat.jpg} \end{minipage} & %
\begin{minipage}[t]{0.45\textwidth} \centering Singular values of
$\widehat{A}$ \\
\includegraphics[width=%
\textwidth]{cumulative_distribution_N100_m66_r66_Ahat.jpg} \end{minipage} \\
\hline
\end{tabular}%
\captionsetup{name=Figure}
\caption{ The blue line shows the cumulative distribution function of
squared singular values in a random realization of the model. The red dashed
line is the limit distribution. The parameters are $N=100$, $p=r=66$.}
\label{table_limit_distr}
\end{table}

\begin{theorem}
\label{Theorem_main}Let $\widehat{A}\df\left( X^{\ast }X\right) ^{-1}X^{\ast
}U$ and $\widehat{Y}\df X\widehat{A}.$
\\
(i)Suppose that Assumption A1 is satisfied, $U$ is a matrix with
i.i.d. standard Gaussian entries, and $X$ is a full-rank matrix independent of $U.$ Then, as $
N\to \infty ,$ the empirical eigenvalue distribution of the $r$-by-$%
r $ matrix  $\frac{1}{r}\widehat{Y}%
^{\ast }\widehat{Y}$ converges weakly to the Marchenko-Pastur probability measure $\mathbb{P}_{MP}^{(\beta) }$ with the density defined in (\ref{density_MP}).
 \\
(ii) In addition, suppose that the entries of $X$ are i.i.d. standard Gaussian variables. Then the empirical eigenvalue distribution of $\widehat{A}^{\ast }\widehat{A}$ weakly converges to the probability measure $\mathbb{P}^{\left(\lambda ,\beta \right) }$ with the density defined in
(\ref{density}).
\end{theorem}

The Marchenko-Pastur measure $\mathbb{P}_{MP}^{(\beta) }$ is supported on the interval
\begin{equation*}
\left[
\left( \sqrt{\beta }-1\right) ^{2},\left( \sqrt{\beta }+1\right) ^{2}\right]
\end{equation*}
 and has the density
\begin{equation}
p_{MP}^{\beta}(x) =\frac{1}{2\pi }\frac{1}{x}\sqrt{4\beta -\left( \beta
+1-x\right) ^{2}}. \label{density_MP}
\end{equation}%
If $0<\beta <1,$ then the measure $\mathbb{P}_{MP}^{\left( \beta \right) }$ has an
additional atom at $0$ with mass $1-\beta .$

The family of probability measures $\mathbb{P}^{\left( \lambda ,\beta \right) }$ is
parameterized by $\lambda \geq 0$ and $\beta >0.$ The continuous part of $%
\mathbb{P}^{\left( \lambda ,\beta \right) }$ is supported on the interval $\left[
x_{1},x_{2}\right] .$ If $\lambda >0,$ then
\begin{equation}
x_{1,2}=\frac{1}{\lambda ^{2}\beta }\left[ \left( 1+\beta \right) \lambda
+2\pm 2\sqrt{\lambda ^{2}\beta +\lambda \left( \beta +1\right) +1}\right] ,
\label{support}
\end{equation}%
where $x_1$ and $x_2$ correspond to the $-$ and $+$ signs before the square root, respectively.
If $\lambda =0,$ then $x_{1}=\left( \beta -1\right) ^{2}/\left( 4\beta
\right) $ and $x_{2}=\infty .$ In both cases, the density is
\begin{equation}
p^{(\lambda ,\beta)}(x) =\frac{1}{2\pi }\frac{1}{x\left( x+1\right) }\sqrt{%
-\left( \beta -1\right) ^{2}+2\beta \left[ \left( 1+\beta \right) \lambda +2%
\right] x-\beta ^{2}\lambda ^{2}x^{2}}.  \label{density}
\end{equation}%
If $0<\beta <1,$ then the measure $\mathbb{P}^{\left( \lambda ,\beta \right) }$ has
an additional atom at $0$ with mass $1-\beta .$

\textbf{Proof:} Both parts immediately follow from Theorem \ref{theorem_fundamental}. The first part uses a property of the Wishart random matrices discovered by Marchenko and Pastur \cite{marchenko_pastur67}
(and independently rediscovered by Jonnson \cite{jonnson82} and Wachter \cite{wachter78}).

 The second part uses a property of the Jacobi ensemble shown by Wachter \cite{wachter80}
and Silverstein \cite{silverstein85}. In appendix, we give another proof of this property which uses the S-transform from free probability.

\textbf{Remarks:} 1. The theorem is illustrated by a numerical example in
Figure \ref{table_limit_distr}.

2. Here is what happens in some special cases:

a) If $\lambda \rightarrow \infty $ and $\beta $ is fixed, then both $x_1$ and $x_2$ converge to $0.$ This is in agreement with the intuitive notion that
the true matrix $A_{N}=0$ can be estimated precisely if the number of
observations is large relative to the number of variables in the model.

b) If $\lambda \to \infty $ and $\beta \to 0$ so that $%
\lambda \beta \to \xi >0,$ then both $x_1$ and $x_2$ converge to $\xi^{-1} .$ Note
that $\xi =\lim_{N\to \infty }\left( N/r\right) .$

c) If $\lambda $ is fixed and $\beta \to \infty $, then both
$x_1$ and $x_2$ converge to $\lambda ^{-1}.$

In both b) and c),  the regression will pick up spurious dependencies. In b), it is because the number of responses is very large, and in c), it is because the number of predictors is comparable to the number of observations.

d) If $p=r$, then $\widehat{A}$ is square and one can ask about the distribution of its eigenvalues.
By using the methods from \cite{guionnet_krishnapur_zeitouni11} and \cite{haagerup_larsen00},
 the limit distribution of eigenvalues of $\widehat{A}$ can be recovered from that of its singular values. It turns out that the limit distribution is supported
on the disc $|z|^2 \leq \frac{1}{\lambda}$ and has the density
			\begin{equation*}
			  \frac{1+\lambda}{\pi}\frac{dm(z)}{(1+|z|^2)^2},
			\end{equation*}
where $dm(z)$ is the Lebesgue measure on the complex plane.
After the stereographic projection this measure maps to the uniform measure on a Riemann sphere's cap.

This is a generalization of the result for the spherical ensemble of random matrices, which occurs when $N=p=r$, and therefore $\widehat{A}=X^{-1}U$. For the eigenvalues of this ensemble, it is known that the limit distribution is uniform on the Riemann sphere after the stereographic projection. (See \cite{krishnapur08}, \cite{bordenave11}, \cite{forrester_mays12}
and \cite{tikhomirov13} for more results about this ensemble).

3. In \cite{tikhomirov13}, Tikhomirov considers the spherical ensemble $X^{-1}U$ and finds the limit
distribution of its singular values under rather weak assumptions on the
distribution of matrix entries. The basis for the results in \cite{tikhomirov13}
is the Gaussian case which is extended to non-Gaussian matrices by
consecutively changing every matrix entry to a Gaussian random variable and
then verifying that the total change in the Stieltjes transform of the
eigenvalue distribution is negligible. We conjecture that the result in
Theorem \ref{Theorem_main} can be extended to non-Gaussian matrices in a
similar way.

\subsubsection{Consistent estimation of singular values}

Now, suppose that the true matrix $A$ is from the reduced-rank model (%
\ref{low_rank_model}). We are interested to know if the singular values $%
\theta _{i}$ can be estimated consistently.

The estimator $\widehat{A}=\left( X^{\ast }X\right) ^{-1}\left( X^{\ast
}Y\right) $ is an $m$-by-$r$ matrix that can be written as follows:
\begin{equation*}
\widehat{A}=A+\left( X^{\ast }X\right) ^{-1}\left( X^{\ast }U\right) .
\end{equation*}%
In other words, $\widehat{A}$ is the sum of the matrix $A$ and a random matrix. This random matrix has a rotationally invariant distribution by the assumption that both $X$ and $U$ are Gaussian  with i.i.d entries. In addition, matrix $A$ has a fixed set of non-zero singular values
$\theta_1\geq \ldots \geq \theta_s > 0$ and we assume that $N\gg s$. In this situation,  one can apply the results by
Benaych-Georges and Nadakuditi from \cite{BenaychGeorges_Nadakuditi12}.

It turns out that the singular values of $\widehat{A}$ do not converge to that of $A$ as $N\to \infty$. We need to define a correction function, $D_A(x),$ that will map the singular values of $\widehat{A}$ to consistent estimates of the singular values of $A$.

 Recall that the
Stieltjes transform of a probability measure $\mathbb{P}$ is an analytic
function defined \ as
\begin{equation*}
G\left( z\right) =\int \frac{\mathbb{P}\left( dt\right) }{z-t},
\end{equation*}%

Let $G_A(z)$ denote the  Stieltjes transform for the measure $\mathbb{P}^{(\lambda,\beta)}$
 (from Theorem \ref{Theorem_main}). It can be computed as
\begin{eqnarray}
G_A(z) &  =& \frac{1}{2z\left( 1+z\right) } \times \label{limit_Stieltjes} \\
       && \left[ 1-\beta +\left(
2+\lambda \beta \right) z-\sqrt{\left( \lambda \beta z\right) ^{2}-2\beta %
\left[ \lambda \left( 1+\beta \right) +2\right] z+\left( \beta -1\right) ^{2}%
}\right] .  \nonumber
\end{eqnarray}
(This follows from formula (\ref{Stieltjes_P_lambda_beta}) in Appendix after rescaling by $\beta^{-1}$.)

\begin{table}
\includegraphics[width=10cm]{D_function_plot.jpg}

\captionsetup{name=Figure}
\caption{ The function $D_A(x)-x$ for various values of parameters. The dashed
lines are for $\protect\lambda=1$ and the solid lines are for $%
\protect\lambda=0.1$. The plot lines start from $x=x_2(\lambda,\beta)$ defined in (\ref{support}).
}
\label{D_function}
\end{table}

Let
\begin{equation*}
\widetilde{G}_A(x) \df\left[ \beta ^{-1}G_A(x) +\left(
1-\beta ^{-1}\right) \frac{1}{x}\right] ,
\end{equation*}%
and
\begin{equation*}
D_A(x) \df\frac{1}{x\sqrt{G_A\left( x^{2}\right) \widetilde{G}_A\left(
x^{2}\right) }}.
\end{equation*}%
This function has no singularities for real $x>\sqrt{x_{2}},$ (with $x_{2}$ defined
in (\ref{support})), and it is increasing on $\left(\sqrt{x_{2}},\infty \right) .$
The behavior of the function $D_A(x)-x$ for various values of parameters $%
\lambda$ and $\beta$ is illustrated in Figure \ref{D_function}.

Let
\begin{equation*}
\overline{\theta}_A\df\lim_{x\downarrow \sqrt{x_{2}}}D_A(x).
\end{equation*}
Note that $\overline{\theta}_A\leq x_{2}$. (The threshold is below the upper edge of the support for the limit measure
 $\mathbb{P}^{(\lambda,\beta)}$.)

\begin{theorem}
\label{theorem_perturbation}
Let $\widehat{\sigma}_1 \geq \widehat{\sigma}_2 \geq \ldots \geq \widehat{\sigma}_s$  be the first $s$
largest singular values of $\widehat{A}=\left( X^{\ast }X\right) ^{-1}\left(
X^{\ast }Y\right) $, where $Y=XA+U$ and $A=\sum_{j=1}^{s}\theta _{j}u_{j}v_{j}^{\ast }$ with $\theta_1\geq \ldots \geq \theta_s > 0$. Suppose Assumptions A1 is satisfied with $\lambda
>0$ and $X$ and $U$ are independent matrices with
i.i.d. standard Gaussian entries.  For each fixed $i$, if $\theta _{i}>\overline{\theta}_A,$ then almost
surely as $N\to \infty ,$
\begin{equation*}
\widehat{\theta}_{i}\df D_A\left( \widehat{\sigma }_{i}\right) \to
\theta _{i}.
\end{equation*}%
If $\theta _{i}\leq \overline{\theta }_A$, then $\widehat{\sigma }%
_{i}\rightarrow \sqrt{x_{2}}.$
\end{theorem}

In other words, $\widehat{\theta }_{i}=D_A\left( \widehat{\sigma }_{i}\right) $
is a consistent estimator of $\theta _{i}$ provided that $\theta _{i}>%
\overline{\theta }_A,$ otherwise $\theta _{i}$ is hidden by spurious eigenvalues of $\widehat{A}$.

There is a similar result that uses $\widehat{Y}$ instead of $\widehat{A}$. Define
\begin{equation*}
D_Y(x) \df\frac{1}{x\sqrt{G_{MP}^{\beta}\left( x^{2}\right) \widetilde{G}_{MP}\left(
x^{2}\right) }},
\end{equation*}%
where $G_{MP}^{\beta}(x)$ is the Stieltjes transform of the Marchenko-Pastur distribution with parameter $\beta$:
\begin{equation}
G_{MP}^{\beta}(z)=\frac{1}{2z}\left[ 1-\beta +z-\sqrt{(1-\beta+z)^2-4z}\right] .
 \label{MP_Stieltjes}
\end{equation}
and $\widetilde{G}_{MP}^{\beta}\df\left[ \beta ^{-1}G_{MP}^{\beta}(x) +\left(1-\beta ^{-1}\right) \frac{1}{x}\right].$

		   Define $x_{u,Y}$ as the upper edge of the support of the Marchenko-Pastur distribution $\mathbb{P}_{MP}^{(\beta)}$ and
    $ \overline{\theta_Y}\df\lim_{x\downarrow \sqrt{x_{u,Y}}}D_Y(x).$

		\begin{theorem}
		\label{theorem_perturbation_Y}
		Let $\widehat{Y}=X\left( X^{\ast }X\right) ^{-1}\left(
X^{\ast }Y\right) ,$  where $Y=XA+U$ and $A=\sum_{j=1}^{s}\theta _{j}u_{j}v_{j}^{\ast }$ with $\theta_1\geq \ldots \geq \theta_s > 0$.  Suppose Assumption A1 is satisfied with $\lambda
>0$, and $X$ and $U$ are independent matrices with
i.i.d. standard Gaussian entries. Let $\widehat{\sigma }_{i,\widehat{Y}}$ be the first $s$ largest singular values of $\widehat{Y}/\sqrt{r}.$
			If $\theta _{i}>\overline{\theta_Y},$ then almost surely as $N\rightarrow \infty ,$
      \begin{equation*}
        \widehat{\theta }_{i,\widehat{Y}}\df \mu^{-1/2}D_Y\left( \widehat{\sigma }_{i,\widehat{Y}}\right) \to
        \theta _{i}.
      \end{equation*}%

      If $\theta _{i}\leq \overline{\theta_Y }$, then $\widehat{\sigma}_{i,\widehat{Y}}\rightarrow \sqrt{x_{u,Y}}.$
		\end{theorem}

\begin{table}[tbph]
\includegraphics[width=9cm]{comparison_of_A_and_Y_estimators.jpg}
\captionsetup{name=Figure}
\caption{ Quality of the estimators for the rank-one perturbation. The
horizontal axis shows the parameter $\protect\theta$. The red dashed line is
the difference $\widehat{\protect\theta}_A - \protect\theta$ and the blue
solid line is the difference $\widehat{\protect\theta}_Y-\protect\theta$.
The thresholds are $\overline{\protect\theta}_A=14.2$ and $\overline{\protect\theta}_Y=0.8$. (Parameters are $%
N=500$, $\protect\lambda=0.2$, $\protect\beta=1/2$.)}
 \label{fig_estimate}
\end{table}

\begin{table}[htbp]
  \begin{tabular}{cc}
    \hline
    \begin{minipage}[t]{0.5\textwidth}
		  \includegraphics[width=\textwidth]{comparison_thresholds.jpg}
		\end{minipage} & %
    \begin{minipage}[t]{0.5\textwidth}
      \includegraphics[width=\textwidth]{comparison_thresholds_surfc.jpg}
	  \end{minipage} \\
    \hline
  \end{tabular}%
	\captionsetup{name=Figure}
  \caption{ Contour and surface plots of the difference between the estimator thresholds
	$\overline{\theta}_A(\lambda,\beta)$ and $\overline{\theta}_Y(\lambda,\beta)$ .}
  \label{table_thresholds}
\end{table}

\textbf{Remarks}: 1. These results are similar in spirit to the results in \cite%
{baik_benarous_peche05}, \cite{baik_silverstein06}, and \cite{paul07}, which
are concerned with the singular values of sample covariance matrices. In
these papers, it was found that if the true covariance matrix has a 'spike'
that exceeds a certain threshold, then it will be observed in the spectrum
of the sample covariance matrix. Otherwise it will be hidden among the
spurious eigenvalues.

2. The quality of the estimators is illustrated in Figures \ref{fig_estimate} and \ref{table_thresholds}. They show that for relatively small values of parameters $\lambda$ and $\beta$, the threshold $\overline{\theta}_Y$ is smaller than $\overline{\theta}_A$ and the estimator  $\widehat{\theta }_{i,\widehat{Y}}$ is preferable to $\widehat{\theta }_{i,\widehat{A}}$. For large $\lambda$ and $\beta$ the threshold $\overline{\theta}_A$ can be smaller than $\overline{\theta}_Y$. However, the difference is small and in this region the correction term $D(x)-x$ is also small.

3. The proofs of Theorem \ref{theorem_perturbation} and Theorem \ref{theorem_perturbation_Y} are essentially by combining Theorem %
\ref{Theorem_main} in this paper and Theorem 2.8 in \cite%
{BenaychGeorges_Nadakuditi12}. We provide a detailed proof in Section \ref%
{subsection_largest_eigenvalue_nonnull} below for the convenience of the
reader. (The proof of Theorem 2.8 in \cite{BenaychGeorges_Nadakuditi12}
lacks some details and has annoying typos.)
However, we defer to \cite{BenaychGeorges_Nadakuditi12} for details about convergence and continuity issues.

\subsubsection{CLT for the estimator of the singular values}

 Let us define the empirical version of the function $G_A(x)$:
\begin{equation}
G_{A,N}\left( x\right) \df\frac{1}{p}\sum\limits_{i=1}^{p}\frac{1}{x-s_{i}^{2}},
\label{def_G_N}
\end{equation}%
where $s_{i}$ are singular values of the matrix $X\backslash U.$ Let
\begin{equation}
\widetilde{G}_{A,N}(x) \df\frac{r}{p}G_{A,N}(x) +\left( 1-\frac{r}{p}\right) \frac{1}{x}  \label{def_G_N_tilda}
\end{equation}%
and
\begin{equation}
D_{A,N}\left( x\right) \df\frac{1}{x\sqrt{G_{A,N}\left( x^{2}\right) \widetilde{G}%
_{A,N}\left( x^{2}\right) }}.  \label{def_D_N}
\end{equation}
Then Theorem \ref{theorem_perturbation} holds with $D_{A,N}\left( x\right) $
instead of $D_A\left( x\right) .$

Next, let $\widehat{\theta }^{\left( N\right) }\df D_{A,N}\left( \widehat{\sigma }%
^{\left( N\right) }\right) ,$ where $\widehat{\sigma }^{\left( N\right) }$
is the largest singular value of the matrix $\widehat{A}_{N}=\left(
X_{N}^{\ast }X_{N}\right) ^{-1}\left( X_{N}^{\ast }Y_{N}\right) $ and $D_{A,N}$
is as defined in (\ref{def_D_N}).

\begin{theorem}
\label{theorem_clt}
Assume  that $Y=XA+U$ with $A=\theta u v^{\ast }$. Suppose Assumption A1 is satisfied with $\lambda
>0$, and $X$ and $U$ are independent matrices with
i.i.d. standard Gaussian entries.

Then
\begin{equation*}
\frac{\sqrt{r}}{\omega}\left( \widehat{\theta }^{\left( N\right) }-\theta
\right),
\end{equation*}%
with $\omega$ as defined below in (\ref{omega}), converges in distribution to a standard zero-mean Gaussian random variable.
\end{theorem}

\begin{table}[tbph]
\includegraphics[width=10cm]{cdf_largest_s_non-null.jpg}
\captionsetup{name=Figure}
\caption{The blue solid line shows the cumulative distribution function for $%
\frac{\protect\sqrt{r}}{\protect\omega}(D(\widehat{\protect\sigma}^{(N)})-%
\protect\theta).$ The red dashed line is the standard Gaussian cumulative
distribution function. The parameters are $N=300$, $\protect\lambda=0.2$, $%
\protect\beta=2$, and $\protect\theta=9.$}
\label{fig:largest_non-null}
\end{table}

Let $\sigma $ be the largest solution of the
equation
\begin{equation*}
D_A(x) =\theta .
\end{equation*}%
Define
\begin{eqnarray*}
\kappa _{1}^{2} &=&-2\sigma ^{2}\left( G_A^{\prime }\left( \sigma ^{2}\right)
+ \left[ G_A\left( \sigma ^{2}\right) \right] ^{2}\right) , \\
\kappa _{2}^{2} &=&-\frac{2}{\beta }\sigma ^{2}\left( \widetilde{G}_A^{\prime
}\left( \sigma ^{2}\right) +\left[ \widetilde{G}_A\left( \sigma ^{2}\right) %
\right] ^{2}\right) , \\
\tau ^{2} &=&-\left[ \sigma ^{2}\widetilde{G}_A^{\prime }\left( \sigma
^{2}\right) +\widetilde{G}_A\left( \sigma ^{2}\right) \right].
\end{eqnarray*}%

Let
\begin{eqnarray}
\kappa ^{2} &=&\sigma ^{2}\left[ \widetilde{G}_A \left( \sigma ^{2}\right) \right]
^{2}\kappa _{1}^{2}+\sigma ^{2}\left[ G_A\left( \sigma ^{2}\right) %
\right] ^{2}\kappa _{2}^{2}+4\theta ^{-2}\tau ^{2},\text{ and} \notag \\
\omega &=&\frac{1}{2}\theta ^{3}\kappa . \label{omega}
\end{eqnarray}

Remark: We conjecture that the statement of the theorem holds true with the estimator $%
D\left( \widehat{\sigma }^{\left( N\right) }\right) $ instead of $%
D_{N}\left( \widehat{\sigma }^{\left( N\right) }\right) .$ However the proof
of this statement runs into some technical difficulties and might require
some additional assumption on the convergence $N/m\rightarrow 1+\lambda $
and $m/r\rightarrow \beta .$

\section{Proofs of the main results}

\label{subsection_largest_eigenvalue_nonnull}

\subsection{Proof of Theorem \ref{theorem_perturbation}}

The basic tool is the following determinantal identity. Let $|X|$ denote the determinant of a matrix $X$.

\begin{lemma}
\label{lemma_perturbation1}Let $\Psi$ and $D$ be an $N$-by-$N$ and $s$-by-$s$
matrices, respectively, and let $W_{1}$ and $W_{2}$ be an $N$-by-$s$ and $s$-by-$N$ matrices, respectively. Assume that $D$ is invertible. Then,

\begin{equation*}
 \left| tI_N-\Psi-W_{1}DW_{2}\right| =|D|\left| tI_N-\Psi\right|
  \left| D^{-1}-W_{2}\left( tI_N-\Psi\right) ^{-1}W_{1}\right| .
\end{equation*}
\end{lemma}

\textbf{Proof: }The proof is through a sequence of elementary determinantal
identities:
\begin{eqnarray*}
 \left| tI_{N}-\Psi-W_{1}DW_{2}\right| &=&\left| tI_{N}-\Psi\right|
\left| I_{N}-\left( tI_{N}-\Psi\right) ^{-1}W_{1}DW_{2}\right| \\
&=& \left| tI_{N}-\Psi\right| \left| I_{r}-DW_{2}\left(
tI_{N}-\Psi\right) ^{-1}W_{1}\right| \\
&=& \left| tI_{N}-\Psi\right| |D| \left| D^{-1}-W_{2}\left(
tI_{N}-\Psi\right) ^{-1}W_{1}\right| .
\end{eqnarray*}%
\hfill $\square $

We apply this lemma to understand how the singular values of a matrix $Z$ are affected if $Z$ is perturbed by a low-rank matrix $A$.

\begin{lemma}
\label{lemma_reduction_to_M}Let $\widehat{A}=A+Z$ with $A=\sum_{j=1}^{s}\theta _{j}u_{j}v_{j}^{\ast }.$ Let $\Theta =\mathrm{diag}\left( \theta _{1},\ldots ,\theta _{s}\right) ,$
and let $U$ and $V$ be an $p$-by-$s$ and an $r$-by-$s$ matrices whose
columns are vectors $u_{i}$ and $v_{i}$,
respectively. Let
\begin{eqnarray}
M\left( t\right) &\df &\left(
\begin{array}{cc}
tV^{\ast }\left( t^{2}I_{r}-Z^{\ast }Z\right) ^{-1}V, & V^{\ast }Z^{\ast
}\left( t^{2}I_{p}-ZZ^{\ast }\right) ^{-1}U \notag \\
U^{\ast }\left( t^{2}I_{p}-ZZ^{\ast }\right) ^{-1}ZV, & tU^{\ast }\left(
t^{2}I_{p}-ZZ^{\ast }\right) ^{-1}U%
\end{array}%
\right) \\
&-&\left(
\begin{array}{cc}
0 & \Theta ^{-1} \\
\Theta ^{-1} & 0%
\end{array}%
\right) .  \label{definition_Mt}
\end{eqnarray}
 If $t>0$ and $\left|
M\left( t\right) \right|=0,$ then $t$ is a singular value of $\widehat{A}.$
Conversely, if a singular value of $\widehat{A}$ is different from the
singular values of $Z$, then it is a zero of the function $\left| M\left( t\right)\right|$ defined in  (\ref{definition_Mt}) below.
\end{lemma}

\textbf{Proof:} We apply Lemma (\ref{lemma_perturbation1}) to $\Psi=\left(
\begin{array}{cc}
0 & Z \\
Z^{\ast } & 0%
\end{array}%
\right) ,$ $D=\left(
\begin{array}{cc}
0 & \Theta \\
\Theta & 0%
\end{array}%
\right) ,$ $W_{1}=\left(
\begin{array}{cc}
0 & U \\
V & 0%
\end{array}%
\right) ,$ and $W_{2}=W_{1}^{\ast },$ and we use the identity:
\begin{equation*}
\left(
\begin{array}{cc}
tI_{p} & -Z \\
-Z^{\ast } & tI_{r}%
\end{array}%
\right) ^{-1}=\left(
\begin{array}{cc}
t\left( t^{2}I_{p}-ZZ^{\ast }\right) ^{-1} & \left( t^{2}I_{p}-ZZ^{\ast
}\right) ^{-1}Z \\
Z^{\ast }\left( t^{2}I_{p}-ZZ^{\ast }\right) ^{-1} & t\left(
t^{2}I_{r}-Z^{\ast }Z\right) ^{-1}%
\end{array}%
\right) .
\end{equation*}
Then we get
\begin{eqnarray*}
 \left| tI_{p+r}-\left(
\begin{array}{cc}
0 & A+Z \\
A^{\ast }+Z^{\ast } & 0%
\end{array}%
\right) \right| &=&\left( -1\right) ^{s}\prod\limits_{i=1}^{s}\theta
_{i}^{2} \left| tI_{p+r}-\left(
\begin{array}{cc}
0 & Z \\
Z^{\ast } & 0%
\end{array}%
\right) \right|
\\
&&\times  \left|M\left( t\right)\right| .
\end{eqnarray*}%
Hence, the positive eigenvalues of the matrix $\left(
\begin{array}{cc}
0 & A+Z \\
A^{\ast }+Z^{\ast } & 0%
\end{array}%
\right)$ are either positive eigenvalues of the matrix $\left(
\begin{array}{cc}
0 & Z \\
Z^{\ast } & 0%
\end{array}%
\right) $ or positive zeros of $\left|M\left( t\right)\right| .$ Recall that the
singular values of matrix $A+Z$ coincide with the positive eigenvalues of
matrix
\begin{equation*}
\left(
\begin{array}{cc}
0 & A+Z \\
(A+Z)^{\ast } & 0%
\end{array}%
\right) ,
\end{equation*}
and similarly for the singular values of matrix $Z.$ This proves both claims
of the Lemma. \hfill $\square $

We will apply this theorem to $Z=X\backslash \mathcal{U}$, where $\mathcal{U}$ denotes the matrix of noise in the regression model \ref{regression_model} and  should not be confused with matrix $U$ in the definition of $M(t)$. The crucial observation is that the rotational invariance of $Z$ and Theorem \ref{Theorem_main} imply that the $s$-by-$s$ matrix $%
tV^{\ast }\left( t^{2}I_{r}-Z^{\ast }Z\right) ^{-1}V\ $ converges to the
scalar matrix $tG_A\left( t^{2}\right) I_{s}$ provided that $t>\sqrt{x_2}$.

By using Lemma \ref{lemma_shift}, one can also check that the matrix \\
 $tU^{\ast }\left( t^{2}I_{p}-ZZ^{\ast }\right) ^{-1}U$ converges to $\left[
\beta ^{-1}tG_A\left( t^{2}\right) +\left( 1-\beta ^{-1}\right) t^{-1}\right]
I_{s}.$

The off-diagonal blocks of $M\left( t\right) $ converge to zero.

Finally, we use the fact that
\begin{equation*}
\left|
\begin{array}{cc}
xI_{s} & -\Theta ^{-1} \\
-\Theta ^{-1} & yI_{s}%
\end{array}%
\right| =\prod\limits_{i=1}^{s}\left( xy-\theta _{i}^{-2}\right) .
\end{equation*}%
and conclude (by continuity properties of the determinantal equation
solutions proved in \cite{BenaychGeorges_Nadakuditi12}) that if $\theta_i>\overline{\theta}_A$ then there is a singular value of $\widehat{A}$ that converge to the positive solution of the
following equation:%
\begin{equation*}
tG_A\left( t^{2}\right) \left[ \beta ^{-1}tG_A\left( t^{2}\right) +\left(
1-\beta ^{-1}\right) t^{-1}\right] =\theta _{i}^{-2}.
\end{equation*}%
\hfill Since the left hand side equals $\left[ D_A\left( t\right) \right]
^{-2}$ by definition, this convergence establishes the first part of Theorem \ref%
{theorem_perturbation}.

In order to establish the second part, let
 $\theta_1  \geq \ldots \geq \theta_k > \overline{\theta_A} \geq \theta_{k+1} \geq \ldots \geq \theta_s$. Then, by monotonicity of $D_A(x)$, if $i>k$ then for every $\varepsilon>0$ and sufficiently large $N$ the equation $D_A(x)=\theta_i$ does not have roots in the interval $(\sqrt{x_2}+\varepsilon,\infty)$. Hence, by Lemma \ref{lemma_reduction_to_M}), the eigenvalues of matrix $\widehat{A}$ can have only $k$ limits greater than $\sqrt{x_2}+\varepsilon$. It can also be checked that the singular values of the matrix $\widehat{A}$ has the same limiting distribution, $\mathbb{P}^{(\lambda,\beta)}$, as in the null case. (Indeed, a deformation by a fixed-rank matrix does not affect the empirical distribution of eigenvalues). It follows that the $(k+1)$-st, $(k+2)$-nd, \ldots, and $s$-th largest singular values of $\widehat{A}$ must converge to  $\sqrt{x_2}$.
         \hfill $\square $

\subsection{Proof of Theorem \ref{theorem_perturbation_Y}}

An analogue of Lemma \ref{lemma_reduction_to_M} holds and establishes the fact that the singular values of $\widehat{Y}/\sqrt{r}$ are zeros of $\left|M_Y(t)\right|$, where
\begin{eqnarray*}
M_Y(t) &=& -\left(
\begin{array}{cc}
0 & \Theta ^{-1} \\
\Theta ^{-1} & 0%
\end{array}%
\right)\\
&+&\left(
\begin{array}{cc}
  tV^{\ast }\left( t^{2}I_{r}-Z^{\ast }Z\right) ^{-1}V, &
  r^{-\frac{1}{2}}V^{\ast }Z^{\ast}\left( t^{2}I_{N}-ZZ^{\ast }\right) ^{-1}XU \\
  r^{-\frac{1}{2}}U^{\ast }X^{\ast}\left( t^{2}I_{N}-ZZ^{\ast }\right) ^{-1}ZV, &
	r^{-1}tU^{\ast }X^{\ast}\left(t^{2}I_{N}-ZZ^{\ast }\right) ^{-1}XU
\end{array}%
\right),  \label{definition_M_Y_t}
\end{eqnarray*}
where $Z= r^{-\frac{1}{2}}X(X^{\ast}X)^{-1}X \mathcal{U}$, and $\mathcal{U}$ is the matrix of noise in the regression model \ref{regression_model}.
As before, the off-diagonal blocks of $M_Y(t)$ converge to $-\Theta ^{-1}$. Besides, if $t$ is sufficiently large, then the first diagonal block converges to $tG_{MP}^{\beta}(t^2)I_s.$

In addition, for sufficiently large $t$, the limit of the second diagonal block is the scalar matrix $a t I_s$, where $a$ is the limit of
\begin{equation}
\frac{N}{r}\frac{1}{p} \mathrm{Tr} \left( N^{-1}X^{\ast}\left(t^{2}I_{N}-ZZ^{\ast }\right) ^{-1}X \right) \label{matrix_trace}
\end{equation}

In order to calculate this limit, we note that the matrices
$\left(\mathcal{U}/\sqrt{r}, \mathcal{U}^{\ast}/\sqrt{r}\right)$ and
$\left(X/\sqrt{N}, X^{\ast}/\sqrt{N}\right)$ converge in distribution (in the sense of free probability theory)
 to pairs of non-commutative random variables
 $\left(u,u^{\ast}\right)$ and $\left(x,x^{\ast}\right)$, which are free from each other. Let $P_x=x (x^{\ast}x)^{-1} x^{\ast}$. This is the orthogonal projection corresponding to $xx^{\ast}$, since $P_x xx^{\ast}=xx^{\ast}P_x=xx^{\ast}$, $P_x^2=P_x$ and $P_x^{\ast}=P_x$. Then, the limit of the expectation of (\ref{matrix_trace}) is equal to the limit of
\begin{equation}
\frac{N}{r}(1+\lambda) \tau \left( \left(t^{2}I-P_x u u^{\ast} P_x\right) ^{-1}x x^{\ast}\right),
\label{R_XX}
\end{equation}
where $\tau$ is the trace in the corresponding free probability space.

In order to handle this expression, we use the following lemma.
\begin{lemma}
Suppose that the pair of variables $\left( a, P_a \right)$ and the variable $b$ are free from each other, and suppose that $P_a$ has the properties $P_a a= a P_a =a$ and $P_a^2=P_a$. Then
\begin{equation*}
\tau \left((tI-P_a b P_a)^{-1} a\right)=\frac{\tau(a)}{\tau(P_a)}
\left[-\frac{1-\tau(P_a)}{t}+\tau \left((tI-P_a b P_a)^{-1}\right)\right]
\end{equation*}
\end{lemma}

\textbf{Proof:} On both sides we have complex-analytic functions in $t$ (aside of singularities), and we can expand these functions in powers of $t^{-1}$. It is enough to  show the equality in the region of the complex plane where the resulting series converge, since for other values of $t$ the equality will hold by analytic continuation. Consequently, it is enough to show that the equality holds term by term in the series. In particular, we need to check that
\begin{equation*}
\tau \left((P_a b P_a)^n a\right)=\tau \left((P_a b P_a)^n\right) \frac{\tau(a)}{\tau(P_a)}
\end{equation*}
for each integer $n \geq 1$.
By using the properties of $P_a$ and $a$, the left hand-side can be written as
\begin{equation*}
\tau \left(\underbrace{P_a b \ldots P_a b}_{n\ \mathrm{times}}a\right)=\tau \left( b P_a b \ldots P_a ba\right).
\end{equation*}
Next we use the property that $\left( a, P_a \right)$ and $b$ are free, and write:
\begin{equation}
\tau \left( b P_a b \ldots P_a ba\right)=\sum_{\pi \in \mathcal{NC}(n)} \kappa_{\pi}(b,b,\ldots,b) \tau_{K(\pi)}(P_a,\ldots,P_a,a),
\label{Nica_Speicher}
\end{equation}
where the sum is extended over all non-crossing partitions of the sequence $\left\{1,2,\ldots,n\right\}$, $K(\pi)$ is the Kreweras complement of the partition $\pi$, $\tau_{K(\pi)}$ is the multiplicative extension of the trace $\tau$, associated with partition
 $K(\pi)$, and $\kappa_{\pi}$ denotes the free cumulant functional associated with partition $\pi$. (For additional details and a proof of formula (\ref{Nica_Speicher}), see Theorem 14.4 in \cite{nica_speicher06}.)

Let $|\pi|$ denote the number of blocks in the partition $\pi$. It is a fact that $|K(\pi)|=n+1-|\pi|$.
 Then, by multiplicativity of $\tau_{K(\pi)}$, the expression in (\ref{Nica_Speicher}) can be written as
\begin{equation*}
\sum_{\pi \in \mathcal{NC}(n)} \kappa_{\pi}(b,b,\ldots,b) \tau(P_a)^{n-|\pi|} \tau(a).
\end{equation*}
We can take $\tau(a)$ outside of the sum sign, and a similar calculation shows that
\begin{equation*}
\tau \left((P_a b P_a)^n\right)=\tau \left(P_a\right)\sum_{\pi \in \mathcal{NC}(n)} \kappa_{\pi}(b,b,\ldots,b) \tau(P_a)^{n-|\pi|}.
\end{equation*}
Hence,
\begin{equation*}
\tau \left((P_a b P_a)^n a\right)=\tau \left((P_a b P_a)^n\right)\tau \left(a\right) /\tau \left(P_a\right).
\end{equation*}
\hfill $\square$

By applying this lemma to the expression in (\ref{R_XX}), we calculate that its limit equals
\begin{equation*}
\mu\widetilde{G}_{MP}^{\beta}\left( t^2 \right),
\end{equation*}
where $ \mu=\lim_{N\to\infty}\frac{N}{r}$.

Hence, for sufficiently large $t$, $\left|M_Y(t)\right|$ converges to
\begin{equation*}
\prod_{i=1}^{s}\left(\mu t {G}_{MP}^{\beta}\left( t^2 \right) \widetilde{G}_{MP}^{\beta}\left( t^2 \right) - \theta_i^{-2}\right).
\end{equation*}
From this, we conclude that for $\theta_i>\overline{\theta}_Y$, there exists a singular value of $\widehat{Y}/\sqrt{r}$,
 denoted by $\widehat{\sigma}_i$, such that
\begin{equation*}
\mu^{-1/2} D_Y(\widehat{\sigma}_i) \to \theta_i
\end{equation*}
in probability. This is equivalent to the claim in the first part of the Theorem. The second part is proved as in Theorem \ref{theorem_perturbation}. \hfill $\square$

\subsection{Proof of Theorem \ref{theorem_clt}}

Now we are going to prove the
central limit theorem for the estimator $\widehat{\theta }=D\left( \widehat{%
\sigma }\right) .$ Recall that we are dealing with the rank-one perturbation
model, in which $\theta $ is the singular value of the perturbation and $%
\widehat{\sigma }$ is the largest singular value of the estimator
 $\widehat{A} =X\backslash Y.$ First, we note that the function $\frac{1}{r}\sum_{i=1}^{r}%
\frac{t}{t^{2}-s_{i}^{2}}$ concentrates better than what could be expected
if $s_{i}$ were independent.

\begin{lemma}
\label{speed_convergence}Let $s_{i},$ $i=1,\ldots ,r$ \ be the singular
values of the $m$-by-$r$ matrix $Z=X\backslash U $ and $x_{2}$ be as defined in (\ref{support}). Then for every $\varepsilon >0$ and
every $t>x_{2},$
\begin{equation*}
r^{1-\varepsilon }\left[ \frac{1}{r}\sum_{i=1}^{r}\frac{t}{t^{2}-s_{i}^{2}}-%
\mathbb{E}\frac{1}{r}\sum_{i=1}^{r}\frac{t}{t^{2}-s_{i}^{2}}\right]
\rightarrow 0
\end{equation*}%
in $L^{2}$.
\end{lemma}

\textbf{Proof:} The claim of this lemma is a consequence of a CLT for linear
eigenvalue statistics. Indeed, the eigenvalues of $Z^{\ast }Z$ are the
transformed eigenvalues from the Jacobi ensemble of random matrices (with a
smooth transformation (\ref{relation_of_eigenvalues})). Hence we can
use the results about the linear eigenvalue statistics of the Jacobi
ensemble. For example, the results of Dumitriu and
Paquette \cite{Dumitriu_Paquette12} (specifically their Theorem 3.1) imply
that $\mathrm{Tr}\left( t^{2}I-Z^{\ast }Z\right) ^{-1}$ converges to a
Gaussian random variable with a finite variance. Hence, $r^{-\varepsilon
}\left( \mathrm{Tr}\left( t^{2}I-Z^{\ast }Z\right) ^{-1}-\mathbb{E}\mathrm{Tr%
}\left( t^{2}I-Z^{\ast }Z\right) ^{-1}\right) \rightarrow 0$ in $L^{2}.$
\hfill $\square $

\textbf{Remark:} The results of Johansson in \cite{Johansson98} (e.g., his
Theorem 2.4) suggest that one can use $tG_A\left( t^{2}\right) $ instead of $%
\mathbb{E}\frac{1}{r}\sum_{i=1}^{r}\frac{t}{t^{2}-s_{i}^{2}}$ in the above
lemma. However, Johansson's results are proved under assumption of a fixed
potential $V,$ while the potential in the pdf of the Jacobi ensemble is changing
with $N.$ Hence, some additional assumptions on the convergence $%
m/N\rightarrow \lambda $ and $m/r\rightarrow \beta $ might be needed.

Next, we prove a CLT for the matrix $M\left( t\right) .$

\begin{lemma}
\label{lemma_M_fluctuations}Assume  that $Y=XA+U$ with $A=\theta u v^{\ast }$. Let $M_{N}\left( t\right) $
be the $2$-by-$2$ matrix defined in (\ref{definition_Mt}), and let $%
G_{N}\left( t\right) $ and $\widetilde{G}_{N}\left( t\right) $ be as defined
in (\ref{def_G_N}) and ($t$\ref{def_G_N_tilda}), respectively. If $t>x_{2}$
(with $x_{2}$ defined in (\ref{support})), then the random matrix
\begin{equation*}
\widehat{M}_{N}\left( t\right) =\sqrt{r}\left[ M_{N}\left( t\right) -\left(
\begin{array}{cc}
tG_{N}\left( t^{2}\right) & -\theta ^{-1} \\
-\theta ^{-1} & t\widetilde{G}_{N}\left( t^{2}\right)%
\end{array}%
\right) \right]
\end{equation*}%
converges in distribution to the matrix
\begin{equation*}
\left(
\begin{array}{cc}
\kappa _{1}X & \tau Z \\
\tau Z & \kappa _{2} Y%
\end{array}%
\right) ,
\end{equation*}%
where $X,Y,$ and $Z$ are independent standard Gaussian random variables, and
\begin{eqnarray*}
\kappa _{1}^{2} &=&-2t^{2}\left( G^{\prime }\left( t^{2}\right) +\left[
G\left( t^{2}\right) \right] ^{2}\right) , \\
\kappa _{2}^{2} &=&-2\beta ^{-1}t^{2}\left( \widetilde{G}^{\prime }\left(
t^{2}\right) +\left[ \widetilde{G}\left( t^{2}\right) \right] ^{2}\right) ,
\\
\tau ^{2} &=&-\left[ t^{2}\widetilde{G}^{\prime }\left( t^{2}\right) +%
\widetilde{G}\left( t^{2}\right) \right] .
\end{eqnarray*}
\end{lemma}

For the proof we rely on Theorem 7.1 in \cite{bai_yao08} that allows one to
compute the distributional limit for the forms $\sum_{i,j=1}^{n}u_{i}\left(
l\right) A_{ij}u_{j}\left( l^{\prime }\right) ,$ as $n\rightarrow \infty ,$
for independent, identically distributed $K$-tuples of real or complex
valued random variables $\left\{ u_{i}\left( 1\right) ,\ldots ,u_{i}\left(
K\right) \right\} ,$ under some additional assumptions on matrices $\left(
A_{ij}\right) $. The theorem shows that the limit
is Gaussian and provides a formula for the covariance matrix of the limit.
This theorem is not directly applicable in our case since we will have
variables $u_{i}$ which are the coordinates of the vector uniformly
distributed on a sphere $S^{n}$ and, therefore, not independent. This can be
overcome either by a suitable modification of Theorem 7.1 in \cite{bai_yao08}%
, or by a trick that represents the uniformly distributed vector $u$ as a
normalization of a Gaussian vector (which is similar to what is done in the
proof of Theorem 6.4 in \cite{BenaychGeorges_Guionnet_Maida11}). In the
following proof we concentrate on explaining how the variance coefficients
are calculated.

\textbf{Proof of Lemma \ref{lemma_M_fluctuations}:} By a suitable
modification of Theorem 7.1 in \cite{bai_yao08}, we find that the matrix $%
\widehat{M}_{N}\left( t\right) $ converges in distribution to a matrix that
consists of independent zero-mean Gaussian entries. It remains to compute
the variance of the entries. Let us start with
the upper-left diagonal entry of $M_{N}\left( t\right) ,$ which is $tv^{\ast
}\left( t^{2}I_{r}-Z^{\ast }Z\right) ^{-1}v.$ Because of the rotational invariance of $Z,$ one can
take $v$ uniformly distributed on the sphere $S^{r}$. In the basis that
diagonalizes $Z^{\ast }Z,$ we get $\,$%
\begin{equation*}
\left[ M_{N}\left( t\right) \right] _{11}=\sum_{i=1}^{r}\frac{t}{%
t^{2}-s_{i}^{2}}v_{i}^{2},
\end{equation*}%
where $s_{i}\,$are singular values of $Z$ and $v_{i}$ are coordinates of
vector $v.$ Hence,
\begin{eqnarray*}
\mathbb{E}\left( \left[ M_{N}\left( t\right) \right] _{11}|Z\right) &=&\frac{%
1}{r}\sum_{i=1}^{r}\frac{t}{t^{2}-s_{i}^{2}},\text{ and } \\
\mathbb{E}\left( \left[ M_{N}\left( t\right) \right] _{11}\right) &=&\mathbb{%
E}\frac{1}{r}\sum_{i=1}^{r}\frac{t}{t^{2}-s_{i}^{2}}.
\end{eqnarray*}%
Next we use the total variance formula:
\begin{equation*}
\mathbb{V}ar\left( \sqrt{r}\left[ M_{N}\left( t\right) \right] _{11}\right) =%
\mathbb{E}\left( \mathbb{V}ar\left( \left[ \sqrt{r}M_{N}\left( t\right) %
\right] _{11}|Z\right) \right) +\mathbb{V}ar\left( \mathbb{E}\left( \sqrt{r}%
\left[ M_{N}\left( t\right) \right] _{11}|Z\right) \right) .
\end{equation*}%
The second term converges to zero by Lemma \ref{speed_convergence}, hence we
need only to compute the limit of the first term. We calculate:
\begin{equation*}
\mathbb{E}\left( r\left[ M_{N}\left( t\right) \right] _{11}^{2}|Z\right) =%
\frac{3}{(r+2)}\sum_{i=1}^{r}\left( \frac{t}{t^{2}-s_{i}^{2}}\right) ^{2}+%
\frac{1}{(r+2)}\sum_{i\neq j}^{r}\left( \frac{t}{t^{2}-s_{i}^{2}}\right)
\left( \frac{t}{t^{2}-s_{j}^{2}}\right) ,
\end{equation*}%
where we used $E\left( v_{i}^{4}\right) =\frac{3}{r(r+2)}$ and $E\left(
u_{i}^{2}u_{j}^{2}\right) =\frac{1}{r(r+2)}.$ Hence,
\begin{eqnarray*}
\mathbb{V}ar\left( \sqrt{r}\left[ M_{N}\left( t\right) \right]
_{11}|Z\right) &=&\frac{2}{r}\sum_{i=1}^{r}\left( \frac{t}{t^{2}-s_{i}^{2}}%
\right) ^{2}-\frac{2}{r^{2}}\sum_{i\neq j}^{r}\left( \frac{t}{t^{2}-s_{i}^{2}%
}\right) \left( \frac{t}{t^{2}-s_{j}^{2}}\right) \\
&+&O\left( \frac{1}{r}\right)
\\
&\rightarrow &-2t^{2}\left( G^{\prime }\left( t^{2}\right) +\left[ G\left(
t^{2}\right) \right] ^{2}\right) .
\end{eqnarray*}%
In addition, Lemma \ref{speed_convergence} shows that
\begin{equation*}
\sqrt{r}\left\{ \left[ M_{N}\left( t\right) \right] _{11}-\frac{1}{r}%
\sum_{i=1}^{r}\frac{t}{t^{2}-s_{i}^{2}}\right\}
\end{equation*}%
converges in distribution to the same variable as
\begin{equation*}
\sqrt{r}\left\{ \left[ M_{N}\left( t\right) \right] _{11}-\mathbb{E}\frac{1}{%
r}\sum_{i=1}^{r}\frac{t}{t^{2}-s_{i}^{2}}\right\} ,
\end{equation*}%
that is, to a zero-mean Gaussian variable with variance $-2t^{2}\left(
G^{\prime }\left( t^{2}\right) +\left[ G\left( t^{2}\right) \right]
^{2}\right) .$

Similar argument holds for $\left[ M_{N}\left( t\right) \right] _{22}$:
\begin{equation*}
\sqrt{r}\left\{ \left[ M_{N}\left( t\right) \right] _{22}-t\widetilde{G}%
_{N}\left( t^{2}\right) \right\}
\end{equation*}%
converges to a zero-mean Gaussian variable with variance
\begin{equation*}
-2\beta ^{-1}t^{2}\left( \widetilde{G}^{\prime }\left( t^{2}\right) +\left[
\widetilde{G}\left( t^{2}\right) \right] ^{2}\right) .
\end{equation*}

Next, $\left[ M_{N}\left( t\right) +\theta ^{-1}\right] _{12}=u^{\ast
}\left( t^{2}I_{p}-ZZ^{\ast }\right) ^{-1}Zv,$ where $u$ and $v$ are unit
vectors, which are independent and uniformly distributed on $S^{p}$ and $S^{r},$
respectively. In appropriate coordinates,
\begin{equation*}
\left[ M_{N}\left( t\right) +\theta ^{-1}\right] _{12}=\sum%
\limits_{i=1}^{p}u_{i}\frac{s_{i}}{t^{2}-s_{i}^{2}}v_{i}
\end{equation*}%
The expectation of this term is zero and for the conditional variance we
have the following sum:%
\begin{eqnarray*}
\mathbb{V}ar\left( \sqrt{r}\left[ M_{N}\left( t\right) +\theta ^{-1}\right]
_{12}|Z\right) &=&\frac{1}{p}\sum_{i=1}^{p}\left( \frac{s_{i}}{%
t^{2}-s_{i}^{2}}\right) ^{2} \\
&\rightarrow &\int \frac{s^{2}}{\left( t^{2}-s^{2}\right) ^{2}}d\widetilde{%
\mu }\left( s^{2}\right) \\
&=&-\left[ t^{2}\widetilde{G}^{\prime }\left( t^{2}\right) +\widetilde{G}%
\left( t^{2}\right) \right] .
\end{eqnarray*}%
\hfill $\square $

\begin{corollary}
\label{lemma_fluctuations_determinant} If $t>x_{2}$,
then the random variable
\begin{equation*}
\sqrt{r}\left( \left| M_{N}\left( t\right)\right| - \left|
\begin{array}{cc}
tG_{N}\left( t^{2}\right) & -\theta ^{-1} \\
-\theta ^{-1} & t\widetilde{G}_{N}\left( t^{2}\right)%
\end{array
\right|
\right).
\end{equation*}%
converges in distribution to a Gaussian random variable with the variance
\begin{equation*}
\kappa ^{2}=t^{2}\left[ \widetilde{G}\left( t^{2}\right) \right] ^{2}\kappa
_{1}^{2}+t^{2}\left[ G\left( t^{2}\right) \right] ^{2}\kappa
_{2}^{2}+4\theta ^{-2}\tau ^{2},
\end{equation*}%
where $\kappa _{1},$ $\kappa _{2},$ and $\tau $ are as in Lemma \ref%
{lemma_M_fluctuations}.
\end{corollary}

Next, by Lemma \ref{lemma_reduction_to_M}, the largest singular value of $%
\widehat{A}$ satisfies the equation
\begin{equation*}
\left| M_{N}\left( \widehat{\sigma }_{1}\right)\right| =0.
\end{equation*}%
Let
\begin{equation*}
f_{N}\left( t\right) \df t^{2}\widetilde{%
G}_{N}\left( t^{2}\right) G_{N}\left( t^{2}\right) -\theta ^{-2},
\end{equation*}%
so that by Corollary \ref{lemma_fluctuations_determinant}
\begin{equation*}
\left| M_{N}\left( t\right)\right| =f_{N}\left( t\right) +\frac{\kappa }{\sqrt{r}}%
W+o\left( r^{-1/2}\right) ,
\end{equation*}%
where $W$ is a standard Gaussian random variable. If $\sigma _{1}$ denote
the largest root of $f_{N}\left( t\right) =0,$ then this implies that
\begin{eqnarray*}
\sqrt{r}\left( \widehat{\sigma }_{1}-\sigma _{1}\right) &\rightarrow &\left[
f_{N}^{\left( -1\right) }\right] ^{\prime }\left( 0\right) \kappa W= \\
&&\frac{1}{f_{N}^{\prime }\left( \sigma _{1}\right) }\kappa W,
\end{eqnarray*}%
Note that our estimator is $\widehat{\theta }=D_{N}\left( \widehat{\sigma }%
_{1}\right) ,$ where
\begin{equation*}
D_{N}\left( t\right) =\sqrt{\frac{1}{t^{2}\widetilde{G_{N}}\left(
t^{2}\right) G_{N}\left( t^{2}\right) }}=\left( f_{N}\left( t\right) +\theta
^{-2}\right) ^{-1/2}.
\end{equation*}%
Hence
\begin{eqnarray*}
\sqrt{r}\left( \widehat{\theta }-\theta \right) &\rightarrow &D_{N}^{\prime
}\left( \sigma _{1}\right) \frac{1}{f_{N}^{\prime }\left( \sigma _{1}\right)
}\kappa W \\
&=&-\frac{1}{2}\left( f_{N}\left( \sigma _{1}\right) +\theta ^{-2}\right)
^{-3/2}\kappa W \\
&=&-\frac{1}{2}\theta ^{3}\kappa W.
\end{eqnarray*}%
\hfill $\square $

\section{Conclusion}

\label{conclusion} This paper is about the reduced-rank regression in the
multivariate response linear model. We found that if the number of responses
and predictors is large relative to the number of observations, then the
singular values of the OLS estimator of the coefficient matrix do not
converge to zero even if the true coefficient matrix is zero. The same
observation is true for the matrix of fitted responses. Instead, the
empirical distributions of singular values are converging to some limit
distributions that depend on how numerous the predictors and responses are
relative to the number of observations.

In addition, under the null hypothesis $A=0$ we found that the scaled
largest singular value for these matrices are distributed in the limit
according to the Tracy-Widom distribution. This fact can be used to test
whether the true coefficient matrix is zero and to estimate the rank of the model. In numerical simulations, we found that
one of these rank-selection algorithms compares favorably with the algorithm
suggested by Bunea, She, and Wegkamp in \cite{bunea_she_wegcamp11}.

In the case of the low-rank $A\neq 0$, we showed that the singular values of
$A$ are detectable if and only if they exceed a certain threshold. If they
do, then the estimated coefficient matrix has singular values outside of the
support of the limit empirical distribution.

Finally, we showed that consistent estimators of the true singular values
can be obtained by shrinking the outlier singular values of $\widehat{A}$ or $\widehat{Y}$
appropriately.
We found that the estimation based on singular values of $\widehat{Y}$ is preferable to the estimation based on that
of $\widehat{A}.$  We have also proved a CLT for the
asymptotic distribution of one of the estimators.

redpajama_set_name: RedPajamaArXiv


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Nothing stops you from helping with the employer's tasks during this application process. For example, you can fill out forms intended to be completed by your employer and simply ask the employer to check them over and sign them. The less your U. S. employer is inconvenienced, the more it will be willing to act as sponsor for your visa.
L-1 visas are available only to employees of companies outside the U. S. that have related u. s. branches, subsidiaries, affiliates, or joint venture partners. There is also a special category for international accounting firms. For visa purposes, these terms have specific definitions.
Branches are simply different operating locations of the same company. The clearest example of this is a single international corporation that has branch offices in many countries.
In a subsidiary relationship, one company must own a controlling percentage of the other company, that is, 50% or more. For L-1 purposes, when two companies are in the same corporate or limited form, and at least 50% of the stock of a company in the U. S. is owned by a non-U. S. company, or vice versa, this is a classic subsidiary relationship.
branches or subsidiaries because there is no direct ownership between the two companies. Instead, they share the fact that both are controlled by a common third entity, either a company, group of companies, individual, or group of people.
There are two methods of ownership that will support an L-1 visa based on an affiliate relationship. The first is for one common person or business entity to own at least 50% of the non-u. s. company and 50% of the u. s. company. If no single entity owns at least 50% of both companies, the second possibility is for each owner of the non-u. s. company to also own the u. s. company, and in the same percentages. For example, if five different people each own 20% of the stock of the non-u. s. company, then the same five people must each own 20% of the u. s. company for an affiliate relationship to exist.
A joint venture exists when there is no common ownership between the two companies, but they have jointly undertaken a common business operation or project. To qualify for L-1 purposes, each company must have veto power over decisions, take an equal share of the profits, and bear the losses on an equal basis.
In a situation where both the u. s. and non-u. s. companies are in the corporate or limited form and the majority of the stock of both is publicly held, unless they are simply branches of the same company that wish to transfer employees between them, the joint venture relationship is the only one that is practical for L-1 qualifying purposes. The ownership of a publicly held company is too vast and diverse to prove any of the other types of qualifying business relationships.
L-1 visas are available to employees and partners of international accounting firms. In the case of big accounting firms, the interests between one country and another are not usually close enough to qualify as affiliates under normal L-1 visa rules. Nevertheless, the law considers the managers of such companies qualified to support L-1 visa petitions for their employees. The firm must be part of an international accounting organization with an internationally recognized name. ultimately, this option applies to a only limited number of very large and prominent firms.
If, prior to graduation, you accept practical training employment that is a required part of your studies, as is frequently done by students on fellowships, only 50% of your employment time will be deducted from the allotted 12-month total. Practical training in this special situation is called curricular practical training, and has the effect of allowing you to work for a total of 24 months prior to graduation instead of only 12. If you work for six months in curricular practical training, you are required to deduct only three months from your allotment, meaning you may still accept nine months of additional practical training employment prior to graduation and the full 12 months after graduation. If, however, you work more than 20 hours per week, you lose the benefit of this special rule. Under these circumstances, once again you must deduct the whole amount of time you worked, meaning you are limited to a maximum of 12 months of practical training prior to graduation.
As an M-1 student, you can apply for permission to take practical training only after you have completed your entire program of studies. Applications for work permission must be filed not more than 60 days before your graduation. m-1 students can be granted permission to work in a practical training position only for a period of one month for each four months of study, with a total overall maximum of six months.
• the expiration date of your status.


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Padmaavat movie protests: Here's what is happening across the countryhttps://indianexpress.com/article/india/padmaavat-movie-protests-heres-what-is-happening-across-the-country-5038438/
Padmaavat movie protests: Here's what is happening across the country
Here's a look at what is happening in states across the country where Padmaavat has stirred protests
By Express Web Desk |New Delhi | Published: January 25, 2018 5:16:39 pm
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Padmaavat released in theatres worldwide today.
Sanjay Leela Bhansali's movie Padmaavat, which has courted controversy and led to protests from the Rajput community, was released in theatres on Thursday. Earlier this week, the Supreme Court rejected petitions from the Rajasthan and Madhya Pradesh government which sought a ban on the film.
However, violent protests continued to take place in several states including Haryana, Rajasthan, Gujarat, Madhya Pradesh, Uttar Pradesh and Maharashtra. The movie, which was initially slated to release on December 1, could not make it to the theatres in Rajasthan, Gujarat, Madhya Pradesh and Goa on Thursday as the Multiplex Association of India (MAI) announced that its members would not screen Padmaavat.
On Thursday, the screening of the film in some halls, such as in Noida and Pune, remained relatively peaceful, a few other decided to cancel shows or stuck to their decision not to screen the movie. Protests by the Karni Sena and other fringe groups continued in several parts of the country today, including in Bihar, Uttar Pradesh and Rajasthan.
Follow Padmaavat movie protest LIVE updates
A day after protesters attacked a school bus ferrying students in Gurgaon, the Haryana police has arrested 18 people. Some schools in Gurgaon remained closed today as a precautionary measure. A video, which went viral on Wednesday, showed students and staff cowering down on the floor of the bus, the windows of which were shattered and cries of children heard in the background. The agitators had torched a Haryana Roadways bus before attacking the school bus.
Screengrab from the video showing school children huddled in a Gurgaon school bus which was attacked by protesters.
Haryana ministers, who had vocally protested the movie's release, remained silent on the issue, a day before its release. The only voice heard was of the Health Minister Anil Vij who said the Supreme Court's order was "supreme' and the government would provide security to cinemas screening the film.
Haryana Chief Minister Manohar Lal Khattar had previously assured security and said, "The Supreme Court's order must be followed… well and good if some theatre owners do not want to screen the film, but those who want to screen it will be provided full security".
On Sohna Road, where protesters set a bus on fire (Express Photo/Manoj Kumar)
Also Read | Padmaavat movie review and reactions: LIVE UPDATES
In Gujarat, 20,000 policemen and paramilitary personnel were deployed across the state to ensure law and order is maintained. The situation remained normal across the state, with educational institutions, offices and markets open in most parts.
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At least 44 arrests were made on Wednesday under charges of conspiracy, attempt to murder and attempt to commit culpable homicide, damaging private properties, among others, in four FIRs. Nearly 70 people have been named accused, along with a mob of over 2,000, in the FIRs, which say the key accused held a meeting in Sanand and accordingly came to Himalaya Mall "armed with woods, stones and bottles filled with inflammable" to stop the film's release. The mob went on a rampage, burnt over 60 motorcycles and damaged hundreds of them.
On Wednesday, mobs in Gujarat held demonstrations and torched vehicles. (Express Photo/Javed Raja)
Police Commissioner Singh said, "The police had permitted a candlelight march. Investigation reveals that a rogue group, originating mostly from Sanand, Sanathal and other areas of Ahmedabad rural, was responsible for the violence."
Earlier, on the day the Supreme Court refused to modify its order staying Gujarat and Rajasthan's notifications prohibiting the screening of Padmaavat, protests against the film turned violent with mobs in Ahmedabad targeting multiplexes and malls, blocking roads and torching parked vehicles. Police had to fire in the air to scatter the mobs.
At least three Ahmedabad malls — Acropolis which has the PVR multiplex, Himalaya which has Carnival Cinema and Ahmedabad One which has Cinepolis — were targeted by mobs which set ablaze at least 25-30 vehicles, mostly two-wheelers outside Himalaya Mall, and 10-12 vehicles outside Acropolis, Additional Chief Fire Officer Rajesh Bhatt said.
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Gujarat has been witnessing sporadic protests as demonstraters previously took to streets and blocked roads by burning tyres and buses in parts of Banaskantha, Mehsana, Surendranagar, Bhuj and Bhavnagar.
Keeping these incidents in mind, Gujarat police has decided to deploy additional forces at "sensitive areas" across the state.
In UP, police personnel have been asked to remain vigilant through the day on the first day of Padmaavat's screening, which remained peaceful in some halls in Noida. However, the others cancelled shows or stuck to their decision not to screen the movie.
Uttar Pradesh also witnessed demonstrations against the movie's release, including Lucknow, against the screening of the movie Padmaavat forcing the cancellation of special screenings on the day. In Meerut, protesters pelted stones on a multiplex and in Kanpur, a group reportedly announced a bounty on actor Deepika Padukone's nose.
Earlier on Sunday, bike-borne protesters smashed glass screens at the DND toll plaza in Noida, while another group gathered at Pari Chowk in Greater Noida and blocked roads, affecting traffic. There also were reports of protesters gathering outside Venice Mall in Greater Noida and tried to meet the management to voice their concerns over the movie.
A damaged toll booth on the DND, Sunday. (Express photo: Gajendra Yadav)
Read | Padmaavat protests: There's freedom to stone school bus with children
In Rajasthan, arterial roads were blocked, shops vandalised and bike rallies taken out on Thursday as protests against 'Padmaavat' continued in parts of state. The theatre owners refused to screen the film in view of the law and order situation. The Kota-Bundi highway was blocked and traffic was reportedly disrupted on Udaipur-Chittorgarh road and Dungarpur-Ahmedabad road.
Incidents of stone pelting were also reported with about two dozen shops being damaged in Udaipur. However, elaborate security arrangements were in place in the state capital Jaipur, where the five-day Jaipur Literature Festival began today.
The Shri Rajput Karni Sena (SRKS) on Wednesday exhorted people to enforce a 'janta curfew' by boycotting the film. The outfit's chief, founder patron Lokendra Singh Kalvi on being told that there are many in Rajasthan who may want to watch the movie, said, "Those who want to watch should leave the state. They are not worthy of living in Rajasthan. Anyone who can't save his sisters and daughters cannot have Rajasthani blood and is not a Rajasthani."
Sporadic violence was reported from various places in Rajasthan. In Chittor, the threat by Jauhar Kshatrani Manch to commit jauhar (self-immolation) at Jauhar Kund in Chittorgarh Fort fizzled out amid a heavy barricade by police at Padal Pole gate, where police screened tourists. "They are not allowing us to live with swabhimaan (pride), and now they are not allowing us to even die with swabhimaan," Jauhar Kshatrani Manch president Manjushree Shaktawat said.
Leading film distributors and some cinema hall and multiplex owners in Rajasthan have already announced that they will not distribute or screen the movie here.
The BJP government in the state had condemned and banned the movie. However, the Supreme Court cleared the decks for the nationwide release of Padmaavat by staying notifications of the governments of Rajasthan that prohibited screening of the film following protests from the Rajput community.
There were several protests across Maharashtra as well. In Mumbai, many special screenings were cancelled and attendance remained low. In Pune, 10 vehicles were torched by mobs near Vadgaon Bridge. The police arrested 15 people in connection with the violence, all of whom are from Rajasthan.
Maharashtra previously witnessed protests against the movie release with incidents leading to the Mumbai police detaining 96 members of Karni Sena and other Rajput fringe groups. These groups were protesting outside the office of the Central Board of Film Certification (CBFC) against clearance given by the board for the release of movie Padmavat.
Protesters near the CBFC office on Pedder Road. (Express photo: Prashant Nadkar)
Theatre owners in Madhya Pradesh on Thursday decided not to screen the Sanjay Leela Bhansali movie, despite the state government ensuring security. The state witnessed protests and bandh in several places against the period drama.
Earlier, three Muslim youths were arrested for an alleged offensive Facebook post and comments against queen Padmini. They were booked under Section 295 A (deliberate and malicious act, intended to outrage religious feelings of any class). The three were produced before a local magistrate who sent them to judicial custody on Wednesday.
In Bihar's Muzaffarpur protesters brandished swords, burnt tires in protest against the release of Padmaavat on Thursday. Earlier, a group of about 30 people, who identified themselves as Karni Sena members, vandalised a cinema hall in Muzaffarpur in Bihar, which was scheduled to screen Padmaavat after its release. "We are studying video footage of the incident to identify the suspects," said DSP Ashish Anand.
A cinema hall was vandalised in Muzaffarpur, Bihar. (PTI Photo)
Meanwhile in Punjab, Rajput leaders withdrew their opposition to the release after watching a special screening of Padmaavat on Wednesday. The Rajput Mahasabha, which was against the movie's release, found "nothing objectionable" against the community. "We will suggest that community members all over the country must watch the movie because it is in fact showing the Rajput community in a good light," one leader said. Read: In Punjab twister, Rajput leaders give Padmaavat a big thumbs-up
The Kolkata Police said on Thursday that it has made elaborate security arrangements across the city, especially around malls and multiplexes, to prevent instances of violence and that anyone found disrupting peace will face strict action. West Bengal is, therefore, expecting a smooth screening of the movie. Chief Minister Mamata Banerjee, who has been vocal in support for the release, had stated that the state would make special arrangements for the screening, if required.
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Padmaavat movie protests: Here's what is happening across the country
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redpajama_set_name: RedPajamaCommonCrawl


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NewsUKUK Politics
'What we could be': Alex Salmond launches blueprint for independent Scotland
Holyrood's First Minister insisted country would 'stand still' if it stayed part of UK
James Cusick
Tuesday 26 November 2013 14:24
Scottish First Minister Alex Salmond presents the White Paper for Scottish independence
It was billed as the Scottish government's long-awaited authoritative blueprint for independence. Brave Heart and tartan-wrapped aspirations would be minimised in favour of hard-edged economics and unchallengeable number-crunching. The undecided, still key to next year's referendum outcome, would be won over by "the greatest document of its kind ever published by a prospective nation."
But on a damp Glasgow morning, inside a spartan steel building resembling a cardboard box factory, Alex Salmond presented a wish-list and a tame marketing manifesto. The birth-of-a-nation moment or a new Bannockburn it wasn't.
A rainbow was faintly visible over Govan and the River Clyde when the First Minister said the 670-page document he held in his hand showed "what we could be" if next year's referendum was won.
But the weighty tome, on examination, was like a reverse of Dr Who's Tardis – big on the outside, but with little of substance inside.
Nevertheless, Mr Salmond warned that Scotland would "stand still" if it remained part of the United Kingdom, and forecast his new state would create "a revolution in employment and social policy" and deliver an economy with Scotland rather than Westminster at its centre.
Without deviance, hesitation, though with much repetition, he offered serial assurances that Scotland would become part of a new Sterling Zone with Westminster's full backing, and would enjoy guaranteed "continuing" membership of the EU.
The White Paper was headlined "Scotland's Future" and claimed to "comprehensively" answer a list of 650 key questions on independence. After an hour of direct interrogation by international and domestic journalists, the document's merits felt thinner and thinner.
Although unable to go beyond aspiration and assumption, Mr Salmond and his deputy, Nicola Sturgeon, both claimed an independent Scotland would be unquestioningly welcomed and given what it wanted by a range of institutions including Westminster, the Bank of England, the EU and Nato.
Regardless of the fact-limited White Paper, Mr Salmond remained confident that Scotland after the referendum next September would regain "its place as an equal member of the family of nations".
He urged Scotland's four million voters to seize a "once-in-a-generation" chance to create a fairer, more prosperous country by voting to leave the UK and taking control of their own destiny.
He promised to cut corporation tax - initially by three per cent, to honour current pensions with a triple-lock, extend free childcare and to increase the minimum wage.
The childcare changes would mean 30 hours per week in school term times for all three- and four-year-olds.
Although seeking a divorce, he sounded like someone still wanting live-in and visitation rights for the kids. Scotland and the remainder of the UK would, he said, still have close economic ties with each other, and both would still be using the same pound. Buckingham Palace, though in the centre of London, would still be home to Scotland's head of state.
A free Scotland would take 90 per cent of North Sea oil revenue; within the first term of a fully independent Holyrood, Trident nuclear weapons would vanish from Scottish territory; the assets and staff of the BBC would to a new state broadcaster, the Scottish Broadcasting Service.
The scale of Scotland's share of the current UK national debt was straightforward. The document simply stated this "will be negotiated and agreed" within 18 months before Scotland becomes officially independent in March 2016.
Without any obligation to adopt UK tax policies, the document nevertheless listed the Westminster pain it would dump.
The bedroom tax "rejected at the ballot box in Scotland" would go. The Royal Mail would be renationalised. The minimum wage would rise; air passenger duty would be reduced. And in advance of independence day, the UK government would be asked, nicely, to immediately cancel the rollout of universal credit in Scotland.
On defence, the new independent state would create a Scottish military force of 15,000 regulars, with a security and intelligence agency directed to work closely with Police Scotland.
Scotland's diplomatic representation overseas would see the creation of 70 to 90 embassies estimated to cost £120m to run.
In his preface to Scotland's Future, the First Minister said: "It will no longer be possible for governments to be elected and pursue policies against the wishes of the Scottish people. Independence will put the people of Scotland in charge of our destiny."
He added "If we vote no, Scotland stands still. A once-in-a-generation opportunity to follow a different path, and choose a new and better direction for our nation, is lost."
The response to the "promised land" document was predictable. The Scottish Secretary, Alistair Carmichael, said what had been offered fell short of its billing and said huge questions remained.
Mr Carmichael said it was "highly unlikely" the Scottish government's plan to keep the pound and retain the services of the Bank of England as part of a "currency union" with the rest of the UK would work.
The former chancellor, Alistair Darling, in charge of pro-union "Better Together" campaign, said the White Paper had "ducked" answering the big questions
He said the promise it contained rendered it a "non-starter", adding: "How can they guarantee the pound? And what's plan B if they can't get that?"
Attacking the White Paper's scattered assumptions and assertions, Mr Darling said: "We waited months for this. And it has failed to give credible answers on fundamentally important questions."
He branded it "a fantasy" and a "work of fiction", insisting that Scotland could not leave the UK and simultaneously keep all the benefits of UK membership.
Scotland's Future: the White Paper's conditional claims
An Independent Scotland would retain the pound
A free Scotland would decide which currency it used. The Fiscal Commission (in Scotland) looked at the continued use of Sterling, a new currency, and the Euro. Monetary union with the remainder of the UK, using the pound, was considered the best option. Salmond said this would be in the best interests of Scotland and the UK, therefore neither Westminster nor the Bank of England would object to a "Sterling Zone" with Scotland as a participatory member.
The Scotland Office threw cold water on the SNP's assumption, saying currency unions need close political and fiscal co-operation, and that "Scottish independence is about disintegration".
An independent Scotland will continue as member of the EU
This is described as a "natural position" with Scotland "not taken out of the EU against our wishes". The document however notes the current uncertainty over the UK's position in the EU and the coming referendum on EU membership.
What are the joining terms? Would the UK's opt-outs continue? The document doesn't say, other than that a "full and constructive role" as a member would follow after discussions with Westminster, EU states and the EU's institutions. It points to EU treaty law as the basis that Scotland wouldn't need to re-apply.
The apportionment of the UK national debt will be negotiated and agreed
The UK's debt level is currently £1.3 trillion. The document mentions Scotland's historic contribution to UK public finances, and the offsetting of its "share of the value of the UK's assets against our inherited debt". The document predicts this to be a "realistic calculation". It won't be. For lawyers north and south of the border, sorting this out will mean a bonaza of expensive cases.
Scotland will negotiate the transition from being a member of NATO as part of the UK, to becoming an independent member of the alliance
NATO will be notified about Scotland's membership. That is as detailed as it gets. As other members of NATO are non-nuclear states, no problem is envisaged with the parallel decision to fast-track the removal of the UK's nuclear weapons. Trident, says the document, will be gone "within the first term of the Scottish Parliament following independence". Faslane will be retained as a "conventional" naval base and joint HQ for a Scottish Defence Force.
A single GB-wide market for electricity and gas will continue
The document claims taxation revenue from oil and gas will support Scottish public services. It claims Scotland's energy assets will help the rest of the UK secure its supply and meet its renewables obligations.
How UK energy markets are working at the moment, and whether state interventon is needed, is a contentious issue. The White Paper however assumes the current UK-wide market continuing.
The White Paper's red herrings
The Royal Mail would be renationalised
What, the entire royal mail, or just the Scottish bit? Westminster took the decision to sell off a state-owned company - which is now in private hands. So this won't be "the return of the Royal Mail" as the document suggests. It will simply be the creation of a new Scottish state-owned mail carrier. Although it signals defiance, and that Scotland can work differently, it is just a conditional promise.
Retain the current TV licence fee.
But surely not send the cash to London and the BBC? No. This is again about keeping the "system" of a universal tax that pays for state broadcasting. The UK licence fee can't be retained in Scotland – unless it wants to take the unusual step of sending revenue south. "Retaining" sounds more comforting than creating something new. The document is full of such risk-free red herrings.
The Court of Session and the Appeal Court in Edinburgh would become Scotland's Supreme Court.
Law was never part of the 1707 union. Independent or not, Scotland can call its law courts what it wants. Just as it can call the Royal Mail the Caledonian Mail, it can choose Supreme Court or Great Court or the Premier Court. Even though there is currently no naming restriction, this sounds like a benefit of independence, instead of the political illusion it is.
The minimum wage will be increased in line with inflation.
What happens to the current UK "minimum wage" is immaterial because how it is legally enforced by Westminster would no longer apply. If an independent Scotland chooses to introduce a minimum wage, in whatever currency is used, it will not be governed by the wage level stipulated by the rest of the UK government. The White Paper could have stated that a free Scotland would adopt a "living wage". That would have suggested a Scottish government leading, not following.
Reduce air passenger duty by 50 per cent
The tax system of a free state is a matter for it alone. Does this mean that whatever air passenger duty is in the rest of the UK, Scotland will have half that rate? Probably not.
Alex SalmondFamily And ParentingFifeScotlandScottish National PartyTax
1/1'What we could be': Alex Salmond launches blueprint for independent
'What we could be': Alex Salmond launches blueprint for independent