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

The rapid development and wide application of computer techniques
permits to collect and store a huge amount data, where the number of
measured variables is usually large. Such high dimensional data
occur in many modern scientific fields, such as micro-array data in
biology, stock market analysis in finance and  wireless
communication networks. Traditional  estimation or test tools are no
more valid, or perform  badly for such high-dimensional data, since
they typically assume a large sample size $n$ with respect to the
number of variables $p$.  A better approach in this high-dimensional
data setting would be based on asymptotic theory which has both $n$
and $p$ approaching infinity. To illustrate this purpose, let us
mention the case of Hotelling's $T^2$-test. The failure of
$T^2$-test for high-dimensional data has been mentioned as early as
by \citet{D58}.  As a remedy, Dempster proposed a so-called
non-exact test. However, the theoretical justification of Dempster's
test arises  much later in \cite{B96} inspired by  modern random
matrix theory (RMT). These authors have found necessary correction
for the $T^2$-test to compensate effects due to high dimension.

In this paper,  we consider two LR tests  concerning covariance
matrices.  We first give a theoretical  explanation for the fail of
these tests in   high-dimensional data context.
Next,  with the aid of random matrix
theory, we provide  necessary corrections to these LR tests to cope
with the   high dimensional effects.

First, we consider the problem of one-sample covariance hypothesis
test. Suppose that $\mathbf x$ follows a $p$-dimensional Gaussian
distribution $N(\mu_p, \Sigma_p)$  and we want to test
\begin{equation}
  H_0 : ~\Sigma_p  = I_p ~,  \label{H0piao}
\end{equation}
where $I_p$ denotes the  $p$-dimensional identity matrix. Note that
testing  $ \Sigma_p= A $  with an arbitrary covariance matrix $A$
can always be reduced to the above null hypothesis by  the
transformation $A^{-\frac{1}{2}}\mathbf{x}$.

Let $(\textbf{x}_1,\cdots,\textbf{x}_n)$ be a
sample from $\mathbf x$, where  we assume  $ p < n $.  The  sample
covariance matrix is
\begin{equation}
  \textbf{S}=\frac{1}{n}\sum\limits_{i=1}^p(\textbf{x}_i-\overline{\textbf{x}})
  (\textbf{x}_i-\overline{\textbf{x}})^*,\label{S}
\end{equation}
and set
\begin{equation}L^*=tr \textbf{S} - \log|\textbf{S}|-p~.   \label{L*}
\end{equation}
The likelihood ratio test statistic  is
\begin{equation}
  T_n=n\cdot L^* .
  \label{singlVclassict}
\end{equation}
Keeping $p$ fixed while letting $n \rightarrow \infty$, then the
classical theory depicts that $T_n$ converges to the
$\chi^2_{\frac{1}{2}p(p+1)}$ distribution
under $H_0$.

However, as it will be shown, this classical  approximation leads to
a test size much higher than the nominal test level in the case of
high-dimensional data,  because $T_n$ approaches infinity for large
$p$.  As seen from Table 1 in \S\ref{sec:test1},  for dimension and
sample sizes $(p, n)= (50, 500)$, the realized size of the test  is
22.5\% instead of the nominal 5\%  level. The result is even worse
for the case $(p, n)= (300, 500)$, with   a 100\% test size.

Based on a recent  CLT for linear spectral statistics (LSS) of
large-dimensional sample covariance matrices \citep{B04}, we
construct a corrected version of $T_n$ in \S\ref{sec:test1}. As
shown by  the simulation results of  \S\ref{sec:simul1}, the
corrected test performs much better in  case of  high dimensions.
Moreover, it also performs correctly   for moderate dimensions like
$p=10$ or  20. For dimension and sample sizes $(p, n)$  cited above,
the sizes of the  corrected test are 5.9\% and 5.2\%, respectively,
both close to the  5\% nominal  level.

The second test problem we consider is about the  equality between
two high-dimensional covariance matrices.
Let ${\bf x}_i = (x_{1i}, x_{2i},\cdots , x_{pi})^{T},
i=1,\cdots,n_1$ and ${\bf y}_j = (y_{1j}, y_{2j}, \cdots,
y_{pj})^{T},$ $ j=1, \cdots, n_2$ be observations from two
$p$-dimensional normal populations $N(\mu_k,\Sigma_k),~ k=1,2$,
respectively. We wish to test the null  hypothesis
\begin{equation}
  H_0: \Sigma_1=\Sigma_2  ~.
  \label{a0}
\end{equation}
The related sample covariance matrices are
$$
\displaystyle{A=\frac{1}{n_1}\sum\limits_{i=1}^{n_1}({\bf
    x}_{i}-\overline{{\bf x}})({\bf x}_{i}-\overline{{\bf x}})^{*},}
\quad \displaystyle{B=\frac{1}{n_2}\sum\limits_{i=1}^{n_2}({\bf
    y}_{i}-\overline{{\bf y}})({\bf y}_{i}-\overline{{\bf y}})^{*},}
$$
where $\overline{{\bf x}}$ , $\overline{{\bf y}}$ are the respective
sample means. Let
\begin{equation}
  L_1=\frac{\left|A\right|^{\frac{n_1}{2}}\cdot
    \left|B\right|^{\frac{n_2}{2}}}
  {\left|c_1A+c_2B\right|^{\frac{N}{2}}},\label{L1AB}
\end{equation}
where  $N=n_1+n_2$  and  $c_k$ denote $\frac{n_k}{N}, k=1,2.$ The
likelihood ratio test statistic  is
$$
T_N=-2\log L_1,
$$
and  when $n_1, n_2 \rightarrow \infty$, we get
\begin{equation}
  T_N=-2\log L_1 \Rightarrow \chi^2_{\frac{1}{2}p(p+1)}\label{classic}
\end{equation}
under $ H_0$. Of cause, in this limit scheme, the data dimension $p$
is held fixed.

However, employing this  $\chi^2$ limit
distribution for  dimensions  like 30 or 40, increases dramatically
the  size of the test. For instance, simulations in
\S\ref{sec:simul2} show that, for dimension and sample sizes $(p,
n_1, n_2)= (40, 800,400)$, the test size  equals 21.2\% instead of
the nominal 5\%  level. The result is worse for the case of $(p,
n_1, n_2)= (80, 1600,800)$, leading to a 49.5\% test size. The
reason for this fail  of  classical LR  test is the following.
Modern RMT indicates that when both dimension and sample size are
large, the likelihood ratio statistic $T_N$ drifts to infinity
almost surely. Therefore, the classical $\chi^2$ approximation leads
to many false rejections of $H_0$ in case of high-dimensional data.

Based on recent CLT for linear spectral statistics of $F$-matrices
from RMT, we propose a correction to this LR  test  in
~\S\ref{sec:test2}. Although this corrected test is constructed
under the asymptotic scheme $n_1 \wedge n_2 \rightarrow +\infty$,
$y_{n_1}={p}/{n_1} \rightarrow y_1 \in (0, 1)$,
$y_{n_2}={p}/{n_2}\rightarrow y_2 \in (0, 1)$, simulations
demonstrate an overall correct behavior including  small or moderate
dimensions  $p$. For example, for the above cited dimension and
sample sizes $(p, n_1, n_2)$, the sizes of the  corrected test equal
5.6\% and 5.2\%, respectively,  both  close to the nominal 5\%
level.

Related works include \citet{LedoitWolf02}, \citet{Sriv05} and
\citet{Schott07}. These authors propose several procedures in the
high-dimensional setting for testing that i) a covariance matrix is
an identity matrix, proportional to an identity matrix (spherecity)
and is a diagonal matrix or ii) several covariance matrices are
equal. These procedures have the following common feature: their
construction involves some well-chosen distance function between the
null and the alternative hypotheses and rely on the first two
spectral  moments, namely the statistics  tr$S_k$ and  tr$S_k^2$
from sample covariance matrices $S_k$. Therefore, the procedures
proposed by these authors are different from the likelihood-based
procedures we consider here. Another  important difference  concerns
the Gaussian assumption on the random variables used in  all these
references. Actually, for testing the equality between  two
covariance matrices,
the correction proposed in
this paper  applies  equally for non-Gaussian and high-dimensional
data leading to a valid  pseudo-likelihood test.

The rest of the paper is organized as following. Preliminary and
useful RMT results are recalled in \S\ref{sec:useful}. In
\S\ref{sec:test1} and \S\ref{sec:test2}, we introduce our results
for the two tests above. Proofs and technical derivations are
postponed to the last section.

\section{Useful results from the random matrix theory }
\label{sec:useful}

We first recall several results from RMT, which will be useful for
our corrections to tests.
For any $p\times p$ square matrix $M$  with real eigenvalues
$\left(\lambda_i^M\right)$, $F_n^M$ denotes the  empirical spectral
distribution (ESD) of $M$, that is,
$$
F_n^M(x) = \frac{1}{p}\sum\limits_{i=1}^{p}\textbf{ 1}_{\lambda_i^M
  \leq x}, \quad \quad x \in \mathbb{R}.
$$
We will consider random matrix $M$ whose ESD $F_n^M$ converges
(in a sense to be precised ) to a limiting spectral distribution
(LSD) $F^M$. To make statistical inference about a parameter $\theta = \int f
(x)dF^M(x)$,   it is natural to use the estimator
$$\widehat{\theta}= \int f (x)dF_n^M(x)=
\frac{1}{p}\sum\limits_{i=1}^p f(\lambda_i^M),
$$
which is a  so-called linear spectral statistic (LSS) of the random
matrix $M$.

\subsection{CLT for LSS of a high-dimensional sample covariance matrix }

Let  $ \{\xi_{ki} \in \mathbb{C}, i, k = 1, 2, \cdots \}$ be a
double array of $i.i.d.$ complex variables with mean 0 and variance
1. Set $\xi_i = (\xi_{1i}, \xi_{2i},\cdots , \xi_{pi})^{T}$ ,  the
vectors $ (\xi_1, \cdots,\xi_{n}) $ is considered as  an $i.i.d$
sample from some $p$-dimensional distribution with mean $0_p$ and
covariance matrix $I_p$. Therefore the sample covariance matrix is
\begin{equation}
  S_n={1\over{n}}\sum\limits_{i=1}^{n}\xi_i\xi_i^*.\label{Sn}
\end{equation}

For $0<\theta\le 1$, let $a(\theta)=(1-\sqrt{\theta})^2$
and $b(\theta)=(1+\sqrt{\theta})^2$. The Mar\v{c}enko-Pastur
distribution of index $\theta$, denoted as $F^\theta$, is the
distribution on $[a(\theta), b(\theta)]$ with the following density
function
\[  g_\theta(x) =\frac{1}{2\pi \theta x} \sqrt{
  [b(\theta)-x][x-a(\theta)]},\quad  a(\theta)\le x\le b(\theta).
\]

Let $$y_n=\frac{p}n \rightarrow y \in (0, 1)$$
and $F^y, F^{y_n}$ be the Mar\v{c}enko-Pastur law of index
$y~ \mbox{and}  ~ y_n$, respectively.
Let  $\mathcal{U}$ be an open set of the complex plane,
including   $[ I_{(0, 1)}(y) a(y),  b(y)] $,
and $\mathcal{A}$   be the set of analytic functions $f :
\mathcal{U} \mapsto \mathbb{C}.$  We consider the empirical process
$G_n : = \{G_n(f)\}$ indexed by $\mathcal{A}$ ,
\begin{equation}
  G_n(f)= p\cdot \int_{-\infty}^{+\infty} f(x)\left[F_n-
    F^{y_n}\right] (dx), ~~~~~~\quad f \in  \mathcal{A},\label{Gdef}
\end{equation}
where $F_n$ is the ESD of $S_n$. The following theorem will play a
fundamental role in next derivations, which is a specialization of a
general theorem from \citet{B04}  (Theorem 1.1).

\begin{theorem}
  Assume that
  $ f_1, \cdots ,f_k \in \mathcal{A}$, and  $\{\xi_{ij}\}$  are  $i.i.d.$
  random variables, such that  $E\xi_{11}=0,~ E{|\xi_{11}|^2}=1,
  ~E{|\xi_{11}|}^4 < \infty. $ Moreover, $\frac{p}{n} \rightarrow y
  \in (0, 1) $
  as  $n, p \rightarrow \infty.$\\
  Then: \\
  \textit{(i)} Real Case. \quad  Assume  $\{\xi_{ij}\}$ are real and
  $E(\xi_{11}^4)=3.$  Then the random vector $\left(G_n(f_1),\cdots
  ,G_n(f_k) \right)$  weakly converges to a $k$-dimensional Gaussian
  vector with mean vector,
  \begin{equation}
    m(f_j)=\frac{f_j\left(a(y)\right)+f_j\left(b(y)\right)}{4} -
    \frac{1}{2\pi}\int_{a(y)}^{b(y)}\frac{f_j(x)}{\sqrt{4y-(x-1-y)^2}}dx,
    \quad j=1,\cdots , k,\label{04mean}
  \end{equation}
  and covariance function
  \begin{equation}
    \upsilon\left(f_j,
    f_\ell\right)=-\frac1{2\pi^2}\oint\oint\frac{f_j(z_1)f_\ell(z_2)}{(\underline{m}(z_1)-\underline{m}(z_2))^2}
    d\underline{m}(z_1)d\underline{m}(z_2),\quad j,\ell \in \{1, \cdots,
    k\}\quad \label{04var}
  \end{equation}\\
  where  ~$\underline{m}(z)\equiv m_{\underline{F}^y}(z)$ is the
  Stieltjes Transform of ~ $\underline{F}^y\equiv (1-y)I_{[0,
      \infty)}+yF^y$. The contours
    in (\ref{04var}) are non overlapping and both contain the support of $F^y$.\\

    \textit{(ii)} Complex Case. \quad Assume $\{\xi_{ij}\}$ are complex
    and $E\xi^2_{11}=0$ ,
    $E(|\xi_{11}|^4)=2$.  Then the conclusion of \textit{(i)} also holds, except the mean vector is zero
    and the covariance function is half of the function given in
    (\ref{04var}).\label{T1.1}
\end{theorem}

It is worth noticing that Theorem 1.1 in \citet{B04} covers more general
sample covariance matrices of  form  $S'_n=T_n^{1/2}S_nT_n^{1/2}$ where $(T_n)$ is a
given sequence of positive-definite Hermitian matrices. In the
``white'' case $T_n\equiv I$ as considered here, in a recent
preprint \citet{PL08}, the authors offer a new extension of the CLT
where the constraints   $E|\xi_{11}|^4=3$ or 2, as stated above,  are
removed.

\subsection{CLT for LSS of high-dimensional F matrix }

Let  $ \{\xi_{ki} \in \mathbb{C}, i, k = 1, 2, \cdots \}$ and
$\{\eta_{kj} \in \mathbb{C}, j, k = 1, 2, \cdots \}$ are two
independent double arrays of $i.i.d.$ complex variables with mean 0
and variance 1. Write $\xi_i = (\xi_{1i}, \xi_{2i},\cdots ,
\xi_{pi})^{T}$ and $\eta_j = (\eta_{1j}, \eta_{2j}, \cdots,
\eta_{pj})^{T}$. Also, for any positive integers $n_1, n_2$, the
vectors $ (\xi_1, \cdots,\xi_{n_1}) $ and $ (\eta_1, \cdots,
\eta_{n_2}) $ can be thought as independent samples of size $ n_1 $
and $ n_2 $, respectively, from some $p$-dimensional distributions.
Let $S_1$ and $S_2$ be the associated sample covariance matrices, ~
$i. e.$
$$
S_{1}={1\over{n_1}}\sum\limits_{i=1}^{n_1}\xi_i\xi_i^* \quad
\mbox{and}\quad
S_{2}={1\over{n_2}}\sum\limits_{j=1}^{n_2}\eta_j\eta_j^*$$ Then, the
following  so-called  F-matrix generalizes  the classical Fisher-statistics for
the present $p$-dimensional case,
\begin{equation}
  V_n=S_{1}S_{2}^{-1}\label{F}
\end{equation}
where $n_2 > p$. Here we use the notation $n=(n_1, n_2)$.

Let
\begin{equation}
  y_{n_1}=\frac{p}{n_1}
  \rightarrow y_1 \in (0, 1),
  ~ y_{n_2}=\frac{p}{n_2} \rightarrow y_2
  \in (0, 1).\label{limitscheme}
\end{equation}
Under suitable moment conditions,  the ESD $F_n^{V_n}$ of $V_n$
has a  LSD  $F_{y_1, y_2}$,
which has a density [See P72 of \cite{B06}], given by
\begin{equation}
  \displaystyle{\ell(x)}=\left\{
  \begin{array}{ll} & \displaystyle{\frac {(1-y_{2})\sqrt{(b-x)(x-a)}}{2\pi x(y_{1}+y_{2}x)},
      \quad ~a \leq x \leq b,}\\[6mm]
    &\displaystyle{0, \quad\quad \quad\quad \mbox{otherwise}.}
  \end{array}
  \right.\label{LSDden}
\end{equation}
where $
a=(1-y_{2})^{-2}\left(1-\sqrt{y_{1}+y_{2}-y_{1}y_{2}}\right)^2
~\mbox{and}~
b=(1-y_{2})^{-2}\left(1+\sqrt{y_{1}+y_{2}-y_{1}y_{2}}\right)^2.$

Similar to previously, let $\widetilde{\mathcal{U}}$ be an open set
of the complex plane,
including  the interval
$$\left[I_{(0,1)}(y_{1})
  \frac{(1-\sqrt{y_{1}})^2}{(1+\sqrt{y_{2}})^2},\quad
  \frac{(1+\sqrt{y_{1}})^2}{(1-\sqrt{y_{2}})^2}\right], $$ and
$\widetilde{\mathcal{A}}$  be the set of analytic functions $f :
\widetilde{\mathcal{U}} \mapsto \mathbb{C}.$  Define the empirical
process $\widetilde{G_n} : = \{\widetilde{G_n}(f)\}$ indexed by
$\widetilde{\mathcal{A}}$
\begin{equation}
  \widetilde{G_n}(f)= p\cdot \int_{-\infty}^{+\infty}
  f(x)\left[F_{n}^{V_n}- F_{y_{n_1}, y_{n_2}}\right] (dx), ~~~~~~f \in
  \widetilde{\mathcal{A}}. \label{Gdef2}
\end{equation}
Here $F_{y_{n_1}, y_{n_2}}$ is the limiting distribution in
(\ref{LSDden}) but with $y_{n_k}$ instead of $y_k, k=1,2.$

Recently, \citet{Z08} establishes  a general  CLT for LSS of
large-dimensional F matrix.  The following theorem is a simplified
one quoted from it, which will play an important role.

\begin{theorem}
  Let  $ f_1, \cdots ,f_k \in \widetilde{\mathcal{A}}$,
  and assume:\\
  For each p,  $(\xi_{ij_1})$  and $(\eta_{ij_2})$ variables are $ i.i.d. $,
  $ 1\leq i\leq p, ~ 1\leq
  j_1 \leq n_1, ~  1\leq j_2 \leq n_2. $ $ E\xi_{11}=E\eta_{11}=0, $
  $E|\xi_{11}|^4=E|\eta_{11}|^4<\infty,$~ $y_{n_1}=\frac{p}{n_1}
  \rightarrow y_1 \in (0, 1),\quad
  y_{n_2}=\frac{p}{n_2} \rightarrow y_2 \in (0, 1).$\\
  Then\\
  \textit{(i)} Real Case. \quad Assume $ (\xi_{ij}) $  and $
  (\eta_{ij}) $ are real, $ E|\xi_{11}|^2=E|\eta_{11}|^2=1 $, then the
  random vector $ \left(\widetilde{G_n}(f_1),
  \cdots,\widetilde{G_n}(f_k)\right) $ weakly converges to a
  k-dimensional Gaussian vector  with the mean vector
  \begin{eqnarray}
    m(f_j)&=&\lim\limits_{r \rightarrow
      1_{+}}\left[(\ref{E1})+(\ref{E1betax})+(\ref{E1betay})\right]\nonumber\\
    &&\quad\frac{1}{4\pi i}\oint_{|\zeta|=1}
    f_j(z(\zeta))\left[\frac{1}{\zeta-{1\over r}}+\frac{1}{\zeta+{1\over
          r}}-\frac{2}{\zeta+{y_2\over
          {hr}}}\right]d\zeta \label{E1}\\
    &&\quad + \frac{\beta\cdot y_1(1-y_2)^2}{2\pi i \cdot
      h^2}\oint_{|\zeta|=1}f_j(z(\zeta))\frac{1}{(\zeta+\frac{y_2}{hr})^3}d\zeta\label{E1betax}\\
    &&\quad + \frac{\beta\cdot y_2(1-y_2)}{2 \pi i \cdot
      h}\oint_{|\zeta|=1}f_j(z(\zeta))\frac{\zeta +
      \frac{1}{hr}}{(\zeta+\frac{y_2}{hr})^3}d\zeta ,\label{E1betay}
    \quad\quad j=1, \cdots, k,
  \end{eqnarray}
  where $
  z(\zeta)=(1-y_2)^{-2}\left[1+h^2+2h\mathcal{R}(\zeta)\right], \quad
  h =\sqrt{y_1+y_2-y_1y_2}, $ $ \beta=E|\xi_{11}|^4-3,  $
  and the covariance function   as $1<r_1<r_2 \downarrow 1$
  \begin{eqnarray}
    \lefteqn{\upsilon(f_j, f_\ell)= \lim\limits_{1 < r_1 < r_2
        \rightarrow
        1_{+}}\left[(\ref{cov1})+(\ref{cov1betax}))\right]}\nonumber\\
    &&  -\displaystyle\frac{1}{2\pi^2}\oint_{|\zeta_2|=1}
    \oint_{|\zeta_1|=1}\frac{f_j(z(r_1\zeta_1))f_\ell(z(r_2\zeta_2))r_1r_2}{(r_2\zeta_2-r_1\zeta_1)^2}
    d\zeta_1d\zeta_2,\label{cov1}\\
    && -\frac{\beta \cdot (y_1+y_2)(1-y_2)^2}{4\pi^2h^2}
    \oint_{|\zeta_1|=1}\frac{f_j\left( z(\zeta_1)
      \right)}{(\zeta_1+\frac{y_2}{hr_1})^2}d\zeta_1
    \oint_{|\zeta_2|=1}\frac{f_\ell\left( z(\zeta_2)
      \right)}{(\zeta_2+\frac{y_2}{hr_2})^2}d\zeta_2
    \label{cov1betax}\\
    && j, \ell \in \{1,\cdots , k\}.\nonumber
  \end{eqnarray}

  \textit{(ii)}  Complex Case. Assume $ (\xi_{ij}) $  and $
  (\eta_{ij}) $ are complex,  $ E(\xi_{11}^{2})=E(\eta_{11}^{2})=0, $
  then the conclusion of
  \textit{(i)}  also holds, except the means are $ \lim\limits_{r
    \rightarrow 1_{+}}\left[(\ref{E1betax})+(\ref{E1betay})\right] $ and
  the covariance function is  $ \lim\limits_{1 < r_1 < r_2 \rightarrow
    1_{+}}\left[\displaystyle\frac{1}{2}
    \cdot(\ref{cov1})+(\ref{cov1betax})\right], $ where $
  \beta=E|\xi_{11}|^4-2. ~
  $
  \label{T2.1}
\end{theorem}

We should point out that  Zheng's CLT
for $F$-matrices covers more general situations then those cited in
Theorem~\ref{T2.1}. In particular,  the fourth-moments
$E|\xi_{11}|^4$ and $E|\eta_{11}|^4$ can be different.

The following lemma will be used in \S\ref{sec:test2} for an
application of Theorem \ref{T2.1}  to obtain the formula
(\ref{testE})  and (\ref{testVar}).

\begin{lemma}
  For the function
  $f(x)=\log(a+bx),\quad x\in \mathbb{R}, \quad a, b>0 $, let
  $(c,d)$ be the unique solution to the equations $$\left\{
  \begin{array}{llll}
    c^2+d^2=a(1-y_2)^2+b(1+h^2),\\
    cd=bh,\\
    0 <d< c.\\
  \end{array} \right.$$
  Analogously,  let  $\gamma, \eta$ be the constants similar to $(c,
  d)$ but for the function $g(x)=\log(\alpha+\beta x), \quad \alpha
  > 0,\quad \beta > 0.$
  Then,  the mean and covariance functions in (\ref{E1}) and
  (\ref{cov1}) equal to
  \begin{eqnarray*}
    m(f)&=&\frac{1}{2}\log\frac{(c^2-d^2)h^2}{(ch-y_2d)^2},\\
    \upsilon(f, g)& =&
    2bhd^{-1}c^{-1}\log{\frac{c\gamma}{c\gamma-d\eta}}.
  \end{eqnarray*}
  \label{lem1}
\end{lemma}

\section{Testing the hypothesis that a high-dimensional covariance
  matrix is equal to a given  matrix } \label{sec:test1}

To test the hypothesis   $H_0 : ~\Sigma_p  =I_p$, let be   the
sample covariance matrix \textbf{S}  and likelihood ratio statistic
$T_n$  as  defined in (\ref{S}) and (\ref{singlVclassict}),
respectively.  For $\xi_i=\textbf{x}_i-\mu_p,$ the array
$\{\xi_i\}_{i=1, \cdots, n}$ contains  $p$-dimensional standard
normal variables under $H_0$. Let
$$
\textbf{S}_n={1\over{n}}\sum\limits_{i=1}^{n}\xi_i\xi_i^*.
$$
and  $$\widetilde{L^*}=\mbox{tr} \textbf{S}_n -
\log|\textbf{S}_n|-p.$$

\begin{theorem}
  Assuming that the conditions of Theorem \ref{T1.1} hold, $L^*$ is
  defined as (\ref{L*})
  and $g(x)= x-\log x-1$.  Then,  under $H_0$ and when $n\rightarrow\infty$
  \begin{equation}
    \widetilde{T_n}=\upsilon(g)^{-\frac{1}{2}}\left[ L^*-p \cdot
      F^{y_n}(g)- m(g)\right] \Rightarrow N \left( 0,
    1\right),
    \label{singlsta}
  \end{equation}
  where $F^{y_n}$ is the Mar\v{c}enko-Pastur law of  index  $ y_n$.
\end{theorem}
\begin{proof}
  Because the difference between   $\textbf{S}$
  and $ \textbf{S}_n$  is a rank-1 matrix,
  $\textbf{S}$ and $\textbf{S}_n$ have the same LSD. So, $L^*$
  and $\widetilde{L^*}$ have the same asymptotic
  distribution. We also have
  \begin{eqnarray*}
    \widetilde{L^*}&=& \mbox{tr}\textbf{S}_n - \log|\textbf{S}_n|-p\\
    &=& \sum\limits_{i=1}^{p} \left(\lambda_i^{\textbf{s}_n}-\log
    \lambda_i^{\textbf{s}_n}-1\right) = p \cdot \int (x- \log x-1)
    dF_n(x)\\
    &=&p \cdot \int g(x) d\left(F_n(x)-F^{y_n}(x)\right) +p \cdot
    F^{y_n}(g),
  \end{eqnarray*}
  so that
  \begin{equation}
    G_n(g)=\widetilde{L^*}-p \cdot F^{y_n}(g).\label{singlESD-pLSD}
  \end{equation}
  By Theorem \ref{T1.1}, $G_n(g)$  weakly converges  to a Gaussian
  vector  with the mean
  \begin{equation}
    m(g)=-\frac{\log{(1-y)}}{2}\label{singlmean}
  \end{equation}
  and variance
  \begin{equation}
    \upsilon(g)=-2\displaystyle\log{(1-y)}-2y.\label{singlvar}
  \end{equation}
  for the real case, which are calculated in  \S\ref{sec:proofs}. For
  the complex case, the mean $m(g)$ is zero and the variance is half
  of $ \upsilon(g). $ Then, by (\ref{singlESD-pLSD}) we arrive at
  \begin{equation}
    \widetilde{L^*}-p\cdot F^{y_n}(g)~\Rightarrow~ N\left(m(g),
    \upsilon(g)\right),\label{1connec}
  \end{equation}
  where
  \begin{equation}F^{y_n}(g)=1-\frac{y_n-1}{y_n}\log{(1-y_n)}\label{singllimit}
  \end{equation}
  can be calculated by the density of LSD of sample covariance matrix
  in \S\ref{sec:proofs}. Because $\widetilde{L^*}$ and $ L^*$ have the
  same asymptotic distribution and (\ref{1connec}), finally we get
  \begin{eqnarray}
    \widetilde{T_n}=\upsilon(g)^{-\frac{1}{2}}\left[L^*-p \cdot
      F^{y_n}(g)- m(g)\right] \Rightarrow N \left( 0, 1\right).\nonumber
  \end{eqnarray}
\end{proof}

\subsection{Simulation study I}
\label{sec:simul1} For different values of $(p, n)$, we compute the
realized sizes of traditional likelihood ratio test (LRT) and the
corrected likelihood ratio test (CLRT) proposed previously. The
nominal test level  is set to be $\alpha= 0.05$, and for each $(p,
n)$, we run  10,000 independent replications with real Gaussian
variables. Results are given in Table~1 and Figure~1 below.

\begin{table}[hp]
  \begin{center}
    \begin{tabular}{|l|ccc|cc|}
      \hline & \multicolumn{3}{c|}{CLRT }&  \multicolumn{2}{c|}{LRT } \\[1mm]
             {(p, $n$ ) } & { Size} &Difference with 5\% & {Power} & {Size}& { Power}
             \\[1mm]\hline
             (5, 500) & 0.0803&0.0303&0.6013 &0.0521&0.5233 \\[2mm]
             (10, 500) & 0.0690&0.0190&0.9517 &0.0555&0.9417  \\[2mm]
             (50, 500) & 0.0594&0.0094&1 &0.2252&1  \\[2mm]
             (100, 500) & 0.0537&0.0037&1 &0.9757&1  \\[2mm]
             (300, 500)  & 0.0515&0.0015&1  &1&1  \\[2mm]
             \hline
    \end{tabular}
  \end{center}
  \caption{Sizes and powers of the traditional LRT and the corrected
    LRT, based on 10,000 independent applications with real Gaussian
    variables. Powers are estimated under the alternative
    $\Sigma_p = \mathop{\mathrm{diag}}(1, 0.05, 0.05, 0.05,\ldots)$.}
\end{table}

\begin{figure}[hp]
  \begin{center}
    \includegraphics[width=9cm]{onesampleTypeIerror.eps}\\
    \caption{Realized sizes of the traditional LRT and the corrected
      LRT for different dimensions $p$ with real Gaussian variables.
      10 000 independent runs with 5\%  nominal level and sample size
      $n=500$.\label{1}}
  \end{center}
\end{figure}

As seen from  Table~1, the traditional LRT always rejects $H_0$ when
$p$ is large, like  $p=100$ or 300, while the sizes produced by the
corrected LRT perfectly matches  the nominal level. For moderate
dimensions like $p=50$,  the corrected LRT still performs correctly
while the traditional  LRT has a size much higher than 5\%.

\section{Testing the equality of two high-dimensional covariance matrices }
\label{sec:test2}

Let $ ({\bf x}_i), ~  i=1, \cdots, n_1 $ and $ ({\bf y}_j) , ~  j=1,
\cdots, n_2$ be observations from two normal populations
$N(\mu_k,\Sigma_k),~ k=1,2$, respectively. We examine the test
defined in (\ref{a0}) and (\ref{L1AB}).  The aim is to find a good
scaling of the LR statistic $T_N$, such that the scaled statistic
weakly converges to some limiting distribution. Let
$$
\xi_i= \Sigma^{-\frac{1}2} (\textbf{x}_i-\mu_1),\quad
\eta_i=\Sigma^{-\frac{1}2} (\textbf{y}_i-\mu_2)
$$
where $\Sigma=\Sigma_1=\Sigma_2$ denotes the common covariance
matrix under $H_0$. Note that in a strict sense, the vectors
$(\textbf{x}_i), (\textbf{y}_i)$ and the matrices $\Sigma, \Sigma_1,
\Sigma_2$ depend on $p$. However we do not  signify this dependence
in  notations for ease of statements. Due to Gaussian assumption,
the arrays $(\xi_i)_{i=1, \cdots, n_1}$  and  $(\eta_j)_{j=1,
  \cdots, n_2}$ contain i.i.d. $N(0, 1)$ variables, for which we can
apply Theorem \ref{T2.1}.

Let
\begin{eqnarray*}
  S_{1}&=&{1\over{n_1}}\sum\limits_{i=1}^{n_1}\xi_i\xi_i^*
  =\Sigma^{-\frac{1}2} C \Sigma^{-\frac{1}2} \\
  S_{2}&=&
  {1\over{n_2}}\sum\limits_{j=1}^{n_2}\eta_j\eta_j^*=\Sigma^{-\frac{1}2}
  D \Sigma^{-\frac{1}2},
\end{eqnarray*}
where
\begin{eqnarray}
  C&=&\frac{1}{n_1}\sum\limits_{i=1}^{n_1}({\bf
    x}_{i}-\mu_1)({\bf x}_{i}-\mu_1)^{*}, \nonumber\\
  D&=&\frac{1}{n_2}\sum\limits_{j=1}^{n_2}({\bf y}_{j}-\mu_2)({\bf
    y}_{j}-\mu_2)^{*}.\nonumber
\end{eqnarray}
Note that
\begin{equation}
  V_n=S_{1}S_{2}^{-1}\nonumber
\end{equation}
forms a random F-matrix and we have
\begin{equation}
  \widetilde{L_1}=\frac{\left|S_{1}\right|^{\frac{n_1}{2}}\cdot
    \left|S_{2}\right|^{\frac{n_2}{2}}}
            {\left|c_1S_{1}+c_2S_{2}\right|^{\frac{N}{2}}}=\frac{\left|C\right|^{\frac{n_1}{2}}\cdot
              \left|D\right|^{\frac{n_2}{2}}}
            {\left|c_1C+c_2D\right|^{\frac{N}{2}}}.\label{S1S2CD}
\end{equation}

\begin{theorem} \label{T4.1}
  Assuming that  the conditions of Theorem \ref{T2.1} hold under $H_0$, $L_1$ as
  defined in (\ref{L1AB}) and
  \[
  f(x)=\log(y_{n_1}+y_{n_2}x)-\frac{y_{n_2}}{y_{n_1}
    +y_{n_2}}\log x-\log(y_{n_1}+y_{n_2}).
  \]
  Then, under $H_0$ and as $n_1\wedge n_2\rightarrow \infty$,
  \begin{equation}
    \widetilde{T_N}=\upsilon(f)^{-\frac{1}{2}}\left[
      -\displaystyle\frac{2\log L_1}{N}-p \cdot F_{y_{n_1},y_{n_2}}(f)-
      m(f)\right] \Rightarrow N \left( 0, 1\right).\label{LST}
  \end{equation}
\end{theorem}
\begin{proof}
  As  $A-C$  and $ B-D$  are rank-1 random matrices, $AB^{-1}$  and
  $CD^{-1}$  have the same LSD. Also by (\ref{S1S2CD}),
  $\widetilde{L_1}$ and $ L_1$  have the same asymptotic distribution.
  Because
  \begin{eqnarray*}
    -\frac{2}{N}\log\widetilde{L_1}
    &=&-\frac{2}{N}
    \log\left(
    \frac{\left|S_{1}\right|^{\frac{n_1}{2}}\cdot
      \left|S_{2}\right|^{\frac{n_2}{2}}}
         {\left|c_1S_{1}+c_2S_{2}\right|^{\frac{N}{2}}}
         \right)
         \\
         &=&\log{|c_1V_n^{-1}+c_2|}- c_1\cdot \log|V_n^{-1}|\\
         &=&\sum\limits_{i=1}^p \log(c_1\lambda^{V_n}_i+c_2)-c_1\cdot
         \log(\lambda^{V_n}_i)\\
         &=& p \cdot \int\left[ \log(c_1x+c_2)-c_1\cdot \log(x)
           \right]dF_n^{V_n}(x).
  \end{eqnarray*}
  Define $f(x)=\log(c_1x+c_2)-c_1\cdot \log(x)$, by
  $c_1=\frac{n_1}{N}=\frac{y_{n_2}}{y_{n_1}+y_{n_2}}$ and
  $c_2=\frac{n_2}{N}=\frac{y_{n_1}}{y_{n_1}+y_{n_2}}$, also it can be
  written as
  \begin{equation} f(x)=\log(y_{n_1}+y_{n_2}x)-\frac{y_{n_2}}{y_{n_1}
      +y_{n_2}}\log x-\log(y_{n_1}+y_{n_2}).\label{f(x)}
  \end{equation}
  From
  \begin{eqnarray}
    -\frac{2\log \widetilde{L_1}}{N}&=&p \cdot \int f(x)dF_n^{V_n}(x)\nonumber\\
    &=& p \cdot \int f(x) d\left(F_n^{V_n}(x)-F_{y_{n_1},
      y_{n_2}}(x)\right) +p \cdot F_{y_{n_1}, y_{n_2}}(f),\nonumber
  \end{eqnarray}
  we get
  \begin{equation}
    \widetilde{G_n}(f)=-\frac{2\log \widetilde{L_1}}{N}
    -p \cdot F_{y_{n_1}, y_{n_2}}(f).\label{ESD-pLSD}
  \end{equation}

  By Theorem \ref{T2.1}, $ \widetilde{G_n}(f)$  weakly converges to a
  Gaussian vector with mean
  \begin{equation}
    m(f)=
    \frac{1}{2}\left[\log\left(\frac{y_1+y_2-y_1y_2}{y_1+y_2}\right)
      -\frac{y_1}{y_1+y_2}\log(1-y_2)-\frac{y_2}{y_1+y_2}\log(1-y_1)\right]\label{testE}
  \end{equation}
  and variance
  \begin{equation}
    \upsilon(f)=-\frac{2y_2^2}{(y_1+y_2)^2}\log(1-y_1)-\frac{2y_1^2}{(y_1+y_2)^2}\log(1-y_2)
    -2\log\frac{y_1+y_2}{y_1+y_2-y_1y_2}\label{testVar}
  \end{equation}
  for the real case, which are  calculated by Lemma \ref{lem1} in
  \S\ref{sec:proofs}. For the complex case, the mean $ m(f) $ is zero
  and the variance is half of $ \upsilon(f) $. In other words,
  \begin{eqnarray}
    -\frac{2\log \widetilde{L_1}}{N}-p\cdot F_{y_{n_1}, y_{n_2}}(f)
    &\Rightarrow & N\left(m(f), \upsilon(f)\right), \label{2connec}
  \end{eqnarray}
  where
  \begin{eqnarray*}
    \displaystyle{F_{y_{n_1}, y_{n_2}}(f)} &=&
    \displaystyle{\frac{-(y_{n_1}+y_{n_2}-y_{n_1}y_{n_2})}{y_{n_1}y_{n_2}}\log{(y_{n_1}+y_{n_2}-y_{n_1}y_{n_2})}}\\[5mm]
    \quad
    &+&\displaystyle{\frac{(y_{n_1}+y_{n_2}-y_{n_1}y_{n_2})}{y_{n_1}y_{n_2}}\log{(y_{n_1}+y_{n_2})}}
    +\displaystyle{\frac{y_{n_1}(1-y_{n_2})}{y_{n_2}(y_{n_1}+y_{n_2})}\log{(1-y_{n_2})}}\nonumber\\
    \quad\quad\quad&+&\displaystyle{\frac{y_{n_2}(1-y_{n_1})}{y_{n_1}(y_{n_1}+y_{n_2})}\log{(1-y_{n_1})}},\nonumber
    \label{limit}
  \end{eqnarray*}
  is derived by use of the density of $F_{y_{n_1}, y_{n_2}}$ in
  \S\ref{sec:proofs}. Because $\widetilde{L_1}$and $ L_1$ have the
  same asymptotic distribution and by (\ref{2connec}),  we get by letting
  $n_1\wedge n_2\rightarrow\infty$,
  $$
  \widetilde{T_N}=\upsilon(f)^{-\frac{1}{2}}\left[
    -\displaystyle\frac{2\log L_1}{N}-p \cdot  F_{y_{n_1}, y_{n_2}}(f)-
    m(f)\right] \Rightarrow N \left( 0, 1\right).
  $$
\end{proof}

\subsection{Simulation study II}
\label{sec:simul2}

For different values of $(p, n_1,n_2)$, we compute the realized
sizes of the traditional LRT and the corrected LRT   with 10,000
independent replications. The nominal  test level is $\alpha=  0.05$
and we use real Gaussian variables. Results are summarized in Table
2  and Figure 2.

\begin{table}
  \begin{center}
    \begin{tabular}{|l|ccc|cc|}
      \hline \multicolumn{6}{|c|}{(y1, y2)=(0.05, 0.05)~~}\\
      \hline & \multicolumn{3}{c|}{CLRT  }&  \multicolumn{2}{c|}{LRT } \\[1mm]
             {(p, $n_1$,  $n_2$ ) } & { Size} &Difference with 5\% &
             {Power} & {Size}& { Power}
             \\[1mm]\hline
             (5, 100, 100)  & 0.0770&0.0270&1  &0.0582&1  \\[2mm]
             (10, 200, 200) & 0.0680&0.0180&1 &0.0684&1  \\[2mm]
             (20, 400, 400) & 0.0593&0.0093&1 &0.0872&1  \\[2mm]
             (40, 800, 800) & 0.0526&0.0026&1 &0.1339&1  \\[2mm]
             (80, 1600, 1600)   &0.0501 &0.0001&1&0.2687&1  \\[2mm]
             (160, 3200, 3200)   &0.0491 &-0.0009&1&0.6488&1  \\[2mm]
             (320, 6400, 6400)   &0.0447 &-0.0053&0.9671&1&1  \\[2mm]
             \hline
    \end{tabular}\\[2mm]
    \begin{tabular}{|l|ccc|cc|}
      \hline \multicolumn{6}{|c|}{(y1, y2)=(0.05, 0.1)~~}\\
      \hline & \multicolumn{3}{c|}{CLRT  }&  \multicolumn{2}{c|}{LRT } \\[1mm]
             {(p, $n_1$,  $n_2$ ) } & { Size} &Difference with 5\% &
             {Power} & {Size}& { Power}
             \\[1mm]\hline
             (5, 100, 50)   & 0.0781&0.0281&0.9925 &0.0640&0.9849  \\[2mm]
             (10, 200, 100) & 0.0617&0.0117&0.9847 &0.0752&0.9904  \\[2mm]
             (20, 400, 200) & 0.0573&0.0073&0.9775 &0.1104&0.9938  \\[2mm]
             (40, 800, 400) & 0.0561&0.0061&0.9765 &0.2115&0.9975  \\[2mm]
             (80, 1600, 800)&0.0521 &0.0021&0.9702 &0.4954&0.9998 \\[2mm]
             (160, 3200, 1600)   &0.0520&0.0020&0.9702&0.9433&1  \\[2mm]
             (320, 6400, 3200)   &0.0510 &0.0010&1&0.9939&1  \\[2mm]
             \hline
    \end{tabular}
  \end{center}
  \caption{Sizes and powers of the traditional LRT  and the corrected
    LRT  based on 10,000 independent replications using real
    Gaussian variables.  Powers are estimated under the
    alternative  $\Sigma_1\Sigma_2^{-1}= \mbox{ diag} (3,1,1,1,\cdots)$.
    Upper:  $y_1=y_2=0.05$.~~
    Bottom:   $y_1=0.05,~ y_2=0.1$.\label{tab:23}}
\end{table}

\begin{figure}[hb]
  \begin{center}
    \includegraphics[width=7cm]{0.050.05TypeIError.eps}\quad
    \includegraphics[width=7cm]{0.050.1TypeIError.eps}
    \caption{Sizes of  the traditional LRT  and the corrected
      LRT  based on 10,000 independent replications using real
      Gaussian variables.
      Left:  $y_1=y_2=0.05$.~~
      Right:   $y_1=0.05,~ y_2=0.1$.\label{fig:2}}
  \end{center}
\end{figure}

As we can see, when the dimension $p$ increases,   the traditional
LRT leads to a dramatically high test size while the corrected LRT
remains accurate. Furthermore,  for  moderate dimensions like $p=20$
or 40, the sizes of  the traditional LRT are much higher than 5\%,
whereas the ones of corrected LRT  are  very close. By a closer look
at the  column showing the difference with  5\%, we note that  this
difference rapidly decrease as $p$ increases for the corrected test.
Figure~2 gives  a vivid sight of these  comparisons between the
traditional LRT and the corrected LRT  in term of test  sizes.

\subsection{A pseudo-likelihood test for high-dimensional non-Gaussian data}

As said in Introduction, previous related works as
\citet{LedoitWolf02}, \citet{Sriv05}  or \citet{Schott07} all assume
Gaussian variables.  In contrast, Theorem~\ref{T4.1} applies for
general distributions having a fourth moment. For these non Gaussian
data, we consider the corrected LRT as generalized
pseudo-likelihood ratio test (or Gaussian  LRT).

Moreover, the methods proposed by these authors all rely  on an appropriate
normalization of
the trace of squared difference between two sample covariances
following the
idea of \citet{B96}.
We believe that their method
would strongly depend on the normality assumption (which was supported
by simulation results below).
On the other hand, based on
general understanding, the LRT contains much higher information from
data and its poor performance observed up to now  is just caused by its large bias
when dimension is large. Thus,
from the intuitive understanding, we
are confined ourselves to modify the LRT.

Let us develop in more details an example. Assume that \textbf{x}
follows  a normalized $t$-distribution with 5 degree of freedom,
that is $ \textbf{x}=\sqrt{\frac{3}{5}} t(5), $ \textbf{x} and
\textbf{y} are i.i.d., hence  $E\textbf{x}=E\textbf{y}=0, $ $
E|\textbf{x}|^2=E|\textbf{y}|^2=1 $ and $
E|\textbf{x}|^4=E|\textbf{y}|^4=9. $  We still  employ the result in
Theorem \ref{T4.1} for the test of equality between two covariance
matrices, where
\begin{eqnarray}
  m_1(f)=&
  \displaystyle{\frac{1}{2}\Big[\log\left(\frac{y_1+y_2-y_1y_2}{y_1+y_2}\right)
      -\frac{y_1}{y_1+y_2}\log(1-y_2)-\frac{y_2}{y_1+y_2}\log(1-y_1)}\nonumber\\[2mm]
    &\quad\quad\quad\quad\quad\quad\quad\quad
    +\displaystyle{\frac{6y_1^2y_2}{(y_1+y_2)^2}+\frac{6y_1y_2^2}{(y_1+y_2)^2}\Big]}\label{m1f}
\end{eqnarray}
and
\begin{equation}
  \upsilon_1(f)=-\frac{2y_2^2}{(y_1+y_2)^2}\log(1-y_1)-\frac{2y_1^2}{(y_1+y_2)^2}\log(1-y_2)
  -2\log\frac{y_1+y_2}{y_1+y_2-y_1y_2}\label{v1f}
\end{equation}
instead of $ m(f) $  and $ \upsilon(f) $ for real case,
respectively. (\ref{m1f}) and (\ref{v1f}) are calculated in
\S\ref{sec:proofs}.

The following Table~\ref{tab:t} summarizes a simulation study where
we compare  this corrected pseudo-LRT with the test proposed in
\citet{Schott07}.  We use  1,000 independent replications with the
above $t$-distributed variables. Again, the nominal  test level is
$\alpha=  0.05$. As we can see, the corrected pseudo-LRT performs
correctly while Schott's test is no more valid here since the
variables are not Gaussian.

\begin{table}[htb]
  \begin{center}
    \begin{tabular}{|l|c|c|}
      \hline \multicolumn{3}{|c|}{(y1, y2)=(0.05, 0.1)~~}\\
      \hline {(p, $n_1$,  $n_2$ ) } & {CLRT Size}  & {Schott's Size}
      \\[1mm]\hline
      (10,100, 200) & 0.067&0.517\\[2mm]
      (20, 200, 400) & 0.065&0.603  \\[2mm]
      (40, 400, 800) & 0.054&0.703 \\[2mm]
      (80, 800, 1600)&0.048 &0.764 \\[2mm]
      (160, 1600, 3200)   &0.045&0.826\\[2mm]
      (320, 3200, 6400)   &0.051&0.854\\[2mm]
      \hline
    \end{tabular}
    \caption{Sizes  of the corrected pseudo-likelihood ration test  and Schott's test for the case
      of $y_1=0.1,~ y_2=0.05$, based on 1,000 independent replications with
      normalized $t$-distributed variables with  5 degrees of freedom. \label{tab:t}}
  \end{center}
\end{table}

\section{Proofs}
\label{sec:proofs}

\subsection*{Proof of (\protect\ref{singlmean})}

By Theorem \ref{T1.1}, for $g(x)= x-\log x-1$, by using the variable
change
$x=1+y-2\sqrt{y}\cos\theta,~ 0\leq \theta \leq \pi$, we have
\begin{eqnarray*}
  m(g)&=&\frac{g\left(a(y)\right)+g\left(b(y)\right)}{4} -
  \frac{1}{2\pi}\int_{a(y)}^{b(y)}\frac{g(x)}{\sqrt{4y-(x-1-y)^2}}dx\\
  &=&\frac{y-\log(1-y)}{2}-\frac{1}{2\pi}\int_{0}^{\pi}
  \left[1+y-2\sqrt{y}\cos\theta-\log(1+y-2\sqrt{y}\cos\theta)-1\right]d\theta\\
  &=&\frac{y-\log(1-y)}{2}-\frac{1}{4\pi}\int_{0}^{2\pi}
  \left[y-2\sqrt{y}\cos\theta-\log|1-\sqrt{y}e^{i
      \theta}|^2\right]d\theta\\
  &=&-\frac{\log(1-y)}{2},
\end{eqnarray*}
where $\displaystyle\int_{0}^{2\pi}\log|1-\sqrt{y}e^{i
  \theta}|^2d\theta=0$ is calculated in \cite{B04}.

\subsection*{Proof of (\ref{singlvar})}

For $g(x)=x-\log x -1$, by Theorem \ref{T1.1}, we have
$$\upsilon(g)=-\frac1{2\pi^2}\oint\oint\frac{g(z_1)g(z_2)}{(\underline{m}(z_1)-\underline{m}(z_2))^2}
d\underline{m}(z_1)d\underline{m}(z_2)$$ and
\begin{eqnarray*}
  g(z_1)g(z_2)&=&z_1z_2-z_1\log z_2-z_2\log z_1+\log z_1\log z_2\\
  \quad\quad &&-z_1+\log z_1 -z_2+\log z_2+1.
\end{eqnarray*}
It is easy to see that  $\upsilon(\textbf{1},\textbf{1})=0$, where
\textbf{1} means constant function equals to 1.  For Stieltjes
transform of $F^y$, the following equation is given in \cite{B04},
for $z \in \mathbb{C}^{+}$,
\begin{equation}
  z=-\frac{1}{\underline{m}(z)}+\frac{y}{1+\underline{m}(z)}.
\end{equation}\\
Let  $m_i=\underline{m}(z_i),~~i=1,2$.  For fixed $m_2$,~we
have on a contour enclosed 1, $(y-1)^{-1}$ and -1, but not 0,
\begin{eqnarray*}
  \displaystyle {\oint \frac{\log \left(z(m_1)\right)}
    {(m_1-m_2)^2}dm_1}
  &=& \displaystyle {\oint \frac{\frac{1}{m_1^2}
      -\frac{y}{(1+m_1)^2}}
    {-\frac{1}{m_1}+\frac{y}{1+m_1}}\frac{1}{( m_1-m_2)}dm_1} \\
  &=&\displaystyle{\oint
    \frac{(1+m_{1})^2-ym_1^2}{ym_1(m_1-m_2)}\left(\frac{-1}{m_{1}+1}
    +\frac{1}{m_1-\frac{1}{y-1}}\right)dm_1}\\
  &=&\displaystyle{2\pi i\cdot\left(
    \frac{1}{m_2+1}-\frac{1}{m_2-\frac{1}{y-1}}\right)}.
\end{eqnarray*}
and
\begin{eqnarray*}
  &\quad&\displaystyle {\oint \frac{-\frac{1}{m_1}
      +\frac{y}{1+m_1}}
    {(m_1-m_2)^2}dm_1}\\
  &=&\displaystyle{y\oint (\frac{1}{1+m_{1}}+\frac{1-y}{y}) \cdot
    [1-(1+m_{1})]^{-1}
    \cdot (m_{2}+1)^{-2} \cdot (1-\frac{m_{1}+1}{m_{2}+1})^{-2}
    dm_{1}}\\
  &=&\displaystyle {y\oint (\frac{1}{1+m_{1}}+\frac{1-y}{y}) \cdot
    \sum\limits^\infty_{j=0}(1+m_1)^j (m_2+1)^{-2}
    \sum\limits^\infty_{\ell=1}\ell(\frac{m_{1}+1}{m_{2}+1})^{\ell-1}dm_1}\\
  &=&\displaystyle{2\pi i\cdot  \frac{y}{(m_2+1)^{2}}}.
\end{eqnarray*}
Then we also get $\upsilon(-z_1+\log z_1, ~\textbf{1})=0$.
Similarly, $\upsilon(\textbf{1}, ~ -z_2+\log z_2)=0$.
Furthermore,
\begin{eqnarray*}
  \upsilon(z_1,z_2)&=&\displaystyle{\frac {y^2}{\pi
      i}\oint\frac{1}{(m_2+1)^{2}}(\displaystyle\frac{1}{1+m_{2}}+\displaystyle\frac{1-y}{y})
    \displaystyle\sum\limits^\infty_{j=0}(1+m_2)^jdm_2}
  =2y,
\end{eqnarray*}
and
\begin{eqnarray*}
  \upsilon(z_1, \log z_2)
  &=&\displaystyle{ \frac{y}{\pi i}\oint( \frac{1}{m_2+1}- \frac
    {1}{m_2-1/(y-1)})
    (\frac{1}{1+m_{2}}+\frac{1-y}{y}) \cdot [1-(1+m_{2})]^{-1} dm_2}\\
  &=&\displaystyle{ \frac{y}{\pi i} \oint( \frac{1}{m_2+1}- \frac
    {1}{m_2-1/(y-1)})(\frac{1}{1+m_{2}}+\displaystyle\frac{1-y}{y})
    \sum\limits^\infty_{j=0}(1+m_2)^j dm_2} \\
  &=&2y.
\end{eqnarray*}
By a computation in  \cite{B04}, we know that $ \upsilon(\log z_1,
\log z_2)= -2\log(1-y)$.  Finally, we obtain
\begin{eqnarray*}
  \upsilon(g)&=&\upsilon(z_1, z_2)+\upsilon(\log z_1, \log
  z_2)-2\upsilon(z_1, \log z_2)\\
  &&\quad+\upsilon(-z_1+\log z_1,
  \textbf{1})+\upsilon(\textbf{1},-z_2+\log z_2)+\upsilon(\textbf{1}, \textbf{1})\\
  &=&-2\log (1-y)-2y.
\end{eqnarray*}

\subsection*{Proof of (\ref{singllimit})}

Since  $F^{y_n}$  is the Mar\v{c}enko-Pastur law of index $y_n$,
by using the variable change $x=1+y_n-2\sqrt{y_n}\cos \theta, ~ 0\leq
\theta \leq \pi$ we have
\begin{eqnarray*}
  F^{y_n}(g)&=&\int^{b(y_n)}_{a(y_n)}\frac {x-\log x-1}{2\pi
    xy_n}\sqrt{(b(y_n)-x)(x-a(y_n))} dx\\
  &=&\frac{1}{2\pi
    y_n}\int_0^{\pi}\left[1-\frac{\log(1+y_n-2\sqrt{y_n}\cos\theta)+1}
    {1+y_n-2\sqrt{y_n}\cos\theta}\right]4y_n\sin^2\theta d\theta\\
  &=&\frac{1}{2\pi}\int_0^{2\pi}\left[2\sin^2\theta-\frac{2\sin^2\theta}{1+y_n-2\sqrt{y_n}\cos\theta}
    \left(\log |1-\sqrt{y_n}e^{i\theta}|^2-1\right)\right]d \theta\\
  &=&1-\frac{y_n-1}{y_n}\log (1-y_n),
\end{eqnarray*}
where
$$\displaystyle{\frac{1}{2\pi}\int_0^{2\pi}\frac{2\sin^2\theta}{1+y_n-2\sqrt{y_n}\cos\theta}
  \log |1-\sqrt{y_n}e^{i\theta}|^2d \theta
}=\displaystyle{\frac{y_n-1}{y_n}\log (1-y_n)-1}$$ is calculated in
\cite{B04}.

\subsection*{ Proof of Lemma  \ref{lem1}}

We use the variable change  $x=(1-y_2)^{-2}(1+h^2-2h\cos \theta)$, where
$h=\sqrt{y_1+y_2-y_1y_2}$.  When $c,d $ satisfy
$c^2+d^2=a(1-y_2)^2+b(1+h^2),~ cd=bh,~ 0 < d < c,$ we have
\begin{equation*}
  f(z(\xi))=\log(a+bz(\xi))=\log\left(\frac{\left|c+d\xi\right|^2}{(1-y_2)^2}\right).
\end{equation*}
Similarly,
$$g(z(\xi))=\log(\alpha+\beta
z(\xi))=\log\left(\frac{\left|\gamma+\eta\xi\right|^2}{(1-y_2)^2}\right).$$
Let
$$
\widetilde{f}(z(\xi))=\log\left(\frac{\left(c+d\xi\right)^2}{(1-y_2)^2}\right)
\quad \mbox{and}
\quad\widetilde{g}(z(\xi))=\log\left(\frac{\left(\gamma+\eta\xi\right)^2}{(1-y_2)^2}\right).
$$
Note that $f(z(\xi))=\Re(\widetilde{f}(z(\xi)))\quad \mbox{and}
\quad g(z(\xi))=\Re(\widetilde{g}(z(\xi)))$. By  Theorem \ref{T2.1},
we have
\begin{eqnarray}
  m(f)&=& \displaystyle{\frac{1}{4\pi
      i}\oint_{|\xi|=1}
    f(z(\xi))\left[\frac{1}{\xi-{1\over r}}+\frac{1}{\xi+{1\over r}}-\frac{2}{\xi+{y_2\over {hr}}}\right]d\xi}\nonumber\\
  &=&\displaystyle{\frac{1}{4\pi}\int^{2\pi}_{0}
    f(z(e^{i\theta}))\left[\frac{1}{e^{i\theta}-{1\over r}}
      +\frac{1}{e^{i\theta}+{1\over r}}-\frac{2}{e^{i\theta}+{y_2\over {hr}}}\right]e^{i\theta}d\theta}\nonumber\\
  &=&\displaystyle{\frac{1}{4\pi}\int^{2\pi}_{0}
    f(z(e^{i\theta}))\left[\frac{1}{e^{-i\theta}-{1\over r}}
      +\frac{1}{e^{-i\theta}+{1\over r}}-\frac{2}{e^{-i\theta}+{y_2\over
          {hr}}}\right]e^{-i\theta}d\theta}\nonumber\\
  &=&\displaystyle\frac{1}{8\pi}\int^{2\pi}_{0}
  f(z(e^{i\theta}))\Bigg\{\left[\frac{1}{e^{i\theta}-{1\over r}}
    +\frac{1}{e^{i\theta}+{1\over r}}-\frac{2}{e^{i\theta}+{y_2\over {hr}}}\right]e^{i\theta}+\nonumber\\
  &&\left[\frac{1}{e^{-i\theta}-{1\over
        r}}+\frac{1}{e^{-i\theta}+{1\over r}}
    -\frac{2}{e^{-i\theta}+{y_2\over {hr}}}\right]e^{-i\theta}\Bigg\}d\theta\nonumber\\
  &=&\frac{1}{8\pi}\Re\Bigg\{\int^{2\pi}_{0}
  \widetilde{f}(z(e^{i\theta}))\Bigg[\left(\frac{1}{e^{i\theta}-{1\over
        r}}
    +\frac{1}{e^{i\theta}+{1\over r}}-\frac{2}{e^{i\theta}+{y_2\over {hr}}}\right)e^{i\theta}+\nonumber\\
    &&\left(\frac{r}{r-e^{i\theta}}+\frac{r}{r+e^{i\theta}}
    -\frac{2hr}{y_2e^{i\theta}+hr}\right)\Bigg]d\theta\Bigg\}\nonumber\\
  &=&\Re\Bigg\{\frac{1}{8\pi i}\oint_{|\xi|=1}
  \widetilde{f}(z(\xi))\Bigg[\left(\frac{1}{\xi-{1\over
        r}}+\frac{1}{\xi+{1\over r}}-\frac{2}{\xi+{y_2\over {hr}}}
    \right)\nonumber\\
    &&+\left(\frac{r}{r-\xi}+\frac{r}{r+\xi}
    -\frac{2hr}{y_2\xi+hr}\right)\xi^{-1}\Bigg]d\xi\Bigg\}\nonumber\\
  &=&\frac{1}{4}\left(\widetilde{f}(z(\frac{1}{r}))+\widetilde{f}(z(-\frac{1}{r}))
  -2\widetilde{f}(z(-\frac{y_2}{hr}))\right)\nonumber\\
  &\rightarrow& ^{r\downarrow1}\frac{1}{4}\left[\widetilde{f}(z(1))+\widetilde{f}(z(-1))-2\widetilde{f}(z(-\frac{y_2}{h}))\right]\nonumber\\
  &=&\frac{1}{2}\log\frac{(c^2-d^2)h^2}{(ch-y_2d)^2}.\nonumber
\end{eqnarray}

Let $ m_j=-\frac{1+hr_j\xi_j}{1-y_2} $,  where $|\xi_j|=1, j=1,2,$ $
r_2\downarrow r_1,$ and $r_1\downarrow 1$. By  Theorem~\ref{T2.1},
we have
$$\upsilon(f, g)=-\frac{1}{2\pi ^2}\oint_{|\xi_2|=1}
\left\{\oint_{|\xi_1|=1}\frac{f(z(r_1\xi_1))}{(r_2\xi_2-r_1\xi_1)^2}\cdot
r_1r_2 d\xi_1\right\}g(z(r_2\xi_2))d\xi_2.
$$
When $ r_1\downarrow 1, \quad -\frac{d}{cr_1} \quad \mbox{and}\quad
0$  are poles.  We can then  choose  $r_1$ so that
 $-\frac{c}{dr_1}$  is a not a  pole. Then we get
\begin{eqnarray*}
  &&\displaystyle{\oint_{|\xi_1|=1}\frac{\log(a+bz(r_1\xi_1))}{(r_2\xi_2-r_1\xi_1)^2}\cdot
    r_1r_2 d\xi_1}\\[3mm]
  &=&\displaystyle{\oint_{|\xi_1|=1}\frac{\left(\log(a+bz(r_1\xi_1))\right)'}{r_1\xi_1-r_2\xi_2}\cdot
    r_2 d\xi_1}\\[3mm]
  &=&\displaystyle{\oint_{|\xi_1|=1}\Bigg[\frac{bhr_1\xi_1}{(r_1\xi_1-r_2\xi_2)(c+dr_1\xi_1)c}\cdot
      \frac{1}{\xi_1+\frac{d}{cr_1}}}\\
    &&\quad-\frac{bhr_1^{-1}}{(r_1\xi_1-r_2\xi_2)(c+dr_1\xi_1)c}\cdot
    \frac{1}{(\xi_1+\frac{d}{cr_1})\xi_1}\cdot r_2\Bigg]d\xi_1\\[5mm]
  &=&\displaystyle{2\pi i}\left(
  \frac{bhd^{-1}c^{-1}}{\xi_2}-\frac{bhd^{-1}r_2}{d+cr_2\xi_2}\right).
\end{eqnarray*}
So,
$$
\upsilon(f, g) =-\frac{i}{\pi}
\oint_{|\xi_2|=1}\left(\frac{bhd^{-1}c^{-1}}{\xi_2}
-\frac{bhd^{-1}r_2}{d+cr_2\xi_2}\right) \log\left(\alpha+\beta
z(r_2\xi_2)\right)d\xi_2.
$$
Since the function $g(x)=\log(\alpha+\beta x) $ is analytic, when $
r_2>1$  but sufficiently close to 1, we have
$$
\left|g(z(r\xi_2))-g(z(\xi_2))\right| \leq K(r-1),
$$
for some constant $K$. Thus we have
$$
\begin{array}{lll}
  &\displaystyle{\left|\oint_{|\xi|_2=1}
    \left[g(z(r_2\xi_2))-g(z(\xi_2))\right]
    \left(\frac{bhd^{-1}c^{-1}}{\xi_2}-\frac{bhd^{-1}r_2}{d+cr_2\xi_2}\right)d\xi_2\right|}\\[4mm]
  \rightarrow & 0 \quad \mbox{as} \quad r_2 \downarrow 1,
\end{array}
$$
where the estimations are done according to$|\arg(\xi_2)|$ or
$|\arg(\xi_2)-\pi| \leq \sqrt{r_2-1}$ or not. Thus,
$$
\upsilon(f, g)=-\frac{i}{\pi} \oint_{|\xi_2|=1}
g(z(\xi_2))\left(\frac{bhd^{-1}c^{-1}}{\xi_2}-\frac{bhd^{-1}r_2}{d+cr_2\xi_2}\right)d\xi_2+R(r_2)
$$
where $ R(r_2) \rightarrow 0 , \quad\mbox{as}\quad  r_2\downarrow1.$
Because $\displaystyle{g(z(\xi_2))=\log\left(\frac{\left|\gamma+\eta\xi_2\right|^2}{(1-y_2)^2}\right),}$
for $\gamma, \eta$ satisfying
$\gamma^2+\eta^2=\alpha(1-y_2)^2+\beta(1+h^2),~ \gamma\eta=\beta h,~
0 < \eta < \gamma,$ and if
$\displaystyle{\widetilde{g}(z(\xi_2))=\log\left(\frac{\left(\gamma+\eta\xi_2\right)^2}{(1-y_2)^2}\right),}$
we have  $g(z(\xi_2))=\Re\left(\widetilde{g}(z(\xi_2))\right) $.
Therefore,
\begin{eqnarray*}
  \lefteqn{    \upsilon(f, g)=-\frac{i}{\pi}
    \oint_{|\xi|_2=1}  g(z(\xi_2))
    \left( \frac{bhd^{-1}c^{-1}}{\xi_2} -
    \frac{bhd^{-1}r_2}{d+cr_2\xi_2}      \right)d\xi_2}\\
  &=&\frac{1} {\pi} \int^{2\pi}_{0} g(z(e^{i\theta}))\left(\frac{bhd^{-1}c^{-1}}{e^{i\theta}}
  -\frac{bhd^{-1}r_2}{d+cr_2e^{i\theta}}\right)e^{i\theta}d\theta\\
  &=&^{\theta \rightarrow 2\pi-\theta} ~~\frac{1}{\pi}
  \int^{2\pi}_{0}g(z(e^{i\theta}))\left(\frac{bhd^{-1}c^{-1}}{e^{-i\theta}}
  -\frac{bhd^{-1}r_2}{d+cr_2e^{-i\theta}}\right)e^{-i\theta}d\theta\\
  &=&\frac{1}{2\pi}\int_{0}^{2\pi}g(z(e^{i\theta}))\Bigg[\left(\frac{bhd^{-1}c^{-1}}{e^{i\theta}}
    -\frac{bhd^{-1}r_2}{d+cr_2e^{i\theta}}\right)
    e^{i\theta}
    +bhd^{-1}c^{-1}-\frac{bhd^{-1}r_2}{de^{i\theta}+cr_2}\Bigg]d\theta\\
  &=&\frac{1}{2\pi}\Re\Bigg\{\int_{0}^{2\pi}\widetilde{g}(z(e^{i\theta}))
  \Bigg[\left(\frac{bhd^{-1}c^{-1}}{e^{i\theta}}-\frac{bhd^{-1}r_2}{d+cr_2e^{i\theta}}\right)
    e^{i\theta}
    +bhd^{-1}c^{-1}-\frac{bhd^{-1}r_2}{de^{i\theta}+cr_2}\Bigg]d\theta\Bigg\}\\
  &=&\Re\Bigg\{\frac{1}{2\pi
    i}\oint_{|\xi|_2=1}\widetilde{g}(z(\xi_2))\Bigg[\left(\frac{bhd^{-1}c^{-1}}{\xi_2}-\frac{bhd^{-1}r_2}{d+cr_2\xi_2}\right)
    +\left(bhd^{-1}c^{-1}-\frac{bhd^{-1}r_2}{d\xi_2+cr_2}\right)\xi_2^{-1}\Bigg]d\xi_2\Bigg\}\\
  &=&bhd^{-1}c^{-1}\left[\widetilde{g}(z(0))-\widetilde{g}(z(-\frac{d}{cr_2}))\right]\\
  &\rightarrow& bhd^{-1}c^{-1}\left[\widetilde{g}(z(0))-\widetilde{g}(z(-\frac{d}{c}))\right]\\
  &=&2bhd^{-1}c^{-1}\log\frac{c\gamma}{c\gamma-d\eta}.
\end{eqnarray*}

\subsection*{Proof of (\ref{testE}) and (\ref{testVar})}
Because \textbf{$\xi$}  and  \textbf{$\eta$}   are Gaussian
variables, for real case, $\beta=E|$\textbf{$\xi$}$|^4-3=0,$ then
(\ref{E1betax}), (\ref{E1betay}) and (\ref{cov1betax})
are all 0. Consider (\ref{E1}) and (\ref{cov1}),
as $y_{n_k} \rightarrow y_k, ~k=1,2,$, by the computations done in the
proof of  Lemma~\ref{lem1},  we see that termes tending to zero
could be neglected in the considered contour integrals.
Hence we can put $ y_{n_k}=y_k, k=1,2 $ and   use
$$
f(x)=\log(y_{1}+y_{2}x)-\frac{y_{2}}{y_{1} +y_{2}}\log
x-\log(y_{1}+y_{2})
$$
instead of
$f(x)=\displaystyle{\log(y_{n_1}+y_{n_2}x)-\frac{y_{n_2}}{y_{n_1}
    +y_{n_2}}\log x-\log(y_{n_1}+y_{n_2})}$.
Consider  the variable change
$x=(1-y_2)^{-2}(1+h^2-2h\cos \theta)$, where
$z(\xi)=(1-y_2)^{-2}\left[1+h^2+2h\mathcal{R}(\xi)\right], \quad h
=\sqrt{y_1+y_2-y_1y_2}$. As
\begin{eqnarray*}
  \log(y_{n_1}+y_{n_2}z(\xi))& = & \log\left(\frac{\left|h+y_{2}\xi\right|^2}{(1-y_2)^2}\right),\\
  \log(z(\xi))& = &
  \log\left(\frac{\left|1+h\xi\right|^2}{(1-y_2)^2}\right),
\end{eqnarray*}
we have by  Lemma~\ref{lem1},
\begin{eqnarray*}
  m(f)&=&\displaystyle{\frac{1}{2}\left[\log\frac{(h^2-y_2^2)h^2}{(h^2-y_2^2)^2}
      -\frac{y_2}{y_1+y_2}\log\frac{(1-h^2)h^2}{(h-y_2h)^2}\right]}\\
  &=&\displaystyle{\frac{1}{2}\left[\log\left(\frac{y_1+y_2-y_1y_2}{y_1+y_2}\right)
      -\frac{y_1}{y_1+y_2}\log(1-y_2)-\frac{y_2}{y_1+y_2}\log(1-y_1)\right]},
\end{eqnarray*}
and
\begin{eqnarray*}
  \upsilon(f)&=&\upsilon\big(\log(y_{n_1}+y_{n_2}x)\big)+\frac{y_2^2}{(y_1+y_2)^2}\upsilon
  \big(\log x\big)-\frac{2y_2}{y_1+y_2}\upsilon\big(\log x, \log
  (y_{n_1}+y_{n_2}x)\big)\\
  &=&2\log
  \frac{h^2}{h^2-y_2^2}+2\frac{y_2^2}{(y_1+y_2)^2}\log\frac{1}{1-h^2}-\frac{4y_2}{y_1+y_2}\log
  \frac{1}{1-y_2}\\
  &=&-\frac{2y_2^2}{(y_1+y_2)^2}\log(1-y_1)-\frac{2y_1^2}{(y_1+y_2)^2}\log(1-y_2)
  -2\log\frac{y_1+y_2}{y_1+y_2-y_1y_2}.
\end{eqnarray*}

\subsection*{Proof of $F_{y_{n_1}, y_{n_2}}(f)$}

By (\ref{f(x)}) and  the density of $F_{y_{n_1}, y_{n_2}}(f)$ (the
limiting distribution in (\ref{LSDden}) but with $y_{n_k}$ in place
of $y_k, k=1,2.$ ), where
$h_n=\sqrt{y_{n_1}+y_{n_2}-y_{n_1}y_{n_2}}$,
$a_n=(1-y_{n_2})^{-2}\left(1-\sqrt{y_{n_1}+y_{n_2}-y_{n_1}y_{n_2}}\right)^2$
and
$b_n=(1-y_{n_2})^{-2}\left(1+\sqrt{y_{n_1}+y_{n_2}-y_{n_1}y_{n_2}}\right)^2.$
Using the  substitution $
x=(1-y_{n_2})^{-2}\left(1+h_n^2-2h_n\cos\theta\right), \quad 0 <
\theta < \pi, $   we have
\begin{eqnarray}
  \sqrt{(b_n-x)(x-a_n)}= \frac{2h_n\sin\theta}{(1-y_{n_2})^2},
  \quad&\quad
  dx=\displaystyle\frac{2h_n\sin\theta d\theta}{(1-y_{n_2})^2};\nonumber\\
  x=\frac{\left|1-h_ne^{i\theta}\right|^2}{(1-y_{n_2})^2},\quad&\quad
  \displaystyle{y_{n_1}+y_{n_2}x}=\displaystyle{\frac{\left|h_n-y_{n_2}e^{i\theta}\right|^2}
    {(1-y_{n_2})^2}}.\nonumber
\end{eqnarray}
Therefore,
\begin{eqnarray*}
  &&{F^{y_{n_1},      y_{n_2}}(f)}\\[5mm]
  &=&{\int_{a_n}^{b_n}f(x)\frac{(1-y_{n_2})\sqrt{(b_n-x)(x-a_n)}}{2\pi
      x(y_{n_1}+y_{n_2}x)}dx}\\[6mm]
  &=&{(1-y_{n_2})\int_{a_n}^{b_n}
    \left[\log\left(y_{n_1}+y_{n_2}x\right)
      -\frac{y_{n_2}}{y_{n_1}+y_{n_2}}\log
      x\right]\frac{\sqrt{(b_n-x)(x-a_n)}}{2\pi x(y_{n_1}+y_{n_2}x)}dx}\\
  &&\quad\quad-{\log\left(y_{n_1}+y_{n_2}\right)}\\[6mm]
  &=&{\frac{2(1-y_{n_2})}{\pi}\int_{0}^{\pi}\left[
      \log\frac{\left|h_n-y_{n_2}e^{i\theta}\right|^2}{(1-y_{n_2})^2}
      -\frac{y_{n_2}}{y_{n_1}+y_{n_2}}\log\frac{\left|1-h_ne^{i\theta}\right|^2}
      {(1-y_{n_2})^2}\right]}\\[6mm] &&\quad\quad
  {\cdot\frac{h_n^2\sin^2\theta}{\left|1-h_ne^{i\theta}\right|^2
      \left|h_n-y_{n_2}e^{i\theta}\right|^2}d\theta}-{
    \log\left(y_{n_1}+y_{n_2}\right)}\\[6mm]
  &=&{\frac{2(1-y_{n_2})}{\pi}\int_{0}^{\pi} \left[
      \log\left|h_n-y_{n_2}e^{i\theta}\right|^2
      -\frac{y_{n_2}}{y_{n_1}+y_{n_2}}\log\left|1-h_ne^{i\theta}\right|^2
      \right]} \\[4mm]
  &&\cdot{\frac{h_n^2\sin^2\theta}
    {\left|1-h_ne^{i\theta}\right|^2\left|h_n-y_{n_2}e^{i\theta}\right|^2}d\theta}
  -2\left(1-\frac{y_{n_2}}{y_{n_1}+y_{n_2}}\right)
  \log(1-y_{n_2})-\log{(y_{n_1}+y_{n_2})}\\
  &=&{\Re\Bigg\{\frac{2(1-y_{n_2})}{\pi}\int_{0}^{2\pi}
    \left[ \log(h_n-y_{n_2}e^{i\theta})
      -\frac{y_{n_2}}{y_{n_1}+y_{n_2}}\log(1-h_ne^{i\theta})\right]}\\
  &&
      {\frac{h_n^2\sin^2\theta}{\left|1-h_ne^{i\theta}\right|^2
          \left|h_n-y_{n_2}e^{i\theta}\right|^2}d\theta\Bigg\}}-\frac{2y_{n_1}}{y_{n_1}+y_{n_2}}
      \log(1-y_{n_2})-\log{(y_{n_1}+y_{n_2})}\\[5mm]
      &=&{\Re\Bigg\{\frac{-(1-y_{n_2})}{2\pi i}\oint_{|z|=1}
        \left[\log(h_n-y_{n_2}z)-\frac{y_{n_2}}{y_{n_1}+y_{n_2}}\log(1-h_nz)\right]}\\[5mm]
      &&\cdot{\frac{h_n^2(z-z^{-1})^2}{z\left|1-h_nz\right|^2
          \left|h_n-y_{n_2}z\right|^2}dz\Bigg\}}-{\frac{2y_{n_1}}{y_{n_1}+y_{n_2}}
        \log(1-y_{n_2})-\log{(y_{n_1}+y_{n_2})}}\\[5mm]
      &=&{\Re\Bigg\{\frac{y_{n_2}-1}{y_{n_2}}\cdot\frac{1}{2\pi
          i}\oint_{|z|=1}\left[
          \log(h_n-y_{n_2}z)-\frac{y_{n_2}}{y_{n_1}+y_{n_2}}\log(1-h_nz)\right]}\\[6mm]
      &&\cdot{\frac{(z^2-1)^2}{z(z-h_n)(z-\frac{1}{h_n})
          (z-\frac{y_{n_2}}{h_n})(z-\frac{h_n}{y_{n_2}})}dz}\Bigg\}
      -{\frac{2y_{n_1}}{y_{n_1}+y_{n_2}}
        \log(1-y_{n_2})-\log{(y_{n_1}+y_{n_2})}}.
\end{eqnarray*}
There are three poles inside the unit circle: 0, $h_n, y_{n_2}/h_n$.
Their corresponding residues are
\begin{eqnarray}
  R(0) &=& \frac{y_{n_2}-1}{y_{n_2}}\log (h_n) ,\nonumber\\
  R(h_n)&=& \frac{(h_n^2-1)}{(h_n^2-y_{n_2})}\left[\log (h_n)+\log
    (1-y_{n_2})-\frac{y_{n_2}}{y_{n_1}+y_{n_2}}\log (1-h^2_n)
    \right],\nonumber\\
  R(\frac{y_{n_2}}{h_n})&=&
  \frac{(y_{n_2}^2-h_n^2)}{y_{n_2}(y_{n_2}-h_n^2)} \left[\log
    (h^2_n-y_{n_2}^2)-\log (h_n)-\frac{y_{n_2}}{y_{n_1}+y_{n_2}}\log
    (1-y_{n_2}) \right].\nonumber
\end{eqnarray}
Therefore,
\begin{eqnarray*}
  F^{y_{n_1}, y_{n_2}}(f)&=&
  R(0)+R(h_n)+R(\frac{y_{n_2}}{h_n})\displaystyle{-\frac{2y_{n_1}}{y_{n_1}+y_{n_2}}
    \log(1-y_{n_2})-\log{(y_{n_1}+y_{n_2})}} \\
  &=&\displaystyle{\frac{-(y_{n_1}+y_{n_2}-y_{n_1}y_{n_2})}{y_{n_1}y_{n_2}}\log{(y_{n_1}+y_{n_2}-y_{n_1}y_{n_2})}}\\[2mm]
  &&\quad
  +\displaystyle{\frac{(y_{n_1}+y_{n_2}-y_{n_1}y_{n_2})}{y_{n_1}y_{n_2}}\log{(y_{n_1}+y_{n_2})}}
  +\displaystyle{\frac{y_{n_1}(1-y_{n_2})}{y_{n_2}(y_{n_1}+y_{n_2})}\log{(1-y_{n_2})}}\nonumber\\
  &&\quad\quad\quad+\displaystyle{\frac{y_{n_2}(1-y_{n_1})}{y_{n_1}(y_{n_1}+y_{n_2})}\log{(1-y_{n_1})}}.\nonumber
\end{eqnarray*}

\subsection*{Proof of (\ref{m1f}) and (\ref{v1f})}

Because \textbf{x} and \textbf{y}  are random variables from
normalized $t$-distribution with 5 degree of freedom, \textbf{x} and
\textbf{y} are $i.i.d.$, $E\textbf{x}=E\textbf{y}=0, $ $
E|\textbf{x}|^2=E|\textbf{y}|^2=1 $ and $
E|\textbf{x}|^4=E|\textbf{y}|^4=9. $   For real case,
$\beta=E|$\textbf{$\xi$}$|^4-3=6,$ (\ref{E1}) and (\ref{cov1}) items
are the same to the Gaussian variables. Consider the items
(\ref{E1betax}), (\ref{E1betay}) and (\ref{cov1betax}). As the same
explanation in Proof of (\ref{testE}) and (\ref{testVar}), we use $
f(x)=\log(y_{1}+y_{2}x)-\frac{y_{2}}{y_{1} +y_{2}}\log
x-\log(y_{1}+y_{2}) $ instead.

For (\ref{E1betax}), we have
\begin{eqnarray*}
  &&\frac{\beta\cdot y_1(1-y_2)^2}{2\pi i \cdot h^2} \oint_{|\xi|=1}
  \left[\log\frac{|h+y_2\xi|^2}{(1-y_2)^2}-\frac{y_2}{y_1+y_2}\log\frac{|1+h\xi|^2}{(1-y_2)^2}-\log(y_1+y_2)\right]\\
  &&\quad\quad\quad\quad\quad\quad \quad\quad\quad\quad\quad\quad
  \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
  \quad\quad\quad\cdot \frac{1}{(\xi+\frac{y_2}{hr})^3} d\xi\\
  &=&\frac{\beta\cdot y_1(1-y_2)^2}{2\pi i \cdot h^2} \oint_{|\xi|=1}
  2
  \mathcal{R}\Big\{\log(h+y_2\xi)-\frac{y_2}{y_1+y_2}\log(1+h\xi)\Big\}\cdot
  \frac{1}{(\xi+\frac{y_2}{hr})^3} d\xi\\
  &=&\frac{\beta\cdot y_1(1-y_2)^2}{2\pi i \cdot h^2}
  \oint_{|\xi|=1}\Bigg\{\log(h+y_2\xi)+\log(h+y_2\overline{\xi})\\
  && \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
  -\frac{y_2}{y_1+y_2}\Big[\log(1+h\xi)+\log(1+h\overline{\xi})\Big]\Bigg\}
  \cdot\frac{1}{(\xi+\frac{y_2}{hr})^3} d\xi\\
  &=&\frac{\beta\cdot y_1(1-y_2)^2}{2\pi i \cdot h^2}
  \oint_{|\xi|=1}\left[\log(h+y_2\xi)-\frac{y_2}{y_1+y_2}\log(1+h\xi)\right]
  \\ &&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
  \quad\quad\quad\quad\quad\quad\quad\quad\quad
  \left[\frac{1}{(\xi+\frac{y_2}{hr})^3}
    +\frac{\left(\frac{hr}{y_2}\right)^3\xi}{(\frac{hr}{y_2}+\xi)^3}
    \right] d\xi\\
  &=&\frac{\beta\cdot y_1(1-y_2)^2}{2\pi i \cdot h^2} \cdot 2\pi \cdot
  \frac{1}{2}
  \left[\log(h+y_2\xi)-\frac{y_2}{y_1+y_2}\log(1+h\xi)\right]''\Bigg|_{\xi=-\frac{y_2}{hr}}\\
  &=&\frac{\beta\cdot y_1(1-y_2)^2}{2 h^2}
  \left[-\frac{y_2^2}{(h+y_2\xi)^2}+\frac{y_2}{y_1+y_2}\frac{h^2}{(1+h\xi)^2}\right]\Bigg|_{\xi=-\frac{y_2}{hr}}\\
  &=&\frac{\beta y_1^2y_2}{2(y_1+y_2)^2}.
\end{eqnarray*}

For (\ref{E1betay}), we have
\begin{eqnarray*}
  &&\frac{\beta\cdot (1-y_2)}{4\pi i} \oint_{|\xi|=1}
  \left[\log\frac{|h+y_2\xi|^2}{(1-y_2)^2}-\frac{y_2}{y_1+y_2}\log\frac{|1+h\xi|^2}{(1-y_2)^2}-\log(y_1+y_2)\right]\\
  &&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\cdot
  \frac{\xi^2-\frac{y_2}{h^2r^2}}{(\xi+\frac{y_2}{hr})^2}\left[
    \frac1{\xi-\frac{\sqrt{y_2}}{hr}}+\frac1{\xi+\frac{\sqrt{y_2}}{hr}}-\frac2{\xi+\frac{y_2}{hr}}\right]d\xi\\
  &=&\frac{\beta\cdot (1-y_2)y_2}{2\pi i \cdot h} \oint_{|\xi|=1}
  \left[\log\frac{|h+y_2\xi|^2}{(1-y_2)^2}-\frac{y_2}{y_1+y_2}\log\frac{|1+h\xi|^2}{(1-y_2)^2}-\log(y_1+y_2)\right]\\
  && \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
  \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\cdot
  \left[\frac{\xi+\frac{1}{hr}}{(\xi+\frac{y_2}{hr})^3}\right]
  d\xi\\
  &=&\frac{\beta\cdot (1-y_2)y_2}{2\pi i \cdot h} \oint_{|\xi|=1}2
  \mathcal{R}\Big\{\log(h+y_2\xi)-\frac{y_2}{y_1+y_2}\log(1+h\xi)\Big\}
  \cdot \left[\frac{\xi+\frac{1}{hr}}{(\xi+\frac{y_2}{hr})^3}\right]
  d\xi\\
  &=&\frac{\beta\cdot (1-y_2)y_2}{2\pi i \cdot h}\oint_{|\xi|=1}\Big[
    \log(h+y_2\xi)+\log(h+y_2\overline{\xi})\\
    &&\quad\quad\quad\quad\quad\quad\quad\quad
    -\frac{y_2}{y_1+y_2}\left(\log(1+h\xi)+\log(1+h\overline{\xi})\right)
    \Big]\cdot
  \left[\frac{\xi+\frac{1}{hr}}{(\xi+\frac{y_2}{hr})^3}\right]d\xi\\
  &=&\frac{\beta\cdot (1-y_2)y_2}{2\pi i \cdot
    h}\oint_{|\xi|=1}\left[\log(h+y_2\xi)-\frac{y_2}{y_1+y_2}\log(1+h\xi)\right]\\
  &&\quad\quad\quad\quad\quad\quad\quad\cdot\Big[\frac{\xi+\frac{1}{hr}}{(\xi+\frac{y_2}{hr})^3}+
    \frac{\frac{h^2r^2}{y_2^3}(\xi+hr)}{(\xi+\frac{hr}{y_2})^3}\Big]d\xi\\
  &=&\frac{\beta\cdot (1-y_2)y_2}{2\pi i \cdot h}2\pi i \cdot
  \frac1{2}\left([\log(h+y_2\xi)-\frac{y_2}{y_1+y_2}\log(1+h\xi)]
  \cdot(\xi+\frac{1}{hr})\right)''\Bigg|_{\xi=-\frac{y_2}{hr}}\\
  &=&\frac{\beta y_1y_2^2}{2(y_1+y_2)^2}.
\end{eqnarray*}
Therefore,
\begin{eqnarray}
  m_1(f)&=&
  \frac{1}{2}\Big[\log\left(\frac{y_1+y_2-y_1y_2}{y_1+y_2}\right)
    -\frac{y_1}{y_1+y_2}\log(1-y_2)-\frac{y_2}{y_1+y_2}\log(1-y_1)\nonumber\\
    \quad &+& \frac{6 y_1^2y_2}{(y_1+y_2)^2}+\frac{6
      y_1y_2^2}{(y_1+y_2)^2}\Big].\nonumber
\end{eqnarray}

For covariance, we have
\begin{eqnarray*}
  &&\oint_{|\xi|=1}
  \frac{f\left(\frac{1+h^2+2h\mathcal{R}(\xi)}{(1-y_2)^2}\right)}{(\xi+\frac{y_2}{hr})^2}d\xi\\
  &=& \oint_{|\xi|=1}
  \left[\log\frac{|h+y_2\xi|^2}{(1-y_2)^2}-\frac{y_2}{y_1+y_2}\log\frac{|1+h\xi|^2}{(1-y_2)^2}-\log(y_1+y_2)\right]
  \cdot \frac{1}{(\xi+\frac{y_2}{hr})^2} d\xi\\
  &=& \oint_{|\xi|=1} 2
  \mathcal{R}\Big\{\log(h+y_2\xi)-\frac{y_2}{y_1+y_2}\log(1+h\xi)\Big\}\cdot
  \frac{1}{(\xi+\frac{y_2}{hr})^2} d\xi\\
  &=& \oint_{|\xi|=1}\Bigg\{\log(h+y_2\xi)+\log(h+y_2\overline{\xi})\\
  && \quad\quad
  -\frac{y_2}{y_1+y_2}\Big[\log(1+h\xi)+\log(1+h\overline{\xi})\Big]\Bigg\}
  \cdot\frac{1}{(\xi+\frac{y_2}{hr})^2} d\xi\\
  &=&
  \oint_{|\xi|=1}\left[\log(h+y_2\xi)-\frac{y_2}{y_1+y_2}\log(1+h\xi)\right]
  \left[\frac{1}{(\xi+\frac{y_2}{hr})^2}
    +\frac{\left(\frac{hr}{y_2}\right)^2}{(\xi+\frac{hr}{y_2})^2}
    \right] d\xi\\
  &=& 2\pi i \cdot \left[\log(h+y_2\xi)-\frac{y_2}{y_1+y_2}\log(1+h\xi)\right]'\Bigg|_{\xi=-\frac{y_2}{hr}}\\
  &=& \pi i \cdot
  \left[\frac{y_2}{h+y_2\xi}+\frac{y_2}{y_1+y_2}\frac{h}{1+h\xi}\right]\Bigg|_{\xi=-\frac{y_2}{hr}}\\
  &=&0.
\end{eqnarray*}

So, (\ref{cov1betax}) becomes,
\begin{eqnarray*}
  &&-\frac{\beta \cdot (y_1+y_2)(1-y_2)^2}{4\pi^2h^2}
  \oint_{|\xi_1|=1}
  \frac{f\left(\frac{1+h^2+2h\mathcal{R}(\xi_1)}{(1-y_2)^2}\right)}{(\xi_1+\frac{y_2}{hr_1})^2}d\xi_1
  \oint_{|\xi_2|=1}
  \frac{f\left(\frac{1+h^2+2h\mathcal{R}(\xi_2)}{(1-y_2)^2}\right)}{(\xi_2+\frac{y_2}{hr_2})^2}d\xi_2=0
\end{eqnarray*}
Finally,
\begin{equation}
  \upsilon_1(f)=-\frac{2y_2^2}{(y_1+y_2)^2}\log(1-y_1)-\frac{2y_1^2}{(y_1+y_2)^2}\log(1-y_2)
  -2\log\frac{y_1+y_2}{y_1+y_2-y_1y_2}.\nonumber
\end{equation}

redpajama_set_name: RedPajamaArXiv


JUAN MATA thinks Manchester United played a "very complete game" against Everton as they avoided suffering a poor start like they did versus Arsenal.
Before the international break, United were swept aside by the Gunners as Arsene Wenger's men scored three goals in 19 minutes.
But against the Toffees, it was United who made the brighter start with Morgan Schneiderlin scoring after 18 minutes and Ander Herrera grabbing a second goal four minutes later.
United eventually round out comfortable 3-0 winners on Saturday to stay in the title hunt, two points behind leaders Manchester City.
And in his weekly blog, Mata revealed United's determination not to repeat the Arsenal display.
He said: "Last Saturday we got a solid victory at Goodison Park, always a tough stadium to get the three points.
"I think the score reflects the good work the team did. We managed to take our chances despite Everton's strong start and pressing.
"The first goal was very important, and with the second one things were going in our favour. I'm really happy to see my friend Ander Herrera back on the pitch, contributing his class and his ability to score goals.
"Everton put again some pressure at the beginning of the second half, but the third goal definitely gave us the three points.
"I think we played a very complete game and, especially, we had a much better start here than against Arsenal in our last Premier League game, because we controlled the ball and played mainly on Everton's half of the pitch."


text:
Henry Baker
Company C, 155th Illinois Volunteer Infantry
Submitted by: John and Cindy McCachern
HENRY BAKER OBITUARY
MARSHALL HERALD
Henry Baker died Sunday at his home in this city, after a long illness. A short funeral service was held at his residence this morning and his body was taken to Auburn for burial.
Mr. Baker was born in Monroe County, Ohio, September 2, 1835, where in 1858, he was married to Miss Polly A. Pennell. In 1861, they came to Illinois and settled on a farm in Clark County where the exception of about 8 years, when they lived in Farmersburg, Indiana, they spent the remainder of their lives. About 25 years of this time, they lived on a farm near Auburn. Near the close of the civil war, Mr. Baker enlisted for service in Company C, 155th Illinois, and served to the close of the conflict. Then he returned to his farm. Three children were born: Mrs. Mary Morgan living north of Auburn, Sarah Jane Murphy who died in 1896 and Mrs. Clarissa Read whose home is in Arkansas. About 9 years ago, Mr. Baker moved to Marshall, and in May 1908, he suffered a stroke of paralysis. For a number of years, he had been able to get about and help himself, but during the past 2 or 3 years he failed gradually, and two weeks ago his condition became serious, which culminated in his death Sunday. He had attained the age of 81 years, 6 months and 2 days. He is the last of a family of 8 children. He leaves a widow, 23 grandchildren, 27 great grandchildren, besides some other relatives and friends. In 1896 he united with the Auburn Church and 9 years ago transferred his membership to Marshall Methodist Church, of which he was a member at the time of his death. He had the distinction of having helped to build a church in every community in which he lived, which fact was a joy to him in his declining years. He was a good friend, a devoted husband and father, always punctual in his business affairs and always tried to deal justly with all men.
Return to Scrapbook page

redpajama_set_name: RedPajamaCommonCrawl


By now, you've probably heard of Bitcoin. Perhaps, you've heard the term blockchain, as well. It's been around for a while, but has exploded in the last couple of years. Even so, the chances are you might not understand exactly what it is.
You wouldn't be alone. In 2017, a HSBC report ranked blockchain as the single least-understood technology in the world. As more and more organisations begin to use it, it is becoming increasingly important that those outside the tech sphere have some understanding of what it is and how it works.
Atlas City Global is a one of the growing number of firms offering blockchain technology to businesses. The company wants to make blockchain accessible and widespread, so is evangelical about its benefits to the average person, not least how it allows both businesses and individuals to control and own their data, and provides a gateway to myriad services being built by people around the world.
Blockchain is, at heart, a ledger – ledgers have been used as a way of storing information for thousands of years. Cruicially , it is a distributed ledger, meaning it is shared among many users at the same time. No single user can change the information on the ledger without the consensus of all the other users. This enables users to interact with each other without a central authority, it's a decentralised system.
One of the biggest functions of blockchain is the ability to store data in a transparent, unalterable and – most importantly – decentralised way. That data could be bank records and financial transactions, or any other information that can be found in a spreadsheet, from medical to marketing.
Billions have been invested in blockchain projects in the past two years alone and the global market for the technology is set to quadruple by 2021, to a valuation of $2.3billion. Amid the boom, Atlas City is using blockchain to help businesses automate processes, and developing the next generation of distributed-ledger technology.
Catalyst Network, Atlas City's core product, lets customers build decentralised applications at minimal cost, which the company hopes will open a global market to the technology, something that has been hampered by the sluggishness of current blockchain offerings, with low functionality and relatively high cost. "We designed a new architecture for blockchain that fulfils enterprise users' needs more completely than any existing infrastructure," says Mr Christopherson.
Like many others who have witnessed the development of blockchain, Atlas City sees a lot of similarities between today's landscape and that of the early internet, and they are at the forefront of this new wave of innovation.
With the huge amount of data generated today across the internet the ability to improve the way it is stored and processed is critical. Data stored is predicted to amount to 163 zettabytes globally by 2025 – that's 10 times what it is today – and the methods used to date need to increase in similarly scaled-up capability.
One particular reason is that the cost of using more traditional, centralised forms of data-storage – the "cloud", for example – will skyrocket. And there are other risks. As Mr Christopherson explains, centralised databases are vulnerable and almost certain to be hacked sooner or later.
Atlas City thinks that risks such as this will inevitably encourage more businesses to consider how blockchain could apply to them and their data. It has constructed the Catalyst Network to perform better as it scales, increasing decentralisation and becoming more resilient as the technology is adopted by more users.
Showcasing British business, The Great British Business campaign is giving the country's most exciting growing businesses a share of the limelight. Here, we met Darren Priestnall of Atlas City Global.


text:
Everything is clear about Orlen and Kubica in F1. We have an official announcement
Asteroid 2023 BU passed near Earth. It was discovered by an amateur astronomer
Representative of the US Congress: Thank you for the leadership of Poland
Sigmund Shortle February 16, 2022 2 min read
– I would like to emphasize the strength of the relationship between the United States and Poland, and at the same time I would like to thank you for that The leadership shown by Poland in the face of the crisis In Belarus and Ukraine – said the Chairman of the Armed Forces Committee of the US House of Representatives, Adam Smith, after meeting with Michai Gach (PiS), Chairman of the National Defense Committee (KON) and Cesary Grabarchik (KO), who is its members.
"This leadership enabled us, together with our NATO allies and European Union allies, to ensure the freedom and sovereignty of Ukraine, which is very important in the context of the entire region," the American politician added.
At the same time, he added, the United States wanted to support the independence and territorial integrity of all Central and Eastern European countries. "Speaking on behalf of the US Congress Armed Services Committee, this is our goal to advance the sovereignty and independence of the region," Adam Smith said.
Jak: Poland has an idea to ease tensions
The talks, which took place in the parliament building, dealt with "security issues in our European region, on the eastern borders of Poland, on the eastern borders of NATO and" NATO. On the Russian-Ukrainian border"Immediately after the meeting," Gat said at a press conference.
– We presented our views on security, the threats that were led by President Putin's policy – explained the KON chief, stressing that "it was an interesting meeting, President Smith showed extraordinary knowledge of the situation in our region."
See also Speculation about Putin's failure? Secret Kremlin propaganda instructions leaked | world News
He stressed that Poland has its own ideas to defuse the current and future tensions related to Russia's aggressive policy and presented to the American politician.
Blinken: There is a difference between what Russia says and what it doesRead also:
Head of PiS on the situation in Ukraine. It shouldn't end in a war.
Sigmund Shortle
"Coffee enthusiast. Troublemaker. Incurable introvert. Subtly charming twitter scholar. Award-winning social mediaholic. Internet buff."
Previous The conflict between Ukraine and Russia. Marcin Wirwal from Kiev: We are glad there are no shots
Next An alarming discovery on the Pripyat River. Only satellites noticed it
An exchange of fire in the Gaza Strip. United States: the need for de-escalation
January 27, 2023 Sigmund Shortle
Where is Lukashenko? The Belarusian dictator has secretly left the country
Russia. Media: Dmitry Pavuchka, a high-ranking manager linked to Roscosmos, died in the fire. Another mysterious death in the Russian elite


Discover New Trends with the latest Exhibitions and Events!
Mining, manufacturing and sales of marble blocks.
Manufacturing and sales of marble slabs and tiles.
The largest marble mining company, based on weight extracted.


shine beyond the limits of one's own soul.
lake …joyousness would be like a mighty river flowing out.
privileged to partake of that joy until we are filled with it.
and humiliation in a world which He had made."].
and joyful man of sorrows—let us try to understand this paradox.
above it and triumph over it.
the very source of it. He found His joy in the cross.
man up to the cross.
through righteousness, to the vilest. J.N.D.].
heart of humanity and lifting it helpfully away—this was the joy of the Lord.
in the great overmastering will which has sent these things.
Then again, what a joy the Lord found in His mission of salvation.
If He suffered the others were spared; there was joy in that.
to announce the kingdom as at hand….they had not made one convert!
image, and of sharing with them the eternal [intense happiness].
compensations and balancing of heaven.
How shall we have the joy of the Lord? [It] is a call to unselfish heights.
of God; the expectation of the things to come.
great emprise of salvation into the uttermost parts of the earth?
In Him [Jesus Christ] there is nothing lacking, nothing in excess.
the character of His people; to be beautiful within.
devoted than we really are.
where the whole world shall see it.


text:
Crash of a Boeing 707-328 in Hamburg
Date & Time: Jul 27, 1961 at 1620 LT
F-BHSA
Scheduled Revenue Flight
Paris – Hambourg – Anchorage – Tokyo
Captain / Total hours on type:
Copilot / Total flying hours:
Copilot / Total hours on type:
Aircraft flight hours:
Aircraft flight cycles:
During the takeoff roll on runway 23, at a speed of 120 knots, the captain noted the aircraft was turning slightly to the left and was unable to counteract. As a result, he decided to abandon the takeoff procedure and attempted an emergency braking maneuver. Unable to stop within the remaining distance, the airplane overran, lost its undercarriage and came to rest in a field located 140 meters the runway end, broken in three. Four crew members and six passengers were seriously injured while 31 other occupants were unhurt. The aircraft was destroyed.
The cause of the accident could not be determined with certainty. However, it was reported that a valve on the engine number two remained closed, causing the reverse system to deploy on that engine and the deviation of path during the takeoff procedure.
Final Report:
F-BHSA.pdf2.55 MB


Drs. Henry and Aurora Cube, Smith Street, are celebrating their 50th wedding anniversary. They were married March 8, 1959,in St. Agustin Cathedral, Manila, Philippines.
The former Aurora P. Villanueva is the daughter of Eustacio and Maxima Villanueva. She was born in Zamboanga City, Philippines. Henry was born in Tarlac City, Philippines, the son of Rogelio and Encarnacion Cube.
As newlyweds the couple lived in Quezon City, Philippines, but have lived 35 years at their present address. They are members of St. Peter's Parish Church.
Aurora was a family physician and had a private practice. She belonged to these organizations and work she has done: assistant professor in clinical pathology, faculty of pharmacy and medical technology, University of Sto. Tomas, Manila, Philippines; long-term member and Fellow of American Academy of Family Physicians; member of Pennsylvania Academy of Family Physicians; member of Somerset Medical Society of Family Physicians, medical director/Somerset Patriot Manor; senior member, Somerset Hospital medical staff; Emeritus status, medical staff, Somerset Hospital; volunteered teaching Spanish, grades one through six for two years, St. Peter's Catholic School.
Henry is a board certified neuro-surgeon in private practice. He belonged to these organizations: senior member, American Association of Neurological Surgeons; member Congress of Neurological Surgeons; member Pennsylvania Neurological Society.
Henry did the following work: instructor experimental psychology and neuro-anatomy, Philippines; Fellows in Neurosurgery, Institute of Neurosurgery, Madrid, Spain and Bridgeport Hospital, Bridgeport, Conn.; residents in neurosurgery, Boston hospitals, Boston, Mass., and Hartford Hospital, Hartford, Conn.; post-graduate, neuro-pathology, Yale University, New Haven, Conn.; diplomate in Neurosurgery USA; professor, neurosurgery, University of Sto. Tomas, Manila, Philippines; pioneered, Department of Neuro-Psychiatry, University of Sto. Tomas Manila, Philippines (shown in Internet); chief neurosurgeon, Metropolitan Hospital, Manila, Philippines; consultant in several Manila hospitals; consultant in neurosurgery, Conemaugh Memorial Hospital, Johnstown, and Somerset Community Hospital; member medical staff of the former Lee Hospital, Johnstown.
They are the parents of Ian V. Cube, D.C., Buena Park, Calif., Rowena V. Cube, Harleysville, and Jocelyn Cube Rhinehelder, Westlake, Ohio; and have two grandchildren.
Henry enjoys fishing, gardening, carpentry and he used to play golf. Aurora enjoys golf, gardening, piano, reading and cooking.
They will celebrate with a trip to Paris, France, with daughter, Rowena and a family reunion in Somerset.


Another hot and sunny day. Just had the Caribbean their lunch, 4.ourses and I'm stuffed, ships orchestra with forties fifties nigh then light dinner. Last sea day tomorrow.
Considering we were in The Bay of Biscay good weather if breezy and sat in Crows Nest in the morning but warm enough to sit out in the aftenoon. Packed now apart from what I am wearing and the first case is out, second one after dinner. Comedian Jeff Stevenson again tonight, and then we get our hour back. Early disembarkation at 9.15 tomorrow. Been a bit tired today as last night was the roughest of the cruise and there seemed to be a large object rolling about above my cabin.


text:
De DKW Junior was een compacte auto van de Duitse fabrikant Auto Union. De auto, met een watergekoelde driecilinder tweetaktmotor en voorwielaandrijving werd geproduceerd in de nieuwe DKW-fabriek in Ingolstadt, waarvoor de eerste steen werd gelegd in juli 1958. Ongeveer 240.000 DKW's Junior en Junior de Luxe werden verkocht in de bijna vierjarige bouwperiode tot 1963.

Geschiedenis
Op de Internationale Automobilausstellung in 1957 toonde DKW een prototype met een 660 cm³ tweecilindermotor onder de naam DKW 600. Behalve de motor leek hij op de latere DKW Junior. De serieproductie van de DKW Junior begon in augustus 1959.

Rond 1960 was de DKW Junior een van de weinige automodellen met een tweetaktmotor. Zijn driecilinder lijnmotor met 750 cm³ en 34 pk (25 kW) was vóór de vooras gemonteerd. De radiateur en de volledig gesynchroniseerde vierversnellingsbak stonden erachter. De voor tweetaktauto's kenmerkende vrijloop was niet beschikbaar op de DKW Junior. De krukas was viermaal gelagerd, de cilinders waren van gietijzer en de motor had thermosifonkoeling met een thermostaat. Elke cilinder had een afzonderlijk ontstekingssysteem, dus de motor had drie contactpunten en drie bobines aan de dynamo die boven de cilinders was geplaatst. De aandrijfassen van de auto hadden homokinetische koppelingen aan de buitenkant van de wielen en kruiskoppelingen aan de binnenkant aan de versnellingsbak.

In vergelijking met andere tweetaktauto's uit die tijd, zoals Wartburg of  Trabant, had de motor erg stille en trillingsarme loop. Een andere tekortkoming van het tweetaktprincipe kwam echter des te duidelijker naar voren bij de DKW Junior: het hoge brandstofverbruik voor een auto van deze klasse, dat bij normaal rijgedrag op 8,5 tot 11,5 liter per 100 km bedroeg.

De carrosserie stond op een chassis. Sommige onderdelen, zoals de spatborden, waren geschroefd en daardoor gemakkelijk te vervangen. De voorwielophanging bestond uit dubbele draagarmen met verstelbare torsieveren onderaan en telescopische schokdempers. Aan de achterzijde was het een aan twee draagarmen en een schuine panhardstang geleide starre as met een dwarsliggende torsieveer en telescopische schokdempers. Om het onafgeveerde gewicht laag te houden, zaten de voorste remtrommels aan de binnenkant van de versnellingsbak. Ribben of kleine bladen rond de zogenaamde turbotrommels moesten de koeling verbeteren. Net als de grotere DKW-modellen had de Junior ook tandheugelbesturing. Het chassis was onderhoudsintensief en moest om de 7.500 km op 40 plaatsen worden gesmeerd.

Met de carrosserie van de Junior in trapeziumvorm verliet Auto Union de afgeronde stroomlijnvorm van de DKW-personenauto's, die nog baseerden op een vooroorlogse constructie. De Junior had duidelijke staartvinnen en zijpanelen die in een bijna rechte lijn tot boven de koplampen waren doorgetrokken. Het model werd alleen geleverd als tweedeurs, bood relatief veel binnenruimte en, niet in de laatste plaats dankzij de smalle raamstijlen, een zeer goed overzicht. Bovendien was er een grote kofferbak waarin aan de rechterkant het reservewiel was geplaatst en niet onder de bagage, zoals bij de oudere modellen. De brandstoftank bevond zich onder de kofferbak achter de achteras. Nadelig was de hoge luchtweerstand.

De uitrusting van de Junior omvatte een stuur met een verzonken naaf, stuurversnelling voor de volledig gesynchroniseerde vierversnellingsbak, een schokabsorberend dashboard en lichtsignaal. In de advertenties werd ook de hoedenplank "van bijna een halve vierkante meter"getoond. Een stalen schuifdak en de Saxomat-automatische koppeling waren tegen meerprijs leverbaar. Nadelig waren de voorstoelen waarvan de rugleuningen niet afzonderlijk konden worden versteld.

Junior de Luxe
Op de IAA 1961 presenteerde Auto Union de herziene DKW Junior de Luxe, die met name verschilde van het basismodel door de duidelijk naar voren geplaatste koplampen. Hij was enerzijds bedoeld als opvolger en anderzijds als een luxere variant van de oorspronkelijke Junior, onder andere met uitzetbare tochtramen in de deuren, een motorkap die van binnenuit kon worden ontgrendeld en een chromen sierlijst aan de zijkant.

De cilinderinhoud van de motor was vergroot tot 796 cm³ (boring × slag = 70,5 × 68 mm), het vermogen bleef ongewijzigd op 34 pk. Verder kreeg hij zoals ook de Auto Union 1000 een mengsmeerautomaat die in de DKW Junior vanaf dat moment tegen meerprijs optioneel verkrijgbaar was. De olie hoefde niet langer in de juiste verhouding aan de benzine te worden toegevoegd, zoals bij tweetaktmotoren gebruikelijk, maar werd vanuit een aparte olietank naar de cilinders geleid, afhankelijk van de belasting van de motor. Hierdoor werd minder olie verbrand en de rook- en geurhinder verminderd.

De Junior de Luxe werd slechts twee jaar gebouwd, in 1963 werd hij vervangen door de F11/F12, een doorontwikkeling op basis van de Junior.

Junior

redpajama_set_name: RedPajamaWikipedia


5 ways you can use…Flip Cameras!
Create a project proposing students to create their own public service announcements or commercials related to class material. This could work for any subject, from history to health!
Have students practice their job-seeking skills by recording them interviewing for a fictitious job. They will be able to play back the video and engage in both peer and self-assessments.
When students are delivering speeches in class, use the flip cam to record their speeches so they can view themselves later on and get instant feedback.
Have students create a mock news segment related to class material.
For an end of the semester project, have students create video diaries, reflecting all they have learned in the course throughout the semester.
Additionally, flip cams are a great way to record demonstrations or experiments (think chemistry or art classes) for students who may have missed class or want to review for an exam or lab practical. They are perfect to record segments of lectures where they need to "see" something to really understand it. Flip cams have the potential to make any project into an exciting adventure. For more ways to use flip cams in your class or how to use these devices, feel free to contact us!


text:
<?php

namespace Intervention\Httpauth;

use Illuminate\Support\ServiceProvider;

class HttpauthServiceProvider extends ServiceProvider
{
    /**
     * Indicates if loading of the provider is deferred.
     *
     * @var bool
     */
    protected $defer = false;

    /**
     * Actual provider
     *
     * @var \Illuminate\Support\ServiceProvider
     */
    protected $provider;

    /**
     * Create a new service provider instance.
     *
     * @param  \Illuminate\Contracts\Foundation\Application  $app
     * @return void
     */
    public function __construct($app)
    {
        parent::__construct($app);

        $this->provider = $this->getProvider();
    }

     /**
     * Return ServiceProvider according to Laravel version
     *
     * @return \Intervention\Httpauth\Provider\ProviderInterface
     */
    private function getProvider()
    {
        $app = $this->app;
        $version = intval($app::VERSION);
        $provider = sprintf(
            '\Intervention\Httpauth\HttpauthServiceProviderLaravel%d', $version
        );

        return new $provider($app);
    }

    /**
     * Bootstrap the application events.
     *
     * @return void
     */
    public function boot()
    {
        return $this->provider->boot();
    }

    /**
     * Register the service provider.
     *
     * @return void
     */
    public function register()
    {
        return $this->provider->register();
    }

    /**
     * Get the services provided by the provider.
     *
     * @return array
     */
    public function provides()
    {
        return array('httpauth');
    }

}

redpajama_set_name: RedPajamaGithub


text:
Ailing Big B back on track with IPL for the first time since his surgery
By Sonali Joshi
Amitabh Bachchan will be back in action for the first time since his abdominal surgery last month when he participates in the star-studded opening ceremony of Indian Premier League season 5, set to take place in Chennai on April 3.
The superstar will recite a special poem that writer-lyricist Prasoon Joshi has penned for the event.
Big B's return apart, a highlight of the gala will be the fact that Salman Khan and Priyanka Chopra - known for their very public animosity - will both perform live at the do, besides Prabhu Deva and the Colonial Cousins.
Amitabh Bachchan will recite a poem at the opening ceremony of IPL5
A couple of international acts are also being lined up at the event, to be held at Chennai's YMCA College of Physical Education Nandanam, although no name has been confirmed yet.
Bachchan, who was recovering from stomach surgery, had been advised to take complete rest by his doctors for a few weeks.
The superstar has not taken up any work since then but will be medically fit by April and the IPL opening ceremony is a good opportunity to resume work.
'Asia Cup over soon and then IPL... opening extravaganza in Chennai... and yours truly a small part of it... nervous as hell !! They do the opening show of IPL a day before they play unlike other years... its a separate performance at prestigious venue... meeting all the Captains and teams will be such an honour... seen all of them on TV, now personally. April 3rd is the day for the opening... preparing for it already... getting back into the grind (sic),' tweeted an excited Big B about returning to work with IPL. Priyanka on her part seems quite thrilled at the idea of performing at the IPL opener.
'So what act do u guys suggest I do for the IPL opening ceremony.. Shud be so fun!! (sic)' she tweeted.
Salman Khan and Priyanka Chopra will also perform at the event
Salman, who is busy shooting for Dabangg 2 in Mumbai right now, is expected to fly to Chennai next month with his family and friends, all of whom are known to be huge cricket fans.
In fact, the news was rife last year that Salman wanted to buy an IPL team.
Things, however, did not work out and Salman launched the Celebrity Cricket League.
The Bodyguard star performing at this year's IPL opening ceremony has raised fresh speculation that he might stake part ownership in one of the teams.
Prabhu Deva, who has been more caught up with filmmaking, directing Wanted and the forthcoming Rowdy Rathore, will be seen performing after a long time.
He is going to perform to his superhit Tamil, Telugu and Hindi numbers. IPL 5 starts on April 4, a day after the opening ceremony, and the final of the tournament this year is scheduled to take place on May 27.
Ailing Amitabh Bachchan back on track with IPL for the first time since his surgery


You can make bone broth using whatever bones you choose. For this demonstration, Dr. Becker used an organic chicken. As the bones simmer on the stove they release all their minerals and marrow into the water, and the broth that results is a healthy source of extra nourishment for pets.
When making broth you should use clean, filtered water to which you add a bit of vinegar (acetic acid) to help leech the minerals from the bones into the stockpot water. For this demonstration, Dr. Becker uses Bragg raw apple cider vinegar, but you can use whatever vinegar is in your kitchen.
Dr. Becker demonstrates two rounds of cooking bone broth. Round 1, which takes about 2.5 to 4.5 hours, removes all the skin and meat from the bones. Round 2 cooks all the bones and skin not broken down in round 1, and takes about 24 hours of simmering. Dr. Becker runs the second round of broth through a strainer to separate any remaining pieces of bone, which she discards. Never feed cooked bones to your pets.
Two important things to remember about bone broth: 1) cooked bones are choking and GI hazards; and 2) bone broth alone is not a balanced diet and should be used only as an extra source of nourishment.
Good morning! I'm filming this video from my kitchen this morning as I make bone broth for my dog, Ada.
I thought it would be a good idea to show you how to make bone broth, in case you ever have a situation in your own life in which bone broth could provide excellent nourishment -- perhaps to help a pet recover from illness, or to provide nutrition for an older animal with a declining appetite.
Bone broth is a source of nourishment that has been around for hundreds of years. Some of you may already be making bone broth for your pets. There are lots of different ways to make it, and there really isn't a wrong way. I'm going to show you the way I make bone broth, but you can also find variations on the Internet.
You've probably heard that chicken noodle soup is kind of a "soul food." When we're sick, we want to eat foods that comfort us. And there's research now that proves the science behind why certain types of foods are nourishing to people and comforting to pets.
When we cook a whole chicken and the meat falls off the bone, of course, that delicate, juicy meat is very good for anorexic pets. I use a lot of bone broth for my hospice patients. When animals are dying and no longer interested in eating or chewing food, often they'll still lap broth.
This morning I'm making bone broth because my healthy eight-year-old pit bull, Ada, is allowed to wander around our property, which is over 40 acres. She doesn't go far, but she does get into grass, poop, and dirt. Ada's been throwing up the last couple of days -- she barfed five or six times yesterday. She's not feeling well. We did confirm there's no blockage in her digestive tract, which means she doesn't need to go to surgery, but her GI tract is inflamed. She's also had some diarrhea. So we fasted her yesterday, and now I'm going to make bone broth and offer it to her as her first food after fasting.
Bone broth is a great first food for animals recovering from GI irritation or illness. And if you have a pet that's finicky, you can also use it as a topper on food. If you have a pet that doesn't want to eat anything, feeding bone broth provides very concentrated nourishment and helps get the GI tract functioning again.
I'm going to use a whole organic chicken today. You can also use beef thighbones (called "soup bones" at the store). You can use whatever bones you wish. But when you cook the bones and the meat falls off, that's a great first food for a sick pet, and of course, most pets love it. As you continue to simmer the bones, they release all their minerals and marrow into the water, and that broth becomes a really healthy, nourishing first food or end food, depending on when you offer it to your pet.
It's important to understand that bone broth alone is not a balanced diet. Even if your pets absolutely love it, they need more than bone broth to be healthy. But it's a great way for you to offer extra nourishment in a really palatable form.
Always start with healthy water for cooking. We have a whole-house water filtration system. Our water comes from a well, but we have an Iron Curtain water filtration system that helps filter out impurities.
Fill up your pot with water – a really big stockpot works well. How much water you'll need depends on how much bone broth you want to make. I prefer to make bone broth fresh and use it right away, but you can also make it and freeze it. I prefer to make it fresh, so I'm going to make this today in a smaller stock pot. I've added my water.
As I mentioned earlier, you can use whatever bones you want. I bought a whole organic chicken from Whole Foods and will be using it for today's bone broth.
I do recommend feeding organic meats to your pets whenever possible, because the pesticide residues are not going to be passed up the food chain. Also, happy healthy food animals produce healthier meats for our pets.
I'm going to rinse off my organic chicken, put it into the water, turn on the heat, and do a nice, low simmer for a couple of hours. Because Ada is hungry and hasn't eaten in 48 hours, once the meat falls off the bones I'm going to offer her the meat and skin. Then I'll continue cooking the carcass for another 24 hours on a very low heat.
One important ingredient to add to your stockpot is vinegar – you need to add a little bit of vinegar to the water. Acetic acid (vinegar) helps leech the minerals from the bones into the stockpot water, which is ultimately what you'll feed to your pet. Important note: you should never feed your pet cooked bones. After boiling the bones down to the point where they disintegrate and dissolve, you should throw out any remaining bone fragments.
Now I'm going to cover my pot, turn on the heat and cook the chicken until the meat falls off the bone, which will be in about two and a half hours. I'll stir it occasionally while it cooks.
Once the bones are separated from the meat, I'll remove the meat and some of the broth. Then I'll add more water and continue cooking to make a second round of broth.
It's been about two and a half hours, and this is what my broth looks like. You might be able to hear my dogs getting excited about the broth in the background! The meat isn't all the way off the bone yet, so I'm going to let the chicken simmer for another couple of hours since I'm cooking at a bit lower temperature than I usually do. But the broth is looking pretty good. You can see that the fat is rising to the top, and the skin is coming off the chicken. Some of the meat has fallen off the bone. I'll check it again in another couple of hours.
Now the chicken has been cooking for about four and a half hours. All the meat has fallen off the bones. I have a second pot here to collect the bones and skin. Once all the bones have been removed, I'll feed the meat and broth to Ada for the first round and cook down the other pot of bones for the second round of broth.
Now I've removed all the extra bones and skin not broken down through this first round. I'll serve up a nice bowl of broth and juice and offer it to Ada as soon as it cools.
For my second round of bones, I'll put everything in the pot that didn't get completely broken down. Then I'll add more pure water, another splash of vinegar, and repeat the process.
For these bones to have the minerals leeched out, I'll probably need to simmer them for the next 24 hours. So we'll see what this looks like tomorrow morning.
And here's the second round of bone broth. I took the bones and skin left after the first round, added more water and vinegar and cooked it on low heat for the last 24 hours. There are still some bones in the broth, but they're very soft.
I prefer to run this second round of broth through a strainer, drain off the fluid, let it cool, and feed it to my pets. Some people like to put their second round of broth in a blender to grind down the remaining fine, soft bones, because cooked bones can be a choking hazard. But I don't feed cooked bones to my pets at all. So what I'm going to do is just drain off the liquid, let it cool, and feed it to them.
All my pets love bone broth. My kitties love it. My dogs love it. It's a great food to offer not only as a natural mineral supplement, but also as a bland and nutritious diet for pets who aren't feeling good, who don't have much of an appetite, or who are finicky.
And now you've seen how I make bone broth!


Gene is unit of heredity in all living organisms. A gene is part of a DNA molecule within the nucleus of all cells. Each gene codes for a particular protein. Thus a gene is a unit of the inheritable characteristics of the organism. Humans have tens of thousands of different genes; these determine the phenotype of the individual.
According to new study our level of happiness throughout life is strongly influenced by the genes with which we were born.
An Edinburgh University study of identical and non-identical twins suggests genes may control half the personality traits keeping us happy. The other half is linked to lifestyle, career and relationships.
So if you are not happy, just blame to your genes ;). On a related note positive thinking may improve overall life.
My forecast is that around 2050, the state of Massachusetts will be the first jurisdiction to legalize marriages with robots," artificial intelligence researcher David Levy at the University of Maastricht in the Netherlands told LiveScience. Levy recently completed his Ph.D. work on the subject of human-robot relationships, covering many of the privileges and practices that generally come with marriage as well as outside of it.
I'm just wondering about emotional communication in such marriage. I strongly believe one can build great relationships with emotions. Let us wait and see.
From movie trailers it looks like a comedy that looks differently at relationships. After all it cast include Naseeruddin Shah, Konkona Sen, Ranvir Shoery and others.
Sunil and Malti are a middleclass family couple who has married for ten years. They have a child. But they are getting boring with each other after 10 years (how the hell they spend 10 years together ;) ) of marriage. They are not happy at all. But one day something happens, Sunil suggests unacceptable idea to Malti? which result into funny stuff.