text:
\section{\bf Introduction-Results.}\label{s1}
Let   $X=\{X_j, j\in {\mathbb Z}\}$ be a centered Gaussian stationary  sequence, and let ${\gamma}(n) = {\mathbb E \,} X_0X_n$, $n \in {\mathbb Z}$.   We assume that  $X$ is strongly mixing or   equivalently, that $\lim_{n\to \infty}  {\gamma}(n)  =0$. When $ {\gamma}(n)$ tends sufficiently quickly to $0$, more independence is naturally gained in the structure of $X$.  This can be quantified under the form of  a decoupling inequality. For instance, if
\begin{eqnarray}\label{hyp.dec} p(X) \ =\ \sum_{n\ge 0} {|{\gamma}(n) |\over {\gamma}(0)}<\infty ,
\end{eqnarray}
then for any finite collection $\{f_j, j\in J\}$ of  complex-valued
Borel-measurable functions,
\begin{eqnarray}\label{dec}
\Big|{\mathbb E \,}  \prod_{j\in J}f_j\big(X_j\big) \Big|\ \le\  \prod_{j\in J}\big\|f_j\big(X_0\big) \big \|_{p(X)}.\end{eqnarray}
  This  remarkable  inequality, which
  so nicely condenses the independence properties of these Gaussian sequences,  is Theorem 3 ($d=1$) in Klein, Landau and Shucker \cite{KLS}.
    \vskip 2 pt
  Clearly $|{\mathbb E \,}  \prod_{j\in J}f_j(X_j)|$ measures the degree of independence between the  random variables $f_j(X_j)$.
 An immediate consequence of \eqref{dec} and of well-known Kathri-Sid\'ak's inequality, is that under assumption \eqref{hyp.dec}, we have  the following sharp two-sided estimate,
\begin{equation}\label{ks} \prod_{j\in J} {\mathbb P} \{  |X_j|\le x \}\le {\mathbb P}\big\{ \sup_{j\in J}  |X_j|\le x\big\}\le \prod_{j\in J} {\mathbb P} \{  |X_j|\le x \}^{1/p(X)},
\end{equation}
where $J$ is any finite index  and  $x$ any non-negative real.  \vskip 3 pt
     However, one is often faced with probabilistic questions where  $p(X)=\infty$,
  or simply,  the process $X$ is not stationary in the sense required in \cite{KLS}.   At our knowledge, no extension of \eqref{dec} beyond condition \eqref{hyp.dec} exists in the literature, and it is naturally interesting to search what form could take a decoupling inequality  when \eqref{hyp.dec} fails to be satisfied.
   \vskip 2 pt
   This is the question we address and study
  in this work, which is also  somehow developing
   the recent paper \cite{W}.
 We

  clarify at this stage that our goal is to obtain results   valid for a broad range of Gaussian processes, and thus  not  (possibly)  quite sharp  estimates concerning specific cases, which is another problematic. Our aim is  also to link the question considered with the general theory of Toeplitz forms, and  draw the attention of the reader to the
 interest of this  connection.
This is in that sense continuing the study made in  section 5 of \cite{W}.
  \vskip 2 pt
The role of the stationarity assumption of $X$ in  \cite{KLS}    is   crucial. The proof of \eqref{dec}  much relies  on an analytic inequality due to Brascamp and Lieb, which is of  relevance in the present work.
  \vskip 2 pt

A first  natural question can be stated as follows.
  What form can take the decoupling inequality \eqref{dec} for an arbitrary Gaussian vector? As the law of a Gaussian vector, or more generally of a Gaussian process is completely characterized by its covariance function,   one can make  the question more consistent by asking which characteristics of the covariance matrix $\{{\mathbb E \,} X_iX_j\}_{i,j=1}^n$ of $X$ should be involved (and are to be evaluated):

  a particular function of its eigenvalues, or simply its determinant? It turns out that only the determinant suffices.  More precisely, we  prove  a general decoupling inequality,

free of stationarity assumption.   \vskip 2 pt
Before stating it, we first  extend the notion of decoupling coefficient introduced in  \cite{KLS} to arbitrary Gaussian vectors.
 \begin{definition} Let $ X=\{X_i, 1\le i\le n\}$ be a centered Gaussian vector with non-degenerated components. The decoupling coefficient $p(X)$ of $X$ is defined by
 \begin{eqnarray*}
 p(X)&=& \max_{i=1}^n\sum_{1\le j\le n} \frac{|{\mathbb E \,} X_iX_j|}{{\mathbb E \,} X_i^2}
\,.
\end{eqnarray*}
\end{definition}

This is a natural characteristic of $X$. When $X$ is stationary,
\begin{eqnarray*}
p(X)&=&\max_{i=1}^n\sum_{1\le j\le n} \frac{|{\gamma}(i-j)|}{{\gamma}(0)},\end{eqnarray*}
and so
$$\sum_{1\le h\le n-1} \frac{|{\gamma}(h)|}{{\gamma}(0)}\le p(X)\le 2\sum_{1\le h\le n-1} \frac{|{\gamma}(h)|}{{\gamma}(0)}.$$ Further $p(X)=1$ if and only if $X$ has independent components. Some classes of examples with $p(X)\ll n$ or $p(X)\asymp n$ are given in section \ref{s4}.
\vskip 3 pt

Our first main result states as follows.\vskip 2 pt
\begin{theorem}\label{dec.ineq} Let $ X=\{X_i, 1\le i\le n\}$ be a centered Gaussian vector such that ${\mathbb E \,} X_i^2={\sigma}_i^2>0$ for each  $1\le i\le n$, and with positive definite  covariance matrix $C$.
 Let $p$ be such that
\begin{eqnarray}\label{cond}
p\, \ge \, 2\, p(X).
\end{eqnarray}
 Then for any complex-valued measurable  functions $f_1, \ldots, f_n$ such that $f_i\in L^p({\mathbb R})$, for all $1\le i\le n$, the following inequality holds true,
\begin{eqnarray}\label{ineq}
\bigg|\,{\mathbb E \,}\Big( \prod_{i=1}^n f_i(X_i)\Big)\,\bigg|
&\le &\frac{ 2^{\frac{n}{2}(1-\frac{1}{p})}\big(\prod_{i=1}^n{\sigma}_i\big)^{\frac{1}{p}}}
 {\det(C)^{\frac{1}{2p}}}\ \prod_{i=1}^n\Big( {\mathbb E \,}   f_i(X_i)^p\Big)^\frac{1}{p}.
\end{eqnarray}
 \end{theorem}

\vskip 6 pt
From Theorem \ref{dec.ineq} and Kathri-Sid\'ak's inequality we also get,
 \begin{corollary}\label{co1}Let $ X=\{X_i, 1\le i\le n\}$ be a centered Gaussian vector such that ${\mathbb E \,} X_i^2={\sigma}_i>0$, $1\le i\le n$, and  with positive definite  covariance matrix $C$. Assume that assumption \eqref{cond} is fulfilled for some $p\ge 2$. Then  for any ${\varepsilon}_i >0$, $i=1,\ldots, n$,
 \begin{eqnarray*}
\prod_{i=1}^n{\mathbb P}\big\{  |X_i|\le {\varepsilon}_i\big\} \ \le \ {\mathbb P}\Big\{ \sup_{i=1}^n \frac{|X_i|}{{\varepsilon}_i}\le 1\Big\}
&\le &
\frac{2^{\frac{n}{2}}}{\det(C)^{\frac{1}{2p}} }  \
\prod_{i=1}^n\Big( \frac{\,{\sigma}_i}{\sqrt 2} \ {\mathbb P}\big\{  |X_i|\le {\varepsilon}_i\big\}\Big)^\frac{1}{p}.
\end{eqnarray*}
\end{corollary}

\vskip 6 pt For estimating $\det(C)$, we  place ourselves in the setting of Toeplitz matrices theory where this important question has been and is still  much investigated. A salient aspect of  this theory is that $\det(C)$ can be computed, sometimes  with high degree of accuracy. We refer to the nice book of  Grenander and
Szeg\"o \cite{GS}, we also refer to \cite{W} for a general presentation of the methods used, except for the Laplace transform method,
essentially  in the setting of stationary Gaussian processes.
Let $f(t)= \sum_{-\infty}^\infty d_n e^{int}$ be a function on the unit circle ${\mathbb T}$.
Let $T_{n-1}(f)$ be the Toeplitz matrix defined by $T_{n-1}(f)= \{ d_{j-i}\}_{i,j=0}^{n-1}$ and let $D_{n-1}(f)=\det(T_{n-1}(f))$.

\vskip 2 pt
This   corresponds to the case when   $X$ has a
spectral density function $f(t)$, summable over $[-\pi,\pi]$, and is thus of relevance in our setting.
 Indeed, as
\begin{eqnarray}\label{te} d_{n}=\frac1{2\pi}\int_{- \pi}^{ \pi} e^{-in t} f(t)   {\rm d} t, {\qquad}  \quad  n\in {\mathbb Z} ,
\end{eqnarray}
$T_{n-1}(f)$ is just the $n$-th finite section
of the infinite Toeplitz matrix given by the covariance matrix of the process $X$.  Further as $f\in L^1([-\pi, \pi])$,  by the Riemann-Lebesgue lemma, we have
 $ \lim_{n\to \infty}  d_n =0$.
 \vskip 3 pt

  In the considerable literature  on Toeplitz operators and determinants, $f$ is usually called a symbol or a generating function (generating $(T_{n-1}(f))_n$) and $f$ needs not being a density function. Toeplitz determinants with rational symbols  occur for instance in statistical mechanics and quantum mechanics, see \cite{BF}. They can be calculated using a formula obtained by Day \cite{D}.
\vskip 2 pt
For Toeplitz matrices generated by a density function, $D_{n-1}(f)$ can also be expressed as an integral over the unitary group $U(n)$, by means of the Heine-Szeg\"o identity,
\begin{equation}\label{hs}D_{n-1}(f) \, =\, \int_{U(n)} \Phi_{n,f}(g) {\rm d} g.
\end{equation}
Here
the integration path is taken with respect to the normalized Haar measure on $U(n)$, and $\Phi_{n,f} (g)$ is defined by $\Phi_{n,f} (g)= f(t_1)\ldots f(t_n)$, where $t_1,\ldots,t_n$ are  the eigenvalues of $g$.
This  identity is the starting point of  the proof of a nice form of  the strong Szeg\"o limit theorem established in   \cite{BD} by Bump and Diaconis.

 \vskip 3 pt Using their result we also prove

\begin{theorem}\label{p2} Let   $X=\{X_j, j\in {\mathbb Z}\}$ be a centered Gaussian stationary  sequence with unit variance and
spectral density function
$f(t)$.   Let  $\log f(t) = \sum_{k\in {\mathbb Z}} c_ke^{ikt}$ where the $c_k$ satisfy the following conditions
\begin{equation}\label{c1} \sum_{k\in {\mathbb Z}} |c_k|<\infty,
\end{equation}
\begin{equation}\label{c2} \sum_{k\in {\mathbb Z}} |k|\,|c_k|^2<\infty.
\end{equation}

  Then there exist  reals ${\delta}_n\downarrow 0$, such that for any integer $n\ge 2$,
 any complex-valued measurable  functions $f_1, \ldots, f_n$ with  $f_i\in L^p({\mathbb R})$, for all $1\le i\le n$,
 where
\begin{eqnarray}\label{cond1a}
p\, \ge \, 2\,p(X),
\end{eqnarray}
the following inequality holds true,
\begin{eqnarray*}
\bigg|{\mathbb E \,}\Big( \prod_{i=1}^n f_i(X_i)\Big)\bigg|
&\le &
 \frac{(1+{\delta}_n)\, 2^{\frac{n}{2}} }{(2b(f)G(f))^{\frac{n }{2p}}}\  \prod_{i=1}^n\Big( {\mathbb E \,} |  f_i(X_i)|^p\Big)^\frac{1}{p},
\end{eqnarray*}
where $b(f)=\exp\big\{   \sum_{k=1}^\infty kc_kc_{-k}\big\}$
 and $G(f)$ is the geometric mean of $f$, namely
\begin{equation}\label{G(f)}
G(f) =\exp\big\{\frac{1}{2\pi}\int_{- \pi}^{ \pi}
\log f(t) {\rm d} t \big\}.
\end{equation}

 \end{theorem}
\vskip 2 pt

\begin{corollary}\label{co2}
Let   $X=\{X_j, j\in {\mathbb Z}\}$ be a centered Gaussian stationary  sequence with unit variance and
spectral density function
$f(t)$ satisfying  conditions \eqref{c1} and \eqref{c2}.
 \vskip 2 pt  Then there exist  reals ${\delta}_n\downarrow 0$, such that for any integer $n\ge 2$,
 any  $p$ satisfying \eqref{cond1a},
we have for any ${\varepsilon}_i >0$, $i=1,\ldots, n$,
 \begin{eqnarray*}
{\mathbb P}\Big\{ \sup_{i=1}^n |X_i|\le {\varepsilon}_i\Big\}
&\le &
 \frac{(1+{\delta}_n)\, 2^{\frac{n}{2}} }{(2b(f)G(f))^{\frac{n }{2p}}}\ \prod_{i=1}^n\big( {\mathbb P}\big\{  |X_0|\le {\varepsilon}_i\big\}\big)^\frac{1}{p}
.
\end{eqnarray*}
\end{corollary}

\section{\bf Proof of Theorem \ref{dec.ineq}.}\label{s2}
We first state the proposition below which follows from Theorem 6 in Brascamp and Lieb \cite{BL}. We also refer to \cite{KLS}. It should be indicated here that all that is required for the application of this Theorem, is that the   matrix be positive definite. This one  is written  in terms of its eigenvectors and eigenvalues, and the eigenvectors are  $a^j$, $j=k+1,\ldots, k+m$, which have nothing to do with the vectors
 $a^j$, $j=1,2,\ldots, k$ of their Theorem 1. This point was clarified to the author by Abel Klein \cite{K}. Introduce some notation. Let    $I$  be the $n\times n$ identity matrix and let $\underline b=(b_1, \ldots ,b_n)\in {\mathbb R}^n$. Then $  I(\underline b)$ will denote throughout the diagonal matrix whose values on the diagonal are the corresponding values of $\underline b$. Also, when $b_i\neq 0$ for each $i=1, \ldots, n$, we will  use the notation $\underline b^{-1}=(b_1^{-1}, \ldots ,b_n^{-1})$.
\begin{proposition} \label{p1}Let $1\le p<\infty$. Let $B$ be a positive definite $n\times n$ matrix. Then for any measurable  functions $g_1, \ldots, g_n$ such that $g_i\ge 0$ and $g_i\in L^p({\mathbb R})$, $1\le i\le n$, the following inequality holds true,
\begin{eqnarray}\label{BL.ineq}
\int_{{\mathbb R}^n}\Big( \prod_{i=1}^n g_i(x_i)\Big) \exp\Big\{ -\frac12\langle \underline x, B\underline x\rangle\Big\}{\rm d} \underline x&\le & E_B\ \prod_{i=1}^n\Big( \int_{{\mathbb R}}    g_i(x)^p{\rm d} x\Big)^\frac{1}{p},
\end{eqnarray}
where
\begin{eqnarray}\label{EB}
E_B\,=\,(2\pi)^{\frac{n}{2}(1-1/p)}p^{\frac{n}{2p}} \, \sup_{b_i>0\atop i=1,\ldots,n}\frac{\prod_{i=1}^nb_i^\frac{1}{2p}}{ \det (B+I(\underline b))^\frac12}.
\end{eqnarray}
\end{proposition}
 \vskip 2 pt
\begin{remark}\label{REB}
 The constant $E_B$  is defined in Theorem 6 by
\begin{eqnarray*}
E_B&=&\sup_{b_i>0\atop i=1,\ldots,n}\frac{\int_{{\mathbb R}^n}\big( \prod_{i=1}^n \exp\{-\frac12 b_ix_i^2\}\big) \exp\{ -\frac12\langle \underline x, B\underline x\rangle\}{\rm d} \underline x}{\prod_{i=1}^n\big( \int_{{\mathbb R}}   \exp\{-\frac{p}{2} b_ix^2\}{\rm d} x\big)^\frac{1}{p}},
\end{eqnarray*}
namely  inequality \eqref{BL.ineq}  is maximal when the $g_i$'s are Gaussian.
\vskip 2 pt
But it is elementary
that (see also \cite{KLS}, p.\,705, after (4))

 \begin{eqnarray*}
\frac{\int_{{\mathbb R}^n}\big( \prod_{i=1}^n \exp\{-\frac12 b_ix_i^2\}\big) \exp\{ -\frac12\langle \underline x, B\underline x\rangle\}{\rm d} \underline x}{\prod_{i=1}^n\big( \int_{{\mathbb R}}   \exp\{-\frac{p}{2} b_ix^2\}{\rm d} x\big)^\frac{1}{p}}
 &=& \frac{\int_{{\mathbb R}^n} \exp\{ -\frac12\langle \underline x, (B+I(\underline b))\underline x\rangle\}{\rm d} \underline x}{\prod_{i=1}^n\big( \frac{2\pi}{pb_i}  \big)^\frac{1}{2p}}
\cr &=&(2\pi)^{\frac{n}{2}(1-\frac{1}{p})}p^{\frac{n}{2p}} \,\frac{\prod_{i=1}^nb_i^\frac{1}{2p}}{ \det (B+I(\underline b))^\frac12}.
\end{eqnarray*}
So that\begin{eqnarray*}
E_B\,=\,(2\pi)^{\frac{n}{2}(1-\frac{1}{p})}p^{\frac{n}{2p}} \, \sup_{b_i>0\atop i=1,\ldots,n}\frac{\prod_{i=1}^nb_i^\frac{1}{2p}}{ \det (B+I(\underline b))^\frac12}.
\end{eqnarray*}
\end{remark}
\vskip 3 pt
\begin{remark}\label{RKLS}
In the proof of Theorem 3 in \cite{KLS}, Klein, Landau and Shucker  apply Theorem 6 under the form of that Proposition, p.\,705, with the choice $B=C^{-1}-\frac{1}{pc}I$, where  $C$ is the covariance matrix of the process $X$, $p$ is the decoupling coefficient of $X$, $c={\mathbb E \,} X_0^2$. Further $g_i(x)= f_i(x) e^{-(1/2pc)x^2}$, for $i=1,\ldots, n$.

This requires that $B$ is positive definite, or equivalenly that $pcI-C$ is positive definite. This is ensured by the choice of $p$ made in \cite{KLS}.
\end{remark}
\vskip 3 pt

\vskip 5 pt
In the next lemma, we establish a general bound of $E_B$.
\begin{lemma}\label{l1}
 \begin{eqnarray*}
E_B&\le&
\frac{(2\pi)^{\frac{n}{2}(1-\frac{1}{p})}}{\det(B)^{\frac12(1-\frac{1}{p})}} .
\end{eqnarray*}

\end{lemma}

\begin{proof}  We use the following Lemma.
\begin{lemma}[\cite{B}, Th. 4, p.\,128] If $U$ and $V$ are positive definite matrices, then
\begin{eqnarray*}
\det ({\lambda} U+(1-{\lambda}) V)\ge \det(U)^{\lambda} \det (V)^{1-{\lambda}},
\end{eqnarray*}
for any $0\le {\lambda} \le 1$. \end{lemma}
Therefore
\begin{eqnarray*}
\det \big( U+(\frac{1-{\lambda}}{{\lambda}}) V\big)\ge {\lambda}^{-n}\det(U)^{\lambda} \det (V)^{1-{\lambda}},
\end{eqnarray*}
if $0< {\lambda} \le 1$. We apply this with the choice $U= B$, $V=(\frac{{\lambda}}{1-{\lambda}} )I(\underline b)$.  We get
\begin{eqnarray*}
\det \big( B +I(\underline b)\big)&\ge& {\lambda}^{-n}\det(B)^{\lambda} \det \big((\frac{{\lambda}}{1-{\lambda}} )I(\underline b)\big)^{1-{\lambda}}
\cr &=& {\lambda}^{-n}\det(B)^{\lambda}\,\big(\frac{{\lambda}}{1-{\lambda}} \big)^{n(1-{\lambda})}  \det (I(\underline b))^{1-{\lambda}}
\cr &=&\Big(\frac1{\lambda}\,\big(\frac{{\lambda}}{1-{\lambda}} \big)^{1-{\lambda}}\Big)^n\det(B)^{\lambda}\   \prod_{i=1}^nb_i^{1-{\lambda}}\, .
\end{eqnarray*}
Consequently,
\begin{eqnarray*}
 \frac{\prod_{i=1}^nb_i^\frac{1}{2p}}{ \det (B+I(\underline b))^\frac12}
 &\le &\frac{1}{\det(B)^{\frac{{\lambda}}{2}}}\Big({\lambda}\big(\frac{1-{\lambda}}{{\lambda}}\big)^{1-{\lambda}}\Big)^{\frac{n}{2}} \prod_{i=1}^nb_i^{\frac{1}{2}(\frac{1}{p}-(1-{\lambda}))}
 \cr  &= &\frac{1}{\det(B)^{\frac{{\lambda}}{2}}}\Big({\lambda}^{\lambda} (1-{\lambda})^{1-{\lambda}}\Big)^{\frac{n}{2}} \prod_{i=1}^nb_i^{\frac{1}{2}(\frac{1}{p}-(1-{\lambda}))}.
\end{eqnarray*}
Take ${\lambda}=1-\frac{1}{p}$ and note that  $(1-{\lambda})^{1-{\lambda}}=p^{-\frac{1}{p}} $. We obtain
\begin{eqnarray*}
\frac{\prod_{i=1}^nb_i^\frac{1}{2p}}{ \det (B+I(\underline b))^\frac12}
 &\le &\frac{1}{\det(B)^{\frac12(1-\frac{1}{p})}}  p^{-\frac{n}{2p}}\Big(1-\frac{1}{p}\Big)^{\frac{n}{2}(1-\frac{1}{p})}
 .
\end{eqnarray*}
Therefore
\begin{eqnarray*}
 (2\pi)^{\frac{n}{2}(1-\frac{1}{p})}p^{\frac{n}{2p}}\frac{\prod_{i=1}^nb_i^\frac{1}{2p}}{ \det (B+I(\underline b))^\frac12}
 &\le &\frac{(2\pi)^{\frac{n}{2}(1-\frac{1}{p})}}{\det(B)^{\frac12(1-\frac{1}{p})}}p^{\frac{n}{2p}}\!\cdot\! p^{-\frac{n}{2p}}\Big(1-\frac{1}{p}\Big)^{\frac{n}{2}(1-\frac{1}{p})}
 \cr &= &\frac{(2\pi)^{\frac{n}{2}(1-\frac{1}{p})}}{\det(B)^{\frac12(1-\frac{1}{p})}}\Big(1-\frac{1}{p}\Big)^{\frac{n}{2}(1-\frac{1}{p})} .
\end{eqnarray*}

Whence
\begin{eqnarray*}
E_B&\le&
\frac{(2\pi)^{\frac{n}{2}(1-\frac{1}{p})}}{\det(B)^{\frac12(1-\frac{1}{p})}} .
\end{eqnarray*}

\end{proof}

\begin{proof}[Proof of Theorem \ref{dec.ineq}] It  suffices to prove  inequality \eqref{ineq} when $f_i$ are real-valued and non-negative, for all $i=1,\ldots,n$.
Let $\underline{{\gamma}}=({\sigma}_1^{2}, \ldots, {\sigma}_n^{2})$. We apply Proposition \ref{p1}
 with $B=C^{-1}-\frac{1}{p}I(\underline{{\gamma}}^{-1})$,    $g_i(x)= |f_i(x)| e^{-x^2/(2p{\sigma}_i^2) }$,   $i=1,\ldots, n$.
\vskip 2 pt
We get by using also Lemma \ref{l1},
\begin{eqnarray*}
\int_{{\mathbb R}^n}\Big( \prod_{i=1}^n g_i(x_i)\Big) \exp\Big\{ -\frac12\langle \underline x, B\underline x\rangle\Big\}{\rm d} \underline x
&\le & E_B\prod_{i=1}^n\Big( \int_{{\mathbb R}}    g_i(x)^p{\rm d} x\Big)^\frac{1}{p}
\cr &\le & \frac{(2\pi)^{\frac{n}{2}(1-\frac{1}{p})}}{\det(B)^{\frac12(1-\frac{1}{p})}}\ \prod_{i=1}^n\Big( \int_{{\mathbb R}}    g_i(x)^p{\rm d} x\Big)^\frac{1}{p}.
\end{eqnarray*}

This  is equivalently rewritten as
\begin{align*}
\int_{{\mathbb R}^n}\Big( \prod_{i=1}^n f_i(x_i)\Big) \exp\Big\{ &-\frac12\langle \underline x, C^{-1}\underline x\rangle\Big\}\frac{{\rm d} \underline x}{(2\pi)^{\frac{n}{2}}\det(C)^{\frac12}}
\cr \le &\  \frac{ \big(\prod_{i=1}^n{\sigma}_i\big)^{\frac{1}{p}}}{\det(B)^{\frac12(1-\frac{1}{p})} \det(C)^{\frac12}}  \prod_{i=1}^n\Big( \int_{{\mathbb R}}    |f_i(x)|^p e^{-x^2/(2{\sigma}_i^2) }\frac{{\rm d} x}{{\sigma}_i\sqrt{2\pi}}\Big)^\frac{1}{p},
\end{align*}
namely
\begin{eqnarray*}
{\mathbb E \,}\Big( \prod_{i=1}^n f_i(X_i)\Big)\ \le \
\frac{ \big(\prod_{i=1}^n{\sigma}_i\big)^{\frac{1}{p}}}{\det(B)^{\frac12(1-\frac{1}{p})} \det(C)^{\frac12}}\
 \prod_{i=1}^n\Big( {\mathbb E \,}   f_i(X_i)^p\Big)^\frac{1}{p}.
\end{eqnarray*}
\vskip 3pt
Writing
$$B=C^{-1}-\frac{1}{p}I(\underline{{\gamma}}^{-1})=\frac{1}{p}\,C^{-1}I(\underline{{\gamma}}^{-1})\Big(p\,I(\underline{{\gamma}})- C\Big),$$
 we have
\begin{eqnarray*}
\det(B)=\frac{1}{p^n}\,\det\big(C^{-1}I(\underline{{\gamma}}^{-1})\big)\det\Big(p\,I(\underline{{\gamma}})-C\Big)=\frac{\det\big(p\,I(\underline{{\gamma}})- C\big) }{p^n\,\det(C)\prod_{i=1}^n{\sigma}_i^2} .
\end{eqnarray*}
So that,
\begin{eqnarray*}
{\mathbb E \,}\Big( \prod_{i=1}^n f_i(X_i)\Big)
&\le &
 \frac{ \big(\prod_{i=1}^n{\sigma}_i\big)^{\frac{1}{p}}}
 {\big(\frac{\det(pI(\underline{{\gamma}})-C)}{p^n\det(C)\prod_{i=1}^n{\sigma}_i^2}\big)^{\frac12(1-\frac{1}{p})} \det(C)^{\frac12}}\ \prod_{i=1}^n\Big( {\mathbb E \,}   f_i(X_i)^p\Big)^\frac{1}{p}
\cr &=&
\frac{p^{\frac{n}{2}(1-\frac{1}{p})} \big(\prod_{i=1}^n{\sigma}_i\big)}
 { \det\big(pI(\underline{{\gamma}})-C\big)^{\frac12(1-\frac{1}{p})} \det(C)^{\frac{1}{2p}}}\ \prod_{i=1}^n\Big( {\mathbb E \,}   f_i(X_i)^p\Big)^\frac{1}{p}
.
\end{eqnarray*}

In order to estimate $ \det\big(pI(\underline{{\gamma}})-C\big)$, we recall a well-known result on Hadamard matrices.
\begin{lemma}[\cite{O},\,(2)]\label{l2}  Let $A=\{a_{i,j}, 1\le i,j\le n\}$
 and assume that
\begin{eqnarray*}
\sum_{j=1\atop j\neq i}^n|a_{i,j}|\,<\,|a_{i,i}|,{\qquad} {\qquad} i=1,\ldots , n.
\end{eqnarray*}
Then
\begin{eqnarray*}\det(A)\,\ge \,\prod_{i=1}^n \Big( |a_{i,i}|-\sum_{j=1\atop j\neq i}^n|a_{i,j}|\Big).
\end{eqnarray*}\end{lemma}

By assumption \ref{cond},
\begin{eqnarray*}
\sum_{1\le j\le n\atop j\neq i} |{\mathbb E \,} X_iX_j|\,\le \,\Big(\frac{p}{2} -1\Big)\, {\sigma}_i^2, {\qquad} {\qquad} i=1,\ldots, n.
\end{eqnarray*}
 Letting $pI(\underline{{\gamma}})-C= \{d_{i,j},1\le i,j\le n\}$, we have $|d_{i,i}|=(p-1){\sigma}_i^2$ and
\begin{eqnarray*}
\sum_{j=1\atop j\neq i}^n|d_{i,j}|\,=\, \sum_{j=1\atop j\neq i}^n|{\mathbb E \,} X_iX_j|\,<\,\Big(\frac{p}{2} -1\Big)\,{\sigma}_i^2\,<\, |d_{i,i}|,{\qquad}  i=1,\ldots , n.
\end{eqnarray*}
Thus
$$|d_{i,i}|-\sum_{j=1\atop j\neq i}^n|d_{i,j}| \,\ge\, (p-1){\sigma}_i^2-\Big(\frac{p}{2} -1\Big)\,{\sigma}_i^2\,=\, \frac{ p{\sigma}_i^2}{2}.$$
By Lemma \ref{l2}, it follows that
\begin{eqnarray*}
\det\big(pI(\underline{{\gamma}})-C\big)\,\ge\, \prod_{i=1}^n\big(p{\sigma}_i^2- \sum_{1\le j\le n\atop j\neq i} |{\mathbb E \,} X_iX_j|\big) \,\ge\, \Big(\frac{ p }{2}\Big)^n\prod_{i=1}^n {\sigma}_i^2 .
\end{eqnarray*}
{\noindent}
Thus

\begin{eqnarray*}
{\mathbb E \,}\Big( \prod_{i=1}^n f_i(X_i)\Big)
 &\le &
\frac{p^{\frac{n}{2}(1-\frac{1}{p})} \big(\prod_{i=1}^n{\sigma}_i\big)}
 { \det\big(pI(\underline{{\gamma}})-C\big)^{\frac12(1-\frac{1}{p})} \det(C)^{\frac{1}{2p}}}\ \prod_{i=1}^n\Big( {\mathbb E \,}   f_i(X_i)^p\Big)^\frac{1}{p}
\cr &\le &
\frac{ p^{\frac{n}{2}(1-\frac{1}{p})}\big(\prod_{i=1}^n{\sigma}_i\big)^{\frac{1}{p}}}
 {\big(\frac{ p }{2}\big)^n\big)^{\frac12(1-\frac{1}{p})} \det(C)^{\frac{1}{2p}}}\ \prod_{i=1}^n\Big( {\mathbb E \,}   f_i(X_i)^p\Big)^\frac{1}{p}
\cr &= &
\frac{ 2^{\frac{n}{2}(1-\frac{1}{p})}\big(\prod_{i=1}^n{\sigma}_i\big)^{\frac{1}{p}}}
 {\det(C)^{\frac{1}{2p}}}\ \prod_{i=1}^n\Big( {\mathbb E \,}   f_i(X_i)^p\Big)^\frac{1}{p}
.
\end{eqnarray*}

\end{proof}

\section{\bf Proof of Theorem \ref{p2}.}\label{s3}
  Recall  Bump and Diaconis \cite[Th.\,4]{BD}  strong Szeg\"o limit theorem. Let $(c_k)_{k\in {\mathbb Z}}$ satisfy conditions \eqref{c1}  and \eqref{c2}. Let ${\sigma}(t)= \exp\{\sum_{k\in {\mathbb Z}} c_ke^{ikt}\}$. Then
\begin{equation}\label{bdsslm}D_{n-1} ({\sigma} )\sim \exp\Big( nc_0 + \sum_{k=1}^\infty kc_kc_{-k}\Big).
\end{equation}
    Let $\Gamma_n={\rm Cov}(X_1, \ldots, X_n)$. By applying
   \eqref{bdsslm} with the choice  ${\sigma}(t) = \exp\{\log f(t)\}$, we have
 \begin{equation}\label{sslm}\det (\Gamma_n) \, \sim\, \exp\Big( nc_0+ \sum_{k=1}^\infty kc_kc_{-k}\Big).
\end{equation}

As
 $$c_0= \frac{1}{2\pi}\int_{- \pi}^{ \pi}
\log f(t) {\rm d} t=G(f),$$
it follows that
  \begin{eqnarray*} \det (\Gamma_n)
 &\ge & \Big(\frac{1}{1+{\delta}_n}\Big) \exp\Big\{n G(f)+ \sum_{k=1}^\infty kc_kc_{-k}\Big\}
,\end{eqnarray*}
where ${\delta} _n\downarrow 0$.
 Whence
 \begin{eqnarray*}
\bigg|{\mathbb E \,}\Big( \prod_{i=1}^n f_i(X_i)\Big)\bigg|
&\le &(1+{\delta}_n)
 2^{\frac{n}{2}(1-\frac{1}{p})}\exp\Big\{ -\frac{1}{2p}\big( nG(f)+ \sum_{k=1}^\infty kc_kc_{-k}\big)\Big\}  \prod_{i=1}^n\Big( {\mathbb E \,} |  f_i(X_i)|^p\Big)^\frac{1}{p}
\cr
&= &(1+{\delta}_n)\,b(f)^{-\frac{1}{2p}}\,
 2^{\frac{n}{2}(1-\frac{1}{p})}

 G(f)^{-\frac{n }{2p}}\  \prod_{i=1}^n\Big( {\mathbb E \,} |  f_i(X_i)|^p\Big)^\frac{1}{p},

\end{eqnarray*}
recalling that $b(f)=\exp\big\{   \sum_{k=1}^\infty kc_kc_{-k}\big\}$
  and $G(f)$ is the geometric mean of $f$.

\section{\bf Examples.}\label{s4}
  We list  some remarkable classes of examples.

\subsection{} Let   $X=\{X_j, j\in {\mathbb Z}\}$ be a centered Gaussian stationary  sequence, and assume that $X$ has spectral density. Recall that we have the following representation
 \begin{equation} \label{sgpr} X_k= \sum_{m\in{\mathbb Z}} c_m\xi_{k-m},
 \end{equation}
where $(c_m)\in\ell_2({\mathbb Z})$ and $(\xi_j)$ are i.i.d. standard Gaussian.
Note that
$$ {\mathbb E \,} X_kX_\ell =\sum_{m\in{\mathbb Z}} c_mc_{m-k+\ell}. $$
Consider the sections  $X^n=\{X_1,\ldots,X_n\}$, $n\ge 1$.
\subsubsection{}Let  $c_m=|m|^{-1}$, $m\in{\mathbb Z}\backslash \{0\}$, $c_0=0$.  \begin{proposition}\label{gs1}  We have  \begin{eqnarray*} p(X^n)
   &\le & 4\, (\log n)^2+ \mathcal O (\log n).\end{eqnarray*}
\end{proposition}
The bound of  $p(X^n)$  is in fact optimal, up to some numerical constant.  As $ p(X^n)\ll n$,  this is making   inequality \eqref{ineq} effective.
 \vskip 2 pt
We first  prove a lemma.
\begin{lemma} We have
\begin{eqnarray*}  {\mathbb E \,} X_kX_\ell
 &=& 4\ \frac{\log |k-\ell| }{|k-\ell|}+ \mathcal O \Big(\frac{1}{|k-\ell|}\Big).\end{eqnarray*}
 \end{lemma}
 \begin{proof} We note that
  $$ {\mathbb E \,} X_kX_\ell
 =\sum_{m\in{\mathbb Z}\atop m\neq0, m\neq |k-\ell|} \frac{1}{|m||m-|k-\ell||} ,$$
if $k\neq \ell$, and ${\mathbb E \,} X_k^2= \frac{\pi^2}{3}$.
Let ${\mu}$ be some positive integer. Then,
 \begin{eqnarray*} \sum_{{m\in {\mathbb Z}\atop m\neq{\mu}}\atop m\neq 0} \frac{1}{|m||m-{\mu}|}
&=&  \sum_{m\ge {\mu} +1} \frac{1}{m(m-{\mu})} + \sum_{m\le {\mu}-1\atop m\neq 0}\frac{1}{|m|({\mu}-m)}
\cr &=& \sum_{m\ge {\mu} +1} \frac{1}{m(m-{\mu})} +\sum_{m\le-1}\frac{1}{-m({\mu}-m)}
+ \sum_{1\le m\le {\mu}-1}\frac{1}{m({\mu}-m)}
\cr &=& 2\ \sum_{\nu\ge 1} \frac{1}{\nu(\nu+{\mu})} + \sum_{m=1}^{{\mu}-1}\frac{1}{m({\mu}-m)}.
\end{eqnarray*}
Recall that
$\sum_{1\le m\le x} \frac{1}{m}= \log x +{\gamma} + \mathcal O (x^{-1})$,
where ${\gamma}$ is Euler's constant.
At first,
\begin{eqnarray*}  \sum_{m=1}^{{\mu}-1}\frac{1}{m({\mu}-m)}=\frac{1}{{\mu}}
 \sum_{m=1}^{{\mu}-1}\Big( \frac{1}{m}+\frac{1}{{\mu}-m}\Big)=  \frac{2}{{\mu}} \,
 \sum_{m=1}^{{\mu}-1} \frac{1}{m}
 =  2\  \frac{\log {\mu}}{{\mu}}+ \mathcal O ({\mu}^{-1}).\end{eqnarray*}
Next
 \begin{eqnarray*}\sum_{\nu\ge 1} \frac{1}{(\nu+{\mu})\nu}  &=& \sum_{1\le \nu\le {\mu} - 1} \frac{1}{(\nu+{\mu})\nu} + \sum_{\nu\ge {\mu}} \frac{1}{(\nu+{\mu})\nu}
 \cr &=& \sum_{1\le \nu\le {\mu} - 1} \frac{1}{(\nu+{\mu})\nu}+ \mathcal O ({\mu}^{-1})
\cr &=&\frac{1}{{\mu}}\,  \sum_{1\le \nu\le {\mu} - 1} \Big( \frac{1}{\nu}-\frac{1}{\nu+{\mu}}\Big) + \mathcal O ({\mu}^{-1})
\cr &=& \frac{\log {\mu}}{{\mu}} - \frac{1}{{\mu}}\,  \sum_{{\mu}\le h\le 2 {\mu} - 1}\frac{1}{h}+ \mathcal O ({\mu}^{-1})
\cr &=& \frac{\log {\mu}}{{\mu}} + \mathcal O ({\mu}^{-1}). \end{eqnarray*}
Consequently,
\begin{eqnarray*} \sum_{{m\in {\mathbb Z}\atop m\neq{\mu}}\atop m\neq 0} \frac{1}{|m||m-{\mu}|}
&=& 2\ \sum_{\nu\ge 1} \frac{1}{\nu(\nu+{\mu})} + \sum_{m=1}^{{\mu}-1}\frac{1}{m({\mu}-m)}
\cr &=& \frac{4\log {\mu} }{{\mu}}+ \mathcal O ({\mu}^{-1}).\end{eqnarray*}
And so,
\begin{eqnarray*}  {\mathbb E \,} X_kX_\ell
 &=& 4\ \frac{\log |k-\ell| }{|k-\ell|}+ \mathcal O \Big(\frac{1}{|k-\ell|}\Big).\end{eqnarray*}
\end{proof}
 \begin{proof}[Proof of Proposition \ref{gs1}]
It follows that
\begin{eqnarray*}  \sum_{\ell= 1\atop \ell\neq k}^n|{\mathbb E \,} X_kX_\ell|
 &=& 4\, \sum_{\ell= 1\atop \ell\neq k}^n\frac{\log |k-\ell| }{|k-\ell|}+ \mathcal O \Big(\sum_{\ell= 1\atop \ell\neq k}^n\frac{1}{|k-\ell|}\Big)
.\end{eqnarray*}
Now,
\begin{eqnarray*}\sum_{\ell= 1\atop \ell\neq k}^n\frac{\log |k-\ell| }{|k-\ell|}&=&
\sum_{\ell= 1}^{k-1}\frac{\log (k-\ell )}{k-\ell}+\sum_{\ell= k+1}^n\frac{\log (\ell -k )}{\ell-k}
\cr &\le &
\int_1^{k-1}\frac{\log t}{t}{\rm d} t+\int_1^{n-k}\frac{\log t}{t}{\rm d} t
\cr &=& \frac{\big(\log (k-1)\big)^2 +\big((\log (n-k)\big)^2}{2}.\end{eqnarray*}
Similarly,
\begin{eqnarray*}\sum_{\ell= 1\atop \ell\neq k}^n\frac{\log |k-\ell| }{|k-\ell|} &\ge & \frac{\big(\log (k-1)\big)^2 +\big((\log (n-k)\big)^2}{2}- (\log 2)^2
.\end{eqnarray*}
Therefore
\begin{eqnarray*}\sum_{\ell= 1\atop \ell\neq k}^n\frac{\log |k-\ell| }{|k-\ell|}&=&
\frac{\big(\log (k-1)\big)^2 +\big((\log (n-k)\big)^2}{2}+ \mathcal O \big(1\big).\end{eqnarray*}
Also,
\begin{eqnarray*}\sum_{\ell= 1\atop \ell\neq k}^n\frac{1 }{|k-\ell|}
&\le &
\int_1^{k-1}\frac{{\rm d} t}{t} +\int_1^{n-k}\frac{{\rm d} t}{t}
\ =\ \log (k-1)+ \log (n-k).\end{eqnarray*}
Whence, \begin{eqnarray*}  \sum_{\ell= 1\atop \ell\neq k}^n|{\mathbb E \,} X_kX_\ell|
   &= & 2\,\Big(\big(\log (k-1)\big)^2 +\big((\log (n-k)\big)^2\Big)+ \mathcal O \big(\log k+ \log (n-k)\big).\end{eqnarray*}
   We consequently get the estimate
 \begin{eqnarray*} p(X)
   &\le & 4\, (\log n)^2+ \mathcal O (\log n).\end{eqnarray*}
 \end{proof}

\subsubsection{}Now let  $c_m=|m|^{-r}$, $m\in{\mathbb Z}\backslash \{0\}$ where $r\ge 1$, $c_0=0$. Then naturally $X$ has stronger asymptotical independence properties. In fact
$p(X)<\infty$, as soon as $r\ge 2$.
\vskip 2 pt
 Indeed, by H\"older's inequality, next Proposition \ref{gs1},
\begin{eqnarray*}0\ \le \ {\mathbb E \,} X_kX_\ell
 &=&\sum_{m\in{\mathbb Z}\atop m\neq0, m\neq |k-\ell|} \frac{1}{|m|^r|m-|k-\ell||^r}
\cr &\le &\Big(\sum_{m\in{\mathbb Z}\atop m\neq0, m\neq |k-\ell|} \frac{1}{|m||m-|k-\ell||}\Big)^r
\cr &\le &C_r \,\Big(\frac{\log |k-\ell| }{|k-\ell|}\Big)^r,\end{eqnarray*}
a bound   from which easily follows that $p(X)<\infty$. So, this is an instance where
Klein, Landau and Shucker's inequality \eqref{dec} directly applies.

\begin{remark}[A pathological example]   Gaussian stationary  sequences with spectral density form a huge class, and may in particular exhibit pathological covariance functions. We provide here  a simple example of which the study relies on additive Number Theory.
\vskip 2 pt
Let $A\subset {\mathbb N}$ and $(b_j)\in\ell^2({\mathbb N})$. Consider the following special case of \eqref{sgpr},
 \begin{equation} \label{sgp1} X_k= \sum_{|m|\in A} b_{|m|}\xi_{k-m},{\qquad} k\in {\mathbb Z}.
 \end{equation}
Then
\begin{equation} \label{ex}{\mathbb E \,} X_kX_\ell=
\begin{cases}{\qquad}\qq  0\quad & \hbox{if $k-\ell\notin A-A$},\cr
\displaystyle{\sum_{m\in A\atop
m-k+\ell\in A} b_{|m|}b_{|m-k+\ell|}}\quad &
  \hbox{otherwise}.
\end{cases}
 \end{equation}
Thus the   covariance function is supported on the difference set $A-A$, making the study of this example depending on additive properties of the set $A$.

\end{remark}
\vskip 5 pt

\subsection{Hilbert type
covariance matrices} Now consider non stationary Gaussian sequences   having  Hilbert type
 matrices.   More precisely, let  $C_n$ be the symmetric matrix defined by
\begin{equation}\label{Polya.Szego}
C_n =\Big\{\frac{1}{a_k+a_\ell}\ ; \ k,\ell=1,\ldots, n\Big\},
\end{equation}
where $A=(a_i)_{i\ge 1}$ is a sequence of positive
 real numbers. That $C_n$ is positive definite (and so is   a Gram matrix)   follows from the fact that
$$\int_0^\infty \big| \sum_{k=1}^n x_k e^{-a_k t}\big|^2 {\rm d} t\ge 0.$$
Thus $C_n$ is  the covariance matrix of a Gaussian vector, which can be described explicitly. Indeed, there
 exist in ${\mathbb R}^n$   vectors $u^1,\ldots, u^n$ with Gram matrix $C_n$, for instance the rows of
$C_n^{1/2}$.
Let $\{g_i\}_{1\le i\le n}$ be independent Gaussian standard random variables, and form the Gaussian vector $X^n= \{X_i\}_{1\le i\le n}$ where
$$X_i=\sum_{k=1}^n  g_ku^i_k, {\qquad} i=1,\ldots , n.$$
We immediately see that $X^n$ has covariance matrix $C_n$. Assume that the sequence $A$ is increasing. One easily to check that $p(X^n)\asymp n$.

\bigskip

\noindent {\bf Acknowledgements.} The author is  grateful to  Abel Klein  for a clarification of a point in Brascamp and Lieb's paper. He also  thank Estelle Basor for friendly and stimulating exchanges around Toeplitz operators and determinants generated
by symbols.

redpajama_set_name: RedPajamaArXiv


A HUGE "welcome back" to everyone after the long (and rather inclement) Summer holidays! Welcome also to our new friends, Chloe, Caitlin, Daniel & Kristian, joining us from our own Beaver Colony and further afield!
The Cubs had an enjoyable evening playing a good selection of different games to get them back into the swing of things after the summer break including: "Beat the Cub" (no members of the pack were permanently hurt during this activity!); "Jump the rope"; "Crab football" & "Snowballs".
The senior Cubs held a Sixers and Seconders council and some great ideas came out of this meeting which we hope to incorporate into the Pack programme over the coming weeks & months.


Here is your first free beginner songwriting tip for those who are just starting out... no matter your age!
Spend 15 or 20 minutes listening to several of your favorite songs.
These should be songs that you would like to have written. As you listen, let the feeling of the song wash over you. Listen to the melody. Listen to the words. Feel how you are taken on an emotional journey.
No need to write anything down. Just "feel" your way through these songs.
Then go to another room where you can be alone and quiet. If you cannot do this in your home, go and sit in the car or in a park. You need quiet time alone to let your mind be free.
- a piece of paper and a pen or pencil; also a recorder of some sort, perhaps an MP3 player that will record your voice, or a cassette recorder.
If your mind wanders away from your goal, gently bring it back.
Whatever comes to you - lyrics or music, write it down and/or record it. But, if you do not "hear" a bit of your melody or lyrics, don't worry. Try again tomorrow. This kind of quiet, creative work takes patience. But it will pay off.
Each time you do this creative exercise, you should find that your song continues to take shape.
IMPORTANT NOTE! For a great resource that takes you deep into the art and craft of songwriting visit World Of Songwriting.com.
Remember - your mind is your friend. Give it some peace and some time, and great ideas will come to you.
Try to repeat this creative process two or three times a week. It doesn't take long to listen to your favorite music and then listen to your own mind. You will find that great ideas come to you when you give your mind space.
If you are a new songwriter with little knowledge of music theory, think simple.
Go to your nearest music store and look for a book that gives you a very basic approach to learning music. Or look on Amazon for a beginner's music theory book.
Spend time with the book every day, if possible. But feel no pressure to learn fast. Just enjoy the process. You have time.
Remember to Subscribe to my Free Newsletter, HOT TIPS FOR SMART SINGERS and get two free Gifts!


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MILAN/LONDON: Italy's Eni reported its first quarterly profit in 18 months and said it expects oil and gas production to rise 3 percent per year over the next four years after a string of high profile discoveries.
Eni, the biggest foreign oil major in Africa which accounts for more than 50 percent of its output, has made major gas finds in Mozambique and Egypt, and said on Wednesday it was looking to add up to 3 billion barrels of new resources out to 2020 even as it cut investments by 13 percent.
Earlier Eni beat expectations with fourth quarter adjusted net profit of 459 million euros ($484 million) helped by lower spending and higher oil prices.
"We've never had such a strong set of projects in our history," Chief Executive Claudio Descalzi said during a conference call with analysts on the group's 2017-2020 plan.
State-owned Eni said its organic reserve replacement ratio -- a measure of its ability to find hydrocarbons -- stood at 193 percent in 2016 compared to a 35 percent peer average.
Exxon Mobil Corp, the world's largest publicly traded oil producer, failed to replace 100 percent of its oil and gas reserves with new projects last year while ConocoPhillips Corp revised down proved reserves.
Eni, which will reap the benefit of a ramp-up at Kazakhstan's giant Kashagan field, expects output this year to jump by 5 percent as key projects in Angola, Ghana, Indonesia and Egypt come on stream ahead of schedule.
To deal with the gas finds, the company said it aimed to develop its liquefied natural gas (LNG) business.
"We can create a strong portfolio... the aim is to reach 10 million tonnes of LNG per year by 2025," Descalzi said.
Since taking over as CEO in 2014, Descalzi has refocused the group on the upstream (E&P) business of finding oil and gas.
The veteran oilman, previously head of Eni's E&P unit, has won plaudits from investors for steering Eni through the oil market slump by discovering prize acreage, cutting costs and selling assets.
But his mandate comes up for renewal in the coming weeks and there is some concern his position could be undermined by a corruption probe over a Nigerian oilfield.
"Upstream growth at Eni is well supported by significant organic resource access and while transformation is still in progress there's lots of value still to unlock," said Santander analyst Jason Kenney.
Eni said it would pay a dividend of 0.8 euros per share in 2017, the same as the previous year.
Under its so-called "dual exploration" strategy, Eni aims to sell down stakes in oil and gas fields it operates to raise cash to fund future development and support dividends.
The company, which will generate cash flow of 47 billion euros from operations over the next 4 years, said it was targeting asset sales of 5-7 billion euros to 2020.
"Around 60 percent of that will be in 2017-2018 and some 3-4 billion euros will come from diluting our stakes in fields we operate," CFO Massimo Mondazzi said.
Eni last year sold an overall stake of 40 percent in its giant Zohr field in Egypt to Rosneft and BP for around $2.1 billion.
Sources have said Exxon Mobil has clinched a deal to buy a stake in the group's Area 4 concession in Mozambique but that an announcement would not be made for some time.
Descalzi said a deal to sell a stake in the Mozambique field was "not far" away, adding Eni would remain operator of a part of the field.


I an my Friend yash are addicted to facebook and other social networking areas. So taking the bubbles window manager concept into consideration, we decided to build a browser which incorporates the bubble tab system of the Facebook Messenger. ## How it works Just take an example where u have been asked to open a link from whatsapp or you found one on youtube.com. Surf Mania enables u to open that link in a small web view, which can be squeezed into a small bubble. As and when the page is loaded completed, you will be notified with a popup from the bubble itself. Apart from this, users can also open multiple tabs (or) bubbles, which in-turn results in the increase in the efficiency of the user to cope with his daily task, which he performs through his phone.
Our first challenge was creating a bubble which could move over the whole display. The second challenge we faced in our voyage was implementing multiple tabs(probably the most difficult part of the project and it took around 12 hours in its R&D). In the end race against time has always been difficult.
The Bubble tabing system hasnt been incorporated in any browser till now. So we r really proud for this fact.
WindowsManager class and its variants, working in a team and many other things as well.
Probably publish it on PlayStore.
I worked on the main app bubbles and multitasking of these bubbles.
We are fixing some of the bugs to make the User Interface more cool !


It was only 6 months ago that I began my website and blog. Wow did that go quickly! Why did I start it? Well I wanted to share my love of home decor (and of course of all things white)! A passion I have had for many years and that I fulfilled one other time in my life when I had my own home decor boutique. I was lucky enough to help my clients decorate their homes with the treasures I had found and I am so pleased to be able to do that again as I move back into the design field.
My biggest surprise this year came from how much I love blogging and the reason is really because of you! I am amazed at the incredible support from everyone and love and read every comment that I receive.
Thank you for making the past 6 months so wonderful and I will continue to try to surprise and delight you through 2013. Let me know how I am doing!
Here are some of my Top posts of 2012!
My favourite post of the year was the Paris Home of Patrick Gilles and Dorothée Boissier .
This Creamy Cottage was a much loved post by me and you!
And of course we all love Nina Hartmann's home and her books Vintage by Nina!
No year would be complete without a post on Shabby Chic!
And finally the most viewed and loved post was this beautiful Edwardian Cottage..


Visitors to the famous Sacre Coeur in Paris had a different reason to stare up at the stunning basilica on Sunday when a group of protestors climbed up onto the church's dome and unfurled banners. Find out what it was all about.
They have more commonly been seen at the top of cranes in recent weeks, but on Sunday, a group of fathers took their protest to the top of one of Paris's most famous landmarks – the Sacre Coeur.
The parents, who were joined by a Belgian mother, scaled the dome of the famous basilica in Montmartre and unfurled banners, demanding greater rights for divorced fathers separated from their children.
"We paid our entrance fee and we settled in," said Stephane Iltis, a 39-year-old father, who told Le Parisien newspaper he longer has access to his five children.
Iltis and his fellow protesters, who planned to stay until they were removed, demanded that automatic alternate residence for children be introduced. For his part he denounced a court decision that had assigned all rights over his children to their mother.
Across town on the Left Bank of Paris hundreds of protesters marched for parental equality between Montparnasse and the Ministry of Families in the seventh arrondissement.
The rally was to fight against "the injustice suffered by the vast majority of fathers and against depriving children of their fathers," said Nicolas Moreno, founder of the Collective for Co-parenting (COPCO).
"This is not about a battle of the sexes or a return to the patriarchal policies of 40 years ago," the 34-year-old told Le Parisien.
Protest stunts like the one at the Sacre Coeur have become more commonplace in recent mount in France, ever since an exasperated father mounted a crane in Nantes in February and refused to come down.
Since then aggrieved fathers have mounted similar protests across the country atop churches, buildings, factories and of course cranes.
However the association SOS mums denounced the campaign and the attention it is given.
According to a 2007 study, only 2 percent if divorces and 6 percent of separations end in disagreement between the parents about the children's residence. The study also showed that custody of children is awarded to the mother in 72 percent of divorces.


text:
AR Workshop Kill Devil HillsWood & Canvas Workshop
Wood & Canvas Workshop
Join us at AR Workshop for a DIY class to make one of our many Projects! Choose from Signs, Trays, Boxes, Pillows, Frames & More! In the workshop, you will have the opportunity to customize your project with your choice of designer paints and stain colors. Find out at the workshop about our DIY RewARds program!
Visit event website for more information
Pre-registration required
$29.50 - $115
This event is happening next on:
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Where is the event?
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AR Workshop Kill Devil Hills
3105 N Croatan Hwy
Unit 22-23

redpajama_set_name: RedPajamaCommonCrawl


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MR.D.I.Y.'s Head of Marketing &E-commerce, Mr Andy Chin: "The new E-commerce platform points to another progression, that of keeping up with the times and also being a pioneer in the way business is conducted in this region."
He believes that the new platform will be able to cater a diversified market of tech savvy individuals, who never had time or had least interest to visit physical stores and also provides a convenient shopping solution to help solve a lot of the customer's problems, such as those who want to buy in bulk, urgent purchases, limited mobility as well as cost-saving concerns when it comes to petrol and parking fees.
On the other hand, MR.D.I.Y.'s brand ambassador Adibah Noor had launched its campaign with the bold tagline "Nasib baik tak beli tadi! MR.D.I.Y. murah lagi!" since 2017. This year, Adibah endorsing MR.D.I.Y.'s new tagline "Habislah… MR.D.I.Y. Boleh Beli Online!" again!
Dear online shoppers, there are lots of promotions and prizes to be won when you shop online at MR.D.I.Y.!
#1 From June to August 2018, shoppers who purchase from the website will enjoy RM5 off and free shipping.on their first purchase with a minimum spend of RM10.
#2 All successful online transactions during the contest period (June, July and August 2018) can enter a lucky draw to stand a chance to win over RM68,888 worth of prizes, include an iPad, an iPhone X, a MacBook Pro, and also MR.D.I.Y. vouchers worth RM50.
#3 For the month of July 2018, MR.D.I.Y. will be collaborating with P&G to offer exclusive P&G products with offers of up to 35% off in all stores and online.
#4 Smashing deals online with discounts up to 40% for brands such as MR.D.I.Y. house brand, as well as other reputable international and local brands like WD-40, Shamood Anakku, Coido, OHOYO, Pritech, Staedler, GAP, IVON and Eveready.
Once again, a big round of applause to the MR.D.I.Y online store!


The worlds of business and finance, healthcare, environmental protection, and others are increasingly shaped by law and regulation. To be effective - let alone a leader - in these fields, today's graduates need a basic foundation in law. Emory Law's academic partnerships will be an invaluable contribution to that need.
The Juris Master is a one-year degree, earned with the successful completion of 30 credit hours as a full-time student. The J.M. degree does not qualify recipients to practice law. Rather, it is designed to supplement a student's professional pursuits in areas beyond legal practice.
Many industries and professional fields are coping with the growing importance of regulatory oversight, intensifying risk and liability concerns, global economic uncertainty, increasingly complex decision environments as well as pressures for integrated design/build and faster movement to market. As a result, there is an increasing demand for the integration of professional knowledge in the substance of the work with legal principles relevant to the work. With the additional preparation of a J.M. degree, students will be better equipped to deal with such scenarios, by gaining insight in these and other areas of the law. A student who supplements his or her disciplinary knowledge with a related grounding in the law may enhance their value in the market and workplace; realistically assess if further legal studies are appropriate to their career aspirations; and advance interdisciplinary understanding between their undergraduate professions and law.


The weapons of God are used not just for defensive purposes, (casting down imaginations) but are also used for offensive purposes to keep satan and his kingdom of darkness retreating (pulling down strongholds). Born-again Christians has the power to always attack the kingdom of darkness by exposing lies with Holy Bible Truth inspired by God The Holy Spirit. The Word of God is more than instructions of righteousness. Showing every born-again Christian how to live Godly in Jesus Christ and showing us what is wrong and what is right according to the standards of God. But more than that, it is THE weapon of spiritual warfare.
The Holy Bible has the power to exhort, encourage, correct, cast down every imaginations and every high thing that exalts itself against the knowledge of God. Pulling down spiritual strongholds of sin through the illumination and convictions through His word. The Holy Bible reveals perverted hearts hidden in darkness. It even has the power to penetrate every hard-heart soul and breaks down every barrier that we put up.
Lies and ignorance is the killer of the soul.


Flyers are straighforward advertising products which can be handed out or even used as small posters in stores or other notice boards. We design flyers in all sizes, for all purposes and always especially for your company's needs!
A small club in Berlin needed flyers for their upcoming events, so we designed them.


To help kick off National Fishing & Boating Week (NFBW), RBFF's Take Me FishingTM and Vamos A PescarTM brands have joined forces with Walmart®, South Bend®, and Fishing League Worldwide (FLW) to host and promote more than 1000 events across the United States. These #ReelFun events, held at Walmart stores across the country June 3-5, 2016, will help new and seasoned anglers alike prepare for a summer of fun on the water.
Expert FLW anglers will be on-hand at these events, providing tips and techniques to anglers of all skill levels. Attendees of the events will have the chance to practice their casting skills, enter to win prizes, purchase a fishing license and play fishing-related games. Additionally, each attendee will go home with a free digital subscription of FLW Bass Fishing magazine and a protective fishing license holder.
On top of these great NFBW #ReelFun events, anglers across the country have the opportunity to enter the "Reel in a Trip to Amelia Island" sweepstakes on South Bend's Facebook page. By uploading a photo along with a few words about why they love fishing or sharing their favorite fishing story, participants will have the chance to win an all-expense paid, three-night vacation for four at an oceanfront resort in Amelia Island, Fla., including a fishing excursion with Mike D, host of "South Bend's Lunkerville" television show. The fishing trip with the winners will be featured in a future broadcast of the show.
To find dates and times of local #ReelFun fishing events at participating Walmart stores, visit TakeMeFishing.org/ReelFun.
RBFF has just announced an exciting new partnership between 2015 American League Rookie of the Year, Carlos Correa and its Vamos A PescarTM brand. Correa will motivate Hispanics to spend time with their families fishing and boating this summer through the brand's Así Vamos A Pescar campaign.
In 2012, Correa was a top draft pick and led Houston to a playoff appearance last year, taking home American League Rookie of the Year honors on the way. As a kid, he grew up in a small fishing village in Puerto Rico where he spent plenty of time on the water with his family and friends. His passions, personal experiences and family values make him an ideal partner for RBFF.
We encourage all of our stakeholders to amplify this partnership with Carlos Correa by sharing social media posts on Vamos A Pescar's Facebook, Twitter and Instagram channels and utilizing the free resources, including Hispanic market research and a Spanish translation guide, in the RBFF Resource Center.
Results from this year's America's Top Family Fishing and Boating Spots Sweepstakes will be revealed and promoted leading up to NFBW. Resources will be provided to our state partners to help spread the wordabout the Top 100 Places to fish and boat across the country.
Events, like the many Free Fishing Days that fall during NFBW, are now highlighted in prominent locations across the site. We encourage all stakeholders to take advantage of this opportunity to extend your reach by listing your events on our site for free.
A new partnership between RBFF's Take Me FishingTM and Vamos A PescarTM brands and Walmart®, Southbend®, and Fishing League Worldwide (FLW) will help kick off NFBW with #ReelFun fishing events at over 1,000 Walmart stores. Attendees will have the chance to practice their casting, enter to win prizes, purchase a fishing license and meet and learn from expert FLW anglers.
National Fishing and Boating Week and Free Fishing Days have already been featured in the latest issue of O, The Oprah Magazine, which has a circulation of 2.38 million. We're already off to a great start and are excited to see the press coverage and great accounts of NFBW resulting from our cumulative efforts.
We have provided several free resources to spread the word about NFBW on our website and even more materials can be found in our Resource Center. Don't forget to use #NFBW on your social media channels to be part of the social conversation!


Under the terms of the proposed Vibram class action settlement, Class Members who submit valid and completed Claim Forms will be eligible for refunds, up to a maximum of $94 per pair of shoes purchased. However, because the settlement will be paid on a pro rata basis, it is expected that Class Members who submit timely and valid Claim Forms will receive a refund in the range of $20 to $50 per pair. Any money that remains after all claims have been distributed will be donated to the American Heart Association to use for research about the health benefits of running.
Vibram has also agreed to discontinue certain aspects of its advertising and marketing campaign. Specifically, the company says it will no longer make claims that the FiveFingers shoes are effective at strengthening muscles or preventing injury unless those claims are supported by reliable scientific evidence. Under the terms of the class action settlement, Vibram will also refrain from making any other claims about the health benefits of the FiveFingers footwear unless they are supported by competent and reliable scientific evidence.
Vibram has denied any wrongdoing but agreed to settle the FiveFingers class action lawsuit to avoid the expense and uncertainty of trial.
Given what the judge said when Vibram filed to get the case dismissed, the writing was on the wall that they would probably have to settle as their chances of succeeding in this were not looking good. All the evidence that has been published since this suit was filed also points against a chance of Vibram successfully defending this case.
All the predictions and claims by the fan boys as to what would happen to this case did not come to fruition. I guess all their despicable online bullying of the lead claimant in this case did not deter her.
How do we get in on this claim?
I have 3 pairs to claim, so I can buy 3 new pair of Vibrams.
How do I get a claim submission form? Any ideas???
I have three pairs of these. Would love to know when the website is open.
I've been running/hiking with in my KSOs for more than three years now without any injuries. Anyone who does their research knows that you need to change your stride when running barefoot or while wearing Vibrams.
I just filled out my claim and as soon as the money is awarded I'll be spending it on a new pair of Vibrams. If the money isn't claimed then it will just go to the scumbag lawyers.
As you were then not mislead by the claims that Vibram made, why are you joining the class action? Are you comfortable committing fraud?
Just because he enjoys Vibrams doesn't mean he wasn't mislead or that he's committing fraud.
For example: if it were found out that Cheerios doesn't help lower cholesterol (as their ads claim), but I liked the way they taste, I would have been mislead about the health benefits, but I could still choose to continue to buy them because they're delicious.
Actually he is! By joining the class action you are admitting you were mislead by the health claims that were made by Vibram and did not get the health gains that they claimed you would.. The analogy you use makes no sense.
I have a pair of Bikila Vibrams and I have never had any problems whatsoever, in fact I would say that I have experienced the stated health benefits.
I no longer have foot pain, my posture is better and I keep getting personal bests on all my running routes.
But that is not what the case was about. It was about Vibram claiming that there was "ample scientific evidence" for those health gains. When, in reality, there was and still is none.


The shop is amazing all year round, with shelves packed full of lovely things; Falcon enamelware jostles for space with the softest of table linens and miniature glass bottles, tables are stacked with copies of wonderfully thick cookbooks and interiors tomes, and walls are festooned with garlands, fanwheels and bunting. There's ladieswear and accessories and beauty, menswear, bags, flowers and partyware. Childrens toys, beautifully illustrated books and the crispest sheets of lovely wrapping paper. It is, at all times, one of my favourite places to shop. But at Christmas, it's something else.
At Christmas, the fairy lights come out. At Christmas, there are buckets of foliage dotted about, and wreaths hung merrily on the wall. There is a brightly coloured display behind the till, made up of gift wrapping and decorations, giant fanwheels and tiny boxes topped with pompoms. At Christmas the upper floor is turned into a twinkly, sparkling wonderland; glittery ornaments hang delicately from a tree, while beautifully wrapped gifts in a rainbow of colours are piled beneath. There are snowglobes and miniature trees, decorations for the tree and decorations for the table. Sugary treats nestle on shelves; sumptuous Turkish Delight dusted gently with icing sugar, soft marshmallows and endless bars of chocolate in a variety of flavours.
Previous Post it's beginning to look a lot like christmas….


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Q: Django : duration in time Currently, I use django 1.4.8 and I have in my model the following two fields:
class Event(models.Model):
...
    start = models.DateTimeField(_(u"départ"))
    end = models.DateTimeField(_('fin'))

I tried to add a field indicating instead the duration in time
     start = models.DateTimeField(_(u"départ"))
     time_delay = models.TimeField(_(u"Time_delay"), auto_now_add=True, blank=True, default=datetime.datetime.utcnow)

but I get the error in the admin
'EventAdmin.fieldsets[0][1]['fields']' refers to field 'time_delay' that is missing from the form

My admin :
class EventAdmin(admin.ModelAdmin):

list_display = ('title', 'start', 'contact_time', 'user', 'fin', 'frequency', 'one_file_',)

fieldsets = (
    (None, {
        'fields': ('title','start', 'contact_time', 'is_cancelled', 'calendar', 'user', 'description', ('frequency', 'fin' ), 'activated', 'one_file',)
    }),
)

....
what is this problem

A: You set auto_now_add = True for time_delay field so django will not add this field to the form
class EventAdmin(admin.ModelAdmin):

    list_display = ('title', 'start', 'contact_time', 'user', 'fin', 'frequency', 'one_file_',)

    fieldsets = (
    (None, {
        'fields': ('title','start', 'contact_time', 'is_cancelled', 'calendar', 'user', 'description', ('frequency', 'fin' ), 'activated', 'one_file',)
    }),
)
    readonly_fields = ("time_delay",)

redpajama_set_name: RedPajamaStackExchange


text:
Home Stock Analysts Mean recommendation for VEREIT Inc. (VER) was 2.50: Is this the...
Analysts Mean recommendation for VEREIT Inc. (VER) was 2.50: Is this the key time?
Nick Little
Let's start up with the current stock price of VEREIT Inc. (VER), which is $7.51 to be very precise. The Stock rose vividly during the last session to $7.805 after opening rate of $7.69 while the lowest price it went was recorded $7.50 before closing at $7.64.
Recently in News on December 2, 2020, VEREIT® Announces November Rent Collection of 97% and Provides Transaction Activity Update. VEREIT, Inc. (NYSE: VER) ("VEREIT" or the "Company"), a full-service real estate operating company which owns and manages one of the largest portfolios of single-tenant commercial properties in the U.S., announced its November rent collection and updated transaction activity. You can read further details here
VEREIT Inc. had a pretty Dodgy run when it comes to the market performance. The 1-year high price for the company's stock is recorded $10.18 on 02/24/20, with the lowest value was $3.56 for the same time period, recorded on 03/19/20.
VEREIT Inc. (VER) full year performance was -17.92%
Price records that include history of low and high prices in the period of 52 weeks can tell a lot about the stock's existing status and the future performance. Presently, VEREIT Inc. shares are logging -26.23% during the 52-week period from high price, and 111.25% higher than the lowest price point for the same timeframe. The stock's price range for the 52-week period managed to maintain the performance between $3.56 and $10.18.
The company's shares, operating in the sector of Real Estate managed to top a trading volume set approximately around 9062917 for the day, which was evidently lower, when compared to the average daily volumes of the shares.
When it comes to the year-to-date metrics, the VEREIT Inc. (VER) recorded performance in the market was -18.72%, having the revenues showcasing 5.92% on a quarterly basis in comparison with the same period year before. At the time of this writing, the total market value of the company is set at 8.10B, as it employees total of 160 workers.
VEREIT Inc. (VER) in the eye of market guru's
During the last month, 0 analysts gave the VEREIT Inc. a BUY rating, 0 of the polled analysts branded the stock as an OVERWEIGHT, 0 analysts were recommending to HOLD this stock, 0 of them gave the stock UNDERWEIGHT rating, and 0 of the polled analysts provided SELL rating.
According to the data provided on Barchart.com, the moving average of the company in the 100-day period was set at 6.84, with a change in the price was noted +1.18. In a similar fashion, VEREIT Inc. posted a movement of +18.64% for the period of last 100 days, recording 10,165,253 in trading volumes.
Total Debt to Equity Ratio (D/E) can also provide valuable insight into the company's financial health and market status. The debt to equity ratio can be calculated by dividing the present total liabilities of a company by shareholders' equity. Debt to Equity thus makes a valuable metrics that describes the debt, company is using in order to support assets, correlating with the value of shareholders' equity The total Debt to Equity ratio for VER is recording 0.91 at the time of this writing. In addition, long term Debt to Equity ratio is set at 0.91.
VEREIT Inc. (VER): Stocks Technical analysis and Trends
Raw Stochastic average of VEREIT Inc. in the period of last 50 days is set at 83.57%. The result represents improvement in oppose to Raw Stochastic average for the period of the last 20 days, recording 62.66%. In the last 20 days, the company's Stochastic %K was 78.17% and its Stochastic %D was recorded 81.53%.
If we look into the earlier routines of VEREIT Inc., multiple moving trends are noted. Year-to-date Price performance of the company's stock appears to be encouraging, given the fact the metric is recording -18.72%. Additionally, trading for the stock in the period of the last six months notably improved by 3.59%, alongside a downfall of -17.92% for the period of the last 12 months. The shares increased approximately by -1.18% in the 7-day charts and went down by 5.03% in the period of the last 30 days. Common stock shares were driven by 5.92% during last recorded quarter.
VER Shares
VER stock
VEREIT Inc.
VEREIT Inc. (NYSE:VER)
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A darkened bronze surround with golden highlights creates the structure for a bold cabinet fabricated with charred pine door and drawer faces. The process to achieve the finish on the cabinet faces highlight the natural character of the wood in an unexpected way. The unit is complimented with custom machined bronze hardware that lines each door and drawer and that also acts to help support this credenza on the wall.
Amuneal's Collectors' Shelving System, including the Heritage Credenza, is infinitely flexible and can be configured to fit the dimensions and look of your space.
All Amuneal products are endlessly customizable both in size and finish options. Please contact us for more details.
Lead Time: 11 - 16 weeks. Shipping: please call 213.928.0997 to receive shipping quotes.
Please note that this pricing applies to the exact dimensions and finish options displayed here. Pricing subject to change due to customization requests.


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Home Sports Brittney Griner Height, Wife (Glory Johnson), Girlfriend, Parents
Brittney Griner Height, Wife (Glory Johnson), Girlfriend, Parents
Brittney Griner is a professional basketball player currently employed by Phoenix Mercury from the Women's National Basketball Association WNBA. She joined the Phoenix in 2013 after being selected by the team in the first overall selection of the 2013 WNBA Draft.
The tall and beautiful player started playing basketball at Nimitz High School in Houston. She also played as a student at Baylor University Waco, Texas, before being drafted into the WNBA in 2013. As a college player, Brittney was the only NCAA basketball player to score 2,000 points and block 500 shots.
In addition to her high school, college, and WNBA activities, she played with several international teams including the Zhejiang Golden Bulls and the UMMC Ekaterinburg. Learn more exciting facts about the basketball star, her height, wife, girlfriend, parents, and more.
Brittney's full name is Brittney Yevette Griner and she was born on October 18, 1990, to African-American parents – Raymond and Sandra Griner. There is no further information about her parents or what they do for a living. She has three older siblings – two sisters, Shkera and Pier, and one brother Decarlo Griner.
She attended the Nimitz Gymnasium, where she began to show her love for sports. First, she was a member of the school's volleyball team as a freshman. In her second year, she began practicing with the boys' basketball team to develop her leg strength in preparation and readiness for dunking. This process was recorded and the video went virally online and was viewed over 2.7 million times, earning her a meeting with Shaquille O'Neal.
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The young woman became so good that when she graduated from Nimitz High School, she led the school team to the Texas 5A Girls Basketball State Championship game. Unfortunately, the team lost to Mansfield Summit High School. In high school, Brittany scored 52 times in 32 games, setting a single-game record of seven dunkings against Aldine High School.
Her outstanding records and performances soon earned her a WBCA All-American and she took part in the 2009 WBCA High School All-America Game where she led the team by 20 points and 9 rebounds and was soon named the nation's number one high school women's basketball player by Rivals.com.
College Career
After high school, Griner played college basketball at Baylor University, a private Baptist university in Waco, Texas. During her college career, Brittney Griner broke amazing records. In her sophomore year, she received her First Team All-American Honors after averaging 23 points per game, including 40 points against Green Bay at Sweet 16, a career highlight.
During her junior years, she averaged 23.2 points, 9.4 rebounds, and 5 blocks per game, blocked more shots than any other Division I women's team and was named AP Player of the Year.
She became the seventh player to score during a women's college basketball game and the only second woman to score twice in a single college game. She also earned her second and third dunk in her college career in a one-sided 99-18 win over Texas State University.
In 2010, Baylor finished fourth in the NCAA tournament, and Griner set a record of 14 blocked shots in a 49-33 win over Georgetown Hoyas. She became the 2012 prime minister of women's college basketball and the most outstanding player in the last four.
WNBA Career
Brittany Griner began her professional career in 2013 after she was drafted by the Phoenix Mercury as the first overall candidate in the 2013 WNBA Draft for an amazing and successful rookie season. In her debut match against Chicago Sky, she was the first player in WNBA history to dunk twice in one game.
Griner was soon named WNBA All-Star but missed the 2013 WNBA All-Star game due to a right knee injury she sustained and was replaced by Tina Thompson. With the combination of Candice Dupree, Brittney Griner, and Diana Taurasi, the Mercury had an amazing 2014 season, setting the WNBA record for the most wins in a regular season.
After their contract as rookies expired, Griner signed a multi-year contract with the Mercury again in 2017 and has remained with the team ever since. Apart from college and the WNBA, Brittney Griner has also played for the US basketball team. In 2011 she played with the US national team for two weeks as part of their European training tour under coach Geno Auriemma.
She was also on the finalist list for the US women's Olympic basketball team in 2012 but was unable to participate in the Olympics due to her school schedule and some family challenges. Four years later (2016) she played with the US Women's Olympic Basketball Team at the Summer Olympics and won her first Olympic medal after defeating Spain 101-72.
Brittney Griner – Wife (Glory Johnson), Girlfriend, Parents
Brittney Griner came out as a lesbian in an interview with SI.com in 2013. According to her, she first told her mother Sandra Griner about her sexuality during her freshman year in high school but kept it hidden from her father Raymond Briner until her senior year, and when he found out, he threw her out of the house.
The father's girl, who is now in a good relationship with her father, remembered the difficulties she was having at the time and said that this experience made her a tough person and prepared her for life. Her story, which she wrote in a memoir with Sue Hovey, is entitled In My Skin: My Life on and Off the Basketball Court.
In 2014 Brittney announced her engagement to her WNBA colleague Glory Johnson. A year later, in April 2015, both were arrested on charges of assault and battery and disorderly conduct. Police had broken into their Arizona home to break up an argument between them.
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The following month – May 2015 – they entered into a marriage in Phoenix, Arizona, and in June the couple announced that Johnson was pregnant with twins. The next day Griner filed for divorce on the grounds of fraud and coercion.
She revealed that she had no biological connection to the twins and that Johnson had become pregnant through IVF treatment with her eggs and a sperm donation. The divorce was finalized in June 2016.
How Tall Is Brittney Griner? – Height
Basketball players are usually very tall, and Brittney Griner is no exception. She stands 6 feet 8 inches off the ground, has a wingspan of 88 inches, wears men's size 17 shoes, and is in fact one of the biggest basketball players there is.
Interestingly, the only international players to exceed her height are Allyssa DeHaan, who is 6'6″ (6 foot 9 inches) tall, Heidi Gillingham, who is 6'8″ (6 foot 10 inches) tall, late Margo Dydek, who is 7″ (7 foot 2 inches) tall, and late Sue Geh, who is 6″ (6 foot 9 inches) tall.
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