text:
\section{Exponentially flat Hessian: Proof of Theorem \ref{thm:nce_hessian_bound}}
\label{s:bounding_nce_hessian}
The proof of \Cref{thm:nce_hessian_bound} consists of three ingredients. First, in \Cref{s:hessianbound}, we will compute an algebraically convenient upper bound for the spectral norm of the Hessian of the loss (\cref{eq:nce_loss}) in the case when $\set{P_\theta}$ belongs to an exponential family, in terms of the TV distance $\TV(P_*,Q)$ between the true data distribution $P_*$ and the noise distribution $Q$, and the Fisher information matrix of the sufficient statistics at $\theta_*$.

Then, in \Cref{s:constructing_hard_p}, we construct a
distribution $P^*$ for which the TV distance between $P^*$ and $Q$ is large. We do this by ``tensorizing'' a univariate distribution. Namely, we construct a univariate distribution with mean 0 and variance 1 that is at a constant TV distance from a standard univariate Gaussian. Then, we use the fact that the \emph{Hellinger} distance tensorizes, along with the relationship between TV and Hellinger distance, to show that $TV(P^*, Q) \geq 1 - \delta^d$ for some constant $\delta < 1$. (We refer the reader to \cite{larrynotes} for a detailed review of these distance measures.) \Cref{s:bounding_fisher} bounds the Fisher information matrix term, completing all the components required to establish \Cref{thm:nce_hessian_bound}.

\subsection{Bounding the Hessian in terms of TV distance}
\label{s:hessianbound}
Suppose $\set{P_\theta}$ is an exponential family of distributions, that is $p_\theta(x) = \exp(\theta^\transpose T(x))$, where $T(x)$ is a known function. Then, a straightforward calculation (see e.g., Appendix A in \cite{liu2021analyzing}) shows that the gradient and the Hessian the NCE loss (\cref{eq:nce_loss}) with respect to $\theta$ have the following form:
\begin{align*}
	\nabla_\theta p_\theta(x) & = p_\theta(x) \cdot T(x), \labeleqn{eq:p_gradient}\\
	\nabla_\theta L(p_\theta) & = \frac{1}{2} \int_x \frac{q}{p_\theta+q} (p_\theta - p_*) T(x) dx, \labeleqn{eq:nce_loss_gradient}\\
	\nabla^2_\theta L(p_\theta) & = \frac{1}{2} \int_x \frac{(p_* + q)p_\theta q}{(p_\theta+q)^2} T(x) T(x)^\transpose dx. \labeleqn{eq:nce_loss_hessian}
\end{align*}

For $\theta = \theta_*$ and $p_\theta = p_*$, we have
\begin{align*}
	\nabla^2_\theta L(p_{\theta_*})
	&= \frac{1}{2} \int_x \frac{p_* q}{p_* + q} T(x) T(x)^\transpose dx \labeleqn{eq:hessian_at_opt} \\
	&\preceq \frac{1}{2} \int_x \min (p_*, q) T(x) T(x)^\transpose dx \labeleqn{eq:hessian_bound_1}.
\end{align*}
The second line holds since $\frac{p_* q}{p_* + q} = \min(p_* ,q) \cdot \frac{\max(p_*,q)}{p_* + q} \le \min(p_*,q)$.
Applying the vector version of the Cauchy-Schwarz inequality (\Cref{l:vector_cs}, \Cref{s:cs_vector}) to \cref{eq:hessian_bound_1} with two parts $\frac{\min(p_*(x), q(x))}{\sqrt{p_*(x)}}$ and $T(x) T(x)^\transpose \sqrt{p_*(x)}$, we obtain
\begin{align*}

	\|\nabla^2_\theta L(P_*)\|_2 \le \norm{\nabla^2_\theta L(P_*)}_F
	&\le \frac{1}{2} \brac{\int_x \frac{\min(p_*,q)^2}{p_*}}^{\frac{1}{2}} \brac{\int_x \norm{T(x)T(x)^\transpose}_F^2 p_*(x) dx}^{\frac{1}{2}}\\
	&\le \frac{1}{2} \brac{\int_x \min(p_*,q)dx}^{\frac{1}{2}} \brac{\int_x \norm{T(x)T(x)^\transpose}_F^2 p_*(x) dx}^{\frac{1}{2}}\\
	\implies \norm{\nabla^2_\theta L(P_*)}_2
	&\le \frac{1}{2} \brac[\bigg]{1 - \TV(P_*,Q)}^{\frac{1}{2}} \brac{\int_x \norm{T(x)T(x)^\transpose}_F^2 p_*(x) dx}^{\frac{1}{2}}. \labeleqn{eq:hessian_bound}
\end{align*}

We bound the two terms in the product above separately.

\subsection{Constructing the hard distribution \texorpdfstring{$P_*$}{}}
\label{s:constructing_hard_p}

The hard distribution $P_*$ over $\R^d$ will have the property that $\E_{P_*}[x] = 0$, $\E_{P_*}[xx^\top]=I_d$, but will still have large TV distance from the standard Gaussian $Q=\mathcal{N}(0, I_d)$. This distribution will simply be a product distribution---the following lemma formalizes our main trick of tensorization to construct a distribution having large TV distance with the Gaussian.
\begin{lemma}
	\label{l:tv_separation}
	Let $d > 0$ be given. Let $Q=\mathcal{N}(0, I_d)$ be the standard Gaussian in $\mathbb{R}^d$. Then, for some $\delta < 1$, there exists a log-concave distribution $P$ (also over $\mathbb{R}^d$) with mean 0 and covariance $I_d$ satisfying $\TV(P,Q) \ge 1-\delta^d$.
\end{lemma}
\begin{proof}
    \newcommand{c_\rho}{c_\rho}
    Let $\hat{Q}$ denote the standard normal distribution over $\R$. Let $\hat{P}$ be any other distribution over $\R$ with mean $0$ and variance $1$ that satisfies
    $ \rho(\hat{P},\hat{Q}) = \delta < 1 $,
    where $\rho(\hat{P}, \hat{Q})=\int_x \sqrt{\hat{p}\hat{q}} \,dx$ is the Bhattacharya coefficient. Since $\rho$ tensorizes \citep{larrynotes}, we have that $\rho(\hat{P}^d, \hat{Q}^d) = \rho(\hat{P},\hat{Q})^d$ for any $d>1$. We can then write the Hellinger distance between $P,Q$ as \cite{}
	\begin{equation}
		\label{eq:hellinger_tensorization}
		H^2(\hat{P}^d, \hat{Q}^d) = 2(1-\rho(\hat{P}, \hat{Q})^d).
	\end{equation}
	Further, we also know that
	\begin{align*}
		&\frac{1}{2}H^2(\hat{P}^d, \hat{Q}^d) \le \TV(\hat{P}^d, \hat{Q}^d) \labeleqn{eq:hellinger_tv_bound}\\
		\implies \qquad &1-\rho(\hat{P}, \hat{Q})^d \le \TV(\hat{P}^d, \hat{Q}^d)\\
		\implies \qquad &1 - \delta^d \le \TV(\hat{P}^d, \hat{Q}^d).
	\end{align*}

	Setting $P=\hat{P}^d$ and noting that $\hat{Q}^d = Q = \mathcal{N}(0, I_d)$, we have
	$ \TV(P, Q) \ge 1 - \delta^d. $
	Finally, if the chosen $\hat{P}$ is a log-concave distribution, then so is $\hat{P}^d$, which completes the proof.
\end{proof}

We will now explicitly define the distribution $P_*$ that we will work with for rest of the paper.
\begin{definition}
\label{d:pstar}
Consider the exponential family $\set{p_\theta(x) = \exp\brac{\theta^\top T(x)}}_{\theta \in \R^{d+1}}$ given by the sufficient statistics $T(x) = (x_1^4, \ldots, x_d^4,1)$. Let $P_* = \hat{P}^d$ where $\hat{P}$ is the distribution on $\R$ with density function $\hat{p}$ given by
\begin{equation*}

    \hat{p}(x) \propto \exp\brac{-\frac{x^4}{\sigma^4}}.
\end{equation*}
We will set the constant of proportionality $C$ and $\sigma$ appropriately to ensure that $\hat{P}$ has mean 0 and variance 1.
Note that $P_* = P_{\theta_*}$ for $\theta_* = - \brac{\frac{1}{\sigma^4}, \ldots , \frac{1}{\sigma^4}, \log C}. $
\end{definition}

Since $\frac{d^2\log \hat{p}}{dx^2} = -\frac{12x^2}{\sigma^4} \le 0$, $\hat{p}$ is log-concave. Further, symmetry of $\hat{p}$ around the origin gives $\ex*{\hat{P}} = 0$, and the choice of $\sigma$ ensures that $\var*{\hat{P}} = 1$.
The normalizing constant $C$ satisfies
	\begin{align*}
		C &= \int_{-\infty}^{\infty}e^{-\frac{x^4}{\sigma^4}}dx = 2\int_{0}^{\infty}e^{-\frac{x^4}{\sigma^4}}dx.
	\end{align*}
	Substituting $t=\frac{x^4}{\sigma^4}$, $dt = \frac{4x^3}{\sigma^4}dx = \frac{4t^{3/4}}{\sigma}dx$ gives
	\begin{align*}
		C &= \frac{\sigma}{2}\int_0^{\infty}t^{-3/4}e^{-t}dt = \frac{\sigma}{2}\Gamma\left(\frac{1}{4}\right) = 2\sigma\Gamma\left(\frac{5}{4}\right).
	\end{align*}
	where $\Gamma(z)$ is the gamma function defined as $\Gamma(z)\int_0^\infty x^{z-1} e^{-x} dx.$
	The variance is given by
	\begin{align*}
		\var{\hat{P}} &= \frac{1}{C}\int_{-\infty}^{\infty}x^2e^{-\frac{x^4}{\sigma^4}}dx = \frac{2}{C}\int_{0}^{\infty}x^2e^{-\frac{x^4}{\sigma^4}}dx.
	\end{align*}
	The same substitution as above gives
	\begin{align*}
		\text{Var}(\hat{P})
		&= \frac{1}{2C}\int_0^\infty t^{1/2}t^{-3/4}\sigma^3e^{-t}dt = \frac{\sigma^3}{2C}\int_0^\infty t^{-1/4}e^{-t}dt = \frac{\sigma^3}{2C}\Gamma\left(\frac{3}{4}\right)= \frac{\sigma^2}{4}\frac{\Gamma(3/4)}{\Gamma(5/4)}.
	\end{align*}
	Thus, setting $\sigma= \sqrt{\frac{4\Gamma(5/4)}{\Gamma(3/4)}}$ results in $\var*{\hat{P}}=1$. Correspondingly, we have $C = \frac{4\Gamma(5/4)^{3/2}}{\sqrt{\Gamma(3/4)}}$.

For this choice of $\hat{P}$, the Bhattacharya coefficient $\rho(\hat{P}, \hat{Q})$ is given by:
	\begin{align*}
		\rho(\hat{P}, \hat{Q})
		&= \int_{-\infty}^{\infty}\sqrt{\hat{p}(x)\hat{q}(x)}dx
		= \frac{1}{\sqrt{C\sqrt{2\pi}}}\int_{-\infty}^{\infty}\exp\left(-\frac{x^2}{4}-\frac{x^4}{2\sigma^4} \right)dx
		\approx 0.9905 \le 0.991 < 1.
	\end{align*}
Thus, in the proof of \Cref{l:tv_separation}, we can use this choice of $\hat{P}$, and we have that for $\delta=0.991$ and $P_* = \hat{P}^d$, $\TV(P_*, Q)\ge 1-\delta^d$, as required.

\subsection{Bounding the Fisher information matrix}
\label{s:bounding_fisher}
In this subsection, we bound the second factor in \cref{eq:hessian_bound}, which is the Fisher information matrix at $\theta_*$.

\newcommand{M}{M}

\begin{lemma}
    \label{l:nce_fisher_bound}
     The Fisher information matrix at $\theta_*$ satisfies
    \begin{equation}
        \label{eq:nce_fisher_bound}
        \int_x \norm{T(x)T(x)^\top}_F^2 p_*(x) dx \le d^2 M,
    \end{equation}
    for some constant $M > 0$.
\end{lemma}

\begin{proof}

Recall that $T(x) = (x_1^4, \ldots, x_d^4,1)$. Then,
\begin{equation}
    \norm{T(x)T(x)^\transpose}_F^2 = \sum_{i} x_i^{16} + \sum_{i \neq j} x_i^8 x_j^8 + 2 \sum_{i} x_i^4 + 1.
\end{equation}

Therefore, by linearity of expectation, and using the fact that $P_*$ is a product distribution,
\begin{align*}
    \int_x \norm{T(x)T(x)^\top}_F^2 p_*(x) dx
    & = d \cdot \exwith{\hat{P}}{x^{16}} + d(d-1) \cdot \brac{\exwith{\hat{P}}{x^8}}^2 + 2d \cdot \exwith{\hat{P}}{x^4} + 1  \le d^2 M,
\end{align*}
for an appropriate choice of constant $M$. This constant exists since all the expectations above are bounded owing to the fact that the exponential density $\hat{p}$ dominates in the integrals.
\end{proof}
We now have all elements ready to prove \Cref{thm:nce_hessian_bound}.

\subsection{Putting things together}
For $P_*$ defined as above, and $Q = \mathcal{N}(0,I_d)$, \Cref{l:tv_separation} ensures that
\[ 1 - \TV(P_*, Q) \le \delta^d, \]
for $\delta=0.991$. From \Cref{l:nce_fisher_bound}, we have that
\[ \int_x \norm{T(x)T(x)^T}_F^2 p_*(x) dx \le d^2 M. \]
Substituting these bounds in \cref{eq:hessian_bound}, we get that
\[ \norm{\nabla^2_\theta L(P_*)}_2 \le \frac{1}{2}\delta^{d/2} d \sqrt{M} = \exp(-\Omega(d)).\]
By construction, $p_*$ is a product distribution with $\exwith{p_*}{x} = 0$ and $\exwith{p_*}{xx^\top} = I_d$, which completes the proof of the theorem.

\section{Introduction}
\label{s:introduction}

Noise contrastive estimation (NCE), introduced in \citep{gutmann2010noise, gutmann2012noise}, is one of several popular approaches for learning probability density functions parameterized up to a constant of proportionality, i.e. $p(x) \propto \exp(E_{\theta}(x))$, for some parametric family $\{E_{\theta}\}_{\theta}$. A recent incarnation of this paradigm is, for example, energy-based models (EBMs), which have achieved near-state-of-the-art results on many image generation tasks \citep{du2019implicit, song2019generative}.

The main idea in NCE is to train a classifier which distinguishes between samples from the data distribution $P_*$ and a known, easy-to-sample distribution $Q$, often called the ``noise'' distribution. It can be shown that for a large choice of losses for the classification problem, the optimal classifier model is a (simple) function of the density ratio $p_* / q$, so an estimate for $p_*$ can be extracted from a good classifier.
Moreover, this strategy can be implemented \emph{while avoiding} calculation of the partition function, which is necessary when using maximum likelihood to learn $p^*$.

The noise distribution $q$ is the most significant ``hyperparameter'' in NCE training, with both strong empirical \citep{rhodes2020telescoping} and theoretical \citep{liu2021analyzing} evidence that a poor choice of $q$ can result in poor statistical and algorithmic complexity. \citep{chehab2022optimal} show that even the optimal $q$ for finite number of samples can have an unexpected form (e.g., it is not equal to the true data distribution $p^*$).
Since $q$ needs to be a distribution that one can efficiently draw samples from, as well as write an expression for the probability density function, the choices are somewhat limited.

A particularly common way to pick $q$ is as a Gaussian that matches the mean and covariance of the input data \citep{gutmann2012noise, rhodes2020telescoping}. Our main contribution in this paper is to formally show that such a choice can result in an objective that is statistically poorly behaved, even for relatively simple data distributions. We show that even if $p^*$ is a \emph{product distribution} and a member of a very simple \emph{exponential family}, the Hessian of the NCE loss, when using a Gaussian noise distribution $q$ with matching mean and covariance has exponentially small (in the ambient dimension) spectral norm. As a consequence, the optimization landscape around the optimum will be exponentially flat, making gradient-based optimization challenging.
Even more seriously, we show the asymptotic sample efficiency of the NCE objective will be \emph{exponentially bad} in the ambient dimension.

\section{Simulations}
\label{s:experiments}

\begin{wrapfigure}{r}{0.3\textwidth}
  \begin{center}
    \includegraphics[width=0.28\textwidth]{img/real_moments.pdf}
  \end{center}
  \caption{Log MSE versus Dimension---\Cref{thm:nce_sample_complexity} suggests this plot should be linear, as is observed.}
  \label{fig:real-moments}
\end{wrapfigure}

We also verify our results with simulations. Precisely, we study the MSE for the empirical NCE loss as a function of the ambient dimension, and recover the dependence from Theorem~\ref{thm:nce_sample_complexity}. For dimension $d \in \{70,72,\dots,120\}$, we generate $n=500$ samples from the distribution $P_*$ (\Cref{d:pstar}) we construct in the theorem. We generate an equal number of samples from the noise distribution $Q=\mathcal{N}(0, I_d)$, and run gradient descent to minimize the empirical NCE loss to obtain $\hat{\theta}_n$. Since we explicitly know what $\theta_*$ is, we can compute the squared error $\|\hat{\theta}_n-\theta_*\|^2$. We run 100 trials of this, where we obtain fresh samples each time from $P_*$ and $Q$, and average the squared errors over the trials to obtain an estimate of the MSE.

\Cref{fig:real-moments} shows the plot of $\log \text{MSE}$ versus dimension - we can see that the graph is linear. This corroborates the bound in \Cref{thm:nce_sample_complexity}, which tells us that as $n \to \infty$, the MSE scales exponentially with $d$. This behavior is robust even when the proportion of noise samples to true data samples is changed to 70:30 (though our theory only addresses the 50:50 case).
Finally, we note that optimizing the empirical NCE loss becomes numerically unstable with increasing $d$ (due to very large ratios in the loss), which is why we used comparatively moderate values of $d$.
\section{Asymptotic Statistical Complexity: Proof of Theorem \ref{thm:nce_sample_complexity}}
\label{s:nce_sample_complexity}
We will bound the error of the optimizer $\hat{\theta}_n$ of the empirical NCE loss (\cref{eq:empirical_nce}) using the bias-variance decomposition of MSE. To do this, we will reason about the random variable $\sqrt{n}(\hat{\theta}_n - \theta_*)$; let $\Sigma$ be its covariance matrix. Since $\hat{\theta}_n$ is an unbiased estimate of $\theta_*$, the MSE decomposes as
\begin{equation}
    \label{eq:mse}
    \ex{\norm{\hat{\theta}_n - \theta_*}_2^2} = \frac{1}{n} \Tr(\Sigma).
\end{equation}
The proof of \Cref{thm:nce_sample_complexity} proceeds as follows. In \Cref{subsec:4_1}, we show that the random variable $\sqrt{n}(\hat{\theta}_n - \theta_*)$ is asymptotically normal with mean $0$ and covariance matrix $\Sigma$ given by
\begin{equation}
    \label{eq:sigma}
    \Sigma = \nabla_\theta^2 L(\theta_*)^{-1} \var{\sqrt{n} \nabla_\theta L^n(\theta_*)} \nabla_\theta^2 L(\theta_*)^{-1}.
\end{equation}
We prove that the Hessian $\nabla^2_\theta L(\theta_*)$ is invertible in \Cref{s:hessian_invertibility}, so that the above expression is well-defined.

Since $\Sigma \succeq 0$ (it is a covariance matrix), to get a lower bound on $\Tr(\Sigma)$, it suffices to get a lower bound on the largest eigenvalue of $\Sigma$. Looking at the factors on the right hand side of \cref{eq:sigma}, we note first that \Cref{thm:nce_hessian_bound} ensures an exponential lower bound on \textit{all} eigenvalues of $\nabla^2_\theta L(\theta_*)^{-1}$. The bulk of the proof towards lower bounding the largest eigenvalue of $\Sigma$ consists of lower bounding the \textit{directional} variance of $\sqrt{n} \nabla_\theta L^n(\theta_*)$ in a suitably chosen direction $v$ in terms of $v^\top \nabla_\theta^2 L(\theta_*)v$. In \Cref{subsec:4_2} and \Cref{subsec:4_3}, we use
anti-concentration bounds to prove such variance lower bounds.

\subsection{Gaussian limit of \texorpdfstring{$\sqrt{n}(\hat{\theta}_n - \theta_*)$}{normalized MSE}}
\label{subsec:4_1}

To begin, we will show that $\sqrt{n}(\hat{\theta}_n - \theta_*)$ behaves as a Gaussian random variable as $n \to \infty$. Recall that the empirical NCE loss is given by \cref{eq:empirical_nce}:
\begin{equation*}
    L^n(\theta) = \frac{1}{n} \sum_{i=1}^n -\frac{1}{2} \ln \frac{p_\theta(x_i)}{p_\theta(x_i) + q(x_i)} + \frac{1}{n} \sum_{i=1}^n -\frac{1}{2} \ln \frac{q(y_i)}{p_\theta(y_i) + q(y_i)},
\end{equation*}
where $x_i \sim P_*$ and $y_i \sim Q$ are i.i.d. Let $\hat{\theta}_n$ be the optimizer for $L^n$. Then, by the Taylor expansion of $\nabla_\theta L^n$ around $\theta_*$, we have
\begin{equation}
    \label{eq:thetan_expression}
    \sqrt{n} \brac{\hat{\theta}_n - \theta_*} = - \nabla^2_\theta L^n(\theta_*)^{-1} \cdot \sqrt{n} \nabla_\theta L^n(\theta_*) - \sqrt{n} \cdot O\left(\norm{\hat{\theta}_n - \theta_*}^2\right)
\end{equation}
by \cite{gutmann2012noise}, who also show in their Theorem 2 that
$\hat{\theta}_n$ is a consistent estimator of $\theta_*$; hence, as $n \to \infty$, $\norm{\hat{\theta}_n - \theta_*}^2\to 0$.
\citet[Lemma 12]{gutmann2012noise} also assert\footnote{Translating notation: $T_d = n, J_{T_d}(\theta)=-2L^{n}(\theta)$ and setting $\nu=1$ gives $\mathcal{I_\nu}=2\nabla^2L(\theta_*)$ as in \cref{eq:hessian_at_opt}.} that the Hessian of the empirical NCE loss (\cref{eq:empirical_nce}) at $\theta_*$ converges in probability to the Hessian of the true NCE loss (\cref{def:nce}) at $\theta_*$, i.e., $\nabla^2_\theta L^n(\theta_*)^{-1} \xrightarrow{P} \nabla_\theta^2 L(\theta_*)^{-1}$.
On the other hand, by the Central Limit Theorem, $\sqrt{n} \nabla_\theta L^n(\theta_*)$ converges to a Gaussian with mean $\ex{\sqrt{n} \nabla_\theta L^n(\theta_*)} = \sqrt{n} \nabla_\theta L(\theta^*) = 0$, and covariance $\var{\sqrt{n} \nabla_\theta L^n(\theta_*)}$. With these considerations, we conclude that the random variable $\sqrt{n}(\hat{\theta}_n - \theta_*)$ in \cref{eq:thetan_expression} is asymptotically a Gaussian with mean $0$ and covariance $\Sigma=\nabla_\theta^2 L(\theta_*)^{-1} \var{\sqrt{n} \nabla_\theta L^n(\theta_*)} \nabla_\theta^2 L(\theta_*)^{-1}$, as defined in \cref{eq:sigma}.

Next, we introduce some quantities which will be useful in the subsequent calculations. As we already have a handle on the spectrum of $\nabla_\theta^2 L(\theta_*)$ from \Cref{thm:nce_hessian_bound}, the main object of our focus in \cref{eq:sigma} is the term $\var{\sqrt{n} \nabla_\theta L^n(\theta_*)}$. In particular, since we are concerned with the directional variance of $\Sigma$, we will reason about $\var{v^\transpose \cdot \sqrt{n} \nabla_\theta L^n(\theta_*)}$ for a fixed vector of ones, i.e., $v=1^{d+1}$. This vector has the property that for all $x, v^\top T(x) \ge 1$, as all non-constant coordinates of $T$ are non-negative, and the remaining coordinate is $1$. Writing out the variance term explicitly, we have
\begin{align*}
    \var{v^\transpose \cdot \sqrt{n} \nabla_\theta L^n(\theta_*)}
    & = n \cdot \frac{1}{4n} \varwith{x \sim p_*}{\frac{q(x) \cdot v^\transpose T(x)}{p_*(x) + q(x)}} + n \cdot \frac{1}{4n} \varwith{y \sim q}{\frac{p_*(y) \cdot v^\transpose T(y)}{p_*(y) + q(y)}} \tag*{(using linearity and independence)}\\
    & = \frac 14 \text{Var}_{x \sim p_*}\underbrace{\left[{\frac{q(x) \cdot v^\transpose T(x)}{p_*(x) + q(x)}}\right]}_{A(x)} + \frac 14 \text{Var}_{y \sim q}\underbrace{\left[\frac{p_*(y) \cdot v^\transpose T(y)}{p_*(y) + q(y)}\right]}_{B(y)}. \labeleqn{eq:variance_expression}
\end{align*}

Define $A(x) = \frac{q(x) \cdot v^\transpose T(x)}{p_*(x) + q(x)} = \frac{R_1(x)}{1 + R_1(x)} v^\transpose T(x)$ where $R_1(x) = \frac{q(x)}{p_*(x)}$ and $B(y) = \frac{p_*(y) \cdot v^\transpose T(y)}{p_*(y) + q(y)} = \frac{R_2(y)}{1 + R_2(y)} v^\transpose T(y)$ where $R_2(y) = \frac{p_*(y)}{q(y)}$. To show that $\varwith{x \sim p_*}{A(x)}$ and $\varwith{y \sim q}{B(y)}$ are large, we will need anti-concentration bounds on $R_1(x)$ and $R_2(y)$.

\subsection{Anti-concentration of \texorpdfstring{$R_1(x), R_2(y)$}{R1(x), R2(y)}}
\label{subsec:4_2}

Next, we show that $R_1$ and $R_2$ satisfy (quantitative) anti-concentration. We show this by a relatively straightforward application of the Berry-Esseen Theorem, and the proof is given in \Cref{s:r_anti_conc_proof}. Precisely, we show:
\begin{restatable}{lemma}{anticonclemma}
    \label{l:r_anti_conc}
    Let $d > 0$ be sufficiently large. Let $p = \hat{p}^d$ and $q = \hat{q}^d$ be any product distributions, and define $R(x) = \frac{q(x)}{p(x)}$. Suppose we have the following third moment bound:  $\exwith{x \sim \hat{p}}{\brac{\log \frac{\hat{q}}{\hat{p}}}^3} < \infty.$ Then, for any $\epsilon$, there exist constants $\alpha = \alpha(\hat{p},\hat{q}, \epsilon)$, $\mu = \mu(\hat{p},\hat{q}, \epsilon) < 0$ such that
    \begin{equation*}
			\prwith{x \sim p}{R(x) \le \exp\brac{\mu d - \alpha \sqrt{d}}} \ge \frac{1}{2} - \epsilon  \text{ and } \prwith{x \sim p}{R(x) \ge \exp\brac{\mu d + \alpha \sqrt{d}}} \ge \frac{1}{2} - \epsilon.
    \end{equation*}
\end{restatable}
Instantiating \Cref{l:r_anti_conc} for the pair $(p_*, q)$ gives us the anti-concentration result for $R_1$, while instantiating it for the reversed pair $(q, p_*)$ gives us the anti-concentration result for $R_2$. We can verify that the third moment condition holds in both instantiations, since in both the cases, $\log(\hat{q}/\hat{p})$ is a polynomial. Crucially, we will also utilize the fact that the constant $\mu$ is negative (as it equals $-\textup{KL}(\hat{p}|| \hat{q})$).
We are now ready to bound the variance of $A(x)$ and $B(y)$.

\subsection{Bounding the variance of \texorpdfstring{$A(x), B(y)$}{A(x), B(y)}}
\label{subsec:4_3}

Recall that $A(x)=\tfrac{R_1(x) \cdot v^\transpose T(x)}{1+R_1(x)}$ and $B(y)=\tfrac{R_2(y) \cdot v^\transpose T(y)}{1+R_2(y)}$.
Let $\mu, \alpha$ be the constants given by \Cref{l:r_anti_conc} for $p_*,q, \epsilon$. Further, let $L_1 = \exp\brac{\mu d - \alpha \sqrt{d}}$ and $L_2 = \exp\brac{\mu d + \alpha \sqrt{d}}$. Since the mapping $x \mapsto \frac{x}{1+x}$ is monotonically increasing in $x$,
\begin{equation}
\label{eq:r1_1}
\prwith{x \sim p_*}{R_1(x) \le L_1} = \prwith{x \sim p_*}{\frac{R_1(x)}{1 + R_1(x)} \le \frac{L_1}{1 + L_1}} \ge \frac{1}{2} - \epsilon
\end{equation}
\begin{equation}
\label{eq:r1_lb}
\prwith{x \sim p_*}{R_1(x) \ge L_2} = \prwith{x \sim p_*}{\frac{R_1(x)}{1 + R_1(x)} \ge \frac{L_2}{1 + L_2}}  \ge \frac{1}{2} - \epsilon.
\end{equation}
Let $T_{\mathrm{up}}$ be such that
\begin{equation}
	\label{eq:r1_2}
	\prwith[\Big]{x \sim p_*}{\norm{T(x)} \le T_{\mathrm{up}}} \ge \frac 78
	\qquad \text{and} \qquad
	\prwith[\Big]{x \sim q}{\norm{T(x)} \le T_{\mathrm{up}}} \ge \frac 78.
\end{equation}
In \Cref{s:tup_tail_bound}, we show that some $T_{\mathrm{up}} = O(\sigma^2 \sqrt{d})$ suffices for this to hold. Then, from \cref{eq:r1_1}, we have
\begin{align*}
    &\prwith{x \sim p_*}{\frac{R_1(x)}{1 + R_1(x)} \le \frac{L_1}{1 + L_1}} \ge \frac{1}{2} - \epsilon \\
    \implies\quad & \prwith{x \sim p_*}{\frac{R_1(x)\cdot v^\transpose T(x)}{1 + R_1(x)} \le \frac{L_1 \sqrt{d+1}\norm{T(x)}}{1 + L_1}} \ge \frac{1}{2} - \epsilon \tag*{(Cauchy-Schwarz)} \\
    \implies\quad & \prwith{x \sim p_*}{\brac{\frac{R_1(x) \cdot v^\transpose T(x)}{1+R_1(x)} \le \frac{L_1 \sqrt{d+1} \norm{T(x)}}{1 + Li_1}} \land \brac[\bigg]{\norm{T(x)} \le T_{\mathrm{up}}}}  \ge \frac{3}{8} - \epsilon \tag*{(union bound with \cref{eq:r1_2})}\\
    \implies\quad & \prwith{x \sim p_*}{\frac{R_1(x) v^\transpose T(x)}{1+R_1(x)} \le \frac{\sqrt{d+1} L_1 T_{\mathrm{up}}}{1 + L_1}} \ge \frac 38 - \epsilon \\
	\implies \quad &\prwith{x \sim p_*}{A(x)\le \frac{\sqrt{d+1} L_1 T_{\mathrm{up}}}{1 + L_1}} \ge \frac 14,
\end{align*}
for $\epsilon \le \tfrac 18$. On the other hand, recall also that $v$ satisfies $v^\transpose T(x) \ge 1$ for all $x$. Therefore, we have
\begin{align*}
	&\prwith{x \sim p_*}{\frac{R_1(x)}{1 + R_1(x)} \ge \frac{L_2}{1 + L_2}}  \ge \frac{1}{2} - \epsilon  \\
	\implies\quad &\prwith{x \sim p_*}{\frac{R_1(x) \cdot v^\transpose T(x)}{1 + R_1(x)} \ge \frac{L_2}{1 + L_2}}  \ge \frac{1}{2} - \epsilon \\
	\implies \quad &\prwith{x \sim p_*}{A(x)\ge \frac{L_2}{1 + L_2}}  \ge \frac{1}{4}.
\end{align*}
Now, consider the event
\[ A_1 = \set{A(x) \in \sbrac{\frac{1}{2} \exwith{x \sim p_*}{A(x)}, \frac{3}{2} \exwith{x \sim p_*}{A(x)} }}. \]
If this event were to intersect both the events $A_2 = \set{A(x) \le \tfrac{\sqrt{d+1} L_1 T_{\mathrm{up}}}{1 + L_1}}$ and $A_3 = \set{A(x) \ge \tfrac{L_2}{1+L_2}}$, then we would have
\begin{align*}
&\frac{1}{2} \exwith{x \sim p_*}{A(x)} \le \frac{\sqrt{d+1}L_1T_{\mathrm{up}}}{1+L_1} \quad \text{ and } \quad  \frac{3}{2} \exwith{x \sim p_*}{A(x)} \ge \frac{L_2}{1+L_2} \\
\implies \quad & \frac{L_2}{L_1} \cdot \frac{1}{T_{\mathrm{up}}\sqrt{d+1}}\cdot \frac{L_1 + 1}{L_2 + 1} \le 3.
\end{align*}
We will show that this cannot be the case. Recall that $\mu < 0$,
which means that $L_2 = \exp(\mu d+ \alpha\sqrt{d}) < 1$ for sufficiently large $d$. This means that for sufficiently large $d$ we have:
\begin{align*}
 &\exp(\mu d+ \alpha\sqrt{d}) < 1 \\
 \implies \quad & \exp(\mu d+ \alpha\sqrt{d}) - 2\exp(\mu d - \alpha\sqrt{d}) < 1 \\
 \implies \quad & 1 + \exp(\mu d+ \alpha\sqrt{d}) < 2 + 2\exp(\mu d - \alpha\sqrt{d}) \\
 \implies \quad & \frac{1+\exp(\mu d- \alpha\sqrt{d})}{1+\exp(\mu d+ \alpha\sqrt{d})} > \frac{1}{2} \\
 \implies \quad & \frac{L_1 + 1}{L_2 + 1} > \frac 12.
\end{align*}
Further, since $\frac{L_2}{L_1}=\exp(2\alpha\sqrt{d})$ and $T_{\mathrm{up}} = O(\sigma^2 \sqrt{d})$, we get that
\begin{align*}
    \frac{L_2}{L_1} \cdot \frac{1}{T_{\mathrm{up}}\sqrt{d+1}}\cdot \frac{L_1 + 1}{L_2 + 1} > \frac{\exp(2\alpha \sqrt{d})}{O(\sigma^2 d)} \cdot \frac{1}{2} > 3,
\end{align*}
where the last inequality follows for large enough $d$ since the numerator grows faster than the denominator. Hence for large enough $d$, $A_1$ cannot intersect both $A_2$ and $A_3$. If the event $A_1$ is disjoint from $A_2$, then
\begin{align*}
&\prwith{x \sim p_*}{A_1 \cup A_2} = \prwith{x \sim p_*}{A_1} + \prwith{x \sim p_*}{A_2} \le 1 \\
\implies \quad &\prwith{x \sim p_*}{A_1} \le 1 - \prwith{x \sim p_*}{A_2} \\
\implies \quad &\prwith{x \sim p_*}{A(x) \in \sbrac{\frac{1}{2} \exwith{x \sim p_*}{A(x)}, \frac{3}{2} \exwith{x \sim p_*}{A(x)} }} \le \frac 34 \\
\implies \quad & \prwith{x \sim p_*}{ \abs{A - \E_{p_*} A} \ge \frac 12 \E_{p_*} A} \ge \frac 14.
\end{align*}
This finally lower-bounds the variance of $A$ as
\begin{align*}
    \varwith{p_*}{A} & = \ex{(A-\E_{p_*} A)^2} \ge \frac{1}{4} (\E_{p_*} A)^2 \cdot \pr{(A - \E_{p_*} A)^2 \ge \frac{1}{4} (\E_{p_*} A)^2}  \ge \frac{1}{16} (\E_{p_*} A)^2.
\end{align*}
and thus $\E_{p_*}(A^2)-(\E_{p_*} A)^2 = \varwith{p_*}{A} \ge \frac{1}{16} (\E_{p_*}A)^2$, so that $(\E_{p_*} A)^2 \le \frac{16}{17} \E_{p_*} (A^2)$.

Altogether, we get $\varwith{p_*}{A} \ge \frac{1}{17} \E_{p_*} (A^2)$. An analogous argument in the case when $A_1$ is disjoint with $A_3$ yields the same bound on the variance.
Using an identical argument for $R_2$ and $B$, we get that for large enough $d$, $\varwith{q}{B} \ge \frac{1}{17} \E_{q} (B^2)$.

\subsection{Putting things together}
Putting together the lower bounds  $\varwith{p_*}{A} \ge \frac{1}{17} \E_{p_*} (A^2)$ and $\varwith{q}{B} \ge \frac{1}{17} \E_{q} (B^2)$ we showed in the previous subsection, and recalling \cref{eq:variance_expression}, we get
\begin{align*}
    \var{v^\transpose \cdot \sqrt{n} \nabla_\theta L^n(\theta_*)}
    & = \frac{1}{4} \varwith{p_*}{A} + \frac{1}{4} \varwith{p_*}{B} \ge \frac{1}{68} \brac{\exwith{p_*}{A^2} + \exwith{q}{B^2}} \\
    & = \frac{1}{68} \brac{\int_x \brac{\frac{q(x)^2p_*(x) + q(x)p_*(x)^2}{(p_*(x) + q(x))^2}} v^\top T(x)T(x)^\top v \,dx} \\
    & = \frac{1}{68} v^\top\cdot \int_x \frac{p_*(x)q(x)}{p_*(x) + q(x)} T(x) T(x)^\top dx \cdot v= \frac{1}{34} v^\top \nabla^2_\theta L(\theta_*) v \tag*{(from \cref{eq:hessian_at_opt}).}
\end{align*}
Finally, since $\nabla^2_\theta L(\theta_*)$ is invertible as claimed earlier (\Cref{l:hessian_invertibility}, \Cref{s:hessian_invertibility}), let $w$ be such that $v = \nabla^2_\theta L(\theta_*)^{-1} w$. Then, recalling the expression for $\Sigma$ in \cref{eq:sigma}, we can conclude that
\begin{align*}
    w^\top \Sigma w
    = v^\top \var{\sqrt{n} \nabla_\theta L^n(\theta)} v &= \var{v^\transpose \cdot \sqrt{n} \nabla_\theta L^n(\theta_*)} \\
    &\ge \frac{1}{34} v^\top \nabla^2_\theta L(\theta_*) v
    =  \frac{1}{34} w^\top \nabla^2_\theta L(\theta_*)^{-1} w \labeleqn{eq:sigma_lb_2},
\end{align*}
which gives us the desired bound on the MSE, namely
\begin{align*}
    \ex{\norm{\hat{\theta}_n - \theta_*}_2^2}
    &\ge \frac{1}{n} \Tr(\Sigma) \ge \frac{1}{n} \sup_{z} \frac{z^\top \Sigma z}{\norm{z}^2} \\
    &\ge \frac{1}{n} \frac{w^\top \Sigma w}{\norm{w}^2} \ge \frac{1}{34n} \frac{w^\top \nabla^2_\theta L(\theta_*)^{-1} w}{\norm{w}^2}\ge \frac{1}{34n} \inf_z \frac{z^\top \nabla^2_\theta L(\theta_*)^{-1} z}{\norm{z}^2} \ge \frac{\exp\brac{\Omega(d)}}{n},
\end{align*}
where the last inequality follows from \Cref{thm:nce_hessian_bound} and the fact that $\lambda_{\max}(\nabla^2_\theta L(\theta_*))^{-1}=\lambda_{\min}(\nabla^2_\theta L(\theta_*)^{-1})$.
This concludes the proof of \Cref{thm:nce_sample_complexity}.

\section{Overview of Results}
\label{s:overview}

Let $P_*$ be a distribution in a parametric family $\{P_\theta\}_{\theta \in \Theta}$. We wish to estimate $P_*$ via $P_\theta$ for some $\theta_* \in \Theta$ by solving a noise contrastive estimation task. To set up the task, we also need to choose a noise distribution $Q$, with the constraint that we can draw samples from it efficiently, and we can evaluate the probability density function efficiently.
We will use $p_{\theta}, p_*, q$ to denote the probability density functions (pdfs) of $P_{\theta}$, $P_*$, and $Q$.
For a data distribution $P_*$ and noise distribution $Q$, the NCE loss of a distribution $P_\theta$ is defined as follows:
\begin{definition}[NCE Loss]
\label{def:nce}
The NCE loss of $P_\theta$ w.r.t. data distribution $P_*$ and noise $Q$ is
\begin{equation}\label{eq:nce_loss}
\begin{split}
    L(P_\theta) = -\frac{1}{2}\E_{P_*} \log\frac{p_\theta}{p_\theta+q} - \frac{1}{2}\E_{Q} \log\frac{q}{p_\theta+q}.
\end{split}
\end{equation}
Moreover, the empirical version of the NCE loss when given i.i.d. samples $(x_1,\dots,x_n)\sim P_*^n$ and $(y_1,\dots,y_n) \sim Q^n$ is given by
\begin{equation}
    \label{eq:empirical_nce}
    L^n(\theta) = \frac{1}{n} \sum_{i=1}^n -\frac{1}{2} \log \frac{p_\theta(x_i)}{p_\theta(x_i) + q(x_i)} + \frac{1}{n} \sum_{i=1}^n -\frac{1}{2} \log \frac{q(y_i)}{p_\theta(y_i) + q(y_i)}.
\end{equation}
\end{definition}
The NCE loss can be interpreted as the binary cross-entropy loss for the classification task of distinguishing the data samples from the noise samples. To avoid calculating the partition function, one considers it as an additional parameter, namely we consider an augmented vector of parameters $\tilde{\theta} = (\theta, c)$ and let
$$p_{\tilde{\theta}}(x) = \exp(E_{\theta}(x) - c).$$
The crucial property of the NCE loss is that it has a unique minimizer:
\begin{lemma}[\citealt{gutmann2012noise}]
\label{lem:nce_opt}
The NCE objective in Definition \ref{def:nce} is uniquely minimized at $\theta = \theta_*$ and $c = \log(\int_x \exp(E_{\theta^*}(x))dx)$  provided that the support of $Q$ contains that of $P_*$.
\end{lemma}

We will be focusing on the Hessian of the loss $L$, as the crucial object governing both the algorithmic and statistical difficulty of the resulting objective. We will show the following two main results:
\begin{theorem}[Exponentially flat Hessian] For $d > 0$ large enough, there exists a distribution $P_*=P_{\theta_*}$ over $\R^d$ such that \begin{itemize}
	\label{thm:nce_hessian_bound}
	\item$\mathbb{E}_{P_*}[x] = 0$ and $\mathbb{E}_{P_*}[xx^\top] = I_d$.
	\item $P_*$ is a product distribution, namely $p_* (x_1, x_2, \dots, x_d) = \prod_{i=1}^d p^*(x_i)$.
	\item
		The NCE loss when using $q = \mathcal{N}(0, I_d)$ as the noise distribution has the property that
		$$ \|\nabla^2 L(\theta_*)\|_2 \leq \exp\left(-\Omega(d)\right).$$
\end{itemize}
\end{theorem}
By way of remarks, note that the above example of a problematic distribution $P^*$ is extremely simple. Namely, $P^*$ is a product distribution, with 0 mean and identity covariance. It actually is also the case that $P^*$ is log-concave---which is typically thought of as an ``easy'' class of distributions to learn due to the fact that log-concave distributions are unimodal.

Also, by way of remarks, the fact that the Hessian is exponentially flat near the optimum means that gradient-descent based optimization without additional tricks (e.g., gradient normalization, second order methods like Newton's method) will fail. (See, e.g., Theorem 4.1 and 4.2 in \citet{liu2021analyzing}.)

More seriously, we use the result above to show that the asymptotic statistical complexity, using the above choice of $P^*, Q$, is exponentially bad in the dimension. This substantially clarifies results in \citet{gutmann2012noise}, who provide an expression for the asymptotic statistical complexity in terms of $P^*, Q$ (Theorem 3, \citet{gutmann2012noise}), but from which it's very difficult to glean quantitatively how bad the dependence on dimension can be for a particular choice of $P^*, Q$.

\begin{theorem}[Asymptotic Statistical Complexity]
	\label{thm:nce_sample_complexity}
	Let $d > 0$ be sufficiently large and $Q = \mathcal{N}(0,I_d)$. Let $\hat{\theta}_n$ be the optimizer for the empirical NCE loss  $L^n(\theta)$ with the data distribution $P_*$ given by \Cref{thm:nce_hessian_bound} above and noise distribution $Q$.
	Then, as $n \to \infty$, the mean-squared error satisfies $$\ex{\norm{\hat{\theta}_n - \theta_*}_2^2} = \frac{\exp(\Omega(d))}{n}.$$
\end{theorem}

\section{Conclusion}
\label{sec:conclusion}
Despite significant interest in alternatives to maximum likelihood---for example NCE (considered in this paper), score matching, etc.---there is little understanding of what there is to ``sacrifice'' with these losses, either algorithmically or statistically. In this paper, we provided formal lower bounds on the asymptotic sample complexity of NCE, when using a common choice for the noise distribution $Q$, a Gaussian with matching mean and covariance. Thus, it is likely that even for moderately complex distributions in practice, more involved techniques like \citet{gao2020flow, rhodes2020telescoping} will have to be used, in which one learns a noise distribution $Q$ simultaneously with the NCE minimization or ``anneals'' the NCE objective. There is very little theoretical understanding of such techniques, and this seems like a very fruitful direction for future work.

\section{Bounding the matrix integral in Equation \ref{eq:hessian_bound_1}}
\label{s:cs_vector}
We prove a variant of the Cauchy-Schwarz inequality that gives us a handle on norms of matrix integrals.
\begin{lemma}
	\label{l:vector_cs}
	Let $f: \R^d \to \R$ and $A: \R^d \to \R^n$ be integrable functions, with $M = \int_x f(x) A(x) dx$. Then we have
	\begin{equation}
		\label{eq:vector_cs}
		\norm{M}_2^2 = \norm{\int_x f(x)A(x)ds}_2^2 \le \brac{\int_x \abs{f(x)}^2 dx} \brac{\int_x \norm{A(x)}_2^2 dx}.
	\end{equation}
	Similarly, if $A:\R^d \to \R^{n \times n}$ is a matrix valued function then
	\begin{equation}
		\label{eq:frob_norm_cs}
		\norm{M}_F^2 = \norm{\int_x f(x)A(x)}_F^2 \le \brac{\int_x \abs{f(x)}^2 dx} \brac{\int_x \norm{A(x)}_F^2 dx}.
	\end{equation}
\end{lemma}
\begin{proof}
	The proof follows from the Cauchy-Schwarz inequality. Since we integrate component-wise, for \cref{eq:vector_cs} we have that
	\[ M_i^2 = \brac{\int_x f(x)A(x)_i dx}^2 \le \brac{\int_x f(x)^2dx} \brac{\int_x A(x)_i^2 dx}. \]
	Summing over $i$, we get the result. The matrix variant \cref{eq:frob_norm_cs} follows by looking at the matrix $M$ as a vector in $\R^{n^2}$.
\end{proof}

\section{Proof of Lemma \ref{l:r_anti_conc}}
\label{s:r_anti_conc_proof}
We restate the lemma for convenience:
\anticonclemma*

\begin{proof}
We will analyze the behaviour of $R(x)$ using the Berry-Esseen theorem. Given that $p_* = \hat{p}^d$ and $q = \hat{q}^d$ are product distributions, let $r(x)$ be the random variable defined by $r(x) = \frac{\hat{q}(x)}{\hat{p}(x)}$, $x \sim \hat{p}$.
Let $y_i(x) = \log r(x)$ for $1 \le i \le d$ be $d$ independent copies of the random variable $r(x)$.
Let $\ex{y_i} = \mu_r$, $\ex{\norm{y_i - \mu_r}^2} = \sigma_r^2$ and $\ex{\norm{y_i - \mu_r}^3} = \gamma_r$, all of which are well defined by the hypothesis of the lemma.
Let $Y = \sum_{i=1}^d y_i$, and $Z$ be the standard Gaussian in $\R$. Then, by the Berry-Esseen Theorem \cite[Theorem 3.4.17]{durrett2019probability},
\begin{equation*}
	\pr{ \frac{Y - \mu_r d}{\sigma_r \sqrt{d}} \le -c} \ge \pr{Z \le -c} - \frac{C_{\textup{BE}} \cdot \gamma_r}{\sigma_r^3 \sqrt{d}},
\end{equation*}
where $C_{\textup{BE}} < 1$ \citep{van1972application} is an absolute constant.
We can now choose $c=c(\epsilon)$ such that $\pr{Z \le c} \ge \tfrac{1 - \epsilon}{2}$. Further, we can choose $d$ large enough so that $\tfrac{C_{\textup{BE}} \cdot \gamma}{\sigma^3 \sqrt{d}} \le \tfrac{\epsilon}{2}$. Then for $\mu = \mu_r$ and $\alpha = c \sigma_r$, we have
\begin{equation*}
	\prwith{x \sim p}{R(x) \le \exp\brac{\mu d - \alpha \sqrt{d}}} \ge \frac{1}{2} - \epsilon.
\end{equation*}
Since $Z$ is symmetric around 0, Berry-Esseen gives us the other inequality for the same choice of $\mu$ and $\alpha$,
\begin{equation*}
	\pr{ \frac{Y - \mu_r d}{\sigma_r \sqrt{d}} \ge c} \ge \pr{Z \ge c} - \frac{C_{\textup{BE}} \cdot \gamma_r}{\sigma_r^3 \sqrt{d}} \ge \frac{1}{2} - \epsilon.
\end{equation*}
Note that the constants $\mu$ and $\alpha$ are independent of $d$. Further, note that $\mu = \mu_r = -\textup{KL}(\hat{p}|| \hat{q}) < 0$.
\end{proof}

\section{Invertibility of the Hessian}
\label{s:hessian_invertibility}

We prove that the Hessian of NCE loss for the exponential family given by $T(x) = (x_1^4, \ldots, x_d^4, 1)$ is invertible. In particular, we have the following lemma:
\begin{lemma}
	\label{l:hessian_invertibility}
	Let $Q = \mathcal{N}(0,I_d)$ be the standard Gaussian in $\R^d$. Let $\hat{P}$ be the log concave distribution defined in \cref{d:pstar}. Let $P = \hat{P}^d$. Let $q$ and $p$ denote the density functions of $Q$ and $P$ respectively. Observe that $P$ is in the exponential family given by $T(x) = (x_1^4, \ldots, x_d^4, 1)$, and equals $P_{\theta_*}$ for some $\theta_*$. Then the hessian of the NCE loss with respect to distribution $P$ and noise $Q$ given by
	\[ H = \nabla_\theta^2 L(\theta_*) = \frac{1}{2} \int_x \frac{p_*q}{p_* + q} T(x)T(x)^\top \]
	is invertible.
\end{lemma}
\begin{proof}
	For any subset $A \subseteq \R^d$, define
	\[ H_A = \frac{1}{2} \int_{x \in A} \frac{p_* q}{p_* + q} T(x)T(x)^\top. \]
	Observe that the density functions $p_*$ and $q$ of $P_*$ and $Q$ respectively are strictly positive over all of $\R^d$. Therefore, for any subset $A \subseteq \R^d$ and any $v \in \R^{d+1}$, we have
	\[ v^\top H v  \ge \frac{1}{2} \int_{x \in A} \frac{p_* q}{p_* + q} v^\top  T(x) T(x)^\top v = v^\top H_A v. \]
	Given a vector $v \in \R^{d+1}$, we will pick $A$ such that $\abs{T(x)^\top v} > 0$ for all $x \in A$. Note that the set $\mathcal{B} = \set{e_1 + e_{d+1}, \ldots, e_d + e_{d+1}, e_{d+1}}$ is a basis. Therefore, if $b^\top v = 0$ for all $b \in \mathcal{B}$, then $v = 0$. Hence, there exists some $x \in \set{e_1, \ldots, e_d}$ such that $\abs{T(x)^\top v} > 0$.
	Since $x \mapsto T(x)^\top v$ is a continuous function, we can find an open set $A$ around $x$ such that
	\[ \abs{T(y)^\top v} > 0, \qquad \forall y \in A.  \]
	It follows that
	\[ v^\top H_A v = \frac{1}{2} \int_{x \in A} \frac{p_* q}{p_* + q} v^\top T(x)T(x)^\top v = \frac{1}{2} \int_{x \in A} \frac{p_* q}{p_* + q} \abs{T(x)^\top v}^2 > 0. \]
	Let $B = \R^d \setminus A$. Since $v^\top H_A v > 0$ and $ v^\top H_B v \ge 0$, we have that $v^\top H v >0$. Since this holds for any arbitrary non-zero vector $v$, the matrix $H$ must be full rank. Since $H$ is an integral of PSD matrices, it is a full rank PSD matrix and hence invertible.
\end{proof}

\section{Tail bounds for Equation \ref{eq:r1_2}}
\label{s:tup_tail_bound}

We prove that some $T_{\mathrm{up}} = O(\sigma^2\sqrt{d})$ suffices to obtain the bounds in \cref{eq:r1_2}. Concretely, we prove tail bounds for $\norm{T(x)}$ using tail bounds for $P_*$ and $Q$.
We will use Lemma 1 from \cite{laurent2002adaptive} which proves a bound for $\chi^2$ distributions:
\begin{lemma*}[Lemma 1, \cite{laurent2002adaptive}]
	\label{t:chi_square_bound}
	If $X$ is a $\chi^2$ random variable with $d$ degrees of freedom, then for any positive $t$,
	\[ \pr{X - d \ge 2 \sqrt{td} + 2t} \le \exp\brac{-t}. \]
\end{lemma*}
Then, for $x \sim Q$, $\norm{x}^2$ is a $\chi^2$ random variable with $d$ degrees of freedom. Observe that for $t,d \ge 4$, we have $d + 2t + 2\sqrt{td} \le 2td$. In particular, we have the weaker bound
\[ \prwith[\Big]{x \sim Q}{\norm{x}^2 \ge 2dt^2} \le \exp\brac{-t^2}, \]
implying that
\[ \prwith[\Big]{x \sim Q}{\norm{x} \ge t}	\le \exp\brac{-\frac{t^2}{2d}}. \]
Further, if $\norm{x} \ge \sigma^2 \sqrt{d}$, $q(x) \ge p_*(x)$, implying that for $t \ge \sigma^2 \sqrt{d}$
\[ \prwith[\Big]{x \sim P_*}{\norm{x} \ge t} \le \exp\brac{-\frac{t^2}{2d}}. \]
In particular, for any $\delta$ such that $\log(1/\delta) \ge \sigma^4$, we have
\begin{equation}
	\label{eq:gaussian_tail}
	\prwith[\Big]{x \sim Q}{\norm{x} \ge \sqrt{2d \log(1/\delta)}}	\le \delta
	\qquad \text{and} \qquad
	\prwith[\Big]{x \sim P_*}{\norm{x} \ge \sqrt{2d \log(1/\delta)}}	\le \delta.
\end{equation}

redpajama_set_name: RedPajamaArXiv


text:
Armstrong, who spent his childhood in San Luis Obispo, "has good feelings about the area," Levenson said.
Graphics Department graduate Pat Waters helped arrange the donation of paintings that range from a few square feet to 7-by-7 feet, and were appraised from a few thousand dollars up to $80,000.
Duane paints landscapes and modern expressionism, while his wife paints sea shells and flowers.
Some paintings have been displayed on campus, and a couple will go into the new Science Building when it's complete.
The majority are on exhibit and for sale at Pierce Modern gallery in Paso Robles.
Levenson had difficulty finding a venue for an exhibit.
"All I heard is a two-year wait, a three-year wait," he said.
While visiting a couple of new galleries that have no waiting lists, Levenson shot and sent photos to Armstrong to let him decide.
"Hands down, he picked Pierce, saying it looked more like a New York Gallery with its separate rooms," said Levenson.
The Armstrongs, who live near Santa Cruz in Boulder Creek, could not be reached for comment, but their website says that they met in 1986 when Duane was already a well established painter. Annie's early work was inspired by Georgia O'Keeffe paintings, which led to her large-scale paintings, with a focus on minute details. Unlike her husband, Annie is more consistent in her work. With such different styles and subject matter, the two artists don't step on each others' toes.
Duane was only 12 when he won his first art award, according to the website. His seven generations of Pond painting series, each featuring from five to 21 paintings, took nine to 18 months to complete, as they contain up to 70 layers of paint.
His landscapes and sunrises are just a small part of his work, which includes the series "The Lord of the Rings," "Hot Air Balloons" and "Sand Dunes" as well as his "Pueblo Series." His recent abstract series is called "Rings."
Some major corporations and banks have Armstrong's paintings in their permanent collections as do about 80 other corporate collectors nationwide and in Asia. The Sierra Club selected his painting "High Sierra" for its 1985 poster.
Sales proceeds from the Cal Poly donation to the Graphics Communication Department will help fund scholarships, program development endowments and Learn By Doing initiatives, Levenson said.
Annie Armstrong's painting 'Muricadea,' focuses on the details of a flower.
Pet Tales: Meet Maple, a playful puppy looking for a forever home
Fly over Heart Hill and Niner Wine Estates in Paso Robles

redpajama_set_name: RedPajamaCommonCrawl


The Ontario Securities Commission has issued a warning against five unlicensed crypto companies involved in schemes to encourage investors to trade or invest in cryptocurrencies. The regulator started gathering information about these platforms after receiving complaints about them.
Btcreal, Bitserial, Hypercube Ventures LP, Cabincoin OÜ, and Baappay Inc. are not registered in Ontario to solicit investments or provide advice on investing in, buying or selling securities.
The Commission started gathering information on several crypto trading platforms last month after it received a number of complaints about them, CBC reported, adding that to date no platform has been recognized by the OSC as an exchange or exempted from recognition. Any platform that offers cryptocurrencies that fit the definition of securities "must determine whether it is a marketplace. Marketplaces are required to comply with the rules governing exchanges or alternative trading systems," the news outlet elaborated.
Last month, the OSC issued an Investor Alert on Bitconnect, Bitconnect Coin and the BCC Exchange. "Bitconnect, the BCC Exchange and representatives of Bitconnect are not registered in Ontario to solicit investments or provide advice on investing in, buying or selling securities," the regulator wrote.
The next company is Bitserial which "claims to offer opportunities to invest in BTE Tokens." The firm also "encourages investors to participate in a lending program where they can exchange bitcoin, litecoin, or ethereum for BTE Tokens that are 'lent out' for high returns," the Commission detailed.
The third company is Hypercube Ventures LP which manages multiple websites to encourage investors to buy "emission pools" in order to generate "VNN cryptocurrency," according to the Commission.
The fourth is Cabincoin OÜ which "is currently advertising an unregistered token sale for Cabincoin Tokens," claiming that "the future value of these tokens will far exceed their initial price," the OSC wrote.
What do you think of the OSC's warning? Let us know in the comments section below.
Images courtesy of Shutterstock, the OSC, Btcreal, Bitserial, and Cabincoin.


After years of debate, Abu-Huraira will be allowed to build an Islamic center in St. Anthony, Minnesota.
Abu-Huraira was planning to open a prayer and meeting center in the old Medtronic headquarters in the Minneapolis suburb in 2012, but because the building was zoned as light industrial the group needed a special zoning permit. But the city council rejected the request saying city zoning wouldn't support the facility. Some residents were also very outspoken against the center.
In the fall of 2012, the U.S. attorney's office in Minneapolis began investigating the matter and in August 2014, the federal government sued the city for rejecting Abu-Huraira's request, alleging it was a case of religious discrimination under the Religious Land Use and Institutionalized Persons Act (RLUIPA).
And on Tuesday, U.S. Attorney Andy Luger and an official in the U.S. Attorney General's office announced they reached a settlement which will allow an Islamic Center to open in the small Minneapolis suburb, reports note.
The agreement also says the city of St. Anthony will not treat Abu-Huraira or any other religious group in a discriminatory way, KSTP reports, requiring elected leaders, city managers and some city employees to be trained about RLUIPA.
The city will also make RLUIPA information available to the public through its website and periodically report to the justice department, FOX 9 notes.
St. Anthony is one of six incidents in recent years where cities opposed Muslim land use in Minnesota, the Star Tribune reports. The others cities include Blaine, Plymouth, Willmar, Bloomington and St. Cloud. All were eventually approved, except for St. Cloud – that proposal was withdrawn.


In a group of 2 you will create a shared Prezi presentation about your Tablet findings.
Page 1 - Your top 2 tablets, side by side for comparison.
Page 2 - Your #1 choice of tablets with facts to back up your arguments.
this tablet will or will not be successful.
Today, tablet computers are taking a bigger and bigger percentage of the desktop PC market.
The tablet category is growing as it expands to include a variety of E-Readers, and many different sizes of Tablets.
.... and this is just barely scratching the surface.
The dominating companies in this industry are currently Apple, Samsung, Amazon, Microsoft among others.
Other companies have attempted to compete with but failed miserably including the HP Slate, and Asus Eee Pad to name a few.
In this 2 day assignment I would like you to compare and contrast 10 pro's and con's to using a tablet compared to a traditional PC or Notebook.
I would like you to focus on what you feel are the top 2 tablets on the market. I would like you research an upcoming tablet that has not been released yet. Please provide information and photos on each tablet.
Create a Match up between your top 2 tablet choices. Compare them side by side based on spec's, performance, and 2 other categories of your choosing.
You must choose 1 overall winner at the end.
The assignment should be at least 4-5 paragraphs with photos, specs and details.
choose an overall winner - why does is this tablet better than all the rest?
Use this Tablet Comparison Chart as your starting point.
Read: 10 Best Tablets in the World Today - Do you agree or disagree?
Due, end of class on Thursday.


July 12 - In her first speech since the Taliban in Pakistan shot her for advocating girls' education, Malala Yousafzai celebrated her 16th birthday at the U.N. to appeal for free education for all children. Rough cut (no reporter narration).
STORY: In her first speech since the Taliban in Pakistan tried to kill her for advocating education for girls, Malala Yousafzai celebrated her 16th birthday on Friday at the United Nations, appealing for compulsory free schooling for all children. Wearing a pink head scarf, Yousafzai told U.N. Secretary-General Ban Ki-moon and nearly 1,000 students from around the world attending a Youth Assembly at U.N. headquarters in New York that education was the only way to improve lives. "Let us pick up our books and pens. They are our most powerful weapons. One child, one teacher, one pen and one book can change the world. Education is the only solution," she said. Yousafzai was shot in the head at close range by gunmen in October as she left school in Pakistan's Swat Valley, northwest of the country's capital Islamabad, after campaigning against the Islamist Taliban efforts to deny women education. She said her attackers thought a bullet would silence her and others like her, but that they had "failed." "The terrorists thought that they would change my aims and stop my ambitions but nothing changed in my life except this: weakness, fear and hopelessness died. Strength, power and courage was born," said Yousafzai to loud applause. U.N. Special Envoy for Global Education, former British Prime Minister Gordon Brown, said Friday's event was not just a celebration of Malala's birthday and her recovery, but of her vision. Yousafzai presented U.N. Secretary General Ban Ki-moon with a petition signed by nearly 4 million people in support of 57 million children who are not able to go to school and demanding that world leaders fund new teachers, schools and books and end child labor, marriage and trafficking.


The Community and Economic Development area of study explores how community and government leaders work together to build critical organizations and partnerships, such as community-based services, arts centers, downtown businesses and a variety of urban renewal and redevelopment projects.
Demonstrate a working knowledge of the issues in CED.
Summarize the policies that influence CED decisions and approaches.
Analyze how CED policies are applied.
Compare and contrast how different tools are applied to solve CED problems.


text:
Kirk Herbstreit: Players who opt out of bowl game 'don't love football' | College Football Enquirer
Yahoo Sports' Dan Wetzel and Pete Thamel, and Sports Illustrated's Pat Forde discuss Kirk Herbstreit and Desmond Howard's comments on ESPN's College GameDay about players who opted out of bowl games to focus on the NFL Draft.
[CROWD CHEERING]
DAN WETZEL: "Isn't that what we do as football players? We compete. I don't know if changing it, expanding it is going to change anything. I really don't," Herbstreit said on "College Gameday" last weekend, setting off a tumultuous fire. "I think this era of player just doesn't love football." Desmond Howard then agreed. Herbstreit later clarified that, of course, some players love the game the same today as ever but some don't. I'll always love the players of this game and sorry if people thought I generalized them.
It's a good controversy because it's fairly harmless. I thought the pile-on was a little much. I will say this, I think we will all just get this out of the way. Desmond Howard and Kirk Herbstreit are both really good guys who work hard and are great for college football. And yes, look, if these guys don't think that, then so be it. I disagree. I don't know. I don't know whether they love football as much. They might not. I really have no-- that's a hard one to quantify, but the Bowl opt-outs and the lack of the Bowl, I just think it's a terrible way to try to figure that out.
And to not mention Brian Kelly bailed on his team while they still could have made the playoffs and Lincoln Riley bailed on them. All these guys bailed on Bowl games and not bring up-- look, the ball games just don't matter that much to anybody I think was their mistake on that. What did you think of the kerfuffle, Pat?
PAT FORDE: Yeah, I wrote about it Saturday night. I really didn't intend to write about it because I thought, kind of like you, Dan, that there was such a kind of almost an overreaction during the day. But when I saw Matt Corral got hurt Saturday night in the Sugar Bowl and he comes out on crutches I was like, OK, I think there's a column. That on the same day that Kirk Herbstreit and Desmond Howard kind of decried today's players not playing in Bowl games, Matt Corral showed why there's a good reason players don't play in Bowl games.
Now, I think Matt Corrall is going to be fine. I don't think his pro stock is probably going to be affected. He should be-- whatever the injury is, and we usually never know for sure, but if the X-rays were negative, if it's something that he will be able to rehab and be able to perform at a Pro day, at a scouting combine, whatever he needs to do.
But that injury, you watch that play-- OK, yeah. Could have been worse. And you watch what happened to Jaylen Smith for Notre Dame in the Fiesta Bowl five, six years ago, and Jake Butt, what happened to him in the Orange Ball, and there are reasons that players don't want to play in games that aren't absolutely materially vital to their season's success. And it doesn't mean they don't love football and it doesn't mean they're entitled.
The coaches and the programs have made it very clear for the last decade or so that the Bowl games are of secondary importance by what they do with hirings and firings, and the players are looking at that saying well, OK, why do I have to put it on the line here when I'm worried and I've got an agent talking to me and I've got an advisor and I've got my mom and dad saying, don't get hurt, you know? So, I really cannot blame players for sitting out any games that are non playoff games.
PETE THAMEL: So, the business has changed and all that we've seen is the players getting wise to the business, right?
PAT FORDE: Yeah.
PETE THAMEL: Like, Lincoln Riley went and got his, Brian Kelly went and got his, and all these players who are headed to the draft, they're just ready to go get theirs. Like, they have followed the example set to them by the coaches and the leaders in the sport. And, you know, if you look back from Herbstreit's time to now and Desmond Howard's time to now, ESPN has went and got theirs too. They've doubled the inventory of Bowls and they make a lot of money off it.
So, everyone's making business decisions here. That's all there is. It's quite simple and linear and I certainly have no problem with the players making the business decisions that they did and really didn't from the start. Because, look, people are still watching, the games are entertaining, everybody's getting what they want out of this. The system isn't radically different.
Now, are there are less people who care about the Peach Bowl because Kenneth Walker and Kenny Pickett aren't there? Sure. You know what? And that's the Peach Bowl's fault and that's the commissioners' faults for not looping that showcase in to matter more where those guys would have definitely played, so. And it's, again, you're not going to expand the playoff and end opt-outs, but if you expand the playoff you make the postseason more vibrant and relevant for longer, and it matters more and more places, like we talked about in the last pod, and that's what really helps the sport.
Footage of fire drill at Bangladesh airport shared as genuine emergency
A video of what appears to be a plane engulfed by flames has been viewed thousands of times in Facebook posts that claim it shows an accident at the Hazrat Shahjalal International Airport that serves the Bangladeshi capital Dhaka. However, the video has been shared in a misleading context. Major media outlets reported that the footage was taken during an emergency fire drill that used a dummy plane.The video was shared here on Facebook on January 3, 2022. It shows flames and smoke billowing from


This is the 16th part of a series of posts adapted from a paper I presented at a New Covenant Theology think tank in upstate New York in July 2010.
Our sanctification is achieved by God through our union with Christ. "He who calls you is faithful; he will surely do it" ( ).
The great existence-altering event that happens in our salvation is our union with Christ through His Spirit.
 Do you not know that all of us who have been baptized into Christ Jesus were baptized into his death?  We were buried therefore with him by baptism into death, in order that, just as Christ was raised from the dead by the glory of the Father, we too might walk in newness of life.  For if we have been united with him in a death like his, we shall certainly be united with him in a resurrection like his ( ).
That union will result in our glorification on the last day.
However, we're still stuck in the tension of the now and not yet.
We have been given the Holy Spirit and we are now positionally sanctified and set apart by God as His chosen redeemed. Yet the holiness we are promised often can be elusive in the Christian walk.
What is Paul's answer to this? Paul encourages us in two things: first, to remember the gospel of Jesus Christ and second, to rely on the Holy Spirit – to "walk in a manner worthy of the calling to which you have been called" ( ). Paul roots his exhortations to believers in the indicative, our position in Christ. Paul's imperative, which can flow only from that indicative, is to walk as those called by Christ.
That indicative and imperative are treated as inseparable by Paul: because of Christ, this is who you now are and this is how you walk as one called. Paul uses exhortation, encouragement, and parenesis – Paul implores us to be who we now are because of the Spirit of Christ who dwells in us.
 Abraham Kuyper, The Work of the Holy Spirit (London: Funk and Wagnall's, 1900), 333.
 Sinclair B. Ferguson, The Holy Spirit (Downers Grove, IL: 1996), 152.
3 Do you not know that all of us who have been baptized into Christ Jesus were baptized into his death? 4 We were buried therefore with him by baptism into death, in order that, just as Christ was raised from the dead by the glory of the Father, we too might walk in newness of life.


Poland is currently experiencing a big turnaround in regard to national housing policy, with the introduction of the Home Plus program as well as the revision of the thermo-modernization policy being an important part of the process. Habitat for Humanity Poland is actively taking part in the consultancy process with the aim of introducing the issue of energy poverty into the government's agenda.
In Poland, the responsibility for energy efficiency-related issues is shared between the Ministry of Energy, the Ministry of Environment and the Ministry of Infrastructure and Construction. Moreover, some instruments of energy poverty prevention (such as fuel allowance) also fall under the responsibility of the Ministry of Family, Labor and Social Policy. However, no interdepartmental body has been established to manage the issue and the relevant stakeholders are only in the very early stages of their advocacy work.
The framework of thermo-modernization in Poland covers the thermal refurbishment of all types of residential and municipal buildings (including schools and hospitals); the local district heating network and local sources of heating; and the installation of renewable energy sources or high efficiency energy equipment. Thermo-modernization has traditionally fallen within the area of competence of the Ministry of Infrastructure that is responsible for the program based on the 1998 Act on Supporting Thermo-modernization and Repair of Exploited Buildings (dedicated mostly to multi-family housing). The program is based on a premium constituting a repayment of part of a commercial loan distributed by the State Development Bank. The premium is only offered after a thorough and costly energy audit and is based on the estimates of the annual energy savings. Therefore, despite having been offered to cooperatives, condominiums and individual owners, it has ultimately only helped to renovate mostly bigger units. The program's evaluators have also pointed out a 'creaming' effect – in practice, the premiums were distributed among the units which were well-managed and inhabited by families with high financial and cultural capital.
After reviewing the results of the program in 2017, the Act on Supporting Thermo-modernization will be amended and adjusted to the needs of contemporary renovation investments. Habitat for Humanity Poland is using the opportunity to share its experience gathered during thermo-modernization and renovation projects. Direct contact with beneficiaries and other stakeholders (such as social workers and property administrators) has helped the organization to understand the obstacles preventing people from taking advantage of the subsidy.
Introducing a premium or subsidy which is not based on the commercial loan but on the amount of savings gathered for the purpose of renovating the unit.
We believe the proposed solutions can help to increase the scale of thermo-modernization and effectively engage private funding in cases where such engagement would benefit inhabitants.
Barbara Audycka is an advocacy officer at Habitat for Humanity Poland. She holds a Ph.D. in social sciences, having written a dissertation on housing-related social movements. Barbara has been working both on Poland's Home Plus program and Housing in the EU Urban Agenda.


Frederck Wiegand Jr. is a world wide consultant in the oil and gas business.
Soliciting investor - partner to bring H-well drilling and UBD well drilling rig and associated equipment to Northern MX. Have verbal agreements to drill for shale gas as Mexico contractor.


As a promethion looks comparable in startup cost with a MiSeq (though, actually, because the machine is free, you get more sequencing for your buck with the promethion), I wonder how much it would cost to sequence a bacterial genome on the promethion? This post involves a bunch of guesswork, extrapolation and the backs of envelopes, so ymmv. Corrections/improvements welcome.
Throughput per flowcell, promethion flowcells have 4x the pores of minion, so i'm assuming 4x the throughput. From a quick scan of the community site, 1 Gb of data from the latest pore/chem seems like a decent balance for optimism/realism. ONT have been spouting about 10Gbp, so I'm going to include an optimistic 5 Gbp per minion flowcell as well. Some people have reported this on the community.
Promethion flow cell throughput = 4 to 20 Gbp.
Cost per Gbp of $95-$475.
Add to that the cost of barcoding (each promethion flowcell is divided into 4 sample chambers, not 100% clear that these are distinguishable from the data, as the literature says that each flowcell is addressable, not each sample chamber). $24 per sample.
Reminder that the $119 is very optimistic, and the $219 is definitely 'on a good day'.
nanopore/lab experts – have i missed anything?
What will be minimum promethion flowcell order size/shelf life?
What are the 'official' promethion flowcell throughput stats?
How may people are in the promethion queue/how long would i have to wait if i ordered one now?
* i have used the term cost here, while Mick (who i like very much) points out that i should have been saying price.
@biomickwatson makes the excellent point that staff costs and compute costs for promethion are probably higher. I decided not to address these as it opens up various cans of worms.


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Virtual Screening Room: L'Innocente
NEW RESTORATION!
OPENS FRIDAY, APRIL 24
Streaming Link Available Friday, April 24
A portion of proceeds benefit The Brattle
This title is no longer available to watch in our Virtual Screening Room
1976 • dir Luchino Visconti w/Giancarlo Giannini, Laura Antonelli, Jennifer O'Neill • 129 min
Gabrielle d'Annunzio's passionate novel is brought to life in the final masterpiece from acclaimed director Luchino Visconti (The Leopard). In late-nineteenth century Italy, Tullio (Giannini), an insatiable aristocrat, grows bored with his timid wife Giuliana (Antonelli) and neglects her for his more exciting mistress, the wealthy widow Countess Teresa Raffo (O'Neill). After learning that Giuliana is having a torrid affair of her own, he becomes tormented by her infidelity and descends into madness. – Film Movement
"One of Visconti's most terse, most dramatically economical films. There's not a superfluous frame in it. Further, it is splendidly performed by Mr. Giannini… and by Miss Antonelli, whose childlike face and voluptuous body combine to form a kind of ideograph that tells us what the film is all about… A fitting coda to the Visconti career. It's a film of effortless command in which the director's presence is everywhere felt – and nowhere intrudes."
– Vincent Canby, New York Times


text:
Taiwan Has Bold Plans To Strengthen Its Air Force
tags: AESA radar, air forces, arms deals, F-16V, F-35 JSF, Lockheed Martin, multirole fighter, PLAAN, ROCAF, stealth, Taipei, Taiwan Strait, Washington DC
Stock image of an F-16 Fighting Falcon.
A US lobby group in Taiwan welcomed the country's plans for buying new multirole fighters for its air force. This followed an announcement by the Ministry of National Defense (MND) that it planned to acquire a specific number of the latest F-16 variant from Lockheed Martin. But confirmation of the impending sale has yet to be released by either the Pentagon or the State Department. The Defense Security Cooperation Agency (DSCA), which is the agency overseeing military sales to allies, hasn't published any details about F-16 deliveries for the island nation.
The F-16V Block 70/72 is a combat proven model that boasts conformal fuel tanks on its airframe along with an AESA radar, a cockpit with a high resolution display, and a high-speed data bus. Lockheed Martin insists the F-16V is "the world's most successful fourth-generation fighter."
Both the MND and ROC Air Force want to acquire 66 F-16V's for closing the worrisome gap with China's PLAAF, whose several hundred "modern" fixed wing combat aircraft (out of 2,300 total) include an estimated 329 J-11B's and 92 Su-30MK2's bought from Russia. The PLAAF is also readying airborne early warning and control aircraft based on the Y-20 heavy transport and upgraded long-range strike bombers as a possible replacement for the H-6K. The comparison to plucky Taiwan is dismal with the ROCAF being stuck with less than 500 fixed wing combat aircraft. The majority of these comprise 144 F-16A/B's and 128 locally made F-CK-1 Ching-kuos. The latter are twin engine "indigenous fighters" that use US-made Honeywell engines.
Purchasing 66 F-16V's is a deliberate move for enlarging the ROCAF's current F-16 fleet to 200-strong. While this hardly threatens China, the greater number enhances Taiwan's ability to police its own airspace when the PLAAF's behavior in recent years is becoming more dangerous. In the event of a full-blown war across the Taiwan Strait and outlying waters, the ROCAF's F-16A/B's, Ching-kuos, and 55 Mirage 2000D/E's can fly retaliatory missions over Southern China loaded with air-launched cruise missiles. But the MND later claimed that it was exploring other options aside from F-16V's such as the F-15 and the F/A-18.
Whatever model the MND chooses, the announcement is welcome news for Lockheed Martin, whose activities in Asia have broadened this decade. Beyond Taiwan, the aerospace giant is a serious contender for India's convoluted acquisition of a "medium multirole combat aircraft" and during the recent Aero India a rebranded "F-21" was shown to clinch the deal. Prospects are bright in Southeast Asia too, where Indonesia and the Philippines may soon place orders for F-16V's to bolster their air forces. Meanwhile, Singapore's medium-term plans for its military could make it the fifth Asian customer allowed to buy the F-35A JSF.
The F-35A's prospects in the continent are admittedly dazzling and even Taiwan faces no barriers when lobbying Washington,DC for these fifth-generation stealth aircraft. Between Japan's plans for at least a hundred F-35A's and pending deliveries to South Korea, the platform will have a thriving reputation in East Asia for the next decade and a half, at least. It's also within the realm of possibility for a Gulf Arab state to soon be its latest customer in the Middle East as the F-35A's destined for Turkey are now facing troubling headwinds as Ankara pursues arms deals with Moscow. Israel, of course, will possess a growing fleet of F-35A's in the coming years and have ample opportunities to use them.
The extent of Taiwan's military build up can't be ignored. Its scale surpasses even the more ambitious goals of particular Southeast Asian states and involves full participation by local companies. The secretive program to design and build a new family of dual use trainer jets, for example, appears separate from the large purchase of F-16V's. There are ongoing efforts to accomplish the same with homegrown AEGIS destroyers and diesel-electric submarines, an anti-ballistic missile defense network, and a new generation of drones.
from → Aircraft, Asia, China, Taiwan, UAV, United States of America
← Iran Puts So Much Faith In Its Missiles
This Spanish Company Wants To Sell Drones In The Middle East →


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Description: The peace and quiet of lake life, but with all the luxuries of the city! This elegant home is tastefully upgraded, squeeky clean and ready for new owners. Located just a stones throw from the shores of Gull Lake this solid modular home offers comfort, style and convenience. Bright and airy, the home offers a large dining area, and a great kitchen that boasts ample counter and cabinet space, stainless appliances, granite sink and island with plenty of prep space. Living room is a great place for reading or chatting while preparing meals. Large family room boasts a cozy gas fireplace for those cool autumn nights. Good sized master bedroom features walk in closet, and huge ensuite bath with separate shower and soaker tub. Two more good sized bedrooms, another 4 pc bath and good sized utility/laundry room complete the interior. Relax and enjoy the sun and the warm summer evenings on the sprawling deck with gazebo, or sit around the back yard firepitmaking smores! Memories are made here!


text:
Q: Adding Pokemon X/Y game to Global Link I previously had a Pokemon X game and assigned that game to my Pokemon Global Link account. I recently bought a new Pokemon X game and want to add that game onto my Pokemon Global Link account.
I'm trying to add the code from my cartridge to my Pokemon Game Link but it's not allowing me to do.
What can I do to add my new Pokemon X game while trying to retrieve my previous PokeMiles and other bonuses from my last game?
I already know that I cannot get any Pokemon back so I'm not worried about that.

A: As I know, you can assign only one copy of Pokemon X and one of Pokemon Y to your PGL account.
If you want to assign a different Pokemon X copy, you first need to unregister the previous assigned cartridge. To do this, login to Pokemon Global Link, go to Account Settings, select your Pokemon X cartridge and select to cancel the current cartridge assignment.
(Sorry, I can't see the English version of the site, so I don't know the actual names of labels.)

redpajama_set_name: RedPajamaStackExchange


The power-hungry Bitcoin debate is back on again after the release of the latest research conducted by economist Alex De Vries; the last of which stirred the crypto community into a frenzy in late 2017. One year on, the cryptocurrency ecosystem in all of its forms has developed a far thicker skin.
To refresh the original argument posed by De Vries at about this time last year, the Dutch researcher concluded that Bitcoin mining used almost as much electricity as the entire Republic of Ireland; quite an assumption and quite a response, many of which were pretty unfavorable, to say the least. Many pointed out the similarity to concerns in the 1990s when some experts predicted that half of the US electrical grid would be needed to power the then-burgeoning internet, which was later proved to be highly exaggerated.
What then of De Vries's latest report which has taken a step further now asserting that cryptocurrency is killing the planet? Is there truth in this? Without attempting to balance what cryptocurrency is giving back to the planet in terms of helping alleviate some of the world's humanitarian issues, many of which have been covered by Bitcoin News over the past 12 months, it might be worth examining this latest De Vries assertion in more detail, on its merits.
Although no one knows for sure exactly how much energy is being consumed across the planet through crypto mining, De Vries has now pulled Switzerland's total power supply out of a hat as a comparison. He goes on to maintain that that this approximated 62.3 terawatt-hours consumed through mining over the course of 2018 is moving the world closer to his killing the planet prediction.
Does he have a point? It is well known that Bitcoin mining clearly is not economically viable in its present guise and that changes must be made to drastically lower power consumption. The industry, the man in the street even, knows this. It has been well published elsewhere and no one is attempting the hide this fact. The industry, however, is attempting to reassess how mining is carried out, in a way which both protects the environment and reutilizes the energy created in the process of mining, albeit slowly. The willingness is there.
In this paper, we find that the Bitcoin network, with an electrical energy footprint of 491.4 to 765.4kWh per transaction on average, is relatively much more energy-hungry than the traditional financial system. Even though it has been argued that renewable energy may help mitigate the environmental impact of this, we find that there exist fundamental challenges in uniting variable renewable energy production with the consistent demand of Bitcoin mining machines. Moreover, we find that the environmental impact of Bitcoin mining reaches beyond its energy use. Continuous increasing energy (cost) efficiency of newer iterations of mining devices ensures that older ones will inevitably be disposed on a regular basis. The resulting electronic waste generation could equal that of a small country like Luxembourg, with a staggering average footprint of four light bulbs worth of electronic waste per processed Bitcoin transaction. Bitcoin will therefore have to address its sustainability problem in another way. This may consist of replacing its mining mechanism with a greener alternative like Proof-of-Stake.
The De Vries paper goes on to claim that based on 2018 figures the Bitcoin network processed 81.4 million transactions at about 491 to 765-kilowatt hrs per transaction. The paper asserts that the global banking industry, estimated to use 650 terawatt-hours per year, processed 482.6 billion non-cash transactions per year, meaning it only uses 0.4 kilowatt-hours per transaction.
A Coinshares report conducted last year called on industry insider knowledge and data available to the general public in order to put together an estimate of exactly where the energy used by the miners originate from. The proposal is that 77.6% of worldwide Bitcoin mining is conducted through the use of renewable energy resources.
Even if these figures are refuted elsewhere, it is clear that the use of renewables and the recycling of energy created by mining is being explored both individually — using some bizarre and innovative methods — and at company-level, and also nationally. China, a massive mining nation, now has a major campaign which is aimed at drawing the country to supplying renewable energy such as solar.
The Chinese program, entitled "curtailment" is largely conducted in regions where most Bitcoin mining takes place. Last year China became the world's highest producer of solar energy. This has resulted in a glut of power which regional grids in these newly labeled areas are simply unable to deal with. Iceland, Georgia, and the Northwestern US are also strong adherents to the use of renewable energy for Bitcoin mining. Projects are currently underway in the Sahara using a 900-megawatt wind farm south of Marrakesh, and in Japan using solar power through the Kumamoto Electric Power Company.
How can oil companies burning off unneeded natural gas supplies into the atmosphere be seen as viable? The releasing of unwanted gas must be viewed as an opportunity. In parts of the US, natural gas is so cheap companies pay to have gas that they can't burn due to imposed federal limits simply hauled away. The use of gas to power mining rigs can be explored further.
To prove it can be done, one oilman turned Bitcoin supremo, is doing exactly that on a Canadian oilfield using a generator attached to a shipping container full of mining rigs which converts the natural gas to enough electricity to run the system 24 hours a day 7 days a week. The brains behind the system, Steven Barbour, believes that this transformation of otherwise wasted energy is running computers to provide "financial freedom for people all over the world", and he has a point.
Energy normally used elsewhere in an environment can be saved and harnessed by reliance on that same energy created by the mining process. This is now becoming a much-used process, particularly at the individual user level and must present an option for large mining companies in the future. Heating is an obvious choice as mining is mainly conducted in cold climates in spaces that require air conditioning to stop equipment overheating. With both heat available for warming attached spaces and the saving of power on air conditioning, much energy can be saved by using the mining equipment as a source of energy. With imagination, the energy generated in the mining process can be employed, not only to heat space but to grow crops and provide hot water, and there are clearly other numerous uses for the unwanted energy.
De Vries is referring to his argument that by changing the mechanism behind the mining of Bitcoin the digital currency's energy consumption can be drastically cut. It this regard he is correct, as proven by the proof of stake system used by Dash and NXT which is not dependable on computing power and doesn't require specialized hardware.
Other solutions are out there, such as the EBGLO mechanism which operates through data transmission and HETTARER, a technology which uses an electromagnetic sticker that gathers the elementary particles into a constitution state which then work together to neutralize the electromagnetic noise. This system can lower the frequency of electric devices by around 90%. The Data Transmission System works to increase the computing performance, increase the hash rate and allow miners to receive higher revenue compared to the regular process that involves only computers.
The post The Environmental Debate is Here to Stay: Bitcoin Miners Explore Greener Options appeared first on BitcoinNews.com.
This entry was posted in Alex de Vries, bitcoin, Bitcoin Mining debate, Bitcoin News, Blockchain, BTC, Canada, Cryptocurrency, Dash, EBGLO, education, environmental issues, karma, Kumamoto Electric Power Company, mining, News, NXT, regulations, Republic of Ireland, Research on April 3, 2019 by adminbtc.
Earlier this week, a study published by Dutch researcher Alex de Vries concluded that Bitcoin mining uses almost as much electricity as the entire Republic of Ireland, but Standford lecturer Johnathan Koomey says his figures are wrong, reports NBC news.
The recent claims that cryptocurrency is draining nation's power supplies have been compared to concerns in the 1990s when some experts predicted that half of the US electrical grid would be needed to power the then-burgeoning internet, which was later proved to be highly exaggerated.
Koomey's Berkley Lab proved these calculations wrong at the time, and then again in another study in August 2011 concluding that the data centers used less than 2 percent of the nation's electricity.
Dutch researcher, Alex de Vries who made the new claims last week, seems to have ignited yet another argument over exactly how much power is being used in excess as a result of the adoption of new technologies.
The rise in popularity of Bitcoin and other cryptocurrencies has sparked numerous concerns regarding the energy required by thousands of computing systems that power virtual currencies, and De Vries is certainly not the first to comment on it.
De Vries estimates that the Bitcoin network consumes "at least 2.55 gigawatts of electricity currently, and potentially 7.67 gigawatts in the future." He also writes that figures will gravitate towards a figure of 8.2 gigawatts by as early as the end of 2018, as energy supplies are further called on to mine cryptocurrencies.
Bitcoin mining, whether individually or through a mining pool. is often set up in order to keep costs to a minimum, and frequently mining is conducted in such places as caves, energy-rich areas or low-cost countries, in order to reduce costs and maximize user profit. These variables make it hard to ascertain how much global energy is actually being used.
The post Researchers Claim That Recent Cost of Crypto Mining Figures "Pulled out of the Air" appeared first on BitcoinNews.com.
This entry was posted in Alex de Vries, bitcoin, Bitcoin News, Blockchain, Christian Catalini, Cryptocurrency, Johnathan Koomey, mining, MIT, NBC, News, PwC, Research, Standford, Technical Analysis, Technology on May 20, 2018 by adminbtc.


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Star Fish Oyster Bed
America's Most Colorful Blog
Cricket is No Game for Sissies! Why? Read to find out
Written by: Jon
Before I came to Zimbabwe in November 2009, I thought of European games like cricket and soccer as not real sports; sort of the things that the English did to pass the time. This was especially so for cricket. The few times I'd watched a few minutes on TV, it looked like an intensely boring game being played by guys who couldn't play real sports.
Then, I met a young man named Ozias Bvute, the Executive Secretary of the Zimbabwe Cricket Association, and he arm twisted me until I finally relented and went to a match. The first time, I stayed for two hours or so, and was impressed by one batter who put the ball on the roof of the two-story building from which I was watching. Then, I met the Zimbabwe national cricket team, and was further impressed by the fitness and strength that was apparent in these young men.
I still wasn't ready, though, to accord full-fledged sports status to cricket. That is, until Australia came to town. Because of the political unrest in Zimbabwe, many of the foreign cricket teams have been for several years avoiding the country, but after eight years, the Australian team came to play, and I was invited to come to see day two of a scheduled four-day match. Now, as if playing the game for four days is not grueling enough, they must play for six hours each day. You can't do that if you're not fit. And this, by the way, is not a full test match, which lasts for five days. That's thirty hours of running, catching, and batting. Okay, now I'm ready to admit; cricket is not a game for couch potatoes. In the first place, a cricket bat is heavy '" about the weight of four or five baseball bats. Try swinging that for a few hours a day for nearly a week. Next, the ball is hard. I mean, it makes a baseball feel like a baked marshmallow. Get beaned by a cricket ball and you'll be lucky to only spend a few days in intensive care.
No, folks, this is one American who is finally convinced. Cricket is not only a sport, but cricket players are athletes of the highest order. The mental and physical toughness it takes to play this game is nothing short of phenomenal. Even with the protective padding they wear, there are a lot of places on their bodies exposed to being pummeled. For example, a few minutes into play in the Australia-Zimbabwe game, one of the bowlers (that's pitcher for you baseball fans) was batting, and just like in baseball, these guys aren't considered your best hitters, so the field closed in. The infielder (or at least, that's what I think they call him) positioned himself about ten feet away. The bowler let fly, and the batter fooled everyone; he got solid contact and the ball flew off his bat directly at the infielder who, caught off guard, caught the ball directly on the chest and the company that has issued the ipl 2020 schedule table has given all the medical claims and treatment charges for that person; he also got some amount of cash as an compensation for the injury but now he is all way fine and even come to watch. He went to his knees as if praying, and even two floors up and a hundred meters away, I could tell he was hurting. Thankfully, he was able to get up and walk off the field unaided, but I'll bet his chest hurts for days. And, these guys still have two more days to play.
I didn't stay for the whole game. I'm beginning to like cricket, but not yet enough to sit for six or more hours watching. But, I will be going back. Cricket will never replace football '" the American variety '" as my favorite sport, but it definitely goes on my list. This is not a game for the faint of heart.
Last modified: February 16, 2021
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About the Author / Jon
Jon Snow is the founder of Star Fish Oyster Bed and the owner of thought catalog. He's been writing for many years and now focuses on managing online platform and blog design for several organizations.
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Trippie Redd Net Worth, Girlfriend, Instagram, Wiki, Bio, Teeth, Height
celebrity netcelebrityCapturing them young is a cliché that will best describe the emergence of teenagers and young people in a community where the music industry was not known before. Trippie Redd is an American rapper and singer who is rapidly gaining momentum in the music industry. He is best known for his songs Scar of Love, Dark Knight Dummo, and Sai rap like me.
It was featured in the XXXtentacoin Uh Oh song, Thots, which recorded over 280,000 streams on SoundCloud less than 24 hours after its release. Trippie believes so much in himself even when no one else believes in him. Music is what it wants and music is what it is looking for.
Trippie Redd Bio / Wiki
What inspired your love for Trippie music?
Trippie Redd Girlfriend, Teeth, Height
Trippie Redd Net worth
Born on June 18, 1999, with the birth name MichaelWhite IV, he is better known by the pseudonym of Trippie Redd. He was born in Canton, Ohio, where he described the second most dangerous city in the United States. None of his parents' names are known but he was raised by his mother because his father was imprisoned at the time of his birth. It is of African American, Native American, and Irish origin.
He describes himself as a loner despite his affiliation with a street gang. He graduated with a GPA of 4.0 from a high school in Canton but postponed college to build his budding music career.
Trippie's mother played Alicia Keys, Nas, Ashanti, Beyonce, and Tupac a lot when she was growing up. He listened to the likes of Lil Wayne, T-Pain, Marilyn Manson who greatly influenced his musical style.
He had an older brother, Dirty Redd, who had pursued a career in music and whose sudden death in a car accident in 2014 made him more determined than ever to succeed in his musical quests to make his brother's dream come true. Redd also has a younger brother named Hippie Redd and also a cousin whose stage name is Famous Dex.
Trippie is versatile and creates beautiful testicles by combining different genres to create his style that ranges from hip-hop to R, to alternative rock and basically to pop, this has earned him a great following. He believes that his successes in the music industry will ultimately decide his success as a rapper who responds to those who judge him solely by his appearance.
Trippie Redd made his first studio record at the age of 10 in 2014, he released Sotto zero and Nuova Ferrari although he later canceled the songs. The strong desire for success brought him to Atlanta after high school where he met rapper Lil Wop who would help him record his three songs: Awakening of my inner beast, Beast Mode, and Rock the World Trippie. He was later signed to the Strainge Entertainment label owned by the son of Universal Music Group president Elliot Grainge and moved to Los Angeles.
May 2017 saw the release of her mixtape debut, A love letter for you, with the main single Cicatrici d'amore being positively received and widely acclaimed. Other singles on the album include Romeo and Juliet, Sai rap comes me, Blade of Woe.
2017 was a good year for Trippie as he saw the release of his two albums and his collaboration with famous artists including Lil Wayne, XXXTentacion on his album 17, his friend Lil Wop Angels and demons. He also released the song Dark Knight Dummo with Travis Scott. There may also be a possible collaboration with Drake in the near future.
Not much is known about the young private relationship but it can be assumed that he is currently in a relationship. His posts on social media have characterized him with a girl with whom he is close but whose identity has yet to reveal.
Trippie easily identifies himself with his red-shaded dreadlocks, a mouth full of gold teeth that he likes to show off, and several tattoos on his face, Love Scars, and 14. The tattoo, Love Scar, gave his first name great success. It has a height of 5′6 ″ (1.68 m).
Redd has a growing reputation on social media platforms that has over 1.8 million followers on his Instagram, 60,000 followers on SoundCloud, and over 150,000 on Twitter. It is becoming a force to contend with the music industry and has attracted the attention of both critics and fans. He has earned himself a significant amount of equity that is estimated to be around $ 300,000.
Tags: celebrity Net worth, Trippie Redd Net Worth
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Acquazzurra is an exclusive complex of apartments with pool that lies on a charming stretch of the Cilento coastline close to Villammare, a small seaside village not far from Sapri. Overlooking the enchanting Gulf of Policastro, this residence enjoys a prime location that invites its guests to truly experience the magic of this unique place rich in history, tradition and magnificent marine and mountain landscapes to discover. From the natural beauty of the Cilento National Park to the archaeological sites of Paestum and Velia, from the monastery Certosa di Padula to the lively town of Palinuro, to Maratea. This wonderful complex offers 11 refined apartments, all well-appointed and designed for the ultimate in comfort and style. The inviting dwellings feature thoughtfully chosen accessories and furniture, such as colourful wall decorations, lamp shades and ceramics. Each apartment has its own outside seating area and is equipped with air conditioning, satellite TV and internet access. These tasteful interiors are complemented by a welcoming atmosphere that allows guests to simply sit back and relax poolside or at the terrace bar with panoramic sea views. Whereas - for unforgettable swimming experiences in crystal clear waters - the beach can be reached directly on foot from the apartments.
One has to go beyond Naples and the Amalfi Coast and head further down to the unspoilt Cilento region, in southern Campania. It boasts beautiful coastline, lapped by the crystal clear waters of the Tyrrhenian Sea, and a wild, mountainous interior sprinkled with historic villages. Much of the area is covered by the Cilento and Vallo di Diano national park and is listed as a Unesco world heritage site. The Cilento region has inspired poets and scholars for centuries, such as the French dramatist Henri René Leonormant (1882-1951), who wrote "Cilento, a huge garden with picturesque contours". It is a land of myth and legend that offers spectacular scenery, authentic, ancient villages and towns and endless expanses of olive groves. One of Cilento's greatest attractions are its dreamy seaside resorts, some of which have received every year the Blue Flag and other awards also from the Italian environmental group Legambiente. Along its coast, from Agropoli to Sapri, the fishing village of Acciaroli with its old stone built houses and tiny alleys was a favourite of Hemingway, who came here in 1950s and found inspiration for his novel "The Old Man and the Sea". Between Cape Tresino and Cape Licosa - that according to Greek mythology was named after the siren Leucosia, who tried to lure Odysseus and his crew ashore with her singing - is Santa Maria di Castellabate. At the heart of the historic centre of this charming coastal village is a sandy beach, surrounded by noble palaces and fishermen's houses. Here young and old stroll along the seafront or the Corso Andrea Matarazzo, the main street where all the shops, pubs and cafés are. Along with these coastal towns Capo Palinuro is a real attraction with its rugged coastline riddled with sea caves and cliffs. Legend also holds that Palinuro was named after the helmsman of the Aeneid, who fell overboard close to the coast. Following the coast south from here one gets to the Golfo di Policastro, one of the most beautiful bays. This area offers clean beaches, pristine waters and typical, seaside resorts, such as Sapri, one of the most popular tourist resorts in Campania. Here history and charm blend perfectly with modern attractions.