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

\subsection{Problem formulation}

Let $N=(N_{t} : t\geq 0)$ be a Poisson process with a constant intensity $\lambda>0$, and let $Y_i$ be a sequence of independent random variables, each with distribution $P$, that are also independent of $N$. By definition, a compound Poisson process (abbreviated CPP) $X=(X_t : t\geq 0)$ is
\begin{equation}
\label{cpp}
X_t=\sum_{j=1}^{N_t} Y_j,
\end{equation}
where here and below the sum over an empty index set is understood to be equal to zero. In particular, $X_0=0$. CPP constitutes a classical model in, e.g., risk theory, see~\cite{embrechts97}.

Assume that the process $X$ is observed at discrete times $0<t_1<t_2<\ldots<t_n=T$, where the grid $\{t_i\}$ is not necessarily uniform on $[0,T]$. Based on the observations $X_{t_1},X_{t_2},\ldots,X_{t_n}$, our goal is to estimate the jump size distribution $P$ and the intensity $\lambda$.  We specifically restrict our attention to the case where $P$ is a discrete distribution, $P(\mathbb{N})=1$, and we will write $p=(p_k)_{k\in\mathbb{N}}$ for the probability mass function corresponding to $P$, where $p_k=P(\{k\})$. A similar notation will be used for any other discrete law. The distribution $P$ is called the base distribution. Abusing terminology, we will at times identify it with the corresponding probability mass function $p.$ An assumption that $P$ has no atom at zero is made for identifiability: otherwise this atom gets confounded with $e^{-\lambda}$, which does not allow consistent estimation of the intensity $\lambda$. For a discussion of applications of this CPP model in risk theory, see \cite{zhang14}.

Define the increments $Z_i = X_{t_i} - X_{t_{i-1}}$, $i=1,\ldots,n$. Then $\mathcal{Z}_n = (Z_i:i = 1,\ldots,n)$ is a sequence of  independent random variables. When $\{t_i\}$ are uniformly spaced on $[0,T]$, the random variables $Z_i$ have in fact a common distribution $Q$ satisfying $Q(\mathbb{N}_0)=1$. As $\mathcal{Z}_n$ carries as much information as $(X_{t_i}:i=1,\dots,n)$ do, we can base our estimation procedure directly on the increments $\mathcal{Z}_n$. Since summing up the jumps $Y_j$'s amounts to compounding their distributions, the inverse problem of recovering $P$ and $\lambda$ from $Z_i$ can be referred to as decompounding; see \cite{buchmann03}.

There are two natural ways to parametrise the CPP model: either in terms of the pair $(\lambda,p)$, or in terms of the L\'evy measure $\nu=(\nu_k)_{k\in\mathbb{N}}$ of the process $X$, see \cite{sato13}. A relationship between the two is $\lambda=\sum_{k=1}^{\infty}\nu_k$ and $p=\nu/\lambda$. Inferential conclusions in one parametrisation can be easily translated into inferential conclusions into another parametrisation. However, for our specific statistical approach the L\'evy measure parametrisation turns out to be more advantageous from the computational point of view.

\subsection{Approach and results}

In this paper, we take a non-parametric Bayesian approach to estimation of the L\'evy measure $\nu$ of $X.$ See \cite{ghosal17} and \cite{mueller15} for modern expositions of Bayesian non-parametrics. A case for non-parametric Bayesian methods has already been made elsewhere in the literature, and will not be repeated here. On the practical side, we implement our procedure via the Gibbs sampler and data augmentation, and show that it performs well under various simulation setups. On the theoretical side, we establish its consistency and derive the corresponding posterior contraction rate, which can be thought of as an analogue of a convergence rate of a frequentist estimator (see \cite{ghosal17}). The posterior contraction rate, up to a practically insignificant $\log n$ factor, turns out to be $1/\sqrt{n}$, which is an optimal rate for non-parametric estimation of cumulative distribution functions. Our contribution thus nicely bridges practical and theoretical aspects of Bayesian non-parametrics.

\subsection{Related literature}

To provide a better motivation for our model and approach, in this subsection we briefly survey the existing literature. A Bayesian approach to non-parametric inference for L\'evy processes is a very recent and emerging topic, with references limited at the moment to \cite{belomestny18}, \cite{gugu15}, \cite{gugu18} and \cite{nickl17}. These deal exclusively with the case when the L\'evy measure is absolutely continuous with respect to the Lebesgue density. At least from the computational point of view, these works are of no help in our present context.

Related frequentist papers for CPP models with discrete base distributions are \cite{buchmann03} and \cite{buchmann04}, which, after earlier contributions dating from the previous century, in fact revived interest in non-parametric techniques for L\'evy processes. To estimate the base distribution $p$, \cite{buchmann03} employ a frequentist plug-in approach relying on the  Panjer recursion (i.e., an empirical cumulative distribution estimate of $q$ is plugged into the Panjer recursion equations to yield an estimate of $p$; see below on the Panjer recursion). The drawback is that the parameter estimates are not guaranteed to be non-negative. \cite{buchmann04} fix this problem by truncation and renormalisation. This works, but looks artificial. As noted in \cite{buchmann04}, in practice the latter approach breaks down if no zero values are observed among $Z_i$'s. \cite{buchmann04} establish weak convergence of their modified estimator, but on the downside its asymptotic distribution is unwieldy to give confidence statements on $p$. Most importantly, the plug-in approaches in \cite{buchmann03} and \cite{buchmann04} do not allow obvious generalisations to non-uniformly spaced observation times $\{t_i\}.$ In \cite{lindo18}, another frequentist estimator of the jump measure is introduced, that is obtained via the steepest descent technique as a solution to an optimisation problem over the cone of positive measures. The emphasis in \cite{lindo18}  is on numerical aspects; again, no obvious generalisation to the case of non-uniform $\{t_i\}$ is available.

Finally, some important, predominantly theoretical references on inference for L\'evy processes are \cite{comte11}, \cite{duval11}, \cite{kappus14}, \cite{neumann09}, \cite{nickl12}, \cite{vanes07} and \cite{trabs15}. We refer to \cite{belomestny15}, \cite{coca18}, \cite{coca18_cpp} and \cite{duval17} for more extensive literature surveys.

\subsection{Outline}
The rest of the paper is organised as follows: in Section \ref{sec:algorithm} we introduce our approach and describe an algorithm for drawing from the posterior distribution. In Sections \ref{sec:examples} and \ref{sec:realdata} we study its performance on synthetic and real examples. Section \ref{sec:asymptotics} is devoted to the examination of asymptotic frequentist properties of our procedure. An outlook on our results is given in Section \ref{sec:outlook}. Finally, in Appendix \ref{app:lemmas} technical lemmas used in the proofs of Section \ref{sec:asymptotics} are collected.

\subsection{Notation}
For two sequences $\{a_n\}$ and $\{b_n\}$ of positive real numbers, the notation $a_n\lesssim b_n$  (or $b_n\gtrsim a_n$) means that there exists a constant $C>0$ that is independent of $n$ and such that $a_n\leq C b_n.$ We write $a_n\asymp b_n$ if both $a_n\lesssim b_n$ and $a_n\gtrsim b_n$ hold.
We denote a prior (possibly depending on the sample size $n$) by $\Pi_n$. The corresponding posterior measure is denoted by  $\Pi_n(\cdot\mid \mathcal{Z}_n).$ The Gamma distribution with shape parameter $a$ and rate parameter $b$ ($a,b >0$) is denoted by $\operatorname{Gamma}(a,b)$. Its density is given by
$x \mapsto \frac{b^a}{\Gamma(a)} x^{a-1}e^{-bx}, \quad x>0$,
where $\Gamma$ is the Gamma function. The inverse Gamma distribution with shape parameter $a$ and scale parameter $b$ is denoted by $\operatorname{IG}(a,b)$. Its density is
$x \mapsto \frac{b^a}{\Gamma(a)} x^{-a-1}e^{-b / x}, \quad x>0$. We use the notation $\operatorname{Exp}(a)$ for an exponential distribution with mean $1/a$.

\section{Algorithm for drawing from the posterior}
\label{sec:algorithm}

A Bayesian statistical approach relies on the combination of the likelihood and the prior on the parameter of interest through Bayes' formula. We start with specifying the prior. As far as the likelihood is concerned, although explicit, it is intractable from the computational point of view for non-parametric inference in CPP models. We will circumvent the latter problem by means of data augmentation, as detailed below.

\subsection{Prior}
\label{subsec:prior}
We define a prior $\Pi$ on $\nu$ through a hierarchical specification
\begin{align*}
\{\nu_k\}_{k=1}^\infty  \mid a, m, \beta_k &  \stackrel{\textrm{i.i.d.}}{\sim} \operatorname{Gamma}(a,1/\beta_k)	\cdot \mathbf{1}_{\{1\le k\le m\}}, \\
	\beta_1,\ldots, \beta_m \mid \gamma & \stackrel{\textrm{i.i.d.}}{\sim}\operatorname{IG}(c,\gamma),\\
	\gamma &\sim \operatorname{Exp}(1).
\end{align*}
 Note that the (fixed) hyperparameters $m\in\mathbb{N}$, $a, c >0$ are denoted by  Latin letters.

The hyperparameter $m$ incorporates our  a priori opinion on the  support of the L\'evy measure $\nu$, or equivalently, the base measure $p$. In applications, the support of $p$ may be unknown, which necessitates the use of a large $m$, e.g.\ $m=\max_{i=1,\ldots,n} Z_i$; this latter is the maximal value suggested by the data $\mathcal{Z}_n$ at hand. Nevertheless, we may simultaneously expect that the `true', data-generating $\nu$ charges full mass only to a proper, perhaps even a small subset of the set $\{1,\ldots,m\}$. In other words, $\nu$ may form a sparse sequence, with many components equal to zero. In fact, there are at least two plausible explanations for an occurrence of a large increment $Z_i$ in the data: either a few large jumps $Y_j$'s occurred, which points towards a large right endpoint of the support of $\nu_0$, or $Z_i$ is predominantly formed of many small jumps, which in turn indicates that the intensity $\lambda$ of the Poisson arrival process $N$ may be large. To achieve accurate estimation results, a prior should take a possible sparsity of $\nu$ into account. This is precisely the reason of our hierarchical definition of the prior $\Pi$: a small $\beta_k$ encourages a priori the shrinkage of the components $\nu_k$ of $\nu$ towards zero.

\subsection{Data augmentation}

Assume temporarily $t_i=i,$ $i=1,\ldots,n$, and write $q=(q_k)_{k\in\mathbb{N}_0}$ for $q_k=q(\{k\})$. Then $Z_i$ have the distribution
\begin{equation}
	\label{eq:comp}
	q = e^{-\lambda} \sum_{j=0}^{\infty} \frac{\lambda^j}{j!}p^{\ast j},
\end{equation}
with $*$ denoting convolution.  The compounding mapping $(\lambda,p)\mapsto q$ can be expressed explicitly via the Panjer recursion (see \cite{panjer81}):
\[
q_0 = e^{-\lambda}, \quad q_k = \frac{\lambda}{k}\sum_{j=1}^{k} j p_j q_{k-j}, \quad k\in\mathbb{N}.
\]
This recursion can be inverted to give the inverse mapping $q\mapsto(\lambda,p)$ via
\[
\lambda = - \log q_0, \quad p_k = - \frac{q_k}{q_0\log q_0} - \frac{1}{kq_0}\sum_{j=1}^{k-1}jp_jq_{k-j}, \quad k\in\mathbb{N}.
\]

In view of \eqref{eq:comp}, the likelihood in the CPP model is explicit. Nevertheless, an attempt to directly use \eqref{eq:comp} or the Panjer recursions in posterior computations results in a numerically intractable procedure. Equally important is the fact that a Panjer recursion based approach would not apply to non-uniformly spaced observation times $\{t_i\}$. Therefore, instead of \eqref{eq:comp} and the Panjer recursion, we will employ  data augmentation, see \cite{tanner87}. We switch back to the case when $\{t_i\}$ are not necessarily uniformly spaced.  The details of our procedure are as follows: when the process $X$ is observed continuously over the time interval $[0,T]$, so that our observations are a full sample path $X^{(T)} = (X_t:t\in[0,T])$ of CPP, the likelihood is tractable and is proportional to
\[
e^{-T\sum_{k=1}^m\nu_k} \prod_{k=1}^m \nu_k^{\mu_k},
\]
see \cite{shreve04}, p.~498. Here
\begin{equation}
\label{eq:muk}
\mu_k = \# \{Y_j=k\},
\end{equation}
i.e.\ the total number of jumps of size $k$.
Then the  prior $\Pi$ from Subsection \ref{subsec:prior} leads to conjugate posterior computations. In fact, the full conditionals are
\begin{gather*}
\nu_k \mid \{\mu_k\},\, \{\beta_k\} \stackrel{\textrm{i.i.d.}}{\sim} \operatorname{Gamma}(a+\mu_k,1/\beta_k+T), \quad k=1,\ldots,m,\\
\beta_k \mid \{\nu_k\},\, \gamma \stackrel{\textrm{i.i.d.}}{\sim} \operatorname{IG}(a + c,\gamma + \nu_k), \quad k=1,\ldots,m,\\
\gamma \mid \{\beta_k\} \sim \operatorname{Gamma}\left(c m+1, 1+\sum_{k=1}^m \beta_k^{-1}\right).
\end{gather*}
Therefore, the Gibbs sampler for posterior inference on $\nu$ can be implemented. The Gibbs sampler cycles through the above conditionals a large number of times, generating approximate (dependent) samples from the posterior. See, e.g.,  \cite{gelfand90} and Section 24.5 in \cite{wasserman04} on the Gibbs sampler and its use in Bayesian statistics.

As we do not observe the process $X$ continuously, we will combine the above with the data augmentation device. First note that we have
\[
Z_i = \sum_{j=1}^m j \mu_{ij},
\]
where $(\mu_{ij}:i=1,\ldots,n, j=1,\ldots,m)$ are independent, and $\mu_{ij} \sim \operatorname{Poisson}(\Delta_i\nu_j)$ for $\nu_j = \lambda p_j$ and $\Delta_i = t_i - t_{i-1}$; see Corollary 11.3.4 in \cite{shreve04}. Furthermore, for $\mu_k$ as in \eqref{eq:muk} we trivially have $\mu_k=\sum_{i=1}^n \mu_{ik}$. Data augmentation iterates the following two steps:
\begin{enumerate}[(i)]
	\item Draw $(\mu_{ij})$ conditional on the data $\mathcal{Z}_n$ and the parameter $\nu$.
	\item Draw $\nu$ conditional on $(\mu_{ij})$.
\end{enumerate}
Once the algorithm has been run long enough, this gives approximate (dependent) samples from the posterior of $\nu$. We already know how to deal with step (ii); now we need to handle step (i).

Thus, keeping $\nu$ fixed, for each $i$ we want to compute the conditional distribution
$(\mu_{ij}:j=1,\ldots,m) \mid Z_i$, and furthermore, we want to be able to simulate from this distribution. In turn, this will immediately allow us to simulate $\mu_k$ conditional on the data $\mathcal{Z}_n$. Now, with $\operatorname{Pr}(\cdot)$ referring to probability under the parameter $\nu$, it holds that
\begin{multline*}
\operatorname{Pr}(\mu_{i1}=k_1,\ldots,\mu_{im}=k_m \mid Z_i = z_i) \\
 = \frac{1}{\operatorname{Pr}(Z_i = z_i)} e^{-\Delta_i \sum_{j=1}^m \nu_j} \prod_{j=1}^m\frac{(\Delta_i\nu_j)^{k_j}}{k_j!} \mathbf{1}\Big\{\sum_{j=1}^m j k_j = z_i\Big\}.
\end{multline*}
Knowledge of the normalising constant $\operatorname{Pr}(Z_i = z_i)$ will not be needed in our approach.

In general, simulation from a discrete multivariate distribution is non-trivial; some general options are discussed in \cite{devroye86}, Chapter XI, Section 1.5, but are unlikely to work easily for a large $m$. We will take an alternative route and use the Metropolis-Hastings algorithm, see, e.g., Section 24.4 in \cite{wasserman04}. We start by observing that for a fixed $i$, the support of $\operatorname{Pr}( \cdot \mid  Z_i = z_i)$ is precisely the set $\mathcal{S}_i$ of non-negative solutions $(k_1,\ldots,k_m)$ of the Diophantine equation
$
\sum_{j=1}^m j k_j = z_i.
$
The {\bf R} package {\bf nilde} (see \cite{nilde}) implements an algorithm from \cite{voinov97} that finds all such solutions for given integers $m$ and $z_i$. By Markovianity of the process $X$, we can simulate the vectors $(\mu_{i1},\ldots,\mu_{im})$  independently for each $i=1,\ldots,n$. If $z_i=0$ or $1$, there is only one solution to the Diophantine equation: the trivial solution $(0,\ldots,0)$ in the first case, and the solution $(1,0,\ldots,0)$ in the second case; for such $z_i$, no simulation is required, as $(\mu_{i1},\ldots,\mu_{im})$ is known explicitly. We thus only need to consider each $i\in\mathcal{I} = \{i:z_i \neq 0 \textrm{ or } 1\}$ in turn, and design a Metropolis-Hastings move on the set of the corresponding solutions $\mathcal{S}_i$. Fix once and for all an ordering of elements in $\mathcal{S}_i$ (this could be, e.g., lexicographic, or reverse lexicographic); we use the notation $|\mathcal{S}_i|$ for the cardinality of $\mathcal{S}_i$. Let $\mu = (\mu_{i1},\ldots,\mu_{im})$ be the current state of the chain, corresponding to the $\ell$th element $s_{\ell}$ of $\mathcal{S}_i$. A proposal $\mu^{\circ} = (\mu_{i1}^{\circ},\ldots,\mu_{im}^{\circ})$ is obtained as follows:

\begin{enumerate}[(i)]
	\item If $\ell = 1$, draw $\mu^{\circ}$ uniformly at random among the elements $\{s_2,s_{|\mathcal{S}_i|}\}$ of $\mathcal{S}_i$.
	\item If $\ell = |\mathcal{S}_i|$, draw  $\mu^{\circ}$ uniformly at random among the elements $\{s_1,s_{ |\mathcal{S}_i| -1 }\}$ of $\mathcal{S}_i$.
	\item If $\ell \neq 1$ or $|\mathcal{S}_i|$, draw  $\mu^{\circ}$ uniformly at random among the elements $\{s_{\ell-1},s_{\ell+1}\}$ of $\mathcal{S}_i$.

\end{enumerate}
Occasionally, one may want to propose another type of a move too.
\begin{enumerate}[(i)]
	\setcounter{enumi}{3}
	\item Draw $\mu^{\circ} = (\mu_{i1}^{\circ},\ldots,\mu_{im}^{\circ})$ uniformly at random from $\mathcal{S}_i$.
\end{enumerate}
The two proposals lead to reversible moves, and one may also alternate them with probabilities $\pi$ and $1-\pi$, e.g.\ $\pi=0.8$. The logarithm of the acceptance probability of a move from $(\mu_{i1},\ldots,\mu_{im})$ to $(\mu_{i1}^{\circ},\ldots,\mu_{im}^{\circ})$ is computed as
\[
\log A = \sum_{k=1}^m (\mu_{ik}^{\circ} - \mu_{ik}) \log(\Delta_i  \nu_k) + \sum_{k=1}^m \left \{ \log(\mu_{ik}!) - \log(\mu_{ik}^{\circ}!) \right\}.
\]
The move is accepted if $\log U \leq  \log A$ for $U$ an independently generated uniform random variate on $[0,1]$, and in that case the current state of the chain is reset to $(\mu_{i1}^{\circ},\ldots,\mu_{im}^{\circ})$. Otherwise the chain stays in $(\mu_{i1},\ldots,\mu_{im})$.

\section{Simulation  examples}
\label{sec:examples}

In this section, we test performance of our approach in a range of representative simulation examples. For benchmarking, we use the frequentist plug-in estimator from \cite{buchmann04}. Two real data examples are given in Section \ref{sec:realdata}. Unless otherwise stated, we took $c=2$ and $a=0.01$ as hyperparameters in our prior specification. As can be seen from the update formulae for the Gibbs sampler, as long as $a$ is not taken too large, its precise value is not very influential on the posterior, given a reasonable sample size. The value $c=2$ ensures that the update step for $\beta_k$ has finite variance. At each step of  updating the imputed data for increment size $z$ we have chosen with probability $0.2$ to propose uniformly from all solutions to the Diophantine equation (for that particular value of $z$).

We implemented our procedure in Julia, see \cite{julia}. The code and datasets for replication of our examples are available on GitHub\footnote{See \url{https://github.com/fmeulen/Bdd}} and Zenodo, see \cite{zenodo_bdd}.

\subsection{Uniform base distribution}\label{subsec:uniform}

This simulation example follows with some extensions that in \cite{buchmann04}. Let $\lambda_0 = 2$, and let $P_0$ be the discrete uniform distribution on $\{1,4,6\}$. We simulated data according to the following settings:
\begin{itemize}
  \item[(a)] $n=100$, $\Delta_i=1$ for $1\le i\le n$;
  \item[(b)] $n=500$, $\Delta_i=1$ for $1\le i\le n$ (the data under (a) are augmented with $400$ extra observations);
  \item[(c)] $n=500$, $\Delta_i=\operatorname{Unif}(0,2)$ for $1\le i\le n$.
\end{itemize}

We set $m= \min(15,Z_{(n)})$, where $Z_{(n)} = \max_{1\leq i \leq n} Z_i.$ In all cases this led to $m=15$, as the value of $Z_{(n)}$ was equal to $30$, $35$ and $40$ for the simulated data under settings (a), (b) and (c), respectively.
The Gibbs sampler was run for $500,000$ iterations, of which the first $250,000$ were discarded as burn-in. From the remaining samples, the posterior mean and $2.5\%$ and $97.5\%$ percentiles were computed for each coefficient $\nu_k$. The results for the first $10$ coefficients are shown in Figure  \ref{fig:example1}. For comparison, the estimator from \cite{buchmann04} is also included in the figure.
\begin{figure}
	\begin{center}
		\includegraphics[scale=0.6]{estimatesA1}
		\includegraphics[scale=0.6]{estimatesA2}
		\includegraphics[scale=0.6]{estimatesA3}
	\end{center}
	\caption{Simulation example from Section \ref{subsec:uniform}. In each figure, the horizontal axis gives the magnitudes of $\nu_k$, $k\in \{1,\ldots, 10\}$. The orange balls denote the true values, the black triangles the Buchmann-Gr\"{u}bel estimator. The blue crosses give the posterior means, whereas the vertical blue line segments represent (pointwise) $95\%$ credible intervals. The settings corresponding to (a), (b) and (c) are explained in the main text. Note the differences in vertical scale across the figures.
  \label{fig:example1}}
\end{figure}
For setting (b), traceplots of every $50$th iteration for a couple of coefficients $\nu_k$ are shown in Figure \ref{fig:example1-traceA2}.
\begin{figure}
	\begin{center}
	\includegraphics[scale=0.6]{traceA2}
		\caption{Traceplots for the simulation example from Section \ref{subsec:uniform} under setting (b). The posterior samples were subsampled, with every $50$th iteration kept. The displayed results are for parameters $\nu_1$, $\nu_6$ and $\nu_9$.
			\label{fig:example1-traceA2}}
	\end{center}
\end{figure}
We measure the error of an estimate $\{\hat\nu_k\}$ by $\operatorname{Err}(\nu, \hat\nu) = \sum_{k=1}^{\infty} |\hat\nu_k - \nu_k|$. The errors are reported in Table \ref{table:unif}. In all  settings, for these particular realisations of the simulated data, the Bayesian procedure outperformed the truncated estimator from \cite{buchmann04}. For setting (c), the latter is completely off, as was to be expected, given that it is derived under the assumption $\Delta_i=1$ for all $i$. An advantage of the Bayesian procedure is the included measure of uncertainty, namely the credible intervals for $\nu_k$. On the other hand, for  the Buchmann-Gr\"{u}bel estimator it is hardly possible to derive confidence intervals, since the limiting distribution of the estimator is fairly complicated.

\begin{table}
\begin{center}
\begin{tabular}{l c c c}
\toprule
Simulation setting & (a) & (b) & (c)\\
\midrule
{Buchmann-Gr\"ubel estimator} & $1.40$ & $0.32$ & $1.44$ \\
{Posterior mean} & $0.15$ & $0.07$ & $0.12$ \\
\bottomrule
\end{tabular}
\caption{Estimation results for scenarios (a)--(c) from Section~\ref{subsec:uniform}. See the text for details.}
\label{table:unif}
\end{center}
\end{table}

\subsection{Geometric base distribution}
\label{subsec:geom}

The setup of this synthetic data example likewise follows that in \cite{buchmann04}. Assume $q$ is a geometric distribution with parameter $\alpha$, i.e.\ $q_k=(1-\alpha)^k \alpha$ for $0 < \alpha <1$. Then $\lambda=-\log\alpha$, and
\[
p_{k} = - \frac{(1-\alpha)^k}{k\log\alpha}, \quad k\in\mathbb{N}.
\]
Hence, $\nu_{k}=(1-\alpha)^k/k$.

We consider two simulation setups:
\begin{itemize}
\item[(a)] $n=500$, $\Delta_i=1$ for $1\le i \le n$ and  $\alpha=1/3$;
\item[(b)] $n=500$, $\Delta_i=1$ for $1\le i \le n$ and $\alpha=1/6$.
\end{itemize}
We set $m= \min(15,Z_{(n)})$ and ran the sampler according to the settings of Section \ref{subsec:uniform}.
The results for both scenarios (a) and (b) are reported in Figure \ref{fig:example2}. In Table \ref{table:geom} we also report estimation errors in one simulation run. For this example and these generated data, the Bayesian procedure gives less precise point estimates than the Buchmann-Gr\"{u}bel method. Note that estimation error for $\alpha = 1/3$ is smaller than that for $\alpha = 1/6.$ This appears intuitive, as a smaller value of $\alpha$ corresponds to a larger value of $\lambda.$ The latter implies that on average each $Z_i$ is a superposition of a larger number of jumps, which renders the decompounding problem more difficult. However, this argument is hard to formalise.
\begin{table}
\begin{center}
	\begin{tabular}{l c c }
		\toprule
		Simulation setting & (a) & (b) \\
		\midrule
		{Buchmann-Gr\"ubel estimator} & $0.28$ & $0.60$ \\
		{Posterior mean} & $0.52$ & $1.05$ \\
		\bottomrule
	\end{tabular}
\caption{Estimation results for scenarios (a)--(b) from Section~\ref{subsec:geom}. See the text for details.}
\label{table:geom}
\end{center}
\end{table}

\begin{figure}
	\begin{center}
		\includegraphics[scale=0.6]{estimatesB1}\\
				\includegraphics[scale=0.6]{estimatesB2}
	\end{center}
	\caption{Simulation example from Section \ref{subsec:geom}.  Settings (a) and (b)  correspond to the true jump distributions $\operatorname{Geom}(1/3)$ and $\operatorname{Geom}(1/6),$ respectively. The horizontal axis gives the magnitudes of $\nu_k$, $k\in
	\{1,\ldots, 15\}$. The orange balls denote the true values, the black triangles the Buchmann-Gr\"{u}bel estimator. The blue crosses give the posterior means, whereas the vertical blue line segments represent (pointwise) $95\%$ credible intervals.
  \label{fig:example2}}
\end{figure}

\subsection{Monte Carlo study}\label{subsec:mc}
For a more thorough comparison of the Buchmann-Gr\"{u}bel estimator and our Bayesian method, we performed a small Monte Carlo experiment. We considered two settings:
\begin{enumerate}
  \item The setting from Section \ref{subsec:uniform} with $n=250$. We took $m=\min(15, Z_{(n)})$.
  \item The setting from Section \ref{subsec:geom} with $\alpha=1/3$. We took $m=\min(20, Z_{(n)})$.
\end{enumerate}
In both cases we assumed $\Delta_i=1$ for all $1\le i \le n$. The number of Monte Carlo repetitions was taken equal to $50$. We took $400.00$ MCMC iterations and discarded the first half of these as burn-in. In Figure \ref{fig:mcresults} we give a graphical display of the results by means of boxplots of the errors. Here, as before, if the true values are denoted by $\nu_k$ and the estimate within a particular simulation run  by $\hat\nu_k$, the error is defined by  $\operatorname{Err}(\nu, \hat\nu) = \sum_{k=1}^{\infty} |\hat\nu_k - \nu_k|$ (we truncated the infinite summation to $50$). The results agree with our earlier findings, in that there is no clear ``winner'' in the comparison. Note that for the setting (ii) we considered both $c=2$ and $c=0.01$ in the prior specification. Both values give similar performance of the Bayesian method. This provides insight into sensitivity of our results with respect to the prior specification. A minor difference between the middle and rightmost panels of Figure \ref{fig:mcresults} may be attributed to Monte Carlo error: the $50$ simulated datasets on which these panels are based on are not the same.  Note that our prior promotes sparsity, and in that respect it is not surprising it does better when the true data-generating L\'evy measure is `sparse'.
\begin{figure}
\includegraphics[scale=0.5]{mcstudy_redone}
\caption{Monte Carlo study from Subsection \ref{subsec:mc} comparing the Buchmann-Gr\"{u}bel estimator and the Bayesian method proposed in this paper. In this figure ``bg'' refers to the Buchmann-Gr\"{u}bel estimator, while  ``bayesmedian'' and ``bayesmean'' refer to the Bayesian method, where either the median or mean was used as a point estimator for each $\nu_i$. The leftmost panel corresponds to the setting (i), whereas the other two panels to the setting (ii). In the latter we used both $c=2$ and $c=0.01$ in the prior specification.  \label{fig:mcresults}}
\end{figure}

\subsection{Computing time}
In terms of computational effort, the time it takes to evaluate the Buchmann-Gr\"{u}bel estimator is negligible compared to our algorithm for sampling from the posterior. This is not surprising, as that frequentist estimator relies on a plug-in approach, whereas in our case an approximation to the posterior is obtained by MCMC simulation. However, if proper uncertainty quantification is desired, then the Bayesian method is advantageous in the sense that it does not solely produce a point estimate.

Note that the proposed MCMC scheme requires determination of the solutions to the Diophantine equation $\sum_{j=1}^m j k_j = z$ for all unique values  $z$ in the observation set. For moderate values of $z$, say $z\le 30,$ this is rather quick, but for large values of $z$ the computing time increases exponentially, as does the amount of the allocated memory. The computing time of each Metropolis-Hastings step is then very small, but we potentially need a very large number of iterations to reach stationarity. The latter is caused firstly  by the fact that at a particular iteration, our proposals for $\mu_{ij}$ do not take  into account the current values of $\nu_1,\ldots, \nu_m$; secondly, the size of the state space that needs to be explored  increases exponentially with $m$.

\section{Real data examples}\label{sec:realdata}
\subsection{Horse kick data}\label{subsec:horse}

To further illustrate our procedure, we will use the von Bortkewitsch data on the number of soldiers in the Prussian cavalry killed by horse kicks (available by year and by cavalry corps); this example was also employed in \cite{buchmann03}. Each observation is an integer from $0$ to $4$, giving the number of deaths for a given year and a given cavalry corps, with overall counts reported in Table \ref{table:bort}. The data are extracted from the table on p.~25 in \cite{bortkewitsch98}. Note that von Bortkewitsch corrects for the fact that the Guards and I, VI and XI cavalry corpses of the Prussian army had a different organisation from other units, and justifiably omits the corresponding counts from consideration.

It has been demonstrated by von Bortkewitsch that the Poisson distribution fits the horse kick data remarkably well. Assuming instead that observations follow a compound Poisson distribution is a stretch of imagination, as that would correspond to a horse running amok and killing possibly more than one soldier in one go. Nevertheless, this example provides a good sanity check for our statistical procedure.

\begin{table}
	\begin{center}
		\begin{tabular}{@{}lrrrrr@{}}
			\toprule
			{Deaths} & { $0$} & { $1$} & { $2$}
			& { $3$} & { $4$}  \\
			\midrule
			{Counts}     & $109$  & $65$  & $22$   & $3$ & $1$ \\
			\bottomrule
		\end{tabular}
		\caption{Data on the number of soldiers in the Prussian cavalry killed by horse kicks. See the table on p.~25 in \cite{bortkewitsch98}.}
		\label{table:bort}
	\end{center}
\end{table}
The estimation results are graphically depicted in Figure \ref{fig:horsekick}. Clearly, point estimates of both methods are in agreement and lend support to the Poisson model for this dataset.

\begin{figure}
	\begin{center}
		\includegraphics[scale=0.6]{estimates_hk}
	\end{center}
	\caption{Estimation for the horse kick data from Subsection \ref{subsec:horse}. The horizontal axis gives the magnitudes of $\nu_k$, $k\in
	\{1,\ldots, 4\}$. The black triangles denote the Buchmann-Gr\"{u}bel estimator, the blue crosses give the posterior means, whereas the vertical blue line segments represent (pointwise) $95\%$ credible intervals.
  \label{fig:horsekick}}
\end{figure}

\subsection{Plant data}\label{subsec:plant}

Our second real example is the one used in \cite{buchmann04}. Consider the data in Table \ref{table:plant}, taken from \cite{evans53}. The data were collected as follows: the area was divided into plots of equal size and in each plot the number of plants was counted; the number of plants in each plot ranges from $0$ to $12$. The second row of Table \ref{table:plant} gives the counts of plots containing a given number of plants; thus, there were $274$ plots that contained no plant, $71$ that contained $1$ plant, etc. It is customary in the ecological literature to model such count data as i.i.d.\ realizations from a compound Poisson distribution. Thus, e.g., \cite{neyman39} advocated the use of a Poisson base distribution in this context; another option here is a geometric base distribution. Given existence of several distinct modelling possibilities, performing an exploratory non-parametric  analysis appears to be a sensible strategy.

The estimation results are graphically depicted in Figure \ref{fig:plant}. There are some small differences between the posterior mean and the Buchmann-Gr\"ubel estimate, but overall they are very similar.
\begin{table}
	\begin{center}
		\begin{tabular}{@{}lrrrrrrrrrrrrr@{}}
			\toprule
			{Plants} & $0$ & $1$ & $2$
			& $3$ & $4$ & $5$ & $6$ & $7$ & $8$ & $9$ & $10$ & $11$ & $12$  \\
			\midrule
			{Counts}     & $274$  & $71$  & $58$   & $36$ & $20$ & $12$ & $10$ & $7$ & $6$ & $3$ & $0$ & $2$ & $1$ \\
			\bottomrule
		\end{tabular}
		\caption{Plant population data from \cite{evans53}.}
		\label{table:plant}
	\end{center}
\end{table}

\begin{figure}
	\begin{center}
		\includegraphics[scale=0.6]{estimates_pp}
	\end{center}
	\caption{Estimation results for the plant data from Subsection \ref{subsec:plant}. The horizontal axis gives the magnitudes of $\nu_k$, $k\in
	\{1,\ldots, 10\}$. The black triangles denote the Buchmann-Gr\"{u}bel estimates, the blue crosses give the posterior means, whereas the vertical blue line segments represent (pointwise) $95\%$ credible intervals.
  \label{fig:plant}}
\end{figure}

\section{Frequentist asymptotics}
\label{sec:asymptotics}

In this section we assume that the observation times $\{t_i\}$ are equidistant: $t_i = i, i=1,\ldots,n$. To evaluate our Bayesian method from a theoretical point of view, we will verify that it is consistent, and we will establish the rate at which the posterior contracts around the `true', data-generating L\'evy measure $\nu_0$; see \cite{ghosal17} for a thorough treatment of Bayesian asymptotics from the frequentist point of view. From now on the subscript $0$ in various quantities will refer to the data-generating distribution.

Our strategy consists in proving that the posterior contraction rate for $\nu_0$, given the sample $\mathcal{Z}_n=(Z_1,\ldots, Z_n)$, can be derived from the posterior contraction rate for $q_0$ given $\mathcal{Z}_n$, which is mathematically easier since $Z_1,\ldots, Z_n$ is a sequence of independent and identically distributed random variables with distribution $q_0$. We therefore effectively avoid dealing directly with the inverse nature of the problem of estimating $p_0$.

The  prior we consider in this section is defined  as follows:
\begin{itemize}
  \item Endow the rate $\lambda$ of the Poisson process with a prior distribution.
  \item Independently, endow the vector $(p_1,\ldots, p_{m})$ with a Dirichlet distribution with parameter $(\alpha_1,\ldots,\alpha_m)$.
  \item Set a priori $p_k=0$ for all $k > m$.
\end{itemize}
This is a somewhat simplified version of the prior we used in Section \ref{sec:algorithm}, which allows us to concentrate on essential features of the problem, without need to clutter the analysis with extra and unenlightening technicalities. Also remember the well-known relationship between the Gamma and Dirichlet distributions: if $\xi_1,\ldots,\xi_m$ are independent Gamma distributed random variables, $\xi_i \sim \operatorname{Gamma}(\alpha_i,1)$, then for $\eta_i = \xi_i / \sum_{j=1}^m \xi_j$, the vector $(\eta_1,\ldots,\eta_m)$ follows the Dirichlet distribution with parameter $(\alpha_1,\ldots,\alpha_m)$; furthermore, we have that $\sum_{j=1}^m \xi_j \sim \operatorname{Gamma}\left(\sum_{j=1}^m \alpha_j,1\right)$, and $\eta_i$ are independent of $\sum_{j=1}^m \xi_j$.

In our asymptotic setting, we will make $m=m_n$ dependent on $n$ and let $m_n\rightarrow\infty$ at a suitable rate as $n\rightarrow\infty$.

Recall that we write $Q=(q_k)_{k\in\mathbb{N}_0}$ for $q_k=Q(\{k\})$. Let $\scr{Q}$ denote the collection of all probability measures supported on $\mathbb{N}$.
\begin{thm}\label{thm:postcontractionrate}
	Suppose there exists  $\underline\alpha$, such that $0<\underline{\alpha} \le \alpha_i \le 1$ for all $1\le i \le m_n$. Suppose $\lambda \sim \operatorname{Gamma}(a,b)$ with $a\in (0,1]$ and that $\nu_0$ has a compact support.
Then, for any $\gamma >1$,
\[  \Pi_n\left( \| \nu - \nu_0 \|_1 \geq \frac{\log^{\gamma} n}{\sqrt{n}} \,\middle\vert\, \mathcal{Z}_n \right) \to 0\] in $Q_0^n$-probability, as $n\to \infty$.
\end{thm}

\begin{rem}
	Note that since the support of $\nu_0$ is not assumed to be known, our CPP model is still naturally non-parametric. The assumption of the compact support of $\nu_0$ does not cover the simulation example of Section \ref{subsec:geom}. However, its relaxation appears to pose very difficult technical challenges and is not attempted in this work.
\end{rem}

The remainder of this section is devoted to the proof of Theorem \ref{thm:postcontractionrate}.

\subsection{Basic posterior inequality via the stability estimate}
A key step of the proof of Theorem~\ref{thm:postcontractionrate} is the stability estimate in Equation \eqref{eq:tv} below, which bounds the total variation distance between the L\'evy measures $\nu, \nu^{\prime}$ in terms of the total variation distance between the corresponding compound distributions $q,q^{\prime}$.

In principle, it is conceivable that the Panjer recursion should allow one to bound probability distances between $P$-probabilities via distances between $Q$-probabilities; we call such a bound a stability estimate. Nevertheless, explicit as the equations of the Panjer recursion are, they are still somewhat unwieldy for that purpose. Hence we will use another inversion formula from \cite{buchmann03}, that will lead to the stability estimate we are after.

First we introduce some notation, and also recall a few useful facts summarised in \cite{buchmann03}. The space of absolutely summable sequences is defined as
$\ell_1 \coloneqq \left\{ a\in\mathbb{R}^{\mathbb{N}_0}: \sum_{j=0}^{\infty} |a_j| < \infty \right\}$,
with a norm given by
$\|a\|_1 = \sum_{j=0}^{\infty} |a_j|$.
For probability vectors $a,b$, the norm $\|a-b\|_1$ is (twice) the total variation distance between $a$ and $b$. For any $a,b\in\ell_1$, we have the inequality
\begin{equation}
\label{eq:ineq:conv}
\|a\ast b\|_1 \leq \|a\|_1 \|b\|_1,
\end{equation}
where $\ast$ denotes convolution of $a$ and $b.$ We define a mapping $a\mapsto\exp(a)$ from $\ell_1$ into $\ell_1$ via
\[
\exp(a)=\sum_{j=0}^{\infty} \frac{a^{\ast j}}{j!}.
\]
The exponential has the following two useful properties:
\[
\exp(a+b)=\exp(a)\ast\exp(b), \quad a,b\in\ell_1,
\]
and
\[
\exp(a)=\exp(b) \Longrightarrow a=b, \quad a,b\in\ell_1.
\]
We define a sequence $\delta_0=(\delta_{0,k})_{k\in\mathbb{N}_0}$, such that $\delta_{0,0}=1$ and its all other entries are equal to zero. Then, using the above properties of the exponential, we can write  concisely the compounding mapping in \eqref{eq:comp} in terms of convolutions of infinite sequences: $q=\exp(\lambda(p-\delta_0))$. Its convolution inverse, i.e.\ $q^{\ast(-1)}$ such that $q^{\ast(-1)} \ast q = \delta_0$, is given by $r=q^{\ast(-1)}=\exp(-\lambda(p-\delta_0))$. Note that $r\in\ell_1$. We have the following recursive expressions
\[
r_0 = \frac{1}{q_0}, \quad r_k = - \frac{1}{q_0} \sum_{j=1}^k q_j r_{k-j}, \quad k\in\mathbb{N}.
\]

\begin{lemma}\label{lemma:tv}
Let $q,q^{\prime}$ correspond to two pairs $(\lambda,p)$ and $(\lambda^{\prime},p^{\prime})$, respectively (and $r$ correspond to $q$, i.e.\ the pair $(\lambda,p)$).
Then, in accordance with the notation introduced above and provided that $\|q^{\prime}-q\|_1<\|r\|_1^{-1}$, it holds that
\begin{equation}
\label{eq:tv}
\| \nu^{\prime} - \nu \|_1 = \| \lambda^{\prime}p^{\prime} - \lambda p \|_1 \leq \frac{\|r\|_1 \| q^{\prime} - q \|_1}{1 - \|r\|_1 \| q^{\prime} - q \|_1 }.
\end{equation}
\end{lemma}
\begin{proof}
The result is a direct consequence of Lemma 3 in \cite{buchmann03}, which states that
\[
(\lambda^{\prime}-\lambda)\delta_0 + \lambda p - \lambda^{\prime}p^{\prime} = \sum_{j=1}^{\infty} \frac{1}{j}(r\ast(q-q^{\prime}))^{\ast j}
\]
whenever $\|q^{\prime}-q\|_1<\|r\|_1^{-1}$.
Taking the $\|\cdot\|_{1}$-norm on both sides and some elementary bounding via \eqref{eq:ineq:conv} imply that
$$|\lambda^{\prime}-\lambda|+ \| \lambda^{\prime}p^{\prime} - \lambda p \|_1 \leq \frac{\|r\|_1 \| q^{\prime} - q \|_1}{1 - \|r\|_1 \| q^{\prime} - q \|_1 },$$
and thus Equation \eqref{eq:tv} follows.
\end{proof}
We will use Equation \eqref{eq:tv} to establish the key inequality for the posterior measure $\Pi(\cdot \mid \mathcal{Z}_n)$. We recall once again that the subscript $0$ refers to `true', data-generating quantities.

\begin{prop}\label{prop:contr}
For any prior $\Pi$ on $\nu$, for any $\varepsilon\in (0,1] $ and for any $n\geq 1$, the following posterior inequality holds:
$$\Pi\left( \| \nu - \nu_0 \|_1 \geq \varepsilon \,\middle\vert\, \mathcal{Z}_n \right)\leq 2 \Pi \left( \| q - q_0 \|_1 \geq \frac{\varepsilon}{2\|r_0\|_1} \,\middle\vert\,  \mathcal{Z}_n \right).$$
\end{prop}

\begin{proof}
Write $\{ \nu: \| \nu - \nu_0 \|_1 \geq \varepsilon \}$ as a union of the sets
\[\left\{ \nu: \| \nu - \nu_0 \|_1 \geq \varepsilon \right\} \cap \left\{ \nu: \|r_0\|_1 \| q - q_0 \|_1 < 1/2 \right\}\]
and
\[ \left\{ \nu: \| \nu - \nu_0 \|_1 \geq \varepsilon \right\} \cap \left\{ \nu: \|r_0\|_1 \| q - q_0 \|_1 \geq  1/2 \right\}.\]
Thanks to Lemma \ref{lemma:tv}, the set
\[
\left\{ \nu: \| \nu - \nu_0 \|_1 \geq \varepsilon \right\} \cap \left\{ \nu: \|r_0\|_1 \| q - q_0 \|_1 < 1/2 \right\}
\]
is a subset of $\{ \nu: \| q - q_0 \|_1 \geq {\varepsilon}/(2\|r_0\|_1) \}$. The proof is concluded by observing that $\left\{ \nu: \| \nu - \nu_0 \|_1 \geq \varepsilon \right\} \cap \left\{ \nu: \|r_0\|_1 \| q - q_0 \|_1 \geq  1/2 \right\}$ is a subset of $\{ \nu: \| q - q_0 \|_1 \geq {\varepsilon}/(2\|r_0\|_1) \}$, too, since $\varepsilon\leq 1$.
\end{proof}
In general, stability estimates like the one in Equation \eqref{eq:tv} are unknown in the literature on L\'evy processes. Consequently, studying Bayesian asymptotics for L\'evy models, even in the CPP case, necessitates the use of very intricate arguments under restrictive assumptions; see, e.g., \cite{nickl17}.

\subsection{Proof of Theorem \ref{thm:postcontractionrate}}
The usefulness of Proposition \ref{prop:contr} above lies in the fact that the posterior contraction rate in the inverse problem of estimating the L\'evy measure $\nu_0$ from indirect observations $\mathcal{Z}_n$ can be now deduced from the posterior contraction rate in the direct problem of estimating the compound distribution $q_0$, which is easier (observe that $r_0$ is determined by $\nu_0$ and is therefore fixed in the proofs). The general machinery developed in \cite{ghosal00} can be applied to handle the latter, and also several inequalities obtained in \cite{gugu15} are useful in that respect.
In particular, we make use of the following inequality for the Hellinger distance,
\begin{equation}
\label{eq:HL}
h(q_{\lambda, p}, q_{\lambda^{\prime}, p^{\prime}}) \le \sqrt{\lambda} h(p,p^{\prime}) +|\sqrt{\lambda} - \sqrt{\lambda^{\prime}}|,
\end{equation}
Cf.\ Lemma 1 in \cite{gugu15}.
To ease our notation, in the sequel we will often write  $q$ and $q^{\prime}$ instead of $q_{\lambda, p}$ and $q_{\lambda^{\prime}, p^{\prime}},$ respectively.

Denote
\[ KL(q_0,q) = Q_0\left(\log \frac{q_0}{q}\right), \quad V(q_0,q) = Q_0\left(\log \frac{q_0}{q}\right)^2. \]
Another two inequalities we will use are the following: let $\lambda, \lambda_0 \in [\underline{\lambda}, \overline{\lambda}].$ Then there exists a positive constant $\overline{C}$, such that
\begin{equation}\label{eq:transfer-qtolambdaandp}
\begin{split}  KL(q_0, q) \le \overline{C}\left(KL(p_0,p) +|\lambda_0- \lambda|^2\right), \\
 V(q_0, q) \le \overline{C}\left(V(p_0,p)+ KL(p_0,p) +|\lambda_0- \lambda|^2\right); \end{split}
 \end{equation}
cf.~equations (14) and (15) in Lemma 1 in \cite{gugu15}.

These three inequalities can be obtained by adjustment of the arguments used in \cite{gugu15}. However, we opted to give their direct proofs in Lemma \ref{KLV} from Appendix ~\ref{app:lemmas} under slightly weaker conditions than required in \cite{gugu15}.

Our proof of Theorem \ref{thm:postcontractionrate} proceeds via verification of the conditions for posterior contraction in Theorem 2.1 in \cite{ghosal00}.  In our setting, the latter result reads  as follows.
\begin{thm}\label{thm:ggvdv}
Assume $\scr{Z}_n =(Z_1,\ldots, Z_n),$ where $Z_1,\ldots, Z_n$ are independent and identically distributed with distribution $q_0$. Let $h$ denote the Hellinger metric on $\scr{Q}$, a collection of all measures with support in $\mathbb{N}$.
Suppose that for a sequence $\{\epsilon_n\}$ with $\epsilon_n \to 0$ and $n\epsilon_n^2\to \infty$, a constant $C$ and sets $\scr{Q}_n\subset \scr{Q}$, we have
\[ \log N(\epsilon_n, \scr{Q}_n, h) \le n\epsilon_n^2,\]
\[ \Pi_n(\scr{Q} \setminus \scr{Q}_n) \le \exp\left(-n\epsilon_n^2(C+4)\right), \]
\[ \Pi_n\left( q\colon KL(q_0,q) \le \epsilon_n^2,\, V(q_0,q) \le \epsilon_n^2\right) \ge \exp\left(-C n\epsilon_n^2\right). \]
Then, for sufficiently large $M>0$, we have that $\Pi_n \left(Q\colon h(q,q_0)\ge M\epsilon_n \mid \scr{Z}_n \right) \to 0$ in $Q_0^n$-probability.
\end{thm}

We will now verify the three conditions of this theorem, which we refer to as the entropy condition, the remaining mass condition, and the prior mass condition, respectively.
To that end,
fix strictly positive sequences $\{\underline\Lambda_n\}$, $\{\overline\Lambda_n\}$, and
define the sieves
\[ \scr{Q}_n =\left\{ q_{\lambda,p} \colon \lambda \in [\underline{\Lambda}_n, \overline\Lambda_n],\: \operatorname{supp} p \subseteq \{1,\ldots,m_n\} \right\}. \]

\subsubsection{Entropy}
We start with bounding the entropy of the sieve $\scr{Q}_n$ for $h$-balls of radius $\epsilon_n$.
\begin{lemma}
	\label{lem:entropy}
Assume that as $n\to\infty,$
\begin{equation}
\label{eq:nlarge}
m_n\rightarrow\infty, \quad \epsilon_n \rightarrow 0, \quad \underline{\Lambda}_n \rightarrow 0, \quad \overline{\Lambda}_n \rightarrow \infty.
\end{equation}
Then
\begin{equation}\label{eq:entropy}
\log N(\epsilon_n, \scr{Q}_n, h) \lesssim m_n \left\{ \log (m_n) + \log (\overline{\Lambda}_n) +  \log\left( \frac{1}{\epsilon_n} \right) \right\} + \log \left(\frac{1}{\underline{\Lambda}_n} \right). \end{equation}
\end{lemma}

\begin{proof}
For $\lambda, \lambda' \ge \underline{\Lambda}_n$,
\[ |\sqrt{\lambda}-\sqrt{\lambda'}| = \frac{|\lambda-\lambda'|}{\sqrt{\lambda}+\sqrt{\lambda'}} \le \frac1{2\sqrt{\underline{\Lambda}_n}} |\lambda -\lambda'|. \]
Furthermore, from Section 3.2 in \cite{pollard_2001},
\[   	h(p, p')   \leq \sqrt{ \| p - p^{\prime}\|_1 }  \leq \sqrt{ m_n \| p-p' \|_{\infty} }. \]
Combining the preceding two displays and Equation \eqref{eq:HL}, we get
\[  h(q_{\lambda, p}, q_{\lambda', p'}) \le  \sqrt{\overline{\Lambda}_n m_n \|p-p'\|_{\infty}} + \frac1{2\sqrt{\underline{\Lambda}_n}} |\lambda -\lambda'|. \]
Hence, if
\[ \|p-p'\|_{\infty} \le \frac{\epsilon_n^2}{4 \overline{ \Lambda}_n m_n}, \quad  |\lambda-\lambda'| \le \sqrt{\underline{\Lambda}_n} \epsilon_n, \]
then  the Hellinger distance between $q_{\lambda, p}$ and $q_{\lambda', p'}$ is bounded by $\epsilon_n$.
To cover $[\underline{\Lambda}_n, \overline\Lambda_n]$, we need at most $\lfloor \frac{\overline{\Lambda}_n}{2\epsilon_n \sqrt{\underline{\Lambda}_n}}\rfloor +1$ intervals of length $2 \sqrt{\underline{\Lambda}_n} \epsilon_n$.
To cover discrete distributions with support in $\{1,\ldots,m_n\}$, we need at most
\[ \left(  \left\lfloor \frac{2\overline{\Lambda}_n m_n }{\epsilon_n^2} \right\rfloor +1  \right)^{m_n}\]
$L_{\infty}$-balls of radius $\epsilon_n^2 / (4\overline{\Lambda}_n m_n)$. Under assumption \eqref{eq:nlarge}, the summand $1$ in the above display is asymptotically negligible and can be omitted. In that case, the number of $h$-balls that we need to cover $\scr{Q}_n$ is of order
\[ \left( \frac{\overline{\Lambda}_n m_n }{\epsilon_n^2} \right)^{m_n}\times  \frac{\overline{\Lambda}_n}{\epsilon_n \sqrt{\underline{\Lambda}_n}} . \]
Taking the logarithm and next a straightforward rearrangement of the terms gives the statement of the lemma. \end{proof}

\subsubsection{Remaining prior mass}
Now we will derive an inequality for the remaining prior mass.

\begin{lemma}\label{lem:remainingmass} For $\lambda\sim \operatorname{Gamma}(a,b)$ with $0<a \leq 1$,
	\[ \Pi_n(\scr{Q}\setminus \scr{Q}_n) \lesssim {\overline{\Lambda}}_n^{a - 1} e^{- b \overline\Lambda_n} +  \underline{\Lambda}_n .\]
\end{lemma}

\begin{proof}
We have (with a slight abuse of notation)
	\[ \Pi_n(\scr{Q}\setminus \scr{Q}_n) = \Pi_n\left( [\overline{\Lambda}_n, \infty)\right) +\Pi_n\left( [0,\underline{\Lambda}_n)\right).\]
Now,
\[
\Pi_n( \lambda \geq \overline{\Lambda}_n ) = \frac{ b^a }{\Gamma(a) } \int_{ \overline{\Lambda}_n }^{\infty} \lambda^{a - 1} e^{-b \lambda }\mathrm{d}\lambda \lesssim \overline{\Lambda}_n^{a - 1} e^{-b \overline\Lambda_n}.
\]
Furthermore,
\[
\Pi_n\left( [0,\underline{\Lambda}_n)\right) = \frac{b^a}{\Gamma(a)  } \int_0^{ \underline{\Lambda}_n } \lambda^{a - 1} e^{- b\lambda}\mathrm{d}\lambda \lesssim \underline{\Lambda}_n^a.
\]
The proof is concluded.
\end{proof}

\subsubsection{Prior mass}
Finally, we lower bound the prior mass in a small Kullback-Leibler neighbourhood of the data-generating compound distribution $q_0$. Define the function  $g\colon (0,\infty) \times (0,1) \to (0,\infty)$ by \[g(\epsilon,c) = C \frac{\epsilon^2}{2[\log(e/c)]^2},\] where $C$ is the constant appearing in the statement of Lemma \ref{lem:KLVtoL1} below.
\begin{lemma}\label{lem:priormass}
Assume that
\begin{enumerate}
  \item there exists  $\underline\alpha$, such that $0<\underline{\alpha} \le \alpha_i \le 1$ for all $1\le i \le m_n$;
  \item strictly positive sequences $\underline{p}_n \rightarrow 0$, $\epsilon_n\rightarrow 0$ and $m_n\rightarrow\infty$ satisfy the inequalities $m_n  g(\epsilon_n,\underline{p}_n)<1$ and $\underline{p}_n <g(\epsilon_n,\underline{p}_n)^2.$
\end{enumerate}
Define
\[ B_n(\epsilon) = \left\{q \in \scr{Q}_n \colon  KL(q_0,q) \le \epsilon^2,\: V(q_0,q) \le \epsilon^2\right\}. \]
Then
\begin{multline*} \Pi_n(B_n(\epsilon_n))  \gtrsim \Pi\left(  |\lambda_0-\lambda| \le \widetilde\epsilon_n\right)\\ \times
	\Gamma\left(\sum_{i=1}^{m_n} \alpha_i\right)  \exp\left(-m_n\log(1/(g(\widetilde\epsilon_n,\underline{p}_n)^2-\underline{p}_n)) - m_n \log(1/\underline{\alpha})\right).
\end{multline*}
Here $\widetilde\epsilon_n = \epsilon_n/\sqrt{3\overline{C}}$, with a constant $\overline{C} > 0 $ not depending on $n$.
\end{lemma}
\begin{proof}
Define
\begin{multline*}
 \widetilde{B}_n(\epsilon) = \Big\{(\lambda, p)\colon \lambda \in [\underline\Lambda_n, \overline\Lambda_n],\: \min_{1\le i \le m_n} p_i \ge \underline{p}_n, \operatorname{supp} p \subseteq\{1,\ldots,m_n\}, \\  KL(p_0,p) \le \epsilon^2,\: V(p_0,p) \le \epsilon^2,\: |\lambda_0-\lambda| \le \epsilon\Big\} .\end{multline*}
For all $n$ large enough and $\epsilon$ small, we have $\{ \lambda: |\lambda_0-\lambda| \le \epsilon  \} \subseteq [\underline\Lambda_n, \overline\Lambda_n].$ Then by inequalities in Lemma \ref{KLV},  $\widetilde{B}_n(\epsilon)\subset B_n\left(\sqrt{3\overline{C}}\epsilon\right)$, with a constant $\overline{C}$ that can be taken the same for all large enough $n$; see the arguments in Section 4.2 in \cite{gugu15}. Hence, using the a priori independence of $p$ and $\lambda$,
\[ \Pi(B_n\left(\epsilon_n\right)) \ge \Pi\left(\widetilde{B}_n\left(\widetilde\epsilon_n\right)\right)  = \Pi\left(   |\lambda_0-\lambda| \le \widetilde\epsilon_n\right) \times U_n, \] where
\[ U_n=  \Pi_n\left( \left\{p \colon  KL(p_0, p) \le \widetilde\epsilon_n^2, \,  \, V(p_0,p) \le \widetilde\epsilon_n^2, \, \min_{1\le i \le m_n} p_i \ge \underline{p}_n\right\}\right).
\]
Now, $\Pi\left(   |\lambda_0-\lambda| \le \widetilde\epsilon_n\right)  \asymp \widetilde\epsilon_n.$ Furthermore, by Lemma \ref{lem:KLVtoL1} from Appendix~\ref{app:lemmas}, we have
\[ U_n \ge \Pi_n\left(\left\{p \colon  \sum_{i=1}^{m_n} |p_{0i}-p_i| \le 2 g(\widetilde\epsilon_n, \underline{p}_n), \, \min_{1\le i \le m_n} p_i \ge \underline{p}_n\right\}\right).\]
The statement of the lemma now follows upon  applying Lemma \ref{lem:dirichlet} from Appendix \ref{app:lemmas} with $\eta=\underline{p}_n$ and $\epsilon =g(\widetilde\epsilon_n, \underline{p}_n)$.
\end{proof}

\subsection{Using bounds in Theorem \ref{thm:ggvdv}}
We take
\begin{gather*}
m_n \asymp \log n, \quad \epsilon_n \asymp \frac{\log^{\gamma} n}{\sqrt{n}},\quad \underline{p}_n \asymp \frac{1}{n^2},\\
\overline{\Lambda}_n \asymp \log^{2\gamma} n, \quad  \underline{\Lambda}_n \asymp \exp(-\operatorname{const} \cdot \log^{2\gamma}n)
\end{gather*}
with appropriately selected proportionality constants, and verify the conditions in Theorem \ref{thm:ggvdv}.

Firstly, condition \eqref{eq:nlarge} is trivially satisfied. Therefore, we can invoke Lemma \ref{lem:entropy} and conclude that the entropy is upper bounded by a multiple of $\log^{2\gamma} n$, since $\gamma > 1.$ Now $\log^{2\gamma} n \lesssim n\epsilon_n^2$, and this verifies the entropy condition in Theorem \ref{thm:ggvdv}.

Be Lemma \ref{lem:remainingmass}, for a suitable choice of the constant $C$ the remaining prior mass condition is likewise satisfied.

Finally, for the prior mass condition in a small Kullback-Leibler neighbourhood to hold, by Lemma \ref{lem:priormass} we need that the term
\[
\Pi\left(  |\lambda_0-\lambda| \le \widetilde\epsilon_n\right)  \exp\left(-m_n\log(1/(g(\widetilde\epsilon_n,\underline{p}_n)^2-\underline{p}_n)) - m_n \log(1/\underline{\alpha})\right)
\]
is lower bounded by
 $ \exp( - C n\epsilon_n^2)$ for some large enough $C>0$. Take  the logarithm on both sides of the above display and note that by our conditions
 \[
 \log \left( \Pi\left(  |\lambda_0-\lambda| \le \widetilde\epsilon_n\right) \right) \gtrsim \log( \widetilde\epsilon_n) \gtrsim -n\epsilon_n^2.
 \]
 Likewise,
 \[
 m_n\log(1/(g(\widetilde\epsilon_n,\underline{p}_n)^2-\underline{p}_n)) + m_n \log(1/\underline{\alpha}) \lesssim n\epsilon_n^2,
 \]
so that the prior mass condition holds.

Thus we have verified all the conditions of Theorem \ref{thm:postcontractionrate}. The resulting posterior contraction rate is $\epsilon_n \asymp {\log^\gamma n}/{\sqrt{n}}.$

\section{Outlook}
\label{sec:outlook}

In this paper we introduced a non-parametric Bayesian approach to estimation of the L\'evy measure $\nu$ of a discretely observed CPP, when the support of $\nu$ is a subset of $\mathbb{N}.$ We constructed an algorithm for sampling from the posterior distribution of $\nu$, and showed that in practice our procedure performs well and measures up to a benchmark frequentist plug-in approach from \cite{buchmann04}. Although computationally more demanding and slower than the latter, our method has an added benefit of providing uncertainty quantification in parameter estimates through the spread of the posterior distribution. On the theoretical side we show that our procedure is consistent, in that  asymptotically, as the sample size $n\rightarrow\infty,$ the posterior concentrates around the `true', data-generating distribution. The corresponding posterior contraction rate is the (nearly) optimal rate $\log^{\gamma}n/\sqrt{n}$ for an arbitrary $\gamma > 1$, if we are to ignore a practically insignificant $\log n$ factor.

Among several generalisations of our results, the one that looks the most promising is extension of our methodology to CPP processes with jump size distributions supported on the set of integers $\mathbb{Z}.$ The corresponding model has garnered substantial interest in financial applications, see \cite{barndorff12}. We leave this extension as a topic of possible future research.

redpajama_set_name: RedPajamaArXiv


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text:
Charles W. Mandeville
Bedle Funeral Home 212 Main Street, Matawan, NJ 07747 732-566-1962 Charles W. Mandeville (97), formerly of Matawan, passed away peacefully Saturday morning, May 30, in his home in Ottsville, Pa, where he resided with his daughter, Carol. Born February 5, 1918, Rahway, NJ, he is predeceased by his wife, Eleanor Wyckoff Mandeville and his older brother, George Donald Mandeville. He is survived by his daughter Carol; two sons Charles (Chip) and Stephen and his husband Victor Aluise, and grandchildren, Tracy Caldwell, Scott Caldwell, Charles Daniel Mandeville, Evangeline Mandeville, and Michelle Mandeville. Charlie spent most of his life in Matawan, NJ, graduating from Matawan High School and Monmouth College, Long Branch, NJ. He served his country during WWII in the Army Signal Corp in both North Africa and Italy where he was awarded the Bronze Star. During his lifetime he was regarded in high esteem for his honesty and fairness in professional positions as Vice President of the Matawan Branch of Central Jersey Bank & Trust Company, Manager of Marlboro Branch of same(simulteously), and Financial Consultant to Lawn Doctor. He was an active and ardent member of the Matawan community, serving as President of the Matawan Chamber of Commerce, Board of Directors of Bayshore Community Hospital, Matawan Board of Health, just to name a few... not to mention a gig as choir director for the First Baptist Church of Matawan. Charlie loved opera, classical music, world travel, reading, learning, thinking... and the list goes on. But most of all he loved the company of his family and friends... and a good party! His memorial service will be held on June 20th at Bedle Funeral Home in Matawan from 11AM-1PM with a celebration of his life beginning at 11:30AM. We encourage all to share a story or two! In lieu of flowers, memorial donations can be made in Charlie's memory to World Wildlife Fund 1250 24th Street, N.W. Washington, DC 20037. Please visit Charlie's memorial website at www.bedlefuneralhome.com
Bedle Funeral Home 212 Main Street, Matawan, NJ 07747 732-566-1962 Charles W. Mandeville (97), formerly of Matawan, passed away peacefully Saturday morning, May 30, in his home in Ottsville, Pa, where he resided with his daughter, Carol.... View Obituary & Service Information
The family of Charles W. Mandeville created this Life Tributes page to make it easy to share your memories.
Bedle Funeral Home 212 Main Street, Matawan, NJ  07747 732-566-1962...
Send flowers to the Mandeville family.

redpajama_set_name: RedPajamaCommonCrawl


First-person singular (yo) future subjunctive form of disecar.
Formal second-person singular (usted) future subjunctive form of disecar.
Third-person singular (él, ella, also used with usted?) future subjunctive form of disecar.


If you've spent time and effort improving your SEO strategy, you might feel – and even fear – that all of your hard work could go to waste with a website redesign. Keyword (no pun intended) = could. If you're not careful with how you navigate the SEO realm before relaunching, that might be an unfortunate truth.
Relaunching your website without a carefully planned search engine optimization strategy could result in devastating consequences. But, with the right knowledge and preparation, there's no need to fear losing your street cred in Google.
It's common for content from the old site to be reused and repurposed for the new site. You're redesigning for a reason though, and a big reason is likely that your old information architecture is too unorganized and confusing to be carried over to the new site.
The point of a redesign is to transform your website into a more appealing, useful experience for visitors. Naturally, you'll want to restructure everything in a way that makes sense for your business structure and aligns with how users expect to find the information they need.
This might mean consolidating and eliminating pages, or spacing out your content across even more pages to make it easily digestible. Either way, there won't be a 1-to-1 match between the old site and the new site.
That's where the 301 redirect comes into play.
A 301 redirect is a permanent detour from one URL to another. For example, if the content from web page www.mywebsite.com/service1 now lives on web page www.mywebsite.com/service-package1, you would need to set up a redirect from /service1 to /service-package1.
Once this is complete, it's also crucial to crawl your site for a complete list of URLs prior to relaunching. This ensures every URL has a home, or redirect, on the new site. Mapping out redirects can be a time-consuming process, but it's well-worth it considering the consequences of skipping right over something that can save time in the long run for you and visitors to your site.
For sustaining SEO – Redirects transfer your ranking power from one page to another, so the power you built up before your redesign isn't thrown completely down the drain.
For improving usability – If someone saves or bookmarks a page on your old website and tries to go back to it without a redirect in place, they'll get a 404 error page – and we all know how frustrating those are. Now, they don't know where to find the information they thought was important enough to save or bookmark in the first place.
Just because you're launching a new website, doesn't mean you need to toss all of the effort you put into the previous one out the window.
Before redesigning, perform a baseline ranking report that will tell you all of the keywords you're ranking for in search engines. Use the report to identify which keywords your domain currently ranks well for.
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Website redesigns are common practice and necessary to keep up with technological advances, business changes, and user expectations. At the same time, there are steps you should to stay in Google's good graces to maintain your ranking power.
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text:
Pappas Capital Earns CEO Cancer Gold Standard Accreditation
The CEO Roundtable on Cancer has accredited Pappas Capital, in Durham, NC, with CEO Cancer Gold StandardTM accreditation, recognizing its efforts to reduce the risk of cancer for its employees and covered family members.
CEO Cancer Gold Standard™ Toolkit Supports University Led Health and Leadership
Through a grant from the Robert Wood Johnson Foundation (RWJF), outreach efforts are underway to encourage academic institutions to adopt the CEO Cancer Gold Standard, a no cost employer-led health and wellness initiative that recognizes exceptional workplaces committed to reducing the burden of cancer for their employees.
University of Iowa Health Care earns workplace wellness recognition for reducing cancer risk
University of Iowa Health Care has earned CEO Cancer Gold Standard™ accreditation from the CEO Roundtable on Cancer, a nonprofit organization that recognizes employers that promote worker health and wellness.
Cree Earns CEO Cancer Gold Standard Accreditation
The CEO Roundtable on Cancer has accredited Cree in Durham, NC with the CEO Cancer Gold StandardTM recognizing its efforts to reduce the risk of cancer for its employees and covered family members.
CEO Roundtable on Cancer recognizes the University of North Dakota for ten years of leadership; lauds Robert Wood Johnson Foundation for supporting university-led, employer wellness initiative
The University of North Dakota has earned CEO Cancer Gold Standard™ re-accreditation for maintaining a strong commitment to the health of its employees and satisfying the comprehensive requirements of the Gold Standard for the past ten years.
LEERINK Partners LLC Earns CEO Cancer Gold Standard Accreditation
The CEO Roundtable on Cancer has accredited LEERINK Partners LLC in Boston, MA with the CEO Cancer Gold StandardTM recognizing its efforts to reduce the risk of cancer for its employees and covered family members.
CVS Health Earns CEO Cancer Gold Standard Accreditation
Woonsocket, RI - February 13, 2017
The CEO Roundtable on Cancer has accredited CVS Health (NYSE: CVS), the nation's largest pharmacy innovation company, with the CEO Cancer Gold Standard, recognizing its efforts to reduce the risk of cancer for its employees and covered family members. The company earned the distinguished Gold Standard accreditation for taking actions in five key areas of health and wellness to fight cancer in the workplace, including: discouraging tobacco use; encouraging physical activity; promoting healthy diet and nutrition; detecting cancer at its earliest stages; and providing access to quality care.


text:
What's The Gayest State In The Union?
Gallup and professor Gary Gates have called up over 200,000 people and asked them, "Do you, personally, identify as lesbian, gay, bisexual, or transgender?"
(By the way, I can never remember what the T in LGBT stands for: Transexual? Transvestite? Presumably, you can make a living on the academic conference circuit having a strong opinion on what the "T" should stand for and why your kind of Ts are most oppressed and therefore most deserving.)
Nationally, 3.5% say "Yes," which isn't too surprising or even interesting. It reminds me of the Onion football headline: "Both teams satisfied with three-and-a-half-yard carry." The Los Angeles Times ran an entire article on these results without mentioning the national average figure, which is a lot less than the Kinsey 10% number that all right-thinking people promoted for decades.
Of the 50 states, the gayest is Hawaii, which also isn't surprising. Hawaii has long been a destination for semi-closeted celebrities downshifting their careers, like Jim "Gomer Pyle" Nabors in 1976.
In general, the term "downshifting" comes to mind when looking at the top of the list: many of the the states at the top look like lower cost, quieter places for homosexuals to head for when the bright lights of San Francisco (Oregon, Nevada), Los Angeles (Hawaii, Nevada), and New York (Vermont, Maine) start to lose their luster.
Gay downshifting/retirement is a pretty big deal economically, as I noticed when I visited Palm Springs last year. The overall Palm Springs area is a vast exurban metropolis? exopolis?, while the actual municipality of Palm Springs, the original core, is now mostly inhabited by older homosexuals downshifting or retired from entertainment industry careers or the like. The whole place has been transformed into a fantasy land of early 1960s Suburban Moderne interior decoration. It's like you are five years old again and your mom is taking you to visit the house of her most upscale social-climbing friend, who has just redecorated in Space Age colors.
A gay male reader pointed out that Palm Springs even has a golf course, Indian Canyons, favored by that most minuscule of all demographics: gay male golfers. (He estimates there are about 30 gay male regulars at Indian Canyons.)
In general, a visit to Palm Springs suggests that the overall national population doesn't seem to be getting particularly gayer in the 21st Century, that there are a huge number of old gays from the 1970s around. The only young males around in downtown Palm Springs are the Mexican streetwalkers.
Other questions inspired by the list: is the big difference between South Dakota (4.4%) and North Dakota (1.7%) just caused by small sample sizes, or is South Dakota just gayer than North Dakota? I am reminded of the various serious attempts made by North Dakota legislators over the last generation to change the name of their state to the more manly-sounding "Dakota."
One important question for these kind of surveys where the goal is to come up with precise estimates of quite small percentages is how large the random error rate is in answering questions. I wouldn't be surprised if, say, one percent of respondents mishear or misunderstand the question, which would tend to falsely narrow the spread among states.


text:
<!DOCTYPE html>

    <meta charset="utf8">
    <title>test</title>

it works!

redpajama_set_name: RedPajamaGithub


Are you a current high school student (or current 1st year college student transferring into Carson-Newman) that is interested in social justice or community engagement?
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Applications for students already enrolled in College(Freshmen and Sophomores) are due on October 29, 2018. Applications for incoming freshman are Jan 1, 2019.


text:
Q: $\pi: X \to Y$ is a morphism of LRS, and $p \mapsto q$. If $g$ is a function vanishing at $q$, then it will pull back to a function vanishing at $p$? From Vakil's Foundations of Algebraic Geometry:

6.3.1. Definition. A locally ringed space is a ringed space $(X,\mathcal O_X)$ such that the stalks $\mathcal O_{X,p}$ are all local rings. A morphism of locally ringed spaces $\pi : X \to Y$ is a morphism of ringed spaces such that the induced map of stalks $π^\# : \mathcal O_{Y,q} \to \mathcal O_{X,p}$ sends the maximal ideal of the former into the maximal ideal of the latter (a "morphism of local rings").

Next, Vakil writes

This means something rather concrete and intuitive: "if $p \mapsto q$, and $g$ is a function vanishing at $q$, then it will pull back to a function vanishing at $p$."

I was confused about this for a while. At first I was thinking that if $g$ is a function vanishing at $q$, then $\pi^\#_Y: \mathcal O_Y(Y) \to \mathcal O_X(X)$ satisfies $\pi^\# (g)(p)=g(\pi(p))=g(q)=0$. But Vakil does not define the map $\pi^\#$ between sections in this way.
I came up with this next:
Suppose $g$ is a function vanishing at $q \in Y$. Then $[g, Y] \in \mathcal O_{Y,q}$ is in the maximal ideal $\mathfrak m_q$. Since $π^\# : \mathcal O_{Y,q} \to \mathcal O_{X,p}$ satisfies $\pi^\#(\mathfrak m_q) \subset \mathfrak m_p$, then $[g,Y] \mapsto [\pi^\#_Y(g), X] \in \mathfrak m_p$. So, $\pi^\#(g)$ vanishes at $p$.
Is this what is really going on?

A: Question: "This means something rather concrete and intuitive: "if p↦q, and g is a function vanishing at q, then it will pull back to a function vanishing at p." Is this what is really going on?"
Answer: If $(f,f^{\#}): (X, \mathcal{O}_X) \rightarrow (Y, \mathcal{O}_Y)$ is a map of locally ringed spaces there is for any point $x\in X$ a well defined induced map at the stalks
$$f^{\#}_x: \mathcal{O}_{Y,f(x)} \rightarrow \mathcal{O}_{X,x}$$
defined as follows: Given s germ $a:=(U,s)\in \mathcal{O}_Y(U)$ with $f(x)\in U$. It follows $a=0$ in the stalk at $f(x)$ iff there is an open set $f(x) \in V \subseteq U$ with $s_V=0$. The image of the element $(U,s)$ in $\mathcal{O}_{X,x}$ is the germ
$$f^{\#}_x((U,s)):= (f^{-1}(U), f^{\#}(U)(s)).$$
By definition (since it is a map of locally ringed spaces) it maps the maximal ideal $\mathfrak{m}_{f(x)}$ to $\mathfrak{m}_{x}$. Hence if a section $s\in \mathfrak{m}_{f(x)}$ it follows
$$f^{\#}_x(s) \in \mathfrak{m}_x.$$
Note: You define the stalk at $x$ as follows: $\mathcal{O}_{X,x}:=lim_{x\in U}\mathcal{O}_X(U)$.

redpajama_set_name: RedPajamaStackExchange


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When Kent Austin was introduced as the head football coach on Jan. 27, 2010, expectations were high for his young squad. The results, however, were less than spectacular as the Red finished the season with a 2-8 record, identical to the mark posted under Jim Knowles '87 the year prior.
Despite finishing the year in the cellar of the Ivy League, several players made lasting impacts on the coaching staff heading into the offseason — giving Austin reason to think that things would be different this time around. Perhaps nobody on the 2011 squad is a better example of that than senior defensive end Zack Imhoff, who moves over from linebacker after leading the Red in tackles last season.
Imhoff leads a defensive line that features seven players rotating in and out of four spots — a feature typical for most college football programs; however, the unit Austin will send onto the field every Saturday will be anything but typical. Joining Imhoff on the line are three freshmen defensive tackles — Shane McManamon, Junior Togiaso and Cadell Williams. While the trio recognizes the need to learn from the team's struggles of a season ago, the emphasis remains on becoming the class synonymous with the future of Cornell football.
"It's definitely good to not be a part of [the losing seasons], but to know about it so we can come in and right the ship and get us back on track towards the Ivy League championship," McManamon added.
While many have focused on the size of the three freshmen — a combined weight of 823 pounds — it is the intellect of these young defensive tackles that has Austin most impressed and offensive lines most in fear.
Throughout training camp and preseason practice, few on the Cornell roster garnered more attention than Togiaso, a 6-1, 288-pound electrical engineer, who turned down scholarships to play at elite FBS schools including Utah. The Encinitas, Calif. native recognized the chance to obtain an Ivy League degree, and has not looked back.
Praise for Togiaso and his impressive level of maturity did not go unnoticed by Austin, who considers the tackle a cornerstone for the future of Cornell football.
Not to be lost in the shuffle are three veterans, who provide this youthful defensive line with the experience necessary to go out and compete at a high level every Saturday. Senior tackle Emile Chang, junior tackle Hugh Stewart and sophomore end Tre' Minor hold a combined 21 starts with the Red over the past three seasons. While victories have been hard to come by over that period, the trio is more confident than ever that the injection of youth into the line will yield improved results.
For Imhoff, being a leader for the younger members of the line has been a priority since the start of training camp.


text:
Home » Review » Movie » Up In the Air
By Dustin Jansick @DJansick on October 12, 2010
Up In The Air has an all-star cast which received three Oscar nominations of the six total for the film, including Best Picture. Often times humorous and other times emotional, it features quick and witty dialogue with a unique storyline. That being said, it also does have its flaws.
Ryan Bingham's (George Clooney) life is air travel and because of it, it is very orderly and systematic. Ryan's job is to fly around the country to inform people that they are fired. He is very much alright with flying 270 days a year, in fact, he enjoys traveling in airports and different cities more than being at home.
Ryan meets Alex (Vera Farmiga) in an airport bar and the two first meet by quickly exchanging the differences between car rental companies as an attempt to impress one another. Then the two compare credit card reward cards before admitting they are two people that get turned on by elite status. They agree to meet each other in a couple weeks when they will be traveling closer to each other in another state.
Ryan returns home from his trip and does not forget to remind you that he does not particularly like being home. He is informed the next day at work that the company has decided to no longer fly workers around the country to fire people. Instead, he will be doing his job via video chat, thus grounding him at home and in his mind grounding his life. Not to mention shutting him down from the date he had just lined up.
So to no one's surprise Ryan speaks out again this to his boss. It quickly involves the person who came up with the new plan, Natalie (Anna Kendrick). Natalie is a newly hired hard worker efficiency expert. She is very set on her way and lives life on deadlines both on and off work hours. She is a go getter and a planner and she blatantly says she does not mind being married to her career. She is not ashamed of it but does not want to settle in life.
Ryan convinces his boss into showing Natalie what he does on the road and tries to make his case that it cannot be done over the internet. She then travels with him to see what he does in person. He is trying to save what he believes his life is while at the same time she is trying to save the company's efficiency.
Up In The Air finally reaches its most important part of the film when he is at his sister's wedding. The groom gets cold feet and he is sent to deal with it. He talks people out of commitment and now has the task of talking the groom into commitment. Which is something he knows little about and you could say he is being a hypocrite because he has not subscribed to that way of thinking.
It was right about then he realizes that commitment is not as daunting and he thought. He has such a way with words; he more or less talks himself into it. Ryan realizes that perhaps he is ready to settle down with Alex after all. The only question left is has Alex turned over a new leaf as well?
Interestingly, the majority of the people we see getting fired at the beginning of Up In The Air are not actors (but obviously Zach Galifianakis is) but actual people who were recently laid off. The filmmakers put ads out and instructed people to treat the camera as if were like the person who fired them. I think that is a nice and creative touch to make it seem more believable.
After watching the film, I began to wonder if Natalie's storyline even mattered. I believe they could have without her part. Although Anna Kendrick does a good job with this role, I don't think it ultimately played any role of importance in the film.
The main idea behind the film is loyalty. The word was frequently in the background on airline advertisements. The airline, credit card and car rental places all have membership clubs that reward loyal customers. Both Ryan and Natalie are loyal to their jobs both have different views on what it means to be loyal in a relationship.
Up In The Air has an unique premise and good dialogue with acting that is beyond average but ultimately the storyline is a little bumpy. I am not sure if it is some ironic play on Ryan's philosophy of empty backpacks and pointlessness but nearly every other character in the film besides himself is pointless. It was a well done film that should have been tied in together a little more.
Up In the Air Movie review
You do not have to be a fan of country music to enjoy this!


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