A blocked Gibbs sampler for NGG-mixture models via a priori truncation

Abstract

We define a new class of random probability measures, approximating the well-known normalized generalized gamma (NGG) process. Our new process is defined from the representation of NGG processes as discrete measures where the weights are obtained by normalization of the jumps of Poisson processes and the support consists of independent identically distributed location points, however considering only jumps larger than a threshold $\varepsilon$. Therefore, the number of jumps of the new process, called $\varepsilon$-NGG process, is a.s. finite. A prior distribution for $\varepsilon$ can be elicited. We assume such a process as the mixing measure in a mixture model for density and cluster estimation, and build an efficient Gibbs sampler scheme to simulate from the posterior. Finally, we discuss applications and performance of the model to two popular datasets, as well as comparison with competitor algorithms, the slice sampler and a posteriori truncation.

Appendices

Appendices

1.1 Appendix 1: Proof of (5)

First observe that, since $N_\varepsilon$ has a Poisson distribution with parameter $\varLambda _\varepsilon$, we have

$$\begin{aligned} p_{\varepsilon }(n_1,\dots ,n_k) =\sum _{N_\varepsilon =0}^{+\infty } p_{\varepsilon }(n_1,\dots ,n_k|N_\varepsilon ) \frac{\varLambda _{\varepsilon }^{N_\varepsilon }}{N_\varepsilon !}{\mathrm{e}}^{-\varLambda _{\varepsilon }}. \end{aligned}$$

(12)

Then, formula (30) in Pitman (1996) yields

$$\begin{aligned} p_{\varepsilon }(n_1,\dots ,n_k|N_\varepsilon )= \mathbb {I}_{ \{1,\dots ,N_\varepsilon +1\} }(k) \sum _{j_1,\dots ,j_k}^{}{\mathbb {E}}\left( \prod _{i=1}^{k}P_{j_i}^{n_i} \right) , \end{aligned}$$

where the vector $(j_1,\dots ,j_k)$ ranges over all permutations of $k$ elements in $\{0,\dots ,N_\varepsilon \}$. Using the gamma function identity,

$$\begin{aligned} \dfrac{1}{T_\varepsilon ^n}=\int _{0}^{+\infty }\frac{1}{\varGamma (n)}u^{n-1} e^{-u T_\varepsilon } du, \end{aligned}$$

(13)

we have:

$$\begin{aligned}&p_{\varepsilon }(n_1,\ldots ,n_k|N_\varepsilon ) \\&\quad = \sum _{j_1,\dots ,j_k} \int _{}^{} \prod _{i=1}^{k}\frac{J_{j_i}^{n_i}}{T_{\varepsilon }^{n_i}} \fancyscript{L}(dJ_0,\dots ,dJ_{N_\varepsilon })\\&\quad {=}\!\! \sum _{j_1,\dots ,j_k} \int _{0}^{+\infty } du \left( \frac{u^{n-1}}{\varGamma (n)} . \prod _{i=1}^{k}\int _{0}^{+\infty }J_{j_i}^{n_i}{\mathrm{e}}^{-J_{j_i}u}\rho _{\varepsilon }(J_j)dJ_{j_i} \right. \\&\qquad \qquad \qquad \qquad \qquad \quad \times \!\!\left. \prod _{j\notin \{j_1,\dots ,j_k\}}^{}\int _{0}^{+\infty }{\mathrm{e}}^{-J_{j}u}\rho _{\varepsilon }(J_j)dJ_j \right) \end{aligned}$$

Substituting the expression of $\rho _{\varepsilon }$ in Sect. 3:

$$\begin{aligned}&p_{\varepsilon }(n_1,\ldots ,n_k|N_\varepsilon ) \\&\quad = \sum _{j_1,\dots ,j_k} \int _{0}^{+\infty } du \Big (\frac{1}{\varGamma (n)}u^{n-1} \\&\qquad \times \prod _{i=1}^{k} \int _{\varepsilon }^{+\infty } \frac{J_{j_i}}{\omega ^\sigma \varGamma (-\sigma ,\omega \varepsilon )}J_{j_i}^{-\sigma -1}{\mathrm{e}}^{-(\omega +u)J_j}dJ_{j_i} \\&\qquad \times \prod _{j\notin \{j_1,\dots ,j_k\}}^{}\int _{0}^{+\infty }\!\!\!\frac{1}{\omega ^\sigma \varGamma (-\sigma ,\omega \varepsilon )} J_{j}^{-\sigma -1}{\mathrm{e}}^{-(\omega +u)J_j}dJ_{j} \Big )\\&\quad = \sum _{j_1,\dots ,j_k} \int _{0}^{+\infty } \Big ( \frac{1}{\varGamma (n)}u^{n-1} \\&\qquad \times \prod _{i=1}^{k} \frac{(u+\omega )^{\sigma -n_i}\varGamma (n_i-\sigma ;(u+\omega )\varepsilon )}{\omega ^{\sigma } \varGamma (-\sigma ,\omega \varepsilon )} \\&\qquad \times \left( \frac{(u+\omega )^{\sigma }\varGamma (-\sigma ;(u+\omega )\varepsilon )}{\omega ^{\sigma }\varGamma (-\sigma ,\omega \varepsilon )} \right) ^ {N_\varepsilon +1-k} \Big ) du. \end{aligned}$$

We have crossed out the indicator function in all previous lines. If we switch the finite sum and the integral, since the integrand function does not depend on the position of the clusters $j_i$’s, $i=1,\dots ,k$, but only on the sizes $n_i$, and there are $(N_\varepsilon +1)(N_\varepsilon )\dots (N_\varepsilon +1-k)={(N_\varepsilon +1)!}/(N_{\varepsilon }+1-k)!$ sequences of $k$ distinct elements from $\{0,\dots ,N_\varepsilon \}$, we get:

$$\begin{aligned}&p(n_1,\dots ,n_k|N_{\varepsilon }) =\mathbb {I}_{ \{1,\dots ,N_{\varepsilon }+1\} }(k) \int _{0}^{+\infty }\Bigg ( \frac{u^{n-1}}{\varGamma (n)} \\&\quad \times \frac{(N_\varepsilon +1)!}{(N_{\varepsilon }+1-k)!} \prod _{i=1}^{k} \frac{(u+\omega )^{\sigma -n_i}\varGamma (n_i-\sigma ;(u+\omega )\varepsilon )}{\omega ^{\sigma } \varGamma (-\sigma ,\omega \varepsilon )} \\&\quad \times \left( \frac{(u+\omega )^{\sigma }\varGamma (-\sigma ;(u+\omega )\varepsilon )}{\omega ^{\sigma }\varGamma (-\sigma ,\omega \varepsilon )} \right) ^{N_{\varepsilon }+1-k}\Bigg ) du. \end{aligned}$$

Observe that, because of the indicator function in the above formula, summation in (12) has to be taken for $N_\varepsilon$ from $k-1$ to $+\infty$. Then, by the change of variable $N_{na}=N_\varepsilon +1-k$ in the summation ($N_\varepsilon +1-k$ is the number of non-allocated jumps), simple calculations give

$$\begin{aligned}\begin{aligned}&\! p_{\varepsilon }(n_1,\dots ,n_k)\\&\!\!\!\quad =\!\sum _{N_{na}=0}^{+\infty } \int _{0}^{+\infty } \!\left( \frac{u^{n-1}}{\varGamma (n)} (u\!+\!\omega )^{k\sigma -n} \prod _{i=1}^{k}\varGamma (n_i\!-\!\sigma ,(u\!+\!\omega )\varepsilon ) \right. \\&\!\!\!\qquad \times \frac{1}{\omega ^\sigma \varGamma (-\sigma ,\omega \varepsilon )} \frac{\kappa ^{k-1}}{\varGamma (1-\sigma )^{k-1}} \frac{N_{na}+k}{N_{na}!}\\&\!\!\!\qquad \times \left. \left( \frac{\kappa (u+\omega )^\sigma }{\varGamma (1-\sigma )} \varGamma (-\sigma ,(u+\omega )\varepsilon ) \right) ^{N_{na}} {\mathrm{e}}^{-\varLambda _{\varepsilon }} \right) du.\\ \end{aligned} \end{aligned}$$

By Fubini’s theorem, we can switch integration and summation, and introduce $\varLambda _{\varepsilon ,u}$ as defined in (6), so that

$$\begin{aligned}&p_{\varepsilon }(n_1,\dots ,n_k)= \int _{0}^{+\infty } \bigg ( \frac{u^{n-1}}{\varGamma (n)} (u+\omega )^{k\sigma -n} \\&\quad \times \prod _{i=1}^{k}\varGamma (n_i-\sigma ,(u+\omega )\varepsilon ) \frac{1}{\omega ^\sigma \varGamma (-\sigma ,\omega \varepsilon )} \frac{\kappa ^{k-1}}{\varGamma (1-\sigma )^{k-1}} \\&\quad \times \sum _{N_{na}=0}^{+\infty } \frac{N_{na} +k}{N_{na}!} \varLambda _{\varepsilon ,u}^{N_{na}} {\mathrm{e}}^{-\varLambda _{\varepsilon }} \bigg ) du, \end{aligned}$$

that is (5), since

$$\begin{aligned} \sum _{N_{na}=0}^{+\infty }\frac{N_{na}+k}{N_{na}!}\varLambda _{\varepsilon ,u}^{N_{na}}= {\mathrm{e}}^{\varLambda _{\varepsilon ,u}}\left( \varLambda _{\varepsilon ,u}+k \right) . \end{aligned}$$

1.2 Appendix 2: Lemma 1

We report this simple lemma for a thorough understanding of Lemma 2.

Lemma 1

Let $(a_n)$ and $(b_n)$ be two sequences of real numbers, such that

$$\begin{aligned} \lim _{n\rightarrow +\infty }(a_n+b_n)=l, \quad \liminf _{n\rightarrow +\infty } a_{n}=a_0, \quad \liminf _{n\rightarrow +\infty } b_{n}=b_0, \end{aligned}$$

where $l,a_0,b_0$ are finite, and $a_0+b_0=l$. Then

$$\begin{aligned} \lim _{n\rightarrow +\infty } a_{n}=a_0, \quad \lim _{n\rightarrow +\infty } b_{n}=b_0. \end{aligned}$$

Proof

By definition of $\liminf$ and $\limsup$ we have:

$$\begin{aligned}&\liminf a_n+\liminf b_n \le \liminf (a_n+b_n)\\&\le \liminf a_n +\limsup b_n \\&\le \limsup (a_n+b_n) \le \limsup a_n +\limsup b_n. \end{aligned}$$

From the hypotheses we have

$$\begin{aligned} a_0+b_0=l&= \liminf (a_n+b_n) \le a_0+\limsup b_n\\&\le \limsup (a_n+b_n)=l=a_0+b_0, \end{aligned}$$

so that $\limsup b_n=b_0$, but by hypothesis $b_0=\liminf b_n$, and consequently

$$\begin{aligned} \lim _{n \rightarrow +\infty } b_n =b_0. \end{aligned}$$

We prove similarly that $\lim _{n\rightarrow +\infty }a_{n}=a_0$.

Of course, this lemma can be generalized to any finite number of sequences.

1.3 Appendix 3: Lemma 2

Lemma 2

Let $p_\varepsilon$ be the eppf of a $\varepsilon -$NGG$(\sigma ,\kappa ,\omega , P_0)$ process. Then for each $n_1,\dots ,n_k\ \in \mathbb {N}$ with $k\ge 0$ and $\sum _{i=1}^{k}n_i$ $=n$, we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}p_{\varepsilon }(n_1,\dots ,n_k)=p_0(n_1,\dots ,n_k), \end{aligned}$$

(14)

where $p_0(\cdot )$ is the eppf of a NGG$(\sigma ,\kappa ,\omega ,P_0)$ process.

Proof

Formula (5) can be restated as

$$\begin{aligned} p_{\varepsilon }(n_1,\dots ,n_k)=\int _{0}^{+\infty }f_{\varepsilon }(u;n_1,\dots ,n_k)du \end{aligned}$$

where $f_{\varepsilon }$ denotes the integrand in the right hand side of (5) itself. In addition, the eppf of a NGG$(\sigma ,\kappa ,\omega , P_0)$ process can be written as

$$\begin{aligned} p_{0}(n_1,\dots ,n_k)=\int _{0}^{+\infty } f_{0}(u;n_1,\dots ,n_k)du \end{aligned}$$

where

$$\begin{aligned}&f_0(u;n_1,\dots ,n_k) = \frac{u^{n-1}}{\varGamma (n)} \left( u+\omega \right) ^{k\sigma -n} \prod _{i=1}^{k} \varGamma (n_i-\sigma )\\&\quad \times \left( \frac{\kappa }{\varGamma (1-\sigma )} \right) ^{k-1}\frac{\kappa }{\varGamma (1-\sigma )} \exp \left\{ \!-\!\kappa \frac{(\omega \!+\!u)^\sigma \!-\!\omega ^\sigma }{\sigma } \right\} ; \end{aligned}$$

see, for instance, Lijoi et al. (2007). We first show that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}f_{\varepsilon }(u;n_1,\dots ,n_k)= f_{0}(u;n_1,\dots ,n_k)\quad \mathrm{for \, any }\, u>0. \end{aligned}$$

This is straightforward by the following remarks:

• 1. $\lim _{\varepsilon \rightarrow 0}\varGamma (n_i-\sigma ,(u+\omega )\varepsilon )=\varGamma (n_i-\sigma )$, for any $i=1,2,\ldots ,k$, by the Dominated Convergence Theorem, since $n_i-\sigma \ge 1-\sigma >0$;

• 2. since $\lim _{\varepsilon \rightarrow 0}\varGamma (-\sigma ,\omega \varepsilon )=+\infty$ and

  $$\begin{aligned} \varGamma (1-\sigma ,x)=-\sigma \varGamma (-\sigma ,x) + x^{-\sigma }{\mathrm{e}}^{-x} \end{aligned}$$

  (Gradshteyn and Ryzhik 2000), we have:

  $$\begin{aligned} \begin{aligned}&\lim _{\varepsilon \rightarrow 0} \frac{\varLambda _{\varepsilon ,u}+k}{\omega ^\sigma \varGamma (-\sigma ,\omega \varepsilon )}= \frac{\kappa }{\varGamma (1-\sigma )}, \\&\lim _{\varepsilon \rightarrow 0} \left( \varLambda _{\varepsilon ,u}-\varLambda _{\varepsilon } \right) = -\kappa \frac{(\omega +u)^\sigma -\omega ^\sigma }{\sigma }. \end{aligned} \end{aligned}$$

Now let $\fancyscript{C}=\{C_1,\dots ,C_k\}$ be a partition of $\{ 1, \dots , n \}$ with group sizes $(n_1,\dots ,n_k)$, and let $\Pi _n$ be the set all the possible partitions of $\{ 1, \dots , n \}$, of any size $k=1,\ldots ,n$. Of course, by definition of eppf,

$$\begin{aligned} \sum _{\fancyscript{C} \in \Pi _n}^{} p(n_1,\dots ,n_k)=1 \end{aligned}$$

and, in particular this holds for either $p_\varepsilon$ and $p_0$. Moreover, by Fatou’s Lemma we have

$$\begin{aligned} p_0(n_1,\dots ,n_k)&=\int _{0}^{+\infty } \lim _{\varepsilon \rightarrow 0} f_{\varepsilon }(u;n_1,\dots ,n_k) du \\&=\int _{0}^{+\infty } \liminf _{\varepsilon \rightarrow 0} f_{\varepsilon }(u;n_1,\dots ,n_k) du \\&\le \liminf _{\varepsilon \rightarrow 0} \int _{0}^{+\infty } f_{\varepsilon }(u;n_1,\dots ,n_k) du \\&= \liminf _{\varepsilon \rightarrow 0} p_{\varepsilon }(n_1,\dots ,n_k). \end{aligned}$$

Suppose now that for a particular sequence $\fancyscript{C}\in \Pi _n$, we have ${\displaystyle p_0(n_1,\dots ,n_k)< \liminf _{\varepsilon \rightarrow 0}p_{\varepsilon }(n_1,\dots ,n_k)}$. In this case

$$\begin{aligned} \begin{aligned}&1 = \sum _{\fancyscript{C} \in \Pi _n}^{}p_0(n_1,\ldots ,n_k) < \sum _{\fancyscript{C} \in \Pi _n}^{} \liminf _{\varepsilon \rightarrow 0} p_{\varepsilon }(n_1,\ldots ,n_k) \\&\quad \le \liminf _{\varepsilon \rightarrow 0} \sum _{\fancyscript{C} \in \Pi _n}^{} p_{\varepsilon }(n_1,\dots ,n_k)=1, \end{aligned} \end{aligned}$$

that is a contradiction. Therefore we conclude that

$$\begin{aligned} p_0(n_1,\dots ,n_k)= \liminf _{\varepsilon \rightarrow 0} p_{\varepsilon }(n_1,\dots ,n_k), \end{aligned}$$

for all $n_1,\ldots ,n_k$, all $k$. Summing up, we have proved so far that:

• $\lim _{\varepsilon \rightarrow 0}\sum _{\fancyscript{C} \in \Pi _n}^{}p_{\varepsilon }(n_1,\dots ,n_k) =1$;

• $\liminf _{\varepsilon \rightarrow 0} p_\varepsilon (n_1,\dots ,n_k)=p_0(n_1,\dots ,n_k)$ for all $\fancyscript{C}=(C_1$, $\dots ,C_k) \in \Pi _n$;

• $\sum _{\fancyscript{C} \in \Pi _n}^{} p_0(n_1,\dots ,n_k)=1$.

By Lemma 1, equation (14) follows.

1.4 Appendix 4: Proof of Proposition 1

As mentioned before, $P_\varepsilon$ is a proper species sampling model, so that $p_{\varepsilon }$ defines a probability law on the sets of all partitions of $\mathbb {N}_n:=\{1,\dots ,n\}$, once that we have set a positive integer $n$. Therefore, we introduce $(N_1^\varepsilon ,\ldots ,N_k^\varepsilon )$, the sizes of the blocks (in order of appearance), of the random partition $C_{\varepsilon ,n}$ defined by $p_\varepsilon$, for any $\varepsilon \ge 0$. The probability distributions of $\{(N_1^\varepsilon ,\ldots ,N_k^\varepsilon ),\varepsilon \ge 0 \}$ are proportional to the values of $p_\varepsilon$ (for any $\varepsilon \ge 0$) in (2.6) in Pitman (2006). Hence, by Lemma 2, for any $k=1,\ldots ,n$ and any $n$,

$$\begin{aligned} (N_1^\varepsilon ,\ldots ,N_k^\varepsilon ) \mathop {\rightarrow }\limits ^{d}(N_1^0,\ldots ,N_k^0) \, \mathrm{as }\, \varepsilon \rightarrow 0. \end{aligned}$$

Here $(N_1^0,\ldots ,N_k^0)$ denotes the sizes of the blocks (in order of appearance), of the random partition $C_{0,n}$ defined by $p_0$, the eppf of a NGG$(\sigma ,\kappa ,\omega , P_0)$ process. By formula (2.30) in Pitman (2006), we have

where $P_j^\varepsilon$ and $\tilde{P}_j$ are the $j$-th weights of a $\varepsilon$-NGG and a NGG process (with parameters $(\sigma ,\kappa ,\omega , P_0)$), respectively. Note that the sequences depending on $n$ have only a finite number of positive weights. Recall that the weak convergence of a sequence of random probability measures is equivalent to the pointwise convergence of the Laplace transforms (see Kallenberg 1983, Theorem 4.2). Let $f(\cdot )$ be a continuous and bounded function on $\varTheta$. If we can invert the order of the limit operations below, then we have:

$$\begin{aligned} \begin{aligned}&\lim _{\varepsilon \rightarrow 0}{\mathbb {E}}\left( {\mathrm{e}}^{-\int _\varTheta f d\mu ^{\varepsilon }} \right) \\&\quad = \lim _{\varepsilon \rightarrow 0}\lim _{n\rightarrow +\infty } {\mathbb {E}}\left( {\mathrm{e}}^{-\int _\varTheta f d \mu _n^{\varepsilon }}\right) \\&\quad =\lim _{n\rightarrow +\infty }\lim _{\varepsilon \rightarrow 0} {\mathbb {E}}\left( {\mathrm{e}}^{-\int _\varTheta fd \mu _n^{\varepsilon } }\right) \\&\quad = \lim _{n\rightarrow +\infty } {\mathbb {E}}\left( {\mathrm{e}}^{-\int _{}^{}fd \mu _n^0 }\right) = {\mathbb {E}}\left( {\mathrm{e}}^{-\int _{}^{}fd \mu ^0}\right) . \end{aligned} \end{aligned}$$

(15)

Here we have introduced notation

$$\begin{aligned} \mu ^\varepsilon _n:=\sum _j \frac{N_j^\varepsilon }{n}\delta _{\tau _j}\quad \mathrm{and } \quad \mu ^\varepsilon :=\sum _j \tilde{P}_j^\varepsilon \delta _{\tau _j}, \,\mathrm{for \, any }\,\varepsilon \ge 0; \end{aligned}$$

thus (15) proves the stated convergence, conditioning on $\{ \tau _0$, $\tau _1,\tau _2,\ldots \}$, which are iid from $P_0$.

To justify the interchange of the two limits above, we must prove that the sequence $\left\{ {\mathbb {E}}\left( {\mathrm{e}}^{-\int _{}^{}fd \mu _n^{\varepsilon } } \right) , n\ge 1\right\}$ converges uniformly. To this end, it is sufficient to show that the difference between two next terms in the sequence does not depend on $\varepsilon$; in fact, for any $M>0$, since

$$\begin{aligned} |{\mathrm{e}}^{-x}-{\mathrm{e}}^{-y} | \le {\mathrm{e}}^M|x-y | \,\mathrm{for \, any } \, x, y \in [-M,M], \end{aligned}$$

we have

$$\begin{aligned}&\left| {\mathbb {E}}\left( {\mathrm{e}}^{-\int _{}^{}fd \mu _{n+1}^{\varepsilon } } \right) -{\mathbb {E}}\left( {\mathrm{e}}^{-\int _{}^{}fd \mu _n^{\varepsilon } } \right) \right| \\&\quad \le {\mathbb {E}}\left( \left| {\mathrm{e}}^{-\int _{}^{}fd \mu _{n+1}^{\varepsilon } } -{\mathrm{e}}^{-\int _{}^{}fd \mu _n^{\varepsilon } }\right| \right) \\&\quad \le {\mathrm{e}}^M{\mathbb {E}}\left( \left| \int _{}^{}fd \mu _{n+1}^{\varepsilon } -\int _{}^{}fd \mu _n^{\varepsilon } \right| \right) , \end{aligned}$$

where $M \ge \sup f$. Let now $C_{\varepsilon ,n+1}$ be a random partition on $\{1,\dots ,n+1\}$ such that its restriction to $\{1,\dots ,n\}$ corresponds to $C_{\varepsilon ,n}$. We distinguish two cases:

• 1. $C_{\varepsilon ,n+1}$ has the same number of clusters of $C_{\varepsilon ,n}$; one of these clusters (the $j$-th for instance) has one more element and, as a consequence, has size equal to $n_j+1$;

• 2. $C_{\varepsilon ,n+1}$ has one more cluster than $C_{\varepsilon ,n}$; this cluster contains only one element.

In both cases, it is not difficult to prove that

$$\begin{aligned} {\mathbb {E}}\left( \left| \int _{\varTheta }fd \mu _{n+1}^{\varepsilon } -\int _{}^{}fd \mu _n^{\varepsilon } \right| \right) \le \frac{2M}{n+1}, \end{aligned}$$

so that we are able to interchange the two limits. Finally, it is straightforward to show that the stated convergence follow from the convergence in distribution conditioning on $\{ \tau _0,\tau _1,\tau _2,\ldots \}$, with an argument on Laplace transforms as before. This ends the first part of the Proposition, i.e. convergence for $\varepsilon \rightarrow 0$.

Convergence as $\varepsilon \rightarrow +\infty$ is straightforward as well. In fact, when $\varepsilon$ increases to $+\infty$, there are no jumps to consider in (4) but the extra $J_0$, so that $P_\varepsilon$ degenerates on $\delta _{\tau _0}$.

1.5 Appendix 5: Proof of Proposition 2

The conditional distribution of $\varvec{\theta }$ is:

$$\begin{aligned} \begin{aligned}&\fancyscript{L}(\theta _1,\ldots ,\theta _n | P_{\varepsilon })= \prod _{i=1}^nP_{\varepsilon }(\theta _i)= \prod _{i=1}^{n} \sum _{j=0}^{N_\varepsilon } \left( P_{j}\delta _{\tau _j}(\theta _i)\right) \\&\quad =\sum _{l_1=0}^{N_\varepsilon }P_{l_1}\delta _{\tau _{l_1}}(\theta _1) \sum _{l_2=0}^{N_\varepsilon }P_{l_2}\delta _{\tau _{l_2}}(\theta _2) \dots \sum _{l_n=0}^{N_\varepsilon }P_{l_n}\delta _{\tau _{l_n}}(\theta _n) \\&\quad = \mathbb {I}_{\{1, \ldots ,N_\varepsilon +1 \}}(k)\frac{1}{(T_\varepsilon )^n}\sum _{l^*_1, \dots , l^*_k}{}J_{l^*_1}^{n_1}\\&\qquad \dots J_{l^*_k}^{n_k}\delta _{\tau _{l^*_1}}(\theta ^*_1)\dots \delta _{\tau _{l^*_k}}(\theta ^*_k), \end{aligned} \end{aligned}$$

where $(\theta ^*_1,\theta ^*_2,\dots , \theta ^*_k)$ is the vector of the unique values in the sample. We omit the indicator $\mathbb {I}_{{\{1, \ldots ,N_\varepsilon +1 \}}}(k)$ till we need it. Introducing the auxiliary variable $U$, by (13) we have:

$$\begin{aligned}&\fancyscript{L}({\varvec{\theta }}, u| P_{\varepsilon })=\dfrac{1}{\varGamma (n)}u^{n-1} e^{-u T_\varepsilon }\\&\quad \times \sum _{l^*_1, \ldots , l^*_k}{} \left( J_{l^*_1}^{n_1} \delta _{\tau ^*_{l^*_1}}(\theta ^*_1) \dots J_{l^*_k}^{n_k}\delta _{\tau ^*_{l^*_k}} (\theta ^*_k) \right) . \end{aligned}$$

Hence we have:

$$\begin{aligned}&\fancyscript{L}({\varvec{\theta }}, u, P_{\varepsilon }) = \fancyscript{L}({\varvec{\theta }}, u| P_\varepsilon )\fancyscript{L}(P_\varepsilon ) \nonumber \\&\quad = \frac{u^{n-1}}{\varGamma (n)} e^{-u T_\varepsilon }\sum _{l^*_1, \ldots , l^*_k}{} \big ( J_{l^*_1}^{n_1} \delta _{\tau ^*_{l^*_1}}(\theta ^*_1) \ldots J_{l^*_k}^{n_k} \delta _{\tau ^*_{l^*_k}} (\theta ^*_k) \big )\fancyscript{L}(P_\varepsilon ) \nonumber \\&\quad = \frac{u^{n-1}}{\varGamma (n)}\prod _{j=0}^{N_\varepsilon }( e^{-u J_j})\sum _{l^*_1, \dots , l^*_k}{}\big ( J_{l^*_1}^{n_1} \delta _{\tau ^*_{l^*_1}}(\theta ^*_1) \ldots J_{l^*_k}^{n_k} \delta _{\tau ^*_{l^*_k}} (\theta ^*_k) \big ) \nonumber \\&\qquad \times \prod _{j=0}^{N_\varepsilon }\big (\rho _\varepsilon (J_j)P_0(\tau _j) \big ) \fancyscript{P}_0(N_\varepsilon ;\varLambda _\varepsilon ) \\&\quad =\frac{1}{\varGamma (n)}u^{n-1} \prod _{j=0}^{N_\varepsilon } \big ( e^{-u J_j}\rho _\varepsilon (J_j)P_0(\tau _j) \big ) \nonumber \\&\qquad \times \sum _{l^*_1, \dots , l^*_k}{}\big ( J_{l^*_1}^{n_1} \delta _{\tau ^*_{l^*_1}}(\theta ^*_1) \ldots J_{l^*_k}^{n_k} \delta _{\tau ^*_{l^*_k}} (\theta ^*_k) \big ) \fancyscript{P}_0(N_\varepsilon ;\varLambda _\varepsilon ) \nonumber \end{aligned}$$

(16)

where, in this proof, $\fancyscript{P}_0(N_\varepsilon ;\varLambda _\varepsilon )$ is the density of the Poisson distribution with parameter $\varLambda _\varepsilon$, evaluated in $N_\varepsilon$, and $P_0(\tau )$ is the density of $P_0$ evaluated in $\tau$.

The conditional distribution of $P_\varepsilon$, given $U=u$ and $\varvec{\theta }$, is as follows:

$$\begin{aligned} \begin{aligned} \fancyscript{L}(P_{\varepsilon }|u,\varvec{\theta })&= \fancyscript{L}({\varvec{\tau }},{\varvec{J}},N_\varepsilon |u,\varvec{\theta })\\&= \fancyscript{L}({\varvec{\tau }},{\varvec{J}}|N_\varepsilon , u,\varvec{\theta }) \fancyscript{L}(N_\varepsilon |u,\varvec{\theta }). \end{aligned} \end{aligned}$$

(17)

The second factor in the right handside is proportional to

$$\begin{aligned}&\fancyscript{L}(N_\varepsilon , u,\varvec{\theta })\\&\quad = \int d J_0 \ldots d J_{N_\varepsilon } d\tau _0\ldots d\tau _{N_\varepsilon } \fancyscript{L}({\varvec{\tau }},{\varvec{J}},N_\varepsilon , u,\varvec{\theta })\\&\quad =\sum _{l^*_1, \dots , l^*_k}{} \Biggl \{ \biggl [ \prod _{i=1}^{k}\int J_{l^*_i}^{n_i} \delta _{\tau _{l^*_i}} (\theta ^*_i) e^{-uJ_{l^*_i}}\rho _\varepsilon (J_{l^*_i})P_0(\tau _{l^*_i}) dJ_{l^*_i}d\tau _{l^*_i} \biggr ] \\&\qquad \times \biggl [ \prod _{j \ne \{ l^*_1, \ldots , l^*_k \} } \int e^{-uJ_{j}}\rho _\varepsilon (J_{j})P_0(\tau _{j}) dJ_{j}d\tau _{j} \biggr ] \Biggr \}\\&\qquad \times \frac{1}{\varGamma (n)}u^{n-1} \fancyscript{P}_0(N_\varepsilon ;\varLambda _\varepsilon ). \end{aligned}$$

Observe that, for any $j \ne \{ l^*_1, \ldots , l^*_k \}$,

$$\begin{aligned}&\int e^{-uJ_{j}}\rho _\varepsilon (J_{j})P_0(\tau _{j})dJ_{j}d\tau _{j}\nonumber \\&\quad = \int _0^{+\infty } e^{-uJ_j}\rho _{\varepsilon }(J_j)dJ_j \nonumber \\&\quad = \frac{1}{\omega ^\sigma \varGamma (-\sigma , \omega \varepsilon )}\int _0^{+\infty } x^{-\sigma -1}{\mathrm{e}}^{(u+\omega )x}\mathbb {I}_{(\varepsilon , +\infty )}(x) dx \nonumber \\&\quad = \dfrac{(\omega +u)^{\sigma }}{\omega ^{\sigma }\varGamma (-\sigma , \omega \varepsilon )} \int _{{(\omega +u)\varepsilon }}^{{+\infty }}e^{-y}y^{-\sigma -1}dy \\&\quad =\dfrac{(\omega +u)^{\sigma }\varGamma (-\sigma , (\omega +u)\varepsilon )}{\omega ^{\sigma }\varGamma (-\sigma ,\nonumber \omega \varepsilon )}. \end{aligned}$$

(18)

The integrand function in the second line of the formula above is the kernel of the mean intensity of a $\varepsilon$-NGG$(\sigma ,\kappa ,\omega +u,P_0)$ process. On the other hand, for $i=1,\ldots ,k$:

$$\begin{aligned}&\int J_{l^*_i}^{n_i} \delta _{\tau _{l^*_i}}(\theta ^*_i) {\mathrm{e}}^{-uJ_{l^*_i}}\rho _\varepsilon (J_{l^*_i}) P_0(\tau _{l^*_i})dJ_{l^*_i}d\tau _{l^*_i} \nonumber \\&\quad =\left( \int J_{l^*_i}{\mathrm{e}}^{-uJ_{l^*_i}}\rho _\varepsilon (J_{l^*_i})dJ_{l^*_i} \right) \left( \int \delta _{\tau _{l^*_i}}(\theta ^*_i) P_0(\theta ^*_i)d\theta ^*_i \right) \nonumber \\&\quad = \dfrac{ P_0(\theta ^*_i)}{\omega ^{\sigma }\varGamma (-\sigma , \omega \varepsilon )} \int _{0}^{+\infty } x^{n_i}{\mathrm{e}}^{-ux}x^{-1-\sigma }{\mathrm{e}}^{-\omega x}\mathbb {I}_{(\varepsilon ,+\infty )}dx\\&\quad =\dfrac{(\omega +u)^{\sigma -n_i}}{\omega ^{\sigma }} \dfrac{\varGamma (n_i-\sigma , (u+\omega )\varepsilon )}{\varGamma (-\sigma , \omega \varepsilon )} P_0(\theta ^*_i)\nonumber . \end{aligned}$$

(19)

The integrand function in (19) is the kernel of a gamma density with parameters $(n_i-\sigma ,u+\omega )$, restricted to $(\varepsilon ,+\infty )$. Summing up, we have

$$\begin{aligned}&\fancyscript{L}(N_\varepsilon | u,\varvec{\theta }) \propto \fancyscript{L}(N_\varepsilon , u,\varvec{\theta })= \dfrac{u^{n-1}}{\varGamma (n)} \fancyscript{P}_0(N_\varepsilon ;\varLambda _\varepsilon ) \nonumber \\&\quad \times \sum _{l^*_1, \ldots , l^*_k}{}\biggl \{ \biggl ( \dfrac{(\omega +u)^{k\sigma -n}\prod _{i=1}^{k}\varGamma (n_i-\sigma , (\omega +u)\varepsilon ) P_0(\theta _i^*)}{\omega ^{\sigma k}\varGamma (-\sigma , \omega \varepsilon )^k} \biggr ) \nonumber \\&\quad \times \biggl ( \dfrac{(\omega +u)^{\sigma (N_\varepsilon +1-k)}\varGamma (-\sigma , (u+\omega )\varepsilon )^{N_\varepsilon +1-k}}{ \omega ^{\sigma (N_\varepsilon +1-k)}\varGamma (-\sigma , \omega \varepsilon )^{N_\varepsilon +1-k}} \biggr ) \biggr \} \nonumber \\&=\dfrac{u^{n-1}}{\varGamma (n)}\fancyscript{P}_0(N_\varepsilon ;\varLambda _\varepsilon )\dfrac{(N_\varepsilon +1)!}{(N_\varepsilon +1-k)!} \dfrac{(\omega +u)^{\sigma k-n}}{\omega ^{\sigma k}\varGamma (-\sigma , \omega \varepsilon )^k} \\&\quad \times \dfrac{(\omega +u)^{\sigma N_{na}}\varGamma (-\sigma , \varepsilon (\omega +u))^{N_{na}}}{\omega ^{\sigma N_{na}} \varGamma (-\sigma , \omega \varepsilon )^ {N_{na}}}\nonumber \\&\quad \times \prod _{i=1}^{k}\biggl ( P_0(\theta _i^*)\varGamma (n_i-\sigma , \varepsilon (\omega +u)) \biggr ) \mathbb {I}_{\{(N_\varepsilon +1)\ge k\}}. \nonumber \end{aligned}$$

(20)

As in the proof of formula (5), $N_{na}=N_\varepsilon +1-k$ is the number of non-allocated jumps. Therefore, since $k$ is given, the conditional distribution $\fancyscript{L}(N_\varepsilon |$ $u,\varvec{\theta })$ is identified by $\fancyscript{L}(N_{na}| u,\varvec{\theta })$; we have

$$\begin{aligned}&\fancyscript{L}(N_{na}| u,\varvec{\theta }) \propto \mathbb {I}_{(N_{na} \ge 0)} \dfrac{(\omega +u)^{\sigma k-n}}{\omega ^{\sigma }\varGamma (-\sigma , \omega \varepsilon )} \dfrac{(N_{na}+k)}{N_{na}!}\\&\quad \times \biggl (\dfrac{\kappa (u+\omega )^{\sigma } }{\varGamma (1-\sigma )}\varGamma (-\sigma , (u+\omega )\varepsilon ) \biggr )^{N_{na}}. \end{aligned}$$

Let $\varLambda _{\varepsilon ,u}$ be as in (6); it easily follows that

$$\begin{aligned}&\fancyscript{L}(N_{na}|\varepsilon , u, {\varvec{\theta }}) \propto \dfrac{N_{na}+k}{N_{na}!}{\mathrm{e}}^{-\varLambda _{\varepsilon ,u}} \varLambda _{\varepsilon ,u}^{N_{na}}\nonumber \\&\quad =\dfrac{\varLambda _{\varepsilon ,u}}{(N_{na}-1)!}\varLambda _{\varepsilon ,u}^{(N_{na}-1)}{\mathrm{e}}^{-\varLambda _{\varepsilon ,u}}+ \dfrac{k}{N_{na}!}\varLambda _{\varepsilon ,u}^{N_{na}}{\mathrm{e}}^{-\varLambda _{\varepsilon ,u}}\\&\quad =\dfrac{\varLambda _{\varepsilon ,u}}{\varLambda _{\varepsilon ,u}+k}\fancyscript{P}_1(N_{na};\varLambda _{\varepsilon ,u})+ \dfrac{k}{\varLambda _{\varepsilon ,u}+k}\fancyscript{P}_0(N_{na};\varLambda _{\varepsilon ,u}).\nonumber \end{aligned}$$

(21)

On the other hand, the first factor in the right handside of (17) can be computed by the following comment. Denote by ${\varvec{l}}^*=(l^*_1,\ldots ,l^*_k)$ is the vector of locations of the allocated jumps. From (16), it is clear that

$$\begin{aligned}&\fancyscript{L}({\varvec{J}},{\varvec{\tau }}, {\varvec{l}}^*|N_{na}, u,\varvec{\theta }) \nonumber \\&\quad = J_{l^*_1}^{n_1} \delta _{\tau ^*_{l^*_1}}(\theta ^*_1) \ldots J_{l^*_k}^{n_k} \delta _{\tau ^*_{l^*_k}} (\theta ^*_k) \nonumber \\&\qquad \times \prod _{j=0}^{N_{na}+k-1}\rho _\varepsilon (J_j) P_0(\tau _j){\mathrm{e}}^{-u J_j}\nonumber \\&\quad =\left( \prod _{i=1}^k J_{l^*_i}^{n_i} \delta _{\tau ^*_{l^*_i}}(\theta ^*_i) {\mathrm{e}}^{-u J_{l^*_i}} \rho _\varepsilon (J_{l^*_i} ) P_0(J_{l^*_i})\right) \nonumber \\&\qquad \times \left( \prod _{j \ne \{ l^*_1, \ldots , l^*_k \} } e^{-uJ_{j}}\rho _\varepsilon (J_{j})P_0(\tau _{j}) \right) . \end{aligned}$$

(22)

The first factor in (22) refers to the unnormalized allocated process: the support is ${\varvec{\theta }}^*$, while the jumps follows independent restricted gamma densities, as clearly observed after (19). This shows point 2. of the Proposition.

On the other hand, the second factor in (22) shows that the non-allocated jumps are indeed the jumps of $\varepsilon$-NGG($\sigma$, $\kappa$, $\omega +u$, $P_0$) process, given that exactly $N_{na}$ jumps of the process were obtained; moreover, the conditional distribution of $N_{na}$ is described in (21). This shows point 1. of the Proposition.

Point 3 follows straightforward from (22).

Normalization of the jumps (allocated and non-allocated) gives 4.

With regard to 5., we need to integrate out $N_\varepsilon$ in $\fancyscript{L}(N_\varepsilon ,u,{\varvec{\theta }})$ displayed in (20). We have already made these computations in the proof of formula (5), and thus $f_{U|\varvec{\theta }^{*}}(u|\varvec{\theta }^{*})$ is proportional to the integrand in (5).

1.6 Appendix 6: Details of the blocked Gibbs sampler

We explicitly derive every step of the Gibbs sampler in Fig. 1, starting from the joint distribution of data and parameters in (11).

• 1. The first step is straightforward, since

  $$\begin{aligned} \fancyscript{L}(u|{\varvec{X}},{\varvec{\theta }},P_\varepsilon ,\varepsilon ,\sigma ,\kappa ) \propto \fancyscript{L}(u, {\varvec{X}},{\varvec{\theta }},P_\varepsilon ,\varepsilon ,\sigma ,\kappa ). \end{aligned}$$

• 2. Thanks to the hierarchical structure of the model, the following relation holds true:

  $$\begin{aligned} \begin{aligned}&\fancyscript{L}(\varvec{\theta }|\varvec{X}, P_\varepsilon , \varepsilon , \sigma , \kappa , u)\propto \prod _{i=1}^{n}k(X_i; \theta _i)\sum _{j=0}^{N_\varepsilon }J_j\delta _{\tau _j}(\theta _i)\\&=\prod _{i=1}^{n}\sum _{j=0}^{N_\varepsilon }J_jk(X_i; \theta _i)\delta _{\tau _j}(\theta _i) =\prod _{i=1}^{n}J_jk(X_i; \tau _j), \end{aligned} \end{aligned}$$

  and this proves Step 2.

• 3. As far as $\fancyscript{L}(P_\varepsilon , \varepsilon , \sigma , \kappa |u, \varvec{\theta }, \varvec{X})$ is concerned, we have

  $$\begin{aligned} \begin{aligned}&\fancyscript{L}(P_\varepsilon , \varepsilon , \sigma , \kappa |u, \varvec{\theta }, \varvec{X})= \fancyscript{L}(P_\varepsilon , \varepsilon , \sigma , \kappa |u, \varvec{\theta })\\&= \fancyscript{L}(P_\varepsilon |\varepsilon , \sigma , \kappa , u, \varvec{\theta }) \fancyscript{L}(\varepsilon , \sigma , \kappa | u, \varvec{\theta }), \end{aligned} \end{aligned}$$

  so that Step 3. can be split into two consecutive substeps. First we simulate from $\fancyscript{L}(\varepsilon , \sigma , \kappa | u, \varvec{\theta })$ as follows: we integrate out $N_\varepsilon$ (or equivalently $N_{na}$) from (20) and obtain

  $$\begin{aligned} \begin{aligned}&\fancyscript{L}(\varepsilon , \sigma ,\kappa | u, \varvec{\theta }, \varvec{X}) \propto \sum _{N_{na}=0}^{+\infty } \fancyscript{L}(N_{na},\varepsilon , \sigma ,\kappa | u, \varvec{\theta }, \varvec{X}) \\&\quad = \dfrac{u^{n-1}}{\varGamma (n)}\left( \dfrac{\kappa }{\varGamma (1-\sigma )}\right) ^{k-1} \prod _{i=1}^{k}\biggl [\varGamma (n_i-\sigma , \varepsilon (u+\omega )) \biggr ] \\&\qquad \times \dfrac{(\omega +u)^{\sigma k-n}}{\omega ^{\sigma }\varGamma (-\sigma , \omega \varepsilon )}e^{\varLambda _{\varepsilon ,u}- \varLambda _{\varepsilon }} \left( \varLambda _{\varepsilon ,u}+k\right) \pi (\varepsilon )\pi (\sigma )\pi (\kappa ). \end{aligned} \end{aligned}$$

  In practical terms, Step 3a can be obtained in three substeps:

  $$\begin{aligned}&\fancyscript{L}(\varepsilon | u, \varvec{\theta }, \varvec{X})\propto \prod _{i=1}^{k}\varGamma (n_i-\sigma , \varepsilon (u+\omega )) {\mathrm{e}}^{(\varLambda _{\varepsilon ,u}-\varLambda _{\varepsilon })}\nonumber \\&\quad \times \dfrac{\varLambda _{\varepsilon ,u}+k}{\varGamma (-\sigma , \omega \varepsilon )} \pi (\varepsilon ), \end{aligned}$$

  (23)
  $$\begin{aligned}&\fancyscript{L}(\sigma |u, \varvec{\theta }, \varvec{X}) \propto \dfrac{(u+\omega )^{k\sigma }}{\omega ^\sigma }\dfrac{\varLambda _{\varepsilon ,u}+k}{\varGamma (-\sigma , \omega \varepsilon )} {\mathrm{e}}^{\left( \varLambda _{\varepsilon ,u}-\varLambda _\varepsilon \right) } \nonumber \\&\quad \times \prod _{i=1}^{k} \varGamma (n_i-\sigma , \varepsilon (u+\omega ))\varGamma (1-\sigma )^{1-k}\pi (\sigma ), \end{aligned}$$

  (24)
  $$\begin{aligned}&\fancyscript{L}(\kappa | u, \varvec{\theta }, \varvec{X})= p_1 gamma(\alpha +k, R+\beta ) \nonumber \\&\quad +\,\, (1-p_1) gamma(\alpha +k-1, R+\beta ), \end{aligned}$$

  (25)

  where

  $$\begin{aligned} R=\dfrac{\omega ^\sigma \varGamma (-\sigma , \varepsilon \omega )}{\varGamma (1-\sigma )}-\dfrac{(\omega +u)^\sigma \varGamma (-\sigma , \varepsilon (\omega +u))}{\varGamma (1-\sigma )} \end{aligned}$$

  and $p_1$ is equal to

  $$\begin{aligned} \frac{(\alpha +k-1)(u+\omega )^\sigma \varGamma (-\sigma , \varepsilon (\omega +u))}{ (\alpha \!+\!k\!-\!1)(u\!+\!\omega )^\sigma \varGamma (\!-\!\sigma , \varepsilon (\omega +u))\!+\!k(R+\beta )\varGamma (1-\sigma )}. \end{aligned}$$

  Here we assume that $\pi (\kappa )$ is $gamma(\alpha , \beta )$. Step 3.b consists in sampling from $\fancyscript{L}(P_\varepsilon |\varepsilon , \sigma , \kappa , u, \varvec{\theta })$ and has already been described in Sect. 4.