# Simulations of multivariate gamma distributions and multifactor gamma distributions

Philippe Bernardoff, Bénédicte Puig

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Philippe Bernardoff, Bénédicte Puig. Simulations of multivariate gamma distributions and multifactor gamma distributions. 2025. hal-04826717v2

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# Simulations of multivariate gamma distributions and multifactor gamma distributions

Philippe Bernardoff *

Bénédicte Puig †

Université de Pau et des Pays de l'Adour

Laboratoire de Mathématiques et de leurs Applications, UMR 5142

avenue de l'Université, 64000 Pau, France

January 10, 2025

## Abstract

This article provides a general expression for infinitely divisible multivariate gamma distributions defined by their Laplace transforms, as well as the conditional Laplace transform of infinitely divisible multivariate gamma distributions. We give algorithms for simulating infinitely divisible gamma distributions and infinitely divisible multifactor gamma distributions in dimension 2, 3, 4 and for all dimensions greater than 2 in the Markovian case. We give examples of simulations in dimension 2, 3, 4 and in dimension 5 in the Markovian case.

**KEY WORDS:** Conditional distribution, Lauricella function, Laplace transform, Markovian distribution

MSC: 60E07, 60E10

## 1 Introduction

The aim of this paper is to extend simulations of bivariate gamma distributions, see [1], to multivariate gamma distributions and multifactor gamma distributions defined by their Laplace transforms. In this paper, we consider the following definitions given in [2]. For more details see also [3], [4] and [5]. We

<footnote>*Corresponding author: philippe.bernardoff@univ-pau.fr</footnote>

<footnote>†benedicte.puig@univ-pau.fr</footnote>
use the extension of the classical univariate definition to $\mathbb{R}^n$ obtained as follows: we consider an affine polynomial $P_n(\theta)$ in $\theta = (\theta_1, \dots, \theta_n)$ where 'affine' means that, for $j = 1, \dots, n$, $\partial^2 P_n / \partial \theta_j^2 = 0$. We also assume that $P_n(0) = 1$. For instance, for $n = 2$, we have $P_2(\theta_1, \theta_2) = 1 + p_{\{1\}}\theta_1 + p_{\{2\}}\theta_2 + p_{\{1,2\}}\theta_1\theta_2$. We denote by $\mathfrak{P}_n = \mathfrak{P}([n])$ the family of all subsets of $[n] = \{1, \dots, n\}$ and $\mathfrak{P}_n^*$ the family of non-empty subsets of $[n]$. For simplicity, if $n$ is fixed and if there is no ambiguity, we denote these families by $\mathfrak{P}$ and $\mathfrak{P}^*$, respectively. Similarly, we denote by $\mathfrak{P}_T = \mathfrak{P}(T)$ the family of all subsets of $T = \{1, \dots, n\}$ and $\mathfrak{P}_T^*$ the family of non-empty subsets of $T$. Similarly, if there is no ambiguity, we denote $P_n$ by $P$. We denote by $\mathbb{N}$ the set of non-negative integers. If $z = (z_1, \dots, z_n) \in \mathbb{R}^n$ and $\alpha = (\alpha_1, \dots, \alpha_n) \in \mathbb{N}^n$, then $\alpha! = \alpha_1! \dots \alpha_n!$, $|\alpha| = \alpha_1 + \dots + \alpha_n$, $a_\alpha = a_{\alpha_1, \dots, \alpha_n}$ and

$$z^{\alpha} = \prod_{i=1}^{n} z_i^{\alpha_i} = z_1^{\alpha_1} \dots z_n^{\alpha_n}. \qquad (1)$$

For $T$ in $\mathfrak{P}_n$, we simplify the above notation by writing $z^T = \prod_{t \in T} z_t$ instead of $z^{1_T}$ where

$$\mathbf{1}_T = (\alpha_1, \dots, \alpha_n) \text{ with } \alpha_i = 1 \text{ if } i \in T \text{ and } \alpha_i = 0 \text{ if } i \notin T. \qquad (2)$$

We also write $z^{-T}$ for $\prod_{t \in T} 1/z_t$ if $z_t \neq 0$, $\forall t \in T$. For a mapping $a : \mathfrak{P} \to \mathbb{R}$, we shall use the notation $a : \mathfrak{P} \to \mathbb{R}$, $T \mapsto a_T$. In this notation, an affine polynomial with constant term equal to 1 is $P(\theta) = \sum_{T \in \mathfrak{P}} p_T \theta^T$, with $p_\emptyset = 1$. Now, if there is no ambiguity, for simplicity, we omit the braces and, if $T = \{t_1, \dots, t_k\}$, we denote $a_{\{t_1, \dots, t_k\}} = a_{t_1, \dots, t_k}$ and $a_\emptyset = a_0$. The indicator function of a set $S$ is denoted by $\mathbf{1}_S$, that is, $\mathbf{1}_S(x) = 1$ for $x \in S$ and $0$ for $x \notin S$. We fix $\lambda > 0$. If a random vector $\mathbf{X} = (X_1, \dots, X_n)$ on $\mathbb{R}^n$ with probability distribution (pd) $\mu_{\mathbf{X}}$ is such that its Laplace transform (Lt) is

$$\mathbb{E}\{\exp[-(\theta_1 X_1 + \dots + \theta_n X_n)]\} = [P(\theta)]^{-\lambda}, \qquad (3)$$

where $\mathbb{E}$ denotes the expectation, for a set of $\theta$ with non-empty interior, then we denote $\mu_{\mathbf{X}} = \gamma_{(P,\lambda)}$, and $\gamma_{(P,\lambda)}$ will be called the multivariate gamma distribution (mgd) associated with $(P, \lambda)$. If $\mathbf{X}$ has pd $\gamma_{(P,\lambda)}$, we denote it by $\mathbf{X} \sim \gamma_{(P,\lambda)}$, and $P, \lambda$ is called respectively the scale parameter, the shape parameter. These mgds occur naturally in the classification of natural exponential families in $\mathbb{R}^n$ [3]. The marginal distributions of the mgd associated with $(P, \lambda)$ are univariate gamma distributions (ugd) of parameters $(p_i, \lambda)$ for $i = 1, \dots, n$, with Lt

$$[P(0, \dots, 0, \theta_i, 0, \dots, 0)]^{-\lambda} = (1 + p_i \theta_i)^{-\lambda}, \qquad (4)$$

and pd

$$\gamma_{(p_i, \lambda)}(dx) = x^{\lambda-1} p_i^{-\lambda} [\Gamma(\lambda)]^{-1} \exp(-x/p_i) \mathbf{1}_{(0, \infty)}(x) dx. \qquad (5)$$
As in [2], we extend the first definition to the multifactor gamma distribution (mfgd) associated with $(P, \Lambda)$ where $\Lambda = (\lambda, \lambda_1, \dots, \lambda_n)$ and $\lambda_i \ge \lambda > 0$ for all $i = 1, \dots, n$ by its Lt

$$ \mathbb{E} \{ \exp[-(\theta_1 X_1 + \dots + \theta_n X_n)] \} = [P(\theta)]^{-\lambda} \prod_{i=1}^{n} (1 + p_i \theta_i)^{-(\lambda_i - \lambda)}. \quad (6) $$

Using (4), the marginal distributions of the mgd associated with $(P, \Lambda)$ are ugds of parameters $(p_i, \lambda_i)$ for $i = 1, \dots, n$, with Lt $(1 + p_i \theta_i)^{-\lambda_i}$, and pd

$\gamma_{(p_i, \lambda_i)}(dx) = x^{\lambda_i - 1} p_i^{-\lambda_i} [\Gamma(\lambda_i)]^{-1} \exp(-x/p_i) \mathbf{1}_{(0, \infty)}(x) dx$. We can denote either $\gamma_{(p_i, \lambda_i)}$ or $\gamma_{(1+p_i \theta_i, \lambda_i)}$.

[2] gives a proposition for building a random vector whose distribution is the mfgd associated with $(P, \Lambda)$.

**Proposition 1** A random vector $\mathbf{X}$ with distribution $\gamma_{P,\Lambda}$ can be obtained in the following way:

Let $\mathbf{Y}$ be a random vector with distribution $\gamma_{(P,\lambda)}$. Let $\mathbf{Z} = (Z_1, \dots, Z_n)$ be a random vector of independent components for which its pds are $\gamma_{(p_i, \lambda_i - \lambda)}$, and such that $\mathbf{Z}$ and $\mathbf{Y}$ are independent random vectors. Then the random vector $\mathbf{X} = \mathbf{Y} + \mathbf{Z}$ has Lt (6), and consequently has the mfgd associated with $(P, \Lambda)$.

**Remark 2** This construction clearly allows us to simulate $\gamma_{(P,\Lambda)}$ by simulating $\mathbf{Z} \sim \gamma_{(P,\lambda)}$ and $\mathbf{Y}$.

For the bivariate case, see also [6]. The Lt of mgd associated with $(P, \lambda)$ and the Lt of mfgd associated with $(P, \Lambda)$ are known by definition. But, its pdfs are unknown, except for the bivariate gamma distribution (bgd) associated with $(P, \lambda)$ and the bivariate mfgd associated with $(P, (\lambda, \lambda_1, \lambda_2))$. Let us recall in the Proposition below, the known results. Let $F_m^p$ be the generalized hypergeometric function (see [7]) defined by

$$ F_m^p (\alpha_1, \dots, \alpha_p; \beta_1, \dots, \beta_m; z) = \sum_{k=0}^{\infty} \frac{(\alpha_1)_k \cdots (\alpha_p)_k z^k}{(\beta_1)_k \cdots (\beta_m)_k k!}, \quad (7) $$

where $(a)_k = \Gamma(a+k)/\Gamma(a)$ for $a > 0$ and $k \in \mathbb{N}$ (or more generally by $\forall n \in \mathbb{N}, \forall a \in \mathbb{R}, (a)_0 = 1$, $(a)_{n+1} = (a+n)(a)_n$ is the Pochhammer's symbol. For simplification, we denote $F_m^0$ by $F_m$. [5] gives the following proposition.

**Proposition 3** Let $P(\theta_1, \theta_2) = 1 + p_1\theta_1 + p_2\theta_2 + p_{12}\theta_1\theta_2$ be an affine polynomial where $p_1, p_2 > 0$ and $p_{1,2} > 0$. Let $\mu = \gamma_{(P,\lambda)}$ be the bgd associated with $(P, \lambda)$. The measure $\mu$ exists if and only if $c = (p_1p_2 - p_{1,2})/p_{1,2}^2 = p_1p_2/p_{1,2}^2\rho_{1,2} > 0$, where $\rho_{1,2} = 1 - p_{1,2}/(p_1p_2)$ is the correlation coefficient between margins. Then we have

$$ \gamma_{(P,\lambda)}(dx_1, dx_2) = \frac{p_{1,2}^{-\lambda}}{\Gamma(\lambda)^2} e^{-\frac{p_2}{p_{12}}x_1 - \frac{p_1}{p_{12}}x_2} (x_1 x_2)^{\lambda-1} F_1(\lambda; cx_1 x_2) \mathbf{1}_{(0,\infty)^2}(\mathbf{x}) d\mathbf{x}. \quad (8) $$
with $F_1(\lambda; z) = \sum_{k=0}^{\infty} \frac{1}{(\lambda)_k} \frac{z^k}{k!} = \Gamma(\lambda) z^{-(\lambda-1)/2} I_{\lambda-1}(2\sqrt{z})$, where $I_\lambda$ is the modified Bessel function of order $\lambda$.

For the case $\mathbf{\Lambda} = (\lambda, \lambda, \lambda_2)$, the mgd associated with $(P, \mathbf{\Lambda})$ is named by [8] the multisensor gamma distribution associated with $(P, \lambda, \lambda_2)$ and they have proved that its pd is given by the equality

$$ \gamma_{(P,\mathbf{\Lambda})}(dx_1, dx_2) = \frac{p_{12}^{-\lambda} p_2^{-(\lambda_2-\lambda)}}{\Gamma(\lambda) \Gamma(\lambda_2)} 1_{\lambda_1-1} x_2^{\lambda_2-1} \mathbf{e}^{-\frac{p_2}{p_{12}} x_1 - \frac{p_1}{p_{12}} x_2} \Phi_3(\lambda_2 - \lambda; \lambda_2; c \frac{p_{12}}{p_2} x_2; c x_1 x_2) \\ \times \mathbf{1}_{(0,\infty)^2}(x_1, x_2) \, dx_1 dx_2, \quad (9) $$

where $\Phi_3(a; b; x, y) = \sum_{m,n \ge 0} \frac{(a)_m}{(b)_{m+n}} \frac{x^m}{m!} \frac{y^n}{n!}$ is the Horn function. For the bivariate general case, we have the following Theorem in [2]. Let $F_I$ be the function defined by

$$ F_I(a, b, c, z_1, z_2, z_3) = \sum_{m_1, m_2, m_3 = 0}^{\infty} \frac{(a)_{m_1} (b)_{m_2} (c)_{m_3}}{(a+c)_{m_1+m_3} (b+c)_{m_2+m_3}} \frac{z_1^{m_1} z_2^{m_2} z_3^{m_3}}{m_1! m_2! m_3!}; \quad (10) $$

it is a particular generalized Lauricella function defined, by example, in [9].

**Theorem 4** The pd of $\gamma_{(P,(\lambda,\lambda_1,\lambda_2))}$ is given by the equality

$$ \gamma_{(P,(\lambda,\lambda_1,\lambda_2))}(dx_1, dx_2) = \frac{p_{12}^{-\lambda} p_1^{-(\lambda_1-\lambda)} p_2^{-(\lambda_2-\lambda)}}{\Gamma(\lambda_1) \Gamma(\lambda_2)} x_1^{\lambda_1-1} x_2^{\lambda_2-1} \mathbf{e}^{-\frac{p_2}{p_{12}} x_1 - \frac{p_1}{p_{12}} x_2} \times \\ F_I(\lambda_1 - \lambda, \lambda_2 - \lambda, \lambda, \frac{p_{12}}{p_1} x_1, \frac{p_{12}}{p_2} x_2, c x_1 x_2) \mathbf{1}_{(0,\infty)^2}(x_1, x_2) \, dx_1 dx_2, \quad (11) $$

If we get $\lambda_1 = \lambda$ in the equality (11), we obtain Chatelain and Tourneret's result (9) because

$$ F_I(0, \lambda_2 - \lambda, \lambda, z_1, z_2, z_3) = \sum_{m_2, m_3 = 0}^{\infty} \frac{(b)_{m_2}}{(b+c)_{m_2+m_3}} \frac{z_2^{m_2} z_3^{m_3}}{m_2! m_3!} = \Phi_3(b; b+c; z_2, z_3). $$

[5] gives the following Proposition:

**Proposition 5** Let $\mu$ be a mgd on $\mathbb{R}^n$ associated with $(P, \lambda)$. Assume that $\mu$ is not concentrated on a linear subspace of $\mathbb{R}^n$ of the form $\{x \in \mathbb{R}^n; x_k = 0\}$ for some $k$ in $[n] = \{1, \dots, n\}$. Then:

(i) For all $i \in [n]$, $p_i \ne 0$.

(ii) If $p_1, \dots, p_k < 0$ and $p_{k+1}, \dots, p_n > 0$, then $\text{Supp}(\mu) \subset (-\infty, 0]^k \times [0, \infty)^{n-k}$.

(iii) If $p_1, \dots, p_n > 0$ then $p_{[n]} \ge 0$.

[5] gives a necessary and sufficient condition for infinite divisibility of the mgd associated with $(P, \lambda)$, in the sense that the Lt of $\gamma_{(P,\lambda)}$ power $t$ for all positive $t$ is still the Lt of a positive measure, by the following theorem using the notation $\tilde{b}_S$ from the notation $b_S$ defined in [10]:
**Theorem 6** Let $\mu = \gamma_{P,\lambda}$ be a mgd associated with $(P, \lambda)$, where $\lambda > 0$ and $P(\theta) = \sum_{T \in \mathfrak{P}_n} p_T \theta^T$ is such that $p_i > 0$, for all $i \in [n]$, and $p_{[n]} > 0$. Let $\tilde{P}(\theta) = \sum_{T \in \mathfrak{P}_n} \tilde{p}_T \theta^T$ be the affine polynomial such that $\tilde{p}_T = -p_{\bar{T}}/p_{[n]}$ for all $T \in \mathfrak{P}_n$, where $\bar{T} = [n] \setminus T$. Let

$$ \tilde{b}_S = b_S(\tilde{P}) = \sum_{k=1}^{|S|} (k-1)! \sum_{\mathcal{T} \in \Pi_S^k} \prod_{T \in \mathcal{T}} \tilde{p}_T, \quad (12) $$

with

$$ b_S(P) = \sum_{k=1}^{|S|} (k-1)! \sum_{\mathcal{T} \in \Pi_S^k} \prod_{T \in \mathcal{T}} p_T, \quad (13) $$

where $|S|$ is the cardinality of the set $S$. Then the measure $\mu$ is infinitely divisible if and only if

$$ \tilde{p}_i = \tilde{b}_{\{i\}} < 0 \text{ for all } i \in [n], \quad (14) $$

and

$$ \tilde{b}_S \ge 0 \text{ for all } S \in \mathfrak{P}_n^* \text{ such that } |S| \ge 2. \quad (15) $$

**Corollary 7** By the properties of infinite divisible distributions we conclude that the necessary and sufficient conditions for infinite divisibility of a mgd associated with $(P, \lambda)$ of theorem (6), are also necessary and sufficient conditions for infinite divisibility of mfgd associated with $(P, \Lambda)$.

To illustrate the difficulty to calculate the mgd associated with $(P, \lambda)$ we recall, for the trivariate gamma distribution (tgd), the following theorem given in [2]. Let $F_{II}$ be the function defined by

$$ F_{II}(\lambda_1, \lambda_2, z_1, z_2, z_3, z_4) = \sum_{m_1, \dots, m_4=0}^{\infty} \frac{1}{(\lambda_1)_{m_1+m_2+m_3} (\lambda_2)_{2m_1+m_2+m_4}} \frac{z_1^{m_1} z_2^{m_2} z_3^{m_3} z_4^{m_4}}{m_1! m_2! m_3! m_4!}; \quad (16) $$

it is still a particular generalized Lauricella function. We note that $c$ in (8) is $\tilde{b}_{1,2}$.

**Theorem 8** In the case $n = 3$, $p_i > 0$ for $i \in [3]$, $p_{ij} > 0$ for $i, j \in [3]$, $\tilde{b}_{ij} = -\frac{b_k}{p_{123}} + \frac{p_j k p_{ik}}{p_{123}^2} \ge 0$ for $i \ne j$ and $\{i, j, k\} = [3]$, $p_{123} > 0$, and $\tilde{b}_{123} = -\frac{1}{p_{123}} + \frac{p_{12} p_1}{p_{123}^3} + \frac{p_{13} p_2}{p_{123}^2} + \frac{p_{23} p_1}{p_{123}^2} + 2 \frac{p_{12} p_{13} p_{23}}{p_{123}^3} \ge 0$, the infinitely divisible tgd $\gamma_{(P,\lambda)}$ associated with $(P, \lambda)$, is given by the formula

$$ \gamma_{(P,\lambda)}(\mathrm{d}\boldsymbol{x}) = \frac{p_{123}^{-\lambda}}{[\Gamma(\lambda)]^3} \exp(\tilde{p}_1 x_1 + \tilde{p}_2 x_2 + \tilde{p}_3 x_3) (x_1 x_2 x_3)^{\lambda-1} \times \\ F_{II}(\lambda, \lambda, \tilde{b}_{13} x_1 x_3 \tilde{b}_{23} x_2 x_3, \tilde{b}_{123} x_1 x_2 x_3, \tilde{b}_{12} x_1 x_2, \tilde{b}_{13} x_1 x_3 + \tilde{b}_{23} x_2 x_3) \mathbf{1}_{(0,\infty)^3}(\boldsymbol{x}) \mathrm{d}\boldsymbol{x}. \quad (17) $$

**Remark 9** The case $p_{123} = 0$ is solved by [11].

**Remark 10** If $\tilde{b}_{12} = \tilde{b}_{13} = \tilde{b}_{23} = 0$, Theorem 8 gives

$$ \gamma_{(P,\lambda)}(\mathrm{d}\boldsymbol{x}) = \frac{p_{123}^{-\lambda}}{[\Gamma(\lambda)]^3} \exp(\tilde{p}_1 x_1 + \tilde{p}_2 x_2 + \tilde{p}_3 x_3) (x_1 x_2 x_3)^{\lambda-1} F_2(\lambda, \lambda; \tilde{b}_{123} x_1 x_2 x_3) \mathbf{1}_{(0,\infty)^3}(\boldsymbol{x}) \mathrm{d}\boldsymbol{x}, \quad (18) $$

and if we put $\lambda = 1$ in (18), we obtain the Kibble and Moran distribution given in [4].
After giving a general expression for the pd of infinitely divisible mgds, this paper provides their conditional Laplace transforms. These results allow us to extend the result of [1] from the simulation of bgds to mgds and mfgds. These results allow us to achieve the aim of this paper for $n = 3, 4$ in the general case and, for two particular cases for any $n$. This paper is organised as follows. Let $n \in \mathbb{N}^* \setminus \{1\}$ and let $\mathbf{X}_n = (X_1, \dots, X_n)$ be a real random vector of infinitely divisible mgd, Section 2 gives a general expression for the pd $\mu_{\mathbf{X}_n} = \gamma_{(P_n, \lambda)}$. Let $k \in \mathbb{N}^* \setminus \{1, n\}$ and $(x_1, \dots, x_k) \in \mathbb{R}^k$, Section 3 gives the conditional Lt of $(X_k, \dots, X_n)$ given $(X_1, \dots, X_k) = (x_1, \dots, x_k)$ denoted by $L_{(X_k, \dots, X_n)}^{(X_1, \dots, X_k)} = (x_1, \dots, x_k)$. An important particular case is given. Section 4 gives a simpler expression of $L_{(X_2, \dots, X_n)}^{X_1=x_1}$. Section 5 apply this last result for $n = 2, 3, 4$, and for the case $n = 2$, we obtain the result of [1], see also [12]. For the case $n = 3, 4$, we obtain a new general result and algorithms for simulating tgds and quadrivariate gamma distributions. For $n = 5$, we could give an expression for $L_{(X_2, \dots, X_n)}^{X_1=x_1}$ and apply the same method as for cases $n = 2, 3, 4$. Unfortunately, the computations seem long and arduous. Therefore, we study the simpler Markovian case in Section 6. Section 7 presents simulations of mgds for $n = 2, 3, 4$, simulations of mfgds for $n = 2, 3$, and a simulation of Mmgd for $n = 5$. All simulations are performed using the R software, [13]. In order to facilitate the fluent presentation of the paper, proofs are collected in Appendix A in order of appearance.

## 2 Probability distributions of multivariate gamma distributions

For a simple example of applying the main results, we will need the following example in dimension $n$.

**Example 11** Let $P_n$ be the affine polynomial defined in [5] by

$$P_n(\theta) = \frac{-q}{p} + \frac{1}{p} \prod_{i=1}^{n} (1 + p\theta_i) \qquad (19)$$

where $0 < p = 1 - q < 1$. Let $\mu = \gamma_{P,1} = \varphi_{n,p,1}$ be the infinitely divisible mgd associated to $(P, 1)$. Let $\gamma_{P,\lambda} = \varphi_{n,p,\lambda}$ the mgd associated to $(P, \lambda)$. For $\mathbf{x} = (x_1, \dots, x_n) \in \mathbb{R}^n$, we have

$$\gamma_{(P_n, \lambda)}(\mathbf{d}\mathbf{x}) = \frac{p^{-(n-1)\lambda}}{(\Gamma(\lambda))^n} e^{-\frac{x_1+\dots+x_n}{p}} (\mathbf{x}^{[n]})^{\lambda-1} F_{n-1}(\lambda, \dots, \lambda; q p^{-n} \mathbf{x}^{[n]}) \mathbf{1}_{(0,\infty)^n}(\mathbf{x}) \mathbf{d}\mathbf{x} \quad (20)$$

We will give an expression of the pd $\gamma_{(P_n, \lambda)}(\mathbf{d}\mathbf{x})$ in the general case. Let us denote $\theta_{[n]} = (\theta_1, \dots, \theta_n)$, $P_n(\theta_{[n]}) = \sum_{T \in \mathfrak{P}_n} p_T(P_n) \theta_{[n]}^T$, for $1 \le k < n$, $\theta_{[k]} = (\theta_1, \dots, \theta_k)$, $P_{[k]}(\theta_{[k]}) = \sum_{T \in \mathfrak{P}_k} p_T(P_k) \theta_{[k]}^T = P_n((\theta_1, \dots, \theta_k, 0, \dots, 0))$, let us define for $T \in \mathfrak{P}_n$, $\tilde{p}_T(P_n) = -\frac{p_{[n] \setminus T}}{p_{\{k\}}}$, and for $T \in \mathfrak{P}_k$, $\tilde{p}_T(P_k) = -\frac{p_{[k] \setminus T}}{p_{\{k\}}}$, and $\theta_{P_n} = (\tilde{p}_1(P_n), \dots, \tilde{p}_n(P_n))$, as well as $\theta_{P_k} = (\tilde{p}_1(P_k), \dots, \tilde{p}_k(P_k))$. If there is no ambiguity, we denote $\theta_{[n]}$ by $\theta$, $\theta_{P_n}$ by $\theta_P$, $\tilde{p}_T(P_n) = \tilde{p}_T$, and $P_n(\theta_n)$ by $P(\theta)$. We also denote $\tilde{p}_i(P_n) = \tilde{p}_i(P) = \tilde{p}_i$, $i \in [n]$. We also denote by $\tilde{\mathbf{p}}$ the vector $\theta_{P_n} = (\tilde{p}_1, \dots, p_n)$. If $n > 1$, let
$\theta_{P_n} = (\tilde{p}_1(P_n), \dots, \tilde{p}_n(P_n))$, so that $(\partial/\partial\theta)^{\{i\}}(P_n)(\theta_{P_n}) = 0, \forall i \in [n]$, since $P_n$ is an affine polynomial, using the Taylor formula in $\theta_{P_n}$, we get the following proposition which define the affine polynomial $R_n$.

**Proposition 12** With the above definition, we have

$$P_n(\theta_n) = p_{[n]}(\theta_{[n]} - \theta_{P_n})^{[n]} \{1 - R_n [(\theta_{[n]} - \theta_{P_n})^{-1}]\}. \quad (21)$$

with

$$R_n(\mathbf{z}_{[n]}) = R_n(z_1, \dots, z_n) = \sum_{T \in \mathfrak{P}_n, |T| \ge 2} r_T \mathbf{z}_{[n]}^T, \quad (22)$$

and

$$r_T = \frac{1}{p_{[n]}} (\frac{\partial}{\partial \theta})^{\bar{T}} (P_n)(\theta_{P_n}), \quad T \in \mathfrak{P}_n, |T| \ge 2 \quad (23)$$

Since $R_n$ depends of $P_n$, if necessary we will write $R_n(P_n)$. If there is no ambiguity, denoting $R_n(\mathbf{z}_{[n]})$ by $R(\mathbf{z})$, we have

$$P(\theta) = p_{[n]}(\theta - \theta_P)^{[n]} \{1 - R[(\theta - \theta_P)^{-1}]\} \quad (24)$$

This last equality is still true for $n = 1$, with $R(z_1) = 0$.

More specifically, let $T \in \mathfrak{P}_n, |T| \ge 2$, we have

$$r_T = \sum_{T' \in \mathfrak{P}_T} \tilde{p}_{T \setminus T'} \tilde{\mathbf{p}}^{T'}, \quad (25)$$

and for any $n \in \mathbb{N} \setminus \{0, 1\}$ if $|T| = 2 \le n$,

$$r_T = \tilde{b}_T, \quad (26)$$

if $|T| = 3 \le n$,

$$r_T = \tilde{b}_T, \quad (27)$$

if $|T| = 4 \le n$,

$$r_T = \tilde{b}_T - \sum_{\{U,V\} \in \Pi_T^2, |U|=2, |V|=2} \tilde{b}_U \tilde{b}_V, \quad (28)$$

if $|T| = 5 \le n$,

$$r_T = \tilde{b}_T - \sum_{\{U,V\} \in \Pi_T^2, |U|=3, |V|=2} \tilde{b}_U \tilde{b}_V, \quad (29)$$

Later, we will need the following definition and results. In the sequel, we suppose that $\forall T \in \mathfrak{P}_n, p_T \ne 0$.

**Definition 13** Let $T \in \mathfrak{P}_n$, and $\bar{T} = [n] \setminus T$, and if there is no ambiguity, for simplicity we denote $S_T(\theta_T)$ by $S_T$, the polynomial defined by: if $p_{\bar{T}} \ne 0$, then

$$S_T(\theta_T) = \frac{1}{p_{\bar{T}}} (\frac{\partial}{\partial \theta})^{\bar{T}} [P_n[\theta]]. \quad (30)$$
If $q_U$ is the numbers such that $S_T(\theta_T) = \sum_{U \in \mathfrak{P}_T} q_U \theta^U$, we have

$$q_U = p_{\bar{T} \cup U} p_{\bar{T}}^{-1}. \qquad (31)$$

Since $S_T$ depends on $P_n$, if necessary we will write $S_T(P_n)$.

For example, if $n = 3$, we have $S_{2,3}(\theta_2, \theta_3) = 1 + \frac{p_{1,2}}{p_1}\theta_2 + \frac{p_{1,3}}{p_1}\theta_3 + \frac{p_{1,2,3}}{p_1}\theta_2\theta_3$, $S_i(\theta_i) = 1 + \frac{p_{1,2,3}}{p[3] \setminus \{i\}}\theta_i$, $i = 1, 2, 3$. We will need the following proposition.

**Proposition 14** For $U \in \mathfrak{P}_T$, we have,

$$S_U(S_T) = S_U(P_n). \qquad (32)$$

We also have the following proposition.

**Proposition 15** With the above definition, we have $\forall U \in \mathfrak{P}_T$

$$\tilde{q}_U = \tilde{p}_U, \qquad (33)$$

therefore, we have

$$\tilde{b}_U(S_T) = \tilde{b}_U, \qquad (34)$$

and

$$r_U(S_T) = r_U(P_n). \qquad (35)$$

If $\gamma_{(P,\lambda)}$ is an infinitely divisible mgd, $\gamma_{(S_T,\lambda)}$ is also an infinitely divisible mgd from Theorem (6). We have the following equalities

$$S_T = (-\tilde{p}_T)^{-1} [(-\tilde{p})^T S^T - \sum_{T' \in \mathfrak{P}_T, |T'|>1} r_{T'}(-\tilde{p})^{T \setminus T'} S^{T \setminus T'}] \qquad (36)$$

or

$$(-\tilde{p})^T S^T = (-\tilde{p}_T)S_T + \sum_{T' \in \mathfrak{P}_T, |T'|>1} r_{T'}(-\tilde{p})^{T \setminus T'} S^{T \setminus T'}. \qquad (37)$$

Now, we can give the following expression of $\gamma_{(P_n,\lambda)}$ by the following theorem.

**Theorem 16** Let $c_{\alpha,\lambda}(R)$ such that

$$[1 - R(\mathbf{z})]^{-\lambda} = \sum_{\alpha \in \mathbb{N}^n} c_{\alpha, \lambda}(R) \mathbf{z}^{\alpha}, \qquad (38)$$

then

$$\gamma_{(P,\lambda)}(\mathbf{d}\mathbf{x}) = \frac{p_{[n]}^{-\lambda}}{[\Gamma(\lambda)]^n} \exp(\boldsymbol{\theta}_P, \mathbf{x}) \mathbf{x}^{(\lambda-1)\mathbf{1}_n} \left[ \sum_{\alpha \in \mathbb{N}^n} \frac{c_{\alpha,\lambda}(R)}{(\lambda)_{\alpha}} \mathbf{x}^{\alpha} \right] \mathbf{1}_{(0,\infty)^n}(\mathbf{x}) (\mathbf{d}\mathbf{x}). \qquad (39)$$

or more specifically

$$\gamma_{(P,\lambda)}(\mathbf{d}\mathbf{x}) = \frac{p_{[n]}^{-\lambda}}{[\Gamma(\lambda)]^n} \exp(\boldsymbol{\theta}_P, \mathbf{x}) \mathbf{x}^{(\lambda-1)\mathbf{1}_n} \left[ \sum_{\alpha \in \mathbb{N}^n, c_{\alpha,\lambda}(R) \neq 0} \frac{c_{\alpha,\lambda}(R)}{(\lambda)_{\alpha}} \mathbf{x}^{\alpha} \right] \mathbf{1}_{(0,\infty)^n}(\mathbf{x}) (\mathbf{d}\mathbf{x}). \qquad (40)$$
We deduce a result given in [5] in the form of the following corollary.

**Corollary 17** For $P_n(\theta_{[n]}) = \frac{-q}{p} + \frac{1}{p} \prod_{i=1}^{n} (1 + p\theta_i)$, for $\mathbf{x} = (x_1, \dots, x_n)$ in $\mathbb{R}^n$, we have the result (20).

For $n = 2$, we can give the following corollary.

**Corollary 18** For $n = 2$, we have

$$R_2(\mathbf{z}) = \tilde{b}_{1,2}z_1z_2, \qquad (41)$$

and

$$c_{\alpha, \lambda}(R_2) = \frac{(\lambda)_l}{l!} \tilde{b}_{1,2}^l, \qquad (42)$$

and $c_{\alpha, \lambda}(R_2) = 0$ otherwise. Hence, we have

$$\gamma_{(P_2, \lambda)}(\mathbf{dx}) = \frac{p_{1,2}^{-\lambda}}{[\Gamma(\lambda)]^2} \exp\left(-\frac{p_2}{p_{1,2}}\mathbf{x}_1 - \frac{p_1}{p_{1,2}}x_2\right) (x_1x_2)^{(\lambda-1)} F_1\left(\lambda, \tilde{b}_{1,2}x_1x_2\right) \mathbf{1}_{(0,\infty)^2}(\mathbf{x})(\mathbf{dx}). \quad (43)$$

We obtain the formula (8).

For $n = 3$, we can give the following corollary.

**Corollary 19** For $n = 3$, we have

$$R_3(\mathbf{z}) = \tilde{b}_{1,2}z_1z_2 + \tilde{b}_{1,3}z_1z_3 + \tilde{b}_{2,3}z_2z_3 + \tilde{b}_{1,2,3}z_1z_2z_3 \qquad (44)$$

If $\alpha = (\alpha_1, \alpha_2, \alpha_3) \in \mathbb{N}^3$, and $\max(\alpha_1, \alpha_2, \alpha_3) = \|\alpha\|_{\infty}$, if $\|\alpha\|_{\infty} \le \frac{|\alpha|}{2}$, we have

$$c_{\alpha, \lambda}(R_3) = \sum_{\|\alpha\|_{\infty} \le k \le \frac{|\alpha|}{2}, k \in \mathbb{N}} \frac{(\lambda)_k \tilde{b}_{1,2}^{k-\alpha_3} \tilde{b}_{1,3}^{k-\alpha_2} \tilde{b}_{2,3}^{k-\alpha_1} \tilde{b}_{1,2,3}^{\alpha_1+\alpha_2+\alpha_3-2k}}{(k-\alpha_3)!(k-\alpha_2)!(k-\alpha_1)!(\alpha_1+\alpha_2+\alpha_3-2k)!}. \qquad (45)$$

and $c_{\alpha, \lambda}(R_3) = 0$ if $\|\alpha\|_{\infty} > \frac{|\alpha|}{2}$, in particular if $|\alpha| = 1$, $c_{\alpha, \lambda}(R_3) = 0$. Thus we have for $\mathbf{x} = (x_1, x_2, x_3)$

$$\gamma_{(P_3, \lambda)}(\mathbf{dx}) = \frac{p_{[3]}^{-\lambda}}{[\Gamma(\lambda)]^3} \exp(\theta_P, \mathbf{x}) \mathbf{x}^{(\lambda-1)} \mathbf{1}_3 \times \left\{ \sum_{\alpha \in \mathbb{N}^3, \|\alpha\|_{\infty} \le \frac{|\alpha|}{2}} \left[ \sum_{\|\alpha\|_{\infty} \le k \le \frac{|\alpha|}{2}, k \in \mathbb{N}} \frac{(\lambda)_k \tilde{b}_{1,2}^{k-\alpha_3} \tilde{b}_{1,3}^{k-\alpha_2} \tilde{b}_{2,3}^{k-\alpha_1} \tilde{b}_{1,2,3}^{\alpha_1+\alpha_2+\alpha_3-2k}}{(k-\alpha_3)!(k-\alpha_2)!(k-\alpha_1)!(\alpha_1+\alpha_2+\alpha_3-2k)!} \right] \frac{\mathbf{x}^{\alpha}}{(\lambda)_{\alpha}} \right\} \mathbf{1}_{(0,\infty)^3}(\mathbf{x})(\mathbf{dx}). \qquad (46)$$

Or, with for $\mathbf{z}_{[4]} = (z_1, z_2, z_3, z_4) \in \mathbb{R}^4$, and $\mathbf{1}_3$ is a generalized multivariate Lauricella function defined by

$$\mathbf{1}_3(\lambda; \mathbf{z}_{[4]}) = \sum_{l \in \mathbb{N}^4} \frac{(\lambda)_{l_1+l_2+l_3+l_4}}{(\lambda)_{l_2+l_3+l_4} (\lambda)_{l_1+l_3+l_4} (\lambda)_{l_1+l_2+l_4}} \frac{\mathbf{z}_{[4]}^l}{l!}. \qquad (47)$$

$$\gamma_{(P_3, \lambda)}(\mathbf{dx}) = \frac{p_{[3]}^{-\lambda}}{[\Gamma(\lambda)]^3} \exp(\theta_{P_3}, \mathbf{x}) \times \mathbf{x}^{(\lambda-1)} \mathbf{1}_3 \mathbf{F}_3(\lambda; \tilde{b}_{1,2}x_1x_2, \tilde{b}_{1,3}x_1x_3, \tilde{b}_{2,3}x_2x_3, \tilde{b}_{1,2,3}x_1x_2x_3) \mathbf{1}_{(0,\infty)^3}(\mathbf{x})(\mathbf{dx}). \qquad (48)$$
We note that the latter result compared with (17) gives with $z_1 = \tilde{b}_{1,2}x_1x_2$, $z_2 = \tilde{b}_{1,3}x_1x_3$, $z_3 = \tilde{b}_{2,3}x_2x_3$, $z_4 = \tilde{b}_{1,2,3}x_1x_2x_3$: $\sum_{l \in \mathbb{N}^4} \frac{(\lambda)_{l_1+l_2+l_3+l_4}}{(\lambda)_{l_2+l_3+l_4}} \frac{(\lambda)_{l_1+l_3+l_4}}{(\lambda)_{l_1+l_2+l_4}} \frac{z_l^l}{l!} = \sum_{l \in \mathbb{N}^4} \frac{1}{(\lambda)_{l_1+l_3+l_4}} \frac{(z_1, z_2+z_3, z_2 z_3, z_4)^l}{l!}$.

In the particular three cases $\tilde{b}_{1,2}, \tilde{b}_{1,3}, \tilde{b}_{2,3}, \tilde{b}_{1,2,3} > 0$, $\tilde{b}_{1,2}, \tilde{b}_{1,3}, \tilde{b}_{2,3} > 0$, $\tilde{b}_{1,2,3} = 0$ and $\tilde{b}_{1,2} = \tilde{b}_{1,3} = \tilde{b}_{2,3} = 0$, $\tilde{b}_{1,2,3} > 0$, we can give the following remarks.

**Remark 20** If $\tilde{b}_{1,2}, \tilde{b}_{1,3}, \tilde{b}_{2,3}, \tilde{b}_{1,2,3} > 0$, if $\|\alpha\|_{\infty} \le \frac{|\alpha|}{2}$,

$$c_{\alpha, \lambda}(R_3) = \sum_{\|\alpha\|_{\infty} \le k \le \frac{|\alpha|}{2}, k \in \mathbb{N}} \frac{(\lambda)_k \tilde{b}_{2,3}^{k-\alpha_1} \tilde{b}_{1,3}^{k-\alpha_2} \tilde{b}_{1,2}^{k-\alpha_3} \tilde{b}_{1,2,3}^{\alpha_1+\alpha_2+\alpha_3-2k}}{(k-\alpha_1)!(k-\alpha_2)!(k-\alpha_3)!(\alpha_1+\alpha_2+\alpha_3-2k)!} > 0 \quad (49)$$

and $c_{\alpha, \lambda}(R_3) = 0$ if $\|\alpha\|_{\infty} > \frac{|\alpha|}{2}$. Thus we have

$$\gamma_{(P_3, \lambda)}(\mathrm{d}\mathbf{x}) = \frac{p_{[3]}^{-\lambda}}{[\Gamma(\lambda)]^3} \exp(\boldsymbol{\theta}_P, \mathbf{x}) \mathbf{x}^{(\lambda-1)\mathbf{1}_3} \times \\ \left\{ \sum_{\alpha \in \mathbb{N}^3, \|\boldsymbol{\alpha}\|_{\infty} \le \frac{|\boldsymbol{\alpha}|}{2}} \left[ \sum_{\boldsymbol{\alpha}\|\boldsymbol{\alpha}\|_{\infty} \le k \le \frac{|\boldsymbol{\alpha}|}{2}, k \in \mathbb{N}} \frac{(\lambda)_k \tilde{b}_{2,3}^{k-\alpha_1} \tilde{b}_{1,3}^{k-\alpha_2} \tilde{b}_{1,2}^{k-\alpha_3} \tilde{b}_{1,2,3}^{\alpha_1+\alpha_2+\alpha_3-2k}}{(k-\alpha_1)!(k-\alpha_2)!(k-\alpha_3)!(\alpha_1+\alpha_2+\alpha_3-2k)!} \right] \frac{\mathbf{x}^{\boldsymbol{\alpha}}}{(\lambda)_{\boldsymbol{\alpha}}} \right\} \mathbf{1}_{(0,\infty)^3}(\mathbf{x}) (\mathrm{d}\mathbf{x}). \quad (50)$$

**Remark 21** If $\tilde{b}_{1,2}, \tilde{b}_{1,3}, \tilde{b}_{2,3} > 0$ and $\tilde{b}_{1,2,3} = 0$, if $\|\alpha\|_{\infty} \le \frac{|\alpha|}{2} = k \in \mathbb{N}$, then

$$c_{\alpha, \lambda}(R_3) = \frac{(\lambda)_k \tilde{b}_{2,3}^{k-\alpha_1} \tilde{b}_{1,3}^{k-\alpha_2} \tilde{b}_{1,2}^{k-\alpha_3}}{(k-\alpha_1)!(k-\alpha_2)!(k-\alpha_3)!} > 0, \quad (51)$$

and $c_{\alpha, \lambda}(R_3) = 0$ if $\|\alpha\|_{\infty} > \frac{|\alpha|}{2} \in \mathbb{N}$ or $\frac{|\alpha|}{2} \notin \mathbb{N}$. Thus we have

$$\gamma_{(P_3, \lambda)}(\mathrm{d}\mathbf{x}) = \frac{p_{[3]}^{-\lambda}}{[\Gamma(\lambda)]^3} \exp(\boldsymbol{\theta}_P, \mathbf{x}) \mathbf{x}^{(\lambda-1)\mathbf{1}_3} \left[ \sum_{\alpha \in \mathbb{N}^3, \|\boldsymbol{\alpha}\|_{\infty} \le \frac{|\boldsymbol{\alpha}|}{2}, k \in \mathbb{N}} \frac{(\lambda)_k \tilde{b}_{2,3}^{k-\alpha_1} \tilde{b}_{1,3}^{k-\alpha_2} \tilde{b}_{1,2}^{k-\alpha_3}}{(k-\alpha_1)!(k-\alpha_2)!(k-\alpha_3)!} \frac{\mathbf{x}^{\boldsymbol{\alpha}}}{(\lambda)_{\boldsymbol{\alpha}}} \right] \mathbf{1}_{(0,\infty)^3}(\mathbf{x}) (\mathrm{d}\mathbf{x}). \quad (52)$$

**Remark 22** If $\tilde{b}_{1,2}, \tilde{b}_{1,3}, \tilde{b}_{2,3} = 0$, $\tilde{b}_{1,2,3} > 0$, then for $\alpha = k\mathbf{1}_3$, $k \in \mathbb{N}$,

$$c_{k\mathbf{1}_3, \lambda}(R_3) = \frac{(\lambda)_k \tilde{b}_{1,2,3}^k}{k!} \quad (53)$$

and $c_{\alpha, \lambda}(R_3) = 0$ otherwise. Thus we have

$$\gamma_{(P_3, \lambda)}(\mathrm{d}\mathbf{x}) = \frac{p_{[3]}^{-\lambda}}{[\Gamma(\lambda)]^3} \exp(\boldsymbol{\theta}_P, \mathbf{x}) \mathbf{x}^{(\lambda-1)\mathbf{1}_3} F_2(\lambda, \lambda; \tilde{b}_{1,2,3}\mathbf{x}^{[3]}) \mathbf{1}_{(0,\infty)^3}(\mathbf{x}) (\mathrm{d}\mathbf{x}) \quad (54)$$

### 3 Conditional Laplace transform

We now assume that $P_n$ is an affine polynomial such that $P_n(\boldsymbol{\theta}) = \sum_{T \in \mathfrak{P}_n} p_T \boldsymbol{\theta}^T = 1 + \sum_{T \in \mathfrak{P}_n^*} p_T \boldsymbol{\theta}^T$, with $p_{[k]} > 0$ for $k = 1, \dots, n$, and is such that the mgd $\gamma_{(P_n, \lambda)}$ associated with $(P_n, \lambda)$ is infinitely divisible. If $k \in [n]$, we denote by $\boldsymbol{\theta}_{[k]} = (\theta_1, \dots, \theta_k)$, $\boldsymbol{\theta}_{[n]\setminus[k]} = (\theta_{k+1}, \dots, \theta_n)$, and $P_k$ the affine polynomial $P_k(\boldsymbol{\theta}_{[k]}) = \sum_{T \in \mathfrak{P}_k} p_T \boldsymbol{\theta}_{[k]}^T = P_n(\boldsymbol{\theta}_{[k]}, \mathbf{0}_{n-k})$. Similarly, if $T = \{t_1, \dots, t_k\} \subset [n]$, $t_1 < \dots < t_k$
$t_k$, we denote by $\theta_T = (\theta_{t_1}, \dots, \theta_{t_k})$, and if $[n] \setminus T = \bar{T} = \{t_{k+1}, \dots, t_n\}$, $\theta_{[n] \setminus T} = (\theta_{t_{k+1}}, \dots, \theta_{t_n})$, and $P_T$ the affine polynomial $P_T(\theta_T) = \sum_{S \in \mathfrak{P}(T)} p_S \theta_T^S = P_n(\theta'_T)$, where $(\theta'_T)_i = \theta_{t_i}$ if $t_i \in T$ and $(\theta'_T)_i = 0$ if $t_i \notin T$. For all $\alpha = (\alpha_1, \dots, \alpha_n) \in \mathbb{N}^n$, we introduce the notation $(\partial/\partial\theta)^\alpha = \partial^{|\alpha|}/\partial\theta_1^{\alpha_1} \cdots \partial\theta_n^{\alpha_n}$. For all $T \in \mathfrak{P}_n$, we also define $(\partial/\partial\theta)^T = (\partial/\partial\theta)^{1_T}$.

Let $\mathbf{X} = (X_1, \dots, X_n)$ be a random real vector such that has distribution $\gamma_{(P_n, \lambda)}$, denoted by $\mathbf{X} \sim \gamma_{(P_n, \lambda)}$, we give a formula for the Lt of $X_{[n] \setminus [k]}$ given $\mathbf{X}_{[k]} = \mathbf{x}_{[k]}$, an important conditional distribution for the simulations of $\mathbf{X}$, in the following main theorem.

**Theorem 23** Let $\mathbf{X} = (X_1, \dots, X_n)$ be a random real vector such that $\mathbf{X} \sim \gamma_{(P_n, \lambda)}$, with the notation of Theorem 16. Let $1 < n \in \mathbb{N}$, $1 \le k < n$, and $Q_{[n] \setminus [k]}$ the affine polynomials with respect to the $n-k$ variables $\theta_{k+1}, \dots, \theta_n$ defined by

$$Q_{[n]\setminus[k]}(\theta_{[n]\setminus[k]}) = \prod_{i=k+1}^{n} [1 + \theta_i (-\theta_{P_n})_i^{-1}] \quad (55)$$

If $\mathbf{x} = (x_1, \dots, x_n) \in \mathbb{R}^n$, we denote $\mathbf{x}_{[k]} = (x_1, \dots, x_k) \in \mathbb{R}^k$. If $\mathbf{y} = (y_1, \dots, y_n) \in \mathbb{R}^n$, let $\mathbf{y}_{[k]}\mathbf{x}_{[k]} = (y_1 x_1, \dots, y_k x_k)$, then the Lt of $\mathbf{X}_{[n] \setminus [k]}$ given $\mathbf{X}_{[k]} = \mathbf{x}_{[k]}$ is

$$L_{\mathbf{X}_{[n]\setminus[k]}}^{\mathbf{x}_{[k]}}(\theta_{[n]\setminus[k]}) = [Q_{[n]\setminus[k]}(\theta_{[n]\setminus[k]})]^{-\lambda} \frac{\mathbf{F}_k(\lambda, R_n, \mathbf{x}_{[k]}, \theta_{[n]\setminus[k]})}{\mathbf{F}_k(\lambda, R_n, \mathbf{x}_{[k]}, \mathbf{0}_{[n]\setminus[k]})} \quad (56)$$

with

$$\mathbf{F}_k(\lambda, R_n, \mathbf{x}_{[k]}, \theta_{[n]\setminus[k]}) = \sum_{\alpha=(\alpha_{[k]}, \alpha_{[n]\setminus[k]}) \in \mathbb{N}^n, c_{\alpha,\lambda}(R_n) \neq 0} \frac{\mathbf{x}_{[k]}^{\alpha_{[k]}}}{(\lambda)_{\alpha_{[k]}}} c_{\alpha,\lambda}(R_n) (\theta_{[n]} - \theta_{P_n})_{[n]\setminus[k]}^{-\alpha_{[n]\setminus[k]}} \quad (57)$$

# 4 Conditional Laplace transform in the particular case $k=1$

Another form of Theorem 23 can be given for $k=1$ by the following theorem.

**Theorem 24** Let $\mathbf{X} = (X_1, \dots, X_n)$ be a random real vector such that $\mathbf{X} \sim \gamma_{(P_n, \lambda)}$, with the notation of Theorem 16. Let $1 < n \in \mathbb{N}$, $k=1$, and $S_{n-1}$, $B_{n-1}$ the affine polynomials with respect to the $n-1$ variables $\theta_2, \dots, \theta_n$ defined by

$$S_{[n]\setminus[1]}(\theta_{[n]\setminus[1]}) = \frac{p_{[n]}}{p_1} (\theta_n - \theta_{P_n})_{[n]\setminus[1]}^{1_{[n]\setminus[1]}} \left[ 1 - R_n \left( 0, (\theta_n - \theta_{P_n})_{[n]\setminus[1]}^{-1} \right) \right], \quad (58)$$

$$B_{n-1}(\theta_{[n]\setminus[1]}) = \frac{p_{[n]}}{p_1} (\theta_n - \theta_{P_n})_{[n]\setminus[1]}^{1_{[n]\setminus[1]}} \frac{\partial}{\partial z_1} R_n \left( 0, (\theta_n - \theta_{P_n})_{[n]\setminus[1]}^{-1} \right), \quad (59)$$

we have

$$S_{[n]\setminus[1]}(\theta_{[n]\setminus[1]}) = \frac{1}{p_1} \frac{\partial}{\partial \theta_1} P_n(\theta_n), \quad (60)$$

$$B_{n-1}(\theta_{[n]\setminus[1]}) = -\frac{1}{p_1} P_n(\tilde{p}_1, \theta_{[n]\setminus[1]}), \quad (61)$$
and

$$S_{[n]\setminus[1]}(\boldsymbol{\theta}_{[n]\setminus[1]}) = \sum_{T \subset [n]\setminus[1]} \frac{p_{[1] \cup T}}{p_{[1]}} \boldsymbol{\theta}_{[n]\setminus[1]}^T. \quad (62)$$

Let $\mathfrak{z}_{n-1}$ the function with respect to the $n-1$ variables $\theta_2, \dots, \theta_n$ defined by

$$\mathfrak{z}_{n-1}(\boldsymbol{\theta}_{[n]\setminus[1]}) = \frac{B_{n-1}(\boldsymbol{\theta}_{[n]\setminus[1]})}{S_{[n]\setminus[1]}(\boldsymbol{\theta}_{[n]\setminus[1]})} = \frac{\frac{\partial}{\partial z_1} R_n \left(0, (\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})_{[n]\setminus[1]}^{-1}\right)}{1 - R_n \left(0, (\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})_{[n]\setminus[1]}^{-1}\right)}. \quad (63)$$

Let us define

$$I_1(R_n) = \{\alpha_1 \in \mathbb{N}, \exists \boldsymbol{\alpha} = (\alpha_1, \alpha_{[n]\setminus\{1\}}) \in \mathbb{N}^n, c_{\boldsymbol{\alpha},\lambda}(R_n) \neq 0\}, \quad (64)$$

then we define $\mathbf{G}$, a function with respect to the variable $u_1$, by

$$\mathbf{G}(R_n, u_1) = \sum_{\alpha_1 \in I_1(R_n)} \frac{u_1^{\alpha_1}}{\alpha_1!} \quad (65)$$

If $\mathbf{x} = (x_1, \dots, x_n) \in \mathbb{R}^n$, we denote $\mathbf{x}_{[k]} = (x_1, \dots, x_k) \in \mathbb{R}^k$. If $\mathbf{y} = (y_1, \dots, y_n) \in \mathbb{R}^n$, let $\mathbf{y}_{[k]}\mathbf{x}_{[k]} = (y_1x_1, \dots, y_kx_k)$, then the $\mathrm{Lt}$ of $\mathbf{X}_{[n]\setminus[1]} = (X_2, \dots, X_n)$ given $X_1 = x_1$ is

$$L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}(\boldsymbol{\theta}_{[n]\setminus[1]}) = [S_{[n]\setminus[1]}(\boldsymbol{\theta}_{[n]\setminus[1]})]^{-\lambda} \frac{\mathbf{G}(R_n, \mathfrak{z}_{n-1}(\boldsymbol{\theta}_{[n]\setminus[1]}) x_1)}{\mathbf{G}(R_n, \mathfrak{z}_{n-1}(\mathbf{0}_{[n]\setminus[1]}) x_1)} \quad (66)$$

Before prove Theorem 24, we give the following remark for the affine polynomial
$$S_{[n]\setminus[1]}(\boldsymbol{\theta}_{[n]\setminus[1]}) = \sum_{T \subset [n]\setminus[1]} \frac{p_{[1] \cup T}}{p_1} (\boldsymbol{\theta}_{[n]\setminus[1]})^T.$$

**Remark 25** For $T \in \mathfrak{P}_{[n]\setminus[1]}$, we have $\tilde{p}_T(S_{[n]\setminus[1]}) = \tilde{p}_T(R_n)$, hence we have $\tilde{b}_T(S_{[n]\setminus[1]}) = \tilde{b}_T(P_n)$. Therefore, if $\gamma_{(P_n, \lambda)}$ is infinitely divisible $\gamma_{(S_{[n]\setminus[1]}, \lambda)}$ is also infinitely divisible by Theorem 6. We also have

$$r_T(S_{[n]\setminus[1]}) = r_T(P_n) = r_T \quad (67)$$

and therefore

$$S_{[n]\setminus[1]}(\boldsymbol{\theta}_{[n]\setminus[1]}) = \frac{p_{[n]}}{p_1} (\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})^{[n]\setminus\{1\}} \{1 - \sum_{T \subset [n]\setminus\{1\}} r_T (\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})^{-T}\} \quad (68)$$

Now, we can prove Theorem 24. If there is no ambiguity, we denote $S_{[n]\setminus[1]}$, $B_{n-1}$ and $\mathfrak{z}_{n-1}$ by $S$, $B$ and $\mathfrak{z}$.

Before giving the main theorem, we prove by finite induction the following lemma.

**Lemma 26** With the notations of Theorem (24), unless $Rn = 0$, we have,

$$I_1(R_n) = \{\alpha_1 \in \mathbb{N}, \exists \boldsymbol{\alpha} = (\alpha_1, \alpha_{[n]\setminus\{1\}}) \in \mathbb{N}^n, c_{\boldsymbol{\alpha},\lambda}(R_n) \neq 0\} = \mathbb{N},$$

therefore $\mathbf{G}$ defined in Theorem (24) by (65) is exp.
Now, we can give the following main Theorem.

**Theorem 27** With the notations of Theorem (24), unless $Rn = 0$, we have

$$L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}(\boldsymbol{\theta}_{[n]\setminus[1]}) = S_{[n]\setminus[1]}^{-\lambda}(\boldsymbol{\theta}_{[n]\setminus[1]}) \exp\{-[\frac{P_n(0, \boldsymbol{\theta}_{[n]\setminus[1]})}{S_{[n]\setminus[1]}(\boldsymbol{\theta}_{[n]\setminus[1]})} - 1]\frac{x_1}{p_1}\} \quad (69)$$

$$= S_{[n]\setminus[1]}^{-\lambda}(\boldsymbol{\theta}_{[n]\setminus[1]}) \exp\{-[\frac{\sum_{T \in \mathfrak{P}([n]\setminus[1]))(p_T - \frac{p_{(1)} \cup T)}{p_1}) \boldsymbol{\theta}_{[n]\setminus[1]}^T] \frac{x_1}{p_1}\} \quad (70)$$

$$= S_{[n]\setminus[1]}^{-\lambda}(\boldsymbol{\theta}_{[n]\setminus[1]}) \times \mathbf{e}^{\left\{\frac{p[n]}{p_1} x_1 \sum_{T \subset [n] \setminus \{1\}, 0 < |T|} r\{1\} \cup T(-\tilde{\mathbf{p}})^{\{[n]\setminus\{1\}\setminus T}[\mathbf{S}^{[n]\setminus\{1\}\setminus T}(\boldsymbol{\theta}_{[n]\setminus[1]})] S_{[n]\setminus[1]}^{-1}(\boldsymbol{\theta}_{[n]\setminus[1]}) - 1\}\right\}} \quad (71)$$

We denote by $\mathbf{S} = (S_1, \dots, S_n)$, with $S_i(\theta_i) = 1 + (-\tilde{p}_i)^{-1} \theta_i$, $i = 1, \dots, n$. For simplicity, we denote $L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}(\boldsymbol{\theta}_{[n]\setminus[1]})$ by $L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}$, $S_{[n]\setminus[1]}(\boldsymbol{\theta}_{[n]\setminus[1]})$ by $S_{[n]\setminus[1]}$, and $\mathbf{S}(\boldsymbol{\theta}_{[n]\setminus[1]})$ by $\mathbf{S}$ and we can write

$$L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1} = S_{[n]\setminus[1]}^{-\lambda} \mathbf{e}^{\{(-\tilde{p}_{[n]\setminus[1]})^{-1} x_1 \sum_{T \subset [n] \setminus \{1\}, 0 < |T|} r\{1\} \cup T(-\tilde{\mathbf{p}})^{\{[n]\setminus\{1\}\setminus T}[\mathbf{S}^{[n]\setminus\{1\}\setminus T} S_{[n]\setminus[1]}^{-1} - 1\}\}} \quad (72)$$

We can also write

$$L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1} = S_{[n]\setminus[1]}^{-\lambda} \mathbf{e}^{\{(-\tilde{p}_{[n]\setminus[1]})^{-1} x_1 [\sum_{T \subset [n] \setminus \{1\}, 0 < |T|} r\{1\} \cup T(-\tilde{\mathbf{p}})^{\{[n]\setminus\{1\}\setminus T}[\mathbf{S}^{[n]\setminus\{1\}\setminus T} S_{[n]\setminus[1]}^{-1} - C\}\}]}, \quad (73)$$

where $C = \sum_{T \subset [n] \setminus \{1\}, 0 < |T|} r\{1\} \cup T(-\tilde{\mathbf{p}})^{\{[n]\setminus\{1\}\setminus T}}$ is such that $L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}(\mathbf{0}_{n-1}) = 1$.

# 5 Applications of the main results

## 5.1 A particular case in the case $k = 1$

Now let us apply the result of Corollary 17 to get the following Corollary.

**Corollary 28** For $P_n(\boldsymbol{\theta}) = \frac{q}{p} + \frac{1}{p} \prod_{i=1}^n (1 + p\theta_i)$, we have

$$L_{\mathbf{X}_{[n]\setminus[k]}}^{\mathbf{X}_{[k]}}(\boldsymbol{\theta}_{[n]\setminus[k]}) = [\prod_{i=k+1}^{n} (1 + p\theta_i)]^{-\lambda} \frac{F_{k-1}(\lambda, \dots, \lambda, qp^{-k}\mathbf{x}_k^{[k]} \prod_{i=k+1}^{n} (1 + p\theta_i)^{-1})}{F_{k-1}(\lambda, \dots, \lambda, qp^{-k}\mathbf{x}_k^{[k]})} \quad (74)$$

The case $k = 1$ is simpler because $F_0 = \exp$, so we can give the following corollary.

**Corollary 29** For $n > 1, k = 1$, we have

$$L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}(\boldsymbol{\theta}_{[n]\setminus[1]}) = [\prod_{i=2}^{n} (1 + p\theta_i)]^{-\lambda} \exp\{qp^{-1}x_1[\prod_{i=2}^{n} (1 + p\theta_i)^{-1} - 1]\}, \quad (75)$$

or

$$L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}(\boldsymbol{\theta}_{[n]\setminus[1]}) = \sum_{k=0}^{\infty} \frac{(qp^{-1}x_1)^k}{k!} \exp(-qp^{-1}x_1) [\prod_{i=2}^{n} (1 + p\theta_i)]^{-(\lambda+k)}, \quad (76)$$
As a result, Formula (76) gives a simulation of $\mathbf{X}_{[n]}$. Let us denote by $\mathcal{P}(\mu)$ the Poisson distribution (Pd) of parameter $\mu$, we derive the following theorem.

**Theorem 30** Let $X_1 \sim \gamma_{(p,\lambda)}$, let $V_1 \sim \mathcal{P}(qp^{-1}X_1)$, let $\mathbf{X}_{[n]\times[1]} = (X_2, \dots, X_n)$, and $\mathbf{X}_{[n]\times[1]}(X_1=x_1) \sim \gamma_{(\prod_{i=2}^n(1+p\theta_i),\lambda+V_1)}$, then $\mathbf{X}_{[n]} = (X_1, X_2, \dots, X_n) \sim \gamma_{(P_n,\lambda)}$, with $P_n(\theta_{[n]}) = \frac{q}{p} + \frac{1}{p}\prod_{i=1}^n (1+p\theta_i)$.

We derive the following algorithm to simulate $\mathbf{X}_{[n]} \sim \gamma_{(P_n,\lambda)}$.

**Algorithm 31** Simulation of an infinitely divisible mgd $\gamma_{(P_n,\lambda)}$, with $P_n(\theta_{[n]}) = \frac{q}{p} + \frac{1}{p}\prod_{i=1}^n (1+p\theta_i)$

1. Simulate $X_1 \sim \gamma_{(p,\lambda)}$

2. Simulate $V_1 \sim \mathcal{P}(qp^{-1}X_1)$

3. Simulate independently $X_i \sim \gamma_{(p,\lambda+V_1)}$

4. Then $\mathbf{X}_{[n]} = (X_1, X_2, \dots, X_n)$ simulate $\gamma_{(P_n,\lambda)}$.

## 5.2 The case $n=2$ and $k=1$

In this case, we give another proof of Theorem 14 in [1]:

**Theorem 32** Let $P_2(\theta_1, \theta_2) = 1+p_1\theta_1+p_2\theta_2+p_{1,2}\theta_1\theta_2$, with $p_1 > 0, p_2 > 0, p_{1,2} > 0$ $\tilde{b}_{1,2} = \tilde{b}_{1,2}(P_2) = p_1p_2/p_{1,2}^2 - 1/p_{1,2} > 0$ Let $P_1(\theta_1) = P_2(\theta_1, 0)$, $X_1 \sim \gamma_{(P_1,\lambda)}$. Let $\alpha_1 = \frac{\tilde{b}_{1,2}}{(-P_2)}$, and $V_1 \sim \mathcal{P}(\alpha_1 X_1)$. We have $S_2(\theta_2) = 1 + \frac{p_{1,2}}{p_1}\theta_2$. Let $X_2 \sim \gamma_{(S_2,\lambda+V_1)}$, then $\mathbf{X}_{[2]} = (X_1, X_2) \sim \gamma_{(P_2,\lambda)}$.

We derive the following algorithm to simulate $\mathbf{X}_{[2]} = (X_1, X_2) \sim \gamma_{(P_2,\lambda)}$, see [1]:

**Algorithm 33** Simulation of an infinitely divisible bgd $\gamma_{(P_2,\lambda)}$

1. Simulate $X_1 \sim \gamma_{(p_1,\lambda)}$

2. Simulate $V_1 \sim \mathcal{P}(\frac{p_{1,2}}{p_1}\tilde{b}_{1,2}X_1)$

3. Simulate $X_2 \sim \gamma_{(\frac{p_{1,2}}{p_1},\lambda+V_1)}$

4. Then $\mathbf{X}_{[2]} = (X_1, X_2)$ simulates $\gamma_{(P_2,\lambda)}$.
## 5.3 The case $n = 3$ and $k = 1$

In this case, we give the following theorem

**Theorem 34** Let $P_3(\theta_1, \theta_2, \theta_3) = 1+p_1\theta_1+p_2\theta_2+p_3\theta_3+p_{1,2}\theta_1\theta_2+p_{1,3}\theta_1\theta_3+p_{2,3}\theta_2\theta_3+p_{1,2,3}\theta_1\theta_2\theta_3$ with $p_i > 0, i = 1, 2, 3, p_{1,2} > 0, p_{2,3} > 0, p_{1,3} > 0, p_{1,2,3} > 0$, and $\tilde{b}_{1,2}(P_3) = p_{1,3}p_{2,3}/p_{1,2,3}^2 - p_3/p_{1,2,3} > 0$, $\tilde{b}_{1,3}(P_3) = p_{1,2}p_{2,3}/p_{1,2,3}^2 - p_2/p_{1,2,3} > 0$, $\tilde{b}_{2,3}(P_3) = p_{1,3}p_{1,2}/p_{1,2,3}^2 - p_1/p_{1,2,3} > 0$ ($\tilde{p}_1 = -p_{2,3}/p_{1,2,3} < 0$, $\tilde{p}_2 = -p_{1,3}/p_{1,2,3} < 0$, $\tilde{p}_3 = -p_{1,2}/p_{1,2,3} < 0$) and $\tilde{b}_{1,2,3}(P_3) = \tilde{p}_{1,2,3} + \tilde{p}_1\tilde{p}_{2,3} + \tilde{p}_2\tilde{p}_{1,3} + \tilde{p}_3\tilde{p}_{1,2} + 2\tilde{p}_1\tilde{p}_2\tilde{p}_3 = -1/p_{1,2,3} + p_{2,3}p_1/p_{1,2,3}^2 + p_{1,3}p_2/p_{1,2,3}^2 + p_{1,2}p_3/p_{1,2,3}^2 - 2p_{2,3}p_{1,3}p_{1,2}/p_{1,2,3}^3 > 0$. Let $P_1(\theta_1) = P_3(\theta_1, 0, 0) = 1+p_1\theta_1$, let $X_1 \sim \gamma_{(P_1, \lambda)}$. Let $S_{2,3}(\theta_2, \theta_3) = 1 + \frac{p_{1,2}}{p_1}\theta_2 + \frac{p_{1,3}}{p_1}\theta_3 + \frac{p_{1,2,3}}{p_1}\theta_2\theta_3$, $S_2(\theta_2) = 1 + (-\tilde{p}_2)^{-1}\theta_2$ and $S_3(\theta_3) = 1 + (-\tilde{p}_3)^{-1}\theta_3$. Let $\alpha_1 = \frac{\tilde{b}_{1,2}}{(-\tilde{p}_2)}$, $\alpha_2 = \frac{\tilde{b}_{1,3}}{(-\tilde{p}_3)}$, $\alpha_3 = \frac{\tilde{b}_{1,2,3}}{(-\tilde{p}_{2,3})}$, $\alpha_4 = \frac{\tilde{b}_{1,2}\tilde{b}_{2,3}}{(-\tilde{p}_2)(-\tilde{p}_{2,3})}$, $\alpha_5 = \frac{\tilde{b}_{1,3}\tilde{b}_{2,3}}{(-\tilde{p}_3)(-\tilde{p}_{2,3})}$, and $V_i \sim \mathcal{P}(\alpha_i X_1), i \in [5]$ independent, with the notation $\mathbf{v} = (v_1, \dots, v_5) \in \mathbb{N}$, we have

$$L_{(X_2, X_3)}^{X_1=x_1} = \sum_{\mathbf{v} \in \mathbb{N}^5} \prod_{i=1}^{5} \mathbf{P}(V_i = v_i) S_2^{-(v_1+v_4)} S_3^{-(v_2+v_5)} S_{2,3}^{-(\lambda+v_3+v_4+v_5)}. \quad (77)$$

Let $Y_2 \sim \gamma_{(S_2, V_1+V_4)}$, $Y_3 \sim \gamma_{(S_3, V_2+V_5)}$, $(Z_2, Z_3) \sim \gamma_{(S_{2,3}, \lambda+V_3+V_4+V_5)}$ independent, and let $(X_2, X_3) = (Y_2 + Z_2, Y_3 + Z_3)$, then $\mathbf{X}_{[3]} = (X_1, X_2, X_3) \sim \gamma_{(P_3, \lambda)}$.

From Theorem 34, we derive the following algorithm for simulating $\mathbf{X}_{[3]} = (X_1, X_2, X_3) \sim \gamma_{(P_3, \lambda)}$.

**Algorithm 35** Simulation of an infinitely divisible tgd $\gamma_{(P_3, \lambda)}$

1. Simulate $X_1 \sim \gamma_{(p_1, \lambda)}$.

2. Compute $\alpha_i, i \in [5]$, defined in Theorem 34. Simulate independently $V_i \sim \mathcal{P}(\alpha_i X_1)$.

3. Simulate independently $Y_2 \sim \gamma_{((-\tilde{p}_2)^{-1}, V_1+V_4)}$, $Y_3 \sim \gamma_{((-\tilde{p}_3)^{-1}, V_2+V_5)}$, $(Z_2, Z_3) \sim \gamma_{(S_{2,3}, \lambda+V_3+V_4+V_5)}$.

4. Then $\mathbf{X}_{[3]} = (X_1, Y_2 + Z_2, Y_3 + Z_3)$ simulates $\gamma_{(P_3, \lambda)}$.

We can notice that $\mathbf{W} = (W_1, W_2, W_3) = (V_1 + V_4, V_2 + V_5, V_3 + V_4 + V_5)$ satisfied for $\mathbf{t} = (t_1, t_2, t_3) \in (0, \infty)^3$, $\mathbb{E}(\mathbf{t}^\mathbf{W} | X_1 = x_1) = \exp[(\alpha_1 t_1 + \alpha_2 t_2 + \alpha_3 t_3 + \alpha_4 t_1 t_3 + \alpha_5 t_2 t_3)x_1 - (\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 + \alpha_5)x_1]$, and $\mathbf{W} | X_1 = x_1$ is a trivariate Poisson distribution, see [14].

From Theorem (32), we derive the following theorem and algorithm.

**Theorem 36** Let $X_1 \sim \gamma_{(p_1, \lambda)}$. Let $\alpha_i, i \in [5]$, defined in Theorem 34, $\alpha_6 = \frac{\tilde{b}_{2,3}}{(-\tilde{p}_3)}$, and $V_i \sim \mathcal{P}(\alpha_i X_1), i \in [5]$. Let $Z'_2 \sim \gamma_{(\frac{p_{1,2}}{p_1}, \lambda+V_3+V_4+V_5)}$ and $V_6 \sim \mathcal{P}(\alpha_6 Z'_2)$, let $Y_2 \sim \gamma_{((-\tilde{p}_2)^{-1}, V_1+V_4)}$, $Y_3 \sim \gamma_{(-\tilde{p}_3)^{-1}, V_2+V_5}$, $Z'_3 \sim \gamma_{(-\tilde{p}_3)^{-1}, \lambda+V_3+V_4+V_5+V_6}$ independent, we have $(Z'_2, Z'_3) \sim \gamma_{(S_{2,3}, \lambda+V_3+V_4+V_5)}$, let $(X_2, X_3) = (Y_2 + Z'_2, Y_3 + Z'_3)$, then $\mathbf{X}_{[3]} = (X_1, X_2, X_3) \sim \gamma_{(P_3, \lambda)}$.

**Algorithm 37** Simulation of an infinitely divisible tgd
1. Simulate $X_1 \sim \gamma(p_{1,\lambda})$.

2. Compute $\alpha_i, i \in [6]$, defined in Theorem 36. Simulate independently $V_i \sim \mathcal{P}(\alpha_i X_i), i \in [5]$.

3. Simulate $Z'_2 \sim \gamma(\frac{p_{1,2}}{p_1}, \lambda+V_3+V_4+V_5)$.

4. Simulate $V_6 \sim \mathcal{P}(\alpha_6 Z'_2)$.

5. Simulate independently $Y_2 \sim \gamma((-\tilde{p}_2)^{-1}, V_4+V_5)$, $Y_3 \sim \gamma((-\tilde{p}_3)^{-1}, V_2+V_5)$, $Z'_3 \sim \gamma((-\tilde{p}_3)^{-1}, \lambda+V_3+V_4+V_5+V_6)$.

6. Then $\mathbf{X}_{[3]} = (X_1, X_2, X_3) = (X_1, Y_2 + Z'_2, Y_3 + Z'_3)$ simulates $\gamma_{(P_3,\lambda)}$.

We can give the following remark.

**Remark 38** In Theorem (34) or Algorithm (35), we use 3 univariate gamma distributions, 5 Poisson distributions and one bivariate distribution. In Theorem (36) or Algorithm (37), we use 5 univariate gamma distributions and 6 Poisson distributions. For the simulations themselves, we can use either method with the R software, [13].

## 5.4 The case $n = 4$ and $k = 1$

In this case, we give the following theorem

**Theorem 39** Let $P_4(\theta_4) = \sum_{T \in \mathfrak{P}_4} p_T \theta_4^T$ with $p_T > 0, T \in \mathfrak{P}_4^*$, and $\tilde{b}_T(P_4) > 0, T \in \mathfrak{P}_4^*, |T| > 1$, Let $P_1(\theta_1) = P_4(\theta_1, 0, 0, 0) = 1 + p_1\theta_1$, let $X_1 \sim \gamma_{(P_1,\lambda)}$. Let $S_T, T \in \mathfrak{P}_4^*$ defined by (30), so that $S_2(\theta_2) = 1 + (-\tilde{p}_2)^{-1}\theta_2$ and $S_3(\theta_3) = 1 + (-\tilde{p}_3)^{-1}\theta_3$, $S_4(\theta_4) = 1 + (-\tilde{p}_4)^{-1}\theta_4$, $S_{2,3}(\theta_2, \theta_3) = 1 + \frac{p_{1,2,4}}{p_{1,4}}\theta_2 + \frac{p_{1,3,4}}{p_{1,4}}\theta_3 + \frac{p_{1,2,3,4}}{p_{1,4}}\theta_2\theta_3$, $S_{2,4}(\theta_2, \theta_4) = 1 + \frac{p_{1,2,3}}{p_{1,3}}\theta_2 + \frac{p_{1,3,4}}{p_{1,3}}\theta_4 + \frac{p_{1,2,3,4}}{p_{1,3}}\theta_2\theta_4$, $S_{3,4}(\theta_3, \theta_4) = 1 + \frac{p_{1,2,3}}{p_{1,2}}\theta_3 + \frac{p_{1,2,4}}{p_{1,2}}\theta_4 + \frac{p_{1,2,3,4}}{p_{1,2}}\theta_3\theta_4$, and $S_{2,3,4}(\theta_2, \theta_3, \theta_4) = 1 + \frac{p_{1,2,4}}{p_1}\theta_2\theta_4 + \frac{p_{1,3,4}}{p_1}\theta_3\theta_4 + \frac{p_{1,2,3,4}}{p_1}\theta_2\theta_3\theta_4$.

$$ \text{Let } \alpha_1 = \frac{\tilde{b}_{1,2}}{(-\tilde{p}_2)}, \alpha_2 = \frac{\tilde{b}_{1,3}}{(-\tilde{p}_3)}, \alpha_3 = \frac{\tilde{b}_{1,4}}{(-\tilde{p}_4)}, \alpha_4 = \frac{\tilde{b}_{1,2,3}}{(-\tilde{p}_{2,3})}, \alpha_5 = \frac{\tilde{b}_{1,2,4}}{(-\tilde{p}_{2,4})}, \alpha_6 = \frac{\tilde{b}_{1,3,4}}{(-\tilde{p}_{3,4})}, \alpha_7 = \frac{\tilde{b}_{1,2}\tilde{b}_{2,3}}{(-\tilde{p}_2)(-\tilde{p}_{2,3})}, $$

$$ \alpha_8 = \frac{\tilde{b}_{1,2}\tilde{b}_{2,4}}{(-\tilde{p}_2)(-\tilde{p}_{2,4})}, \alpha_9 = \frac{\tilde{b}_{1,3}\tilde{b}_{2,3}}{(-\tilde{p}_3)(-\tilde{p}_{2,3})}, \alpha_{10} = \frac{\tilde{b}_{1,3}\tilde{b}_{3,4}}{(-\tilde{p}_3)(-\tilde{p}_{3,4})}, \alpha_{11} = \frac{\tilde{b}_{1,4}\tilde{b}_{2,4}}{(-\tilde{p}_4)(-\tilde{p}_{2,4})}, \alpha_{12} = \frac{\tilde{b}_{1,4}\tilde{b}_{3,4}}{(-\tilde{p}_4)(-\tilde{p}_{3,4})}, $$

$$ \alpha_{13} = \frac{\tilde{b}_{1,2,3,4}}{(-\tilde{p}_{2,3,4})}, \alpha_{14} = \frac{\tilde{b}_{1,2}\tilde{b}_{2,3,4} + \tilde{b}_{2,3}\tilde{b}_{1,2,4} + \tilde{b}_{2,4}\tilde{b}_{1,2,3}}{(-\tilde{p}_2)(-\tilde{p}_{2,3,4})}, \alpha_{15} = \frac{\tilde{b}_{1,3}\tilde{b}_{2,3,4} + \tilde{b}_{2,3}\tilde{b}_{1,3,4} + \tilde{b}_{3,4}\tilde{b}_{1,2,3}}{(-\tilde{p}_3)(-\tilde{p}_{2,3,4})}, \alpha_{16} = \frac{\tilde{b}_{1,4}\tilde{b}_{2,3,4} + \tilde{b}_{2,4}\tilde{b}_{1,3,4} + \tilde{b}_{3,4}\tilde{b}_{1,2,4}}{(-\tilde{p}_4)(-\tilde{p}_{2,3,4})}, $$

$$ \alpha_{17} = \frac{2\tilde{b}_{1,2}\tilde{b}_{2,3}\tilde{b}_{2,4}}{(-\tilde{p}_2)^2(-\tilde{p}_{2,3,4})}, \alpha_{18} = \frac{2\tilde{b}_{1,3}\tilde{b}_{2,3}\tilde{b}_{3,4}}{(-\tilde{p}_3)^2(-\tilde{p}_{2,3,4})}, \alpha_{19} = \frac{2\tilde{b}_{1,4}\tilde{b}_{2,4}\tilde{b}_{3,4}}{(-\tilde{p}_4)^2(-\tilde{p}_{2,3,4})}, \alpha_{20} = \frac{\tilde{b}_{2,3}(\tilde{b}_{1,2}\tilde{b}_{3,4} + \tilde{b}_{1,3}\tilde{b}_{2,4})}{(-\tilde{p}_2)(-\tilde{p}_3)(-\tilde{p}_{2,3,4})}, $$

$$ \alpha_{21} = \frac{\tilde{b}_{2,4}(\tilde{b}_{1,2}\tilde{b}_{3,4} + \tilde{b}_{1,4}\tilde{b}_{2,3})}{(-\tilde{p}_2)(-\tilde{p}_4)(-\tilde{p}_{2,3,4})}, \alpha_{22} = \frac{\tilde{b}_{3,4}(\tilde{b}_{1,3}\tilde{b}_{2,4} + \tilde{b}_{1,4}\tilde{b}_{2,3})}{(-\tilde{p}_3)(-\tilde{p}_4)(-\tilde{p}_{2,3,4})}, \alpha_{23} = \frac{\tilde{b}_{1,2,3}\tilde{b}_{2,3,4}}{(-\tilde{p}_{2,3})(-\tilde{p}_{2,3,4})}, \alpha_{24} = \frac{\tilde{b}_{1,2,4}\tilde{b}_{2,3,4}}{(-\tilde{p}_{2,4})(-\tilde{p}_{2,3,4})}, $$

$$ \alpha_{25} = \frac{\tilde{b}_{1,3,4}\tilde{b}_{2,3,4}}{(-\tilde{p}_3)(-\tilde{p}_{2,3,4})}, \alpha_{26} = \frac{\tilde{b}_{2,3}(\tilde{b}_{1,2}\tilde{b}_{2,3,4} + \tilde{b}_{2,4}\tilde{b}_{1,2,3})}{(-\tilde{p}_2)(-\tilde{p}_{2,3})(-\tilde{p}_{2,3,4})}, \alpha_{27} = \frac{\tilde{b}_{2,4}(\tilde{b}_{1,2}\tilde{b}_{2,3,4} + \tilde{b}_{2,3}\tilde{b}_{1,2,4})}{(-\tilde{p}_2)(-\tilde{p}_{2,4})(-\tilde{p}_{2,3,4})}, \alpha_{28} = \frac{\tilde{b}_{2,3}(\tilde{b}_{1,3}\tilde{b}_{2,3,4} + \tilde{b}_{2,4}\tilde{b}_{1,2,3})}{(-\tilde{p}_3)(-\tilde{p}_{2,3})(-\tilde{p}_{2,3,4})}, $$

$$ \alpha_{29} = \frac{\tilde{b}_{3,4}(\tilde{b}_{1,3}\tilde{b}_{2,3,4} + \tilde{b}_{2,3}\tilde{b}_{1,3,4})}{(-\tilde{p}_3)(-\tilde{p}_{3,4})(-\tilde{p}_{2,3,4})}, \alpha_{30} = \frac{\tilde{b}_{2,4}(\tilde{b}_{1,4}\tilde{b}_{2,3,4} + \tilde{b}_{3,4}\tilde{b}_{1,2,4})}{(-\tilde{p}_4)(-\tilde{p}_{2,4})(-\tilde{p}_{2,3,4})}, \alpha_{31} = \frac{\tilde{b}_{3,4}(\tilde{b}_{1,4}\tilde{b}_{2,3,4} + \tilde{b}_{2,4}\tilde{b}_{1,3,4})}{(-\tilde{p}_4)(-\tilde{p}_{3,4})(-\tilde{p}_{2,3,4})}, \alpha_{32} = \frac{\tilde{b}_{1,2}\tilde{b}_{2,3}\tilde{b}_{2,4}}{(-\tilde{p}_2)^2(-\tilde{p}_{2,3})(-\tilde{p}_{2,3,4})}, $$

$$ \alpha_{33} = \frac{\tilde{b}_{1,2}\tilde{b}_{2,3}\tilde{b}_{2,4}}{(-\tilde{p}_2)^2(-\tilde{p}_{2,4})(-\tilde{p}_{2,3,4})}, \alpha_{34} = \frac{\tilde{b}_{1,3}\tilde{b}_{2,3}\tilde{b}_{3,4}}{(-\tilde{p}_3)^2(-\tilde{p}_{2,3})(-\tilde{p}_{2,3,4})}, \alpha_{35} = \frac{\tilde{b}_{1,3}\tilde{b}_{2,3}\tilde{b}_{3,4}^2}{(-\tilde{p}_3)^2(-\tilde{p}_{3,4})(-\tilde{p}_{2,3,4})}, \alpha_{36} = \frac{\tilde{b}_{1,4}\tilde{b}_{3,4}\tilde{b}_{2,4}}{(-\tilde{p}_4)^2(-\tilde{p}_{2,4})(-\tilde{p}_{2,3,4})}, $$

$$ \alpha_{37} = \frac{\tilde{b}_{1,4}\tilde{b}_{2,4}\tilde{b}_{3,4}^2}{(-\tilde{p}_4)^2(-\tilde{p}_{3,4})(-\tilde{p}_{2,3,4})}, \alpha_{38} = \frac{\tilde{b}_{2,3}^2(\tilde{b}_{1,2}\tilde{b}_{3,4} + \tilde{b}_{1,3}\tilde{b}_{2,4})}{(-\tilde{p}_2)(-\tilde{p}_3)(-\tilde{p}_{2,3})(-\tilde{p}_{2,3,4})}, \alpha_{39} = \frac{\tilde{b}_{2,4}^2(\tilde{b}_{1,2}\tilde{b}_{3,4} + \tilde{b}_{1,4}\tilde{b}_{2,3})}{(-\tilde{p}_2)(-\tilde{p}_4)(-\tilde{p}_{2,4})(-\tilde{p}_{2,3,4})}, $$

$$ \alpha_{40} = \frac{\tilde{b}_{3,4}^2(\tilde{b}_{1,3}\tilde{b}_{2,4} + \tilde{b}_{1,4}\tilde{b}_{2,3})}{(-\tilde{p}_3)(-\tilde{p}_4)(-\tilde{p}_{3,4})(-\tilde{p}_{2,3,4})}. \text{ Let } V_i \sim \mathcal{P}(\alpha_i X_i), i \in [40], \text{ be independent Poisson distributions, we} $$
have with the above definitions, with $\mathbf{v} = (v_1, \dots, v_{40}) \in \mathbb{N}^{40}$, and $z_1 = v_1 + v_7 + v_8 + v_{14} + 2v_{17} + v_{20} + v_{21} + v_{26} + v_{27} + 2v_{32} + 2v_{33} + v_{38} + v_{39}$, $z_2 = v_2 + v_9 + v_{10} + v_{15} + 2v_{18} + v_{20} + v_{22} + v_{28} + v_{29} + 2v_{34} + 2v_{35} + v_{38} + v_{40}$, $z_3 = v_3 + v_{11} + v_{12} + v_{16} + 2v_{19} + v_{21} + v_{22} + v_{30} + v_{31} + 2v_{36} + 2v_{37} + v_{39} + v_{40}$, $z_4 = v_4 + v_7 + v_9 + v_{23} + v_{26} + v_{28} + v_{32} + v_{34} + v_{38}$, $z_5 = v_5 + v_8 + v_{11} + v_{24} + v_{27} + v_{30} + v_{33} + v_{36} + v_{39}$, $z_6 = (v_6 + v_{10} + v_{12} + v_{25} + v_{29} + v_{31} + v_{35} + v_{37} + v_{40}$, $z_7 = \sum_{i=13}^{40} v_i$.

$$L_{(X_2, X_3, X_4)}^{X_1=x_1} = \sum_{\mathbf{v} \in \mathbb{N}^{40}} \left[ \prod_{i=1}^{40} \mathbf{P}(V_i = v_i) \right] S_2^{-z_1} S_3^{-z_2} S_4^{-(z_3)} S_{2,3}^{-z_4} S_{2,4}^{-z_5} S_{3,4}^{-z_6} S_{2,3,4}^{-(\lambda+z_7)} \quad (78)$$

Let the following infinitely divisible independent random vectors defined by

$$X_1 \sim \gamma_{(P_1, \lambda)}, \quad (79)$$

$$Y_2 \sim \gamma_{(S_2, V_1 + V_7 + V_8 + V_{14} + 2V_{17} + V_{20} + V_{21} + V_{26} + V_{27} + 2V_{32} + 2V_{33} + V_{38} + V_{39})}, \quad (80)$$

$$Y_3 \sim \gamma_{(S_3, V_2 + V_9 + V_{10} + V_{15} + 2V_{18} + V_{20} + V_{22} + V_{28} + V_{29} + 2V_{34} + 2V_{35} + V_{38} + V_{40})}, \quad (81)$$

$$Y_4 \sim \gamma_{(S_4, V_3 + V_{11} + V_{12} + V_{16} + 2V_{19} + V_{21} + V_{22} + V_{30} + V_{31} + 2V_{36} + 2V_{37} + V_{39} + V_{40})}, \quad (82)$$

$$(U_{1,2}, U_{1,3}) \sim \gamma_{(S_{2,3}, V_4 + V_7 + V_9 + V_{23} + V_{26} + V_{28} + V_{32} + V_{34} + V_{38})}, \quad (83)$$

$$(U_{2,2}, U_{2,4}) \sim \gamma_{(S_{2,4}, V_5 + V_8 + V_{11} + V_{24} + V_{27} + V_{30} + V_{33} + V_{36} + V_{39})}, \quad (84)$$

$$(U_{3,3}, U_{3,4}) \sim \gamma_{(S_{3,4}, V_6 + V_{10} + V_{12} + V_{25} + V_{29} + V_{31} + V_{35} + V_{37} + V_{40})}, \quad (85)$$

$$(W_2, W_3, W_4) \sim \gamma_{(S_{2,3,4}, \lambda + \sum_{i=13}^{40} v_i)} \quad (86)$$

Let $(X_2, X_3, X_4)$ defined by

$$X_2 = Y_2 + U_{1,2} + U_{2,2} + W_2, \quad (87)$$

$$X_3 = Y_3 + U_{1,3} + U_{3,3} + W_3, \quad (88)$$

$$X_4 = Y_4 + U_{2,4} + U_{3,4} + W_4, \quad (89)$$

then

$$\mathbf{X}_{[4]} = (X_1, X_2, X_3, X_4) \sim \gamma_{(P_{[4]}, \lambda)}. \quad (90)$$

The following algorithm is derived from Theorem 39 to simulate $\mathbf{X}_{[4]} \sim \gamma_{(P_4, \lambda)}$

**Algorithm 40** Simulation of an infinitely divisible quadrivariate gamma distribution $\gamma_{(P_4, \lambda)}$

1. Simulate $X_1 \sim \gamma_{(P_1, \lambda)}$;

2. Compute $\alpha_i, i \in [40]$, defined in Theorem 39 and simulate independently $V_i \sim P(\alpha_i X_1)$;

3. Simulate independently $Y_2, Y_3, Y_4, (U_{1,2}, U_{1,3}), (U_{2,2}, U_{2,4}), (U_{3,3}, U_{3,4}), (W_2, W_3, W_4)$ defined in Theorem 39.
4. Compute $X_2, X_3$, and $X_4$, respectively defined by (87), (88) and (89), then $\mathbf{X}_{[4]}$ simulates $\gamma_{(P_4,\lambda)}$.

**Remark 41** We need to simulate $\gamma_{(S_{2,3}, V_4+V_7+V_9+V_{23}+V_{26}+V_{28}+V_{32}+V_{34}+V_{38})}$, $\gamma_{(S_{2,4}, V_5+V_8+V_{11}+V_{24}+V_{27}+V_{30}+V_{33}+V_{36}+V_{39})}$, $\gamma_{(S_{3,4}, V_6+V_{10}+V_{12}+V_{25}+V_{29}+V_{31}+V_{35}+V_{37}+V_{40})}$, and $\gamma_{(S_{2,3,4}, \lambda+\sum_{i=13}^{40} V_i)}$.

To do this, we use 40 Pds and 4 ugds, 3 bgds and 1 tgd. In each time for bgd, we can use 1 Pd and 2 ugds, and for tgd we can use 6 Pds and 5 ugds. Finally, we can simulate $\gamma_{(P_{[4]},\lambda)}$ with 49 Pds and 15 ugds.

We see that it is possible to simulate a mgd by induction. Unfortunately, the complexity of the computations seems enormous from $n = 5$ upwards. This is why we study the Markovian case, which is simpler.

# 6 Markovian multivariate gamma distributions

In this section, we use the results given in [15]. We suppose that $\mathbf{X}_{[n]} = (X_1, \dots, X_n) \sim \gamma_{(P_n, \lambda)}$, where $P_n$ is an affine polynomial and $\lambda > 0$. We assume that $\gamma_{(P_n, \lambda)}$ is infinitely divisible and it satisfies the following first-order Markov property $\mathbf{P}(X_{i+1} \in B | X_i = x_i, \dots, X_1 = x_1) = \mathbf{P}(X_{i+1} \in B | X_i = x_i)$, for any $1 \le i \le n-1$ and for any bounded set all $B \subset \mathbb{R}$. Such a distribution is called a Markovian mgd (Mmgd). Let $f_{\mathbf{X}_{[n]}}$ be the probability density function (pdf) of $\mathbf{X}_{[n]}$ on $(0, \infty)^n$. See [15] for the expression of the probability density of a Mmgd. [15] also give the following Theorem in the case $p_i = 1, i \in [n]$.

**Theorem 42** Let $\mathbf{X}_{[n]}$ be a random vector distributed according to a Mmgd with shape parameter $\lambda > 0$. The Lt of $\mathbf{X}_{[n]}$ can be expressed for $\theta = (\theta_1, \dots, \theta_n) \in \mathbb{R}^n$ as $L_{\mathbf{X}_{[n]}}(\theta) = \det (\mathbf{I}_n + \mathbf{D}_{\theta}\mathbf{R}_{1/2})^{-\lambda}$ where $\mathbf{I}_n$ is the $n \times n$ identity matrix, $\mathbf{D}_{\theta}$ is the diagonal matrix whose diagonal entries are the components of vector $\theta$, and $\mathbf{R}_{1/2} = (a_{i,j})_{1\le i,j\le n}$ is a correlation matrix such that $a_{i,i} = 1, \forall 1 \le i \le n; a_{i,i+1} = \sqrt{\rho_{i,i+1}}, \forall 1 \le i \le n-1; a_{i,j} = \sqrt{\rho_{i,j}} = \sqrt{\rho_{i,k}}\sqrt{\rho_{k,j}}, \forall 1 \le i < k < j \le n$.

We also have $(X_i, X_{i+l}) \sim \gamma_{(P_{[i,i+l]}, \lambda)}$ with $P_{[i,i+l]}(\theta_i, \theta_{i+1}) = 1 + \theta_i + \theta_{i+1} + (1 - a_{i,i+l}^2) \theta_i \theta_{i+l}$, and we have $\rho_{i,i+l} = \prod_{j=i}^{i+l-1} \rho_{j,j+1}$, for all $1 \le i < i+l \le n$. A straightforward consequence is that $(X_i, X_{i+1}) \sim \gamma_{(P_{[i,i+1]}, \lambda)}$ with $P_{[i,i+1]}(\theta_i, \theta_{i+1}) = 1 + \theta_i + \theta_{i+1} + (1 - \rho_{i,i+1}) \theta_i \theta_{i+1}$. We immediately derive the algorithm for simulating $\mathbf{X}_{[n]}$ from Algorithm 33 and the computations $\frac{p_{i,i+1}}{p_i} \tilde{b}_{i,i+1} = \frac{\rho_{i,i+1}}{1-\rho_{i,i+1}}$ and $\frac{p_{i,i+1}}{p_i} = 1 - \rho_{i,i+1}$.

**Algorithm 43** Simulation of $\gamma_{(P_n, \lambda)}$, with $P_n = \det (\mathbf{I}_n + \mathbf{D}_{\theta}\mathbf{R}_{1/2})$ under the conditions of Theorem 42
1. Simulate $X_1 \sim \gamma_{(1,\lambda)}$;

2. For $i=1$ to $n-1$, do

• simulate $V_i \sim P\left(\frac{\rho_{i,i+1}}{1-\rho_{i,i+1}} X_i\right)$,

• simulate $X_{i+1} \sim \gamma_{(1-\rho_{i,i+1},\lambda+V_i)}$;

3. Then $\mathbf{X}_{[n]} = (X_1, \dots, X_n)$ simulate $\gamma_{(P_n,\lambda)}$.

We note that if we consider $\mathbf{Y}_{[n]} = (p_1X_1, \dots, p_nX_n)$, then $\mathbf{Y}_{[n]} = (Y_1, \dots, Y_n) \sim \gamma_{(Q_n,\lambda)}$, where $Q_n$ is an affine polynomial with $Q_n(\theta_{[n]}) = P_n(\mathbf{p}_{[n]}\theta_{[n]}) = \det(\mathbf{I}_n + \mathbf{D}_{\mathbf{p}_{[n]}}\theta_{[n]})R_{1/2})^{-\lambda}$.

# 7 Simulations

We present simulations for examples of mgds for $n \in \{2, 3, 4\}$, for examples of mfgds for $n \in \{2, 3\}$, and for an example of Mmgd for $n = 5$.

## 7.1 Simulations in dimension 2

Let $P_2(\theta_1, \theta_2) = 1 + 3\theta_1 + 3\theta_2 + \theta_1\theta_2$ and $Q_2(\theta_1, \theta_2) = 1 + 15/13\theta_1 + 3/13\theta_2 + 1/13\theta_1\theta_2$, let $\mathbf{X}_{[2]} = (X_1, X_2) \sim \gamma_{(P_2,2)}$ and $\mathbf{Y}_{[2]} = (Y_1, Y_2) \sim \gamma_{(Q_2,2)}$ for which the correlation coefficients $\rho_{X_1,X_2}$ and $\rho_{Y_1,Y_2}$ are respectively, $\rho_{X_1,X_2} = 1 - \frac{p_1 p_2}{9} = 0.889$ and $\rho_{Y_1,Y_2} = \frac{32}{45} = 0.711$.

Simulations for samples of size 1,000 of bgd $\gamma_{(P_2,2)}$ and $\gamma_{(Q_2,2)}$ are illustrated respectively by the graphical representations given in Figure 1.

Let $\mathbf{X}'_{[2]} = (X'_1, X'_2) \sim \gamma_{(P_2,(2,3,4))}$ and $\mathbf{Y}'_{[2]} = (Y'_1, Y'_2) \sim \gamma_{(Q_2,(2,3,4))}$ for which the correlation coefficients $\rho'_{X'_1,X'_2}$ and $\rho'_{Y'_1,Y'_2}$ are respectively (by the formula $\rho'_{X'_1,X'_2} = \frac{\lambda}{\sqrt{\lambda_1\lambda_2}}\rho_{X_1,X_2}$) $\rho'_{X'_1,X'_2} = \frac{2}{\sqrt{12}}\frac{8}{9} = \frac{8}{27}\sqrt{3} \approx 0.513$ and $\rho'_{Y'_1,Y'_2} = \frac{2}{\sqrt{12}}\frac{32}{45} = \frac{32}{135}\sqrt{3} \approx 0.411$.

Simulations for samples of size 1,000 of mfgd $\gamma_{(P_2,(2,3,4))}$ and $\gamma_{(Q_2,(2,3,4))}$ are illustrated by the graphical representations given in Figure 2.

## 7.2 Simulations in dimension 3

Let $P_3(\theta_1, \theta_2, \theta_3) = 1 + \theta_1 + \theta_2 + \theta_3 + 0.55\theta_1\theta_2 + 0.45\theta_1\theta_3 + 0.5\theta_2\theta_3 + 0.2\theta_1\theta_2\theta_3$ and $Q_3(\theta_1, \theta_2, \theta_3) = 1 + \theta_1 + 4\theta_2 + 5\theta_3 + 2.2\theta_1\theta_2 + 2.25\theta_1\theta_3 + 10\theta_2\theta_3 + 4\theta_1\theta_2\theta_3$, let $\mathbf{X}_{[3]} = (X_1, X_2, X_3) \sim \gamma_{(P_3,2)}$ and $\mathbf{Y}_{[3]} = (Y_1, Y_2, Y_3) \sim \gamma_{(Q_3,2)}$ with $(Y_1, Y_2, Y_3) = (X_1, 4X_2, 5X_3)$.
<figure><img src="image_2.png" /><figcaption>Figure 1: Distributions and simulations of $X_{[2]}$ and $Y_{[2]}$</figcaption></figure>

Simulations for samples of size 1,000 of mgd $\gamma_{(P_{3,2})}$ and $\gamma_{(Q_{3,2})}$ are illustrated by the graphical representations given in Figures 3, 4 and Figures 5, 6.

Let $\mathbf{X}'_{[3]} = (X'_1, X'_2, X'_3) \sim \gamma_{(P_{[3]},(2,3,4,5))}$ and $\mathbf{Y}'_{[3]} = (Y'_1, Y'_2, Y'_3) \sim \gamma_{(Q_{[3]},(2,3,4,5))}$.

Simulations for samples of size 1,000 of mfgd $\gamma_{(P_{3,(2,3,4,5)})}$ and $\gamma_{(Q_{3,(2,3,4,5)})}$ are illustrated by the graphical representations given in Figure 7 and Figure 8.
<figure><img src="image_3.png" /><figcaption>Figure 2: Distributions and simulations of $X'_{[2]}$ and $Y'_{[2]}$</figcaption></figure>

<figure><img src="image_4.png" /><figcaption>Figure 3: Distributions of $X_1, X_2, X_3$</figcaption></figure>
<figure><img src="image_5.png" /><figcaption>Figure 4: Distribution and simulation of $\mathbf{X}_{[3]}$</figcaption></figure>

<figure><img src="image_6.png" /><figcaption>Figure 5: Distributions of $Y_1, Y_2, Y_3$</figcaption></figure>

## 7.3 Simulations in dimension 4

First, we look for the symmetric case where $P_4(\theta_1, \theta_2, \theta_3, \theta_4, \theta_4) = 1 + s_1(\theta_1 + \theta_2 + \theta_3 + \theta_4) + s_2(\theta_1\theta_2 + \theta_1\theta_3 + \theta_1\theta_4 + \theta_2\theta_3 + \theta_2\theta_4 + \theta_3\theta_4) + s_3(\theta_1\theta_2\theta_3 + \theta_1\theta_2\theta_4 + \theta_1\theta_3\theta_4 + \theta_2\theta_3\theta_4) + s_4\theta_1\theta_2\theta_3\theta_4$. Ac-
<figure><img src="image_7.png" /><figcaption>Figure 6: Distribution and simulation of $\mathbf{Y}_{[3]}$</figcaption></figure>

cording to formulas (3.13a) or (3.13b) in [5] and [16] pp. 307-8, we respectively have for $n = 4$, and $|S| = 1, 2, 3, 4$, $\tilde{b}_S = -s_3/s_4, (-s_4s_2 + s_3^2)/s_4^2, -(s_1s_4^2 + 2s_3^3 - 3s_2s_3s_4)/s_4^3, (6s_3^4 - s_4^3 + 3s_2^2s_4^2 + 4s_1s_3s_4^2 - 12s_2s_3^2s_4)/s_4^4$. Without loss of generality, we can assume that $s_4 = 1$, so we respectively have for $n = 4$, and $|S| = 1, 2, 3, 4$, $\tilde{b}_S = -s_3, -s_2+s_3^2, -s_1+3s_3s_2-2s_3^3, 6s_3^4-1+3s_2^2+4s_1s_3-12s_2s_3^2$. We must simultaneously check the conditions $-s_3 < 0, -s_2+s_3^2 \ge 0, -s_1+3s_3s_2-2s_3^3 \ge 0$ and $6s_3^4-1+3s_2^2+4s_1s_3-12s_2s_3^2 \ge 0$ for the gamma distribution $\gamma_{(P_{[4]},\lambda)}$ to be indefinitely divisible. These conditions are equivalent to

$$s_3 > 0, s_2 \le s_3^2, \frac{1-3s_2^2}{4s_3} - \frac{3}{2}s_3^3 + 3s_2s_3 \le s_1 \le 3s_3s_2 - 2s_3^3. \quad (91)$$

The last condition (91) is equivalent to $\frac{1-3s_2^2}{4s_3} + \frac{1}{2}s_3^3 + (3s_2s_3 - 2s_3^3) \le s_1 \le 3s_3s_2 - 2s_3^3$, which is only possible for $\frac{1-3s_2^2}{4s_3} + \frac{1}{2}s_3^3 \le 0$. This gives us the following condition $\sqrt{\frac{1+2s_3^4}{3}} \le s_2$ and (91) becomes

$$s_3 > 0, \sqrt{\frac{1+2s_3^4}{3}} \le s_2 \le s_3^2, \frac{1-3s_2^2}{4s_3} - \frac{3}{2}s_3^3 + 3s_2s_3 \le s_1 \le 3s_3s_2 - 2s_3^3. \quad (92)$$

For $s_3 = 2$, we get $3.3166248 \le s_2 \le 4$, and $s_2 = 3.5$ matches. For $s_3 = 2, s_2 = 3.5$, we get $4.53125 \le s_1 \le 5.0$ and $s_1 = 4.75$ is a possible value. We check that for $|S| = 1, 2, 3, 4$, we have
<figure><img src="image_8.png" /><figcaption>Figure 7: Distribution and simulation of $\mathbf{X}'_{[3]}$</figcaption></figure>

<figure><img src="image_9.png" /><figcaption>Figure 8: Distribution and simulation of $\mathbf{Y}'_{[3]}$</figcaption></figure>

respectively $\tilde{b}_S = -2, 0.5, 0.25, 1.75$. Let $\mathbf{X}_{[4]} = (X_1, X_2, X_3, X_4) \sim \gamma_{(P_{4,2})}$. Simulations for samples of size 1,000 of mgd $\gamma_{(P_{4,2})}$ are illustrated by the graphical representations given in Figure 9 by four
one-dimensional projections and Figure 10 by various three-dimensional projections.

<figure><img src="image_10.png" /><figcaption>Figure 9: Distribution and simulation of $X_1, X_2, X_3, X_4$</figcaption></figure>
<figure><img src="image_11.png" /><figcaption>Figure 10: Distribution and simulation of $\mathbf{X}_{[4]}$</figcaption></figure>

Next, we search for the general case by slightly modifying the values of $s_1, s_2, s_3, s_4$ checking that the indefinite divisibility conditions of $\gamma_{(P_{[4]},\lambda)}$ remain verified. For example, we obtain
$$ (p_1, p_2, p_3, p_4, p_{1,2}, p_{1,3}, p_{1,4}, p_{2,3}, p_{2,4}, p_{3,4}, p_{1,2,3}, p_{1,2,4}, p_{1,3,4}, p_{2,3,4}, p_{1,2,3,4}) = $$
(4.75, 4.8, 4.85, 4.7, 3.5, 3.55, 3.6, 3.65, 3.45, 3.4, 2, 1.99, 2.02, 2.01, 1) with
($\tilde{b}_1, \tilde{b}_2, \tilde{b}_3, \tilde{b}_4, \tilde{b}_{1,2}, \tilde{b}_{1,3}, \tilde{b}_{1,4}, \tilde{b}_{2,3}, \tilde{b}_{2,4}, \tilde{b}_{3,4}, \tilde{b}_{1,2,3}, \tilde{b}_{1,2,4}, \tilde{b}_{1,3,4}, \tilde{b}_{2,3,4}, \tilde{b}_{1,2,3,4}) = (-2.01, -2.02, -1.99, -2, 0.6602, 0.5499, 0.37, 0.4198, 0.49, 0.48, 0.111404, 0.2177, 0.3989, 0.5053, 1.590846)

, and $Q_4 (\theta_1, \theta_2, \theta_3, \theta_4, \theta_4) = 1 + 4.75\theta_1 + 4.8\theta_2 + 4.85\theta_3 + 4.7\theta_4 + 3.5\theta_1\theta_2 + 3.55\theta_1\theta_3 + 3.6\theta_1\theta_4 + 3.65\theta_2\theta_3 + 3.45\theta_2\theta_4 + 3.4\theta_3\theta_4 + 2\theta_1\theta_2\theta_3 + 1.99\theta_1\theta_2\theta_4 + 2.02\theta_1\theta_3\theta_4 + 2.01\theta_2\theta_3\theta_4 + \theta_1\theta_2\theta_3\theta_4$. Let $\mathbf{Y}_{[4]} = (Y_1, Y_2, Y_3, Y_4) \sim \gamma_{(Q_{4,2})}$. Simulations for samples of size 1,000 of mgd $\gamma_{(Q_{4,2})}$ are illustrated by the graphical representations given in Figure 11 by four one-dimensional projections and Figure 12 by various three-dimensional projections.

<figure><img src="image_12.png" /><figcaption>Figure 11: Distribution and simulation of $Y_1, Y_2, Y_3, Y_4$</figcaption></figure>
<figure><img src="image_13.png" /><figcaption>Figure 12: Distribution and simulation of $\mathbf{Y}_{[4]}$</figcaption></figure>
7.4 Simulations of Mmgd in dimension 5

For example, we obtain for $\rho_{1,2} = 0.9^2$, $\rho_{2,3} = 0.8^2$, $\rho_{3,4} = 0.7^2$, $\rho_{4,5} = 0.6^2$,

$$
\mathbf{R}_{1/2} = \begin{pmatrix} 1 & 0.9 & 0.72 & 0.504 & 0.3024 \\ 0.9 & 1 & 0.8 & 0.56 & 0.336 \\ 0.72 & 0.8 & 1 & 0.7 & 0.42 \\ 0.504 & 0.56 & 0.7 & 1 & 0.6 \\ 0.3024 & 0.336 & 0.42 & 0.6 & 1 \end{pmatrix} \quad \text{and}
$$

$$
\begin{align*}
P_5 (\theta_1, \theta_2, \theta_3, \theta_4, \theta_5) = {}& 1 + \theta_1 + \theta_2 + \theta_3 + \theta_4 + \theta_5 + 0.19\theta_1\theta_2 + 0.4816\theta_1\theta_3 + 0.745984\theta_1\theta_4 + \\
& 0.90855424\theta_1\theta_5 + 0.36\theta_2\theta_3 + 0.6864\theta_2\theta_4 + 0.887104\theta_2\theta_5 + 0.51\theta_3\theta_4 + 0.8236\theta_3\theta_5 + 0.64\theta_4\theta_5 + \\
& 0.0684\theta_1\theta_2\theta_3 + 0.130416\theta_1\theta_2\theta_4 + 0.16854976\theta_1\theta_2\theta_5 + 0.245616\theta_1\theta_3\theta_4 + 0.39664576\theta_1\theta_3\theta_5 + \\
& 0.47742976\theta_1\theta_4\theta_5 + 0.1836\theta_2\theta_3\theta_4 + 0.296496\theta_2\theta_3\theta_5 + 0.439296\theta_2\theta_4\theta_5 + 0.3264\theta_3\theta_4\theta_5 + \\
& 0.034884\theta_1\theta_2\theta_3\theta_4 + 0.05633424\theta_1\theta_2\theta_3\theta_5 + 0.08346624\theta_1\theta_2\theta_4\theta_5 + 0.15719424\theta_1\theta_3\theta_4\theta_5 + \\
& 0.117504\theta_2\theta_3\theta_4\theta_5 + 0.02232576\theta_1\theta_2\theta_3\theta_4\theta_5.
\end{align*}
$$

Let $\mathbf{X}_{[5]} = (X_1, X_2, X_3, X_4, X_5) \sim \gamma_{(P_{5,2})}$. A simulation for a sample of size 1,000 of Mmgd $\gamma_{(P_{5,2})}$ is illustrated by the graphical representations given in Figure 13 by various one-dimensional and two-dimensional projections.

<figure><img src="image_14.png" /><figcaption>Figure 13: Distribution and simulation of $\mathbf{X}_{[5]}$</figcaption></figure>
# Appendix A Proofs

**Proof of Proposition 12.** Using the Taylor formula in $\theta_{P_n}$, we get

$$ P_n(\theta_{[n]}) = p_{[n]}(\theta_{[n]} - \theta_{P_n})^{[n]} \left(1 - \sum_{T \in \mathfrak{P}_n, |T| \ge 2} -\frac{1}{p_{[n]}} \left(\frac{\partial}{\partial \theta}\right)^T (P_n)(\theta_{P_n}) (\theta_{[n]} - \theta_{P_n})^{-T}\right), \text{ then by} \\ (23) \text{ and (22), we obtain (21). Now we compute } r_T \text{ for } T \in \mathfrak{P}_n, |T| \ge 2, \text{ from (23), we have} \\ r_T = \sum_{S \in \mathfrak{P}_n} \tilde{p}_S \theta_n^{S \setminus \bar{T}} (\theta_{P_n}) = \sum_{S \in \mathfrak{P}_n, S=\bar{T} \cup T', T' \subset [n] \setminus \bar{T}} \tilde{p}_S^{S \setminus \bar{T}} \tilde{p}^{T'} \sum_{T' \in \mathfrak{P}_T} \tilde{p}_{T' \setminus T'}^{T'} \tilde{p}^{T'}. $$

If $|T| = 2$, without loss of generality we compute $r_T$ for $T = \{1, 2\}$: $r_T = \tilde{p}_{1,2} + \tilde{p}_1\tilde{p}_2 + \tilde{p}_2\tilde{p}_1 - \tilde{p}_1\tilde{p}_2 = \tilde{p}_{1,2} + \tilde{p}_1\tilde{p}_2 = \tilde{b}_{\{1,2\}} = \tilde{b}_T$.

Similarly, if $|T| = 3$, we compute $r_T$ for $T = \{1, 2, 3\}$: $r_T = \tilde{p}_{1,2,3} + \tilde{p}_{1,2}\tilde{p}_3 + \tilde{p}_{1,3}\tilde{p}_2 + \tilde{p}_{2,3}\tilde{p}_1 + \tilde{p}_1\tilde{p}_2\tilde{p}_3 + \tilde{p}_2\tilde{p}_1\tilde{p}_3 + \tilde{p}_3\tilde{p}_1\tilde{p}_2 + \tilde{p}_1\tilde{p}_2\tilde{p}_3 - \tilde{p}_1\tilde{p}_2\tilde{p}_3 = \tilde{p}_{1,2,3} + \tilde{p}_{1,2}\tilde{p}_3 + \tilde{p}_{1,3}\tilde{p}_2 + \tilde{p}_{2,3}\tilde{p}_1 + 2\tilde{p}_1\tilde{p}_2\tilde{p}_3 = \tilde{b}_{1,2,3} = \tilde{b}_T$.

Similarly, if $|T| = 4$, we compute $r_T$ for $T = \{1, 2, 3, 4\}$. We prove that

$$ r_{\{1,2,3,4\}} = \tilde{b}_{1,2,3,4} - (\tilde{b}_{1,2}\tilde{b}_{3,4} + \tilde{b}_{1,3}\tilde{b}_{2,4} + \tilde{b}_{1,4}\tilde{b}_{2,3}) = \tilde{b}_T - \sum_{\{U,V\} \in \Pi_T^2, |U|=2, |V|=2} \tilde{b}_U \tilde{b}_V. $$

Similarly, if $|T| = 5$, we compute $r_T$ for $T = \{1, 2, 3, 4, 5\}$, we prove that

$$ r_{\{1,2,3,4,5\}} = \tilde{b}_{1,2,3,4,5} - \tilde{b}_{1,2,3}\tilde{b}_{4,5} - \tilde{b}_{1,2,4}\tilde{b}_{3,5} - \tilde{b}_{1,2,5}\tilde{b}_{3,4} - \tilde{b}_{1,3,4}\tilde{b}_{2,4} - \tilde{b}_{1,3,5}\tilde{b}_{2,4} - \tilde{b}_{1,4,5}\tilde{b}_{2,3} - \tilde{b}_{2,3,4}\tilde{b}_{1,5} - \tilde{b}_{2,3,5}\tilde{b}_{1,4} - \tilde{b}_{2,4,5}\tilde{b}_{1,3} - \tilde{b}_{3,4,5}\tilde{b}_{1,2} = \tilde{b}_T - \sum_{\{U,V\} \in \Pi_T^2, |U|=3, |V|=2} \tilde{b}_U \tilde{b}_V. $$

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**Proof of Proposition 14.** Indeed, we have

$$ S_U(S_T) = \frac{p_{\bar{T} \cup U}}{p_{\bar{T} \cup \bar{U}}} \bar{U} \{ \frac{1}{p_{\bar{T}}} (\frac{\partial}{\partial \theta})^{\bar{T}} [P_n(\theta)] \} = \frac{1}{p_{\bar{T} \cup U}} (\frac{\partial}{\partial \theta})^{\bar{U} \cup \bar{T}} [P_n(\theta)] = \frac{1}{p_{\bar{U} \cap \bar{T}}} (\frac{\partial}{\partial \theta})^{\bar{U} \cap \bar{T}} [P_n(\theta)] = S_U(P_n). $$

■

**Proof of Proposition 15.** Equality (33) results from the following computation $\tilde{q}_U = \frac{p_{\bar{T} \cup T \setminus U}}{p_{\bar{T}}} / \frac{p_{\bar{T} \cup T}}{p_{\bar{T}}} = \frac{p_{\bar{U}}}{p_{[n]}}$. Equality (34) comes from (12) and (33). Equality (35) comes from (25) and (33). According to (21) for $S_T$ and (35), we get $S_T(\theta_T) = (-\tilde{p}_T)^{-1} (\theta_{[n]} - \theta_{P_n})^T (1 - \sum_{T' \in \mathfrak{P}_T, |T'|>1} r_{T'} (\theta_{[n]} - \theta_{P_n})^{-T'}) = (-\tilde{p}_T)^{-1} [(\theta_{[n]} - \theta_{P_n})^T - \sum_{T' \in \mathfrak{P}_T, |T'|>1} r_{T'} (\theta_{[n]} - \theta_{P_n})^{T \setminus T'}]$ and the equalities $\theta_i - \tilde{p}_i = (-\tilde{p}_i) S_i, i = 1, \dots, n$, gives (36). Equality (37) is a rewriting of (36). ■

**Proof of Theorem 16.** Let us remember some definitions introduced in [5]. We construct certain measures on $[0, \infty)^n$ indexed by $I \in \mathfrak{P}_n^*$. For $i \in [n]$, define $l_i(\mathrm{d}x_i) = \mathbf{1}_{(0,\infty)}(x_i) \mathrm{d}x_i$ if $i \in I$ and $l_i(\mathrm{d}x_i) = \delta_0(\mathrm{d}x_i)$ if $i \notin I$. We define the following measure on $[0, \infty)^n$: $h_I(\mathrm{d}\mathbf{x}) = \bigotimes_{i=1}^n l_i(\mathrm{d}x_i)$. For instance, if $n = 3$ and $I = \{2,3\}$, then $h_{\{2,3\}}(\mathrm{d}x_1, \mathrm{d}x_2, \mathrm{d}x_3) = \delta_0(\mathrm{d}x_1) \mathbf{1}_{\{0,\infty\}}^2(x_2, x_3) \mathrm{d}x_2 \mathrm{d}x_3$. We denote by $\mathbf{1}_n$ the vector $(1, \dots, 1) \in \mathbb{R}^n$, and by $\mathbf{0}$ if there is no ambiguity, and by $\mathbf{0}_n$ the vector $(0, \dots, 0) \in \mathbb{R}^n$, and by $\mathbf{0}$ if there is no ambiguity. For $I \in \mathfrak{P}_n^*$, we write $\mathbb{N}_i^I = \mathbb{N}$ if $i \in I$, $\mathbb{N}_i^I = \{0\}$ if $i \notin I$, and $\mathbb{N}^I = \times_{i=1}^n \mathbb{N}_i^I$. For $\theta = (\theta_1, \dots, \theta_n) \in \mathbb{R}^n$ with $\theta_i \neq 0$ for all $i \in [n]$, recall the notations $\theta^{-1} = (\theta_1^{-1}, \dots, \theta_n^{-1})$ and for $\alpha \in \mathbb{N}^n$, $\theta^{-\alpha} = (\theta^{-1})^{\alpha}$. For all $I \in \mathfrak{P}_n^*$, let

$$ \mu_{\alpha, I}(\mathrm{d}\mathbf{x}) = \frac{\mathbf{x}^{\alpha-1}_I}{\alpha - 1_I!} h_I(\mathrm{d}\mathbf{x}) \qquad (93) $$
Thus, for $\theta_1 > 0, \dots, \theta_n > 0$, the Lt of $\mu_{\alpha,I}$ is $L_{\mu_{\alpha,I}}(\theta) = \theta^{-\alpha}$. More generally, for $-a_1+\theta_1 > 0, \dots, -a_n + \theta_n > 0$, if $\mathbf{a} = (a_1, \dots, a_n) \in \mathbb{R}^n$, then we have

$$L_{\exp(\mathbf{a},\mathbf{x})\mu_{\alpha,I}}(\theta) = (-\mathbf{a} + \theta)^{-\alpha}. \qquad (94)$$

The latter is still true if we replace $(\alpha - \mathbf{1}_I)!$ in (93) by $\Gamma(\alpha) = \prod_{i \in I} \Gamma(\alpha_i)$ if $\alpha_i > 0$, $i \in [n]$.
Using (24) and (38), we write $[P(\theta)]^{-\lambda} = p_{[n]}^{-\lambda} \sum_{\alpha \in \mathbb{N}^n} c_{\alpha,\lambda}(R) (-\theta_P + \theta)^{-(\alpha+\lambda\mathbf{1})}$. Using (94) we get
$[P(\theta)]^{-\lambda} = L_{p_{[n]}^{-\lambda}} \exp(\theta_P, \mathbf{x}) \mathbf{x}^{(\lambda-1)1} (\sum_{\alpha \in \mathbb{N}^n} c_{\alpha,\lambda}(R) \frac{\mathbf{x}^{\alpha}}{\Gamma(\alpha+\lambda\mathbf{1})} h_{[n]})(\theta)$ and the equality $(\lambda)_{\alpha} = \Gamma(\alpha + \lambda\mathbf{1}) / [\Gamma(\lambda)]^n$ gives (39). ■

**Proof of Corollary 17.** We apply formula (40) of Theorem (16). We have $p_i = 1$, $i \in [n]$, $p_T = p^{|T|-1}$, $\tilde{p}_T = -p^{-|T|}$, for all $T \in \mathfrak{P}_n^*$. Then $\theta_{P_n} = (-p^{-1}, \dots, -p^{-1}) = -p^{-1}\mathbf{1}_n$. For $(\theta_{P_n} + \theta_{[n]})^{[n]} > qp^{-n}$ and $\theta_i > -p^{-1}$, $\forall i \in [n]$, we obtain $P_n(\theta_{[n]}) = p^{n-1} \prod_{i=1}^n (p^{-1} + \theta_i)[1 - qp^{-n} \prod_{i=1}^n (p^{-1} + \theta_i)^{-1}]$, and if $\mathbf{z} = (z_1, \dots, z_n)$, then $R_n(\mathbf{z}) = qp^{-n}\mathbf{z}^{[n]}$. Since $[1 - R_n(\mathbf{z})]^{-\lambda} = \sum_{l=0}^{\infty} \frac{(\lambda)_l}{l!} (qp^{-n})^l \mathbf{z}^{l1_n}$, we have $c_{l\mathbf{1}_n,\lambda}(R_n) = \frac{(\lambda)_l}{l!} (qp^{-n})^l$ if $l \in \mathbb{N}$, and $c_{\alpha,\lambda}(R_n) = 0$ if $\alpha \neq l\mathbf{1}_n$, $l \in \mathbb{N}$. Therefore (40) gives
$$
\begin{aligned}
\gamma_{(P,\lambda)}(\mathbf{d}\mathbf{x}) &= \frac{(p^{n-1})^{-\lambda}}{[\Gamma(\lambda)]^n} \exp\left(-\frac{x_1+\cdots+x_n}{p}\right) (\mathbf{x}^{[n]})^{(\lambda-1)} \left[\sum_{l=0}^{\infty} \frac{1}{[(\lambda)_l]^{n-1}l!} (qp^{-n}\mathbf{x}^{[n]})^l\right] \mathbf{1}_{(0,\infty)^n}(\mathbf{x})(\mathbf{d}\mathbf{x}) \\
&= \frac{(p^{n-1})^{-\lambda}}{[\Gamma(\lambda)]^n} \exp\left(-\frac{x_1+\cdots+x_n}{p}\right) (\mathbf{x}^{[n]})^{(\lambda-1)} F_{n-1}(\lambda, \dots, \lambda; qp^{-n}\mathbf{x}^{[n]}) \mathbf{1}_{(0,\infty)^n}(\mathbf{x})(\mathbf{d}\mathbf{x}), \text{ and Definition (7) of } F_{n-1} \text{ gives (20).} \quad \blacksquare
\end{aligned}
$$

**Proof of Corollary 18.** From (26), we have (41), and $[1 - R_2(\mathbf{z})]^{-\lambda} = \sum_{l=0}^{\infty} \frac{(\lambda)_l}{l!} \tilde{b}_{1,2}^l \mathbf{z}^{l/2}$. Hence, if $\alpha = l\mathbf{1}_2 = (l,l)$, $l \in \mathbb{N}$, $c_{\alpha,\lambda}(R_2) = \frac{(\lambda)_l}{l!} \tilde{b}_{1,2}^l$, and $c_{\alpha,\lambda}(R_2) = 0$ otherwise. Formula (40) therefore gives $\gamma_{(P_2,\lambda)}(\mathbf{d}\mathbf{x}) = \frac{p_{1,2}^{-\lambda}}{[\Gamma(\lambda)]^2} \exp(-\frac{p_2}{p_{1,2}}\mathbf{x}_1 - \frac{p_1}{p_{1,2}}\mathbf{x}_2) (x_1x_2)^{(\lambda-1)} (\sum_{l \in \mathbb{N}} \frac{\tilde{b}_{1,2x_1x_2}}{(\lambda)_l l!}) \mathbf{1}_{(0,\infty)^2}(\mathbf{x})(\mathbf{d}\mathbf{x})$, and definition (7) of $F_1$ gives (43). ■

**Proof of Corollary 19.** From (26) and (27), we have, $r_T = \tilde{b}_T$ for $T \in \mathfrak{P}_3$, $|T| = 2, 3$. This proves Formula (44).

Now, for $\mathbf{z}_3 = (z_1, z_2, z_3) \in \mathbb{R}^3$, and $\lambda > 0$, we develop $[1 - R_3(\mathbf{z})]^{-\lambda}$ by
$$[1 - R_3(\mathbf{z}_3)]^{-\lambda} = \sum_{l=(l_1,l_2,l_3,l_4) \in \mathbb{N}^4} \frac{(\lambda)_{l_1+l_2+l_3+l_4}}{l_1!l_2!l_3!l_4!} (\tilde{b}_{1,2}z_1z_2)^{l_1} (\tilde{b}_{1,3}z_1z_3)^{l_2} (\tilde{b}_{2,3}z_2z_3)^{l_3} (\tilde{b}_{1,2,3}z_1z_2z_3)^{l_4}.$$

Now, the conditions $k \in \mathbb{N}$, $l_1 + l_2 + l_3 + l_4 = k$; $\alpha_1 = l_1 + l_2 + l_4$, $\alpha_2 = l_1 + l_3 + l_4$, $\alpha_3 = l_2 + l_3 + l_4$ give $l_1 = k - \alpha_3 \ge 0$, $l_2 = k - \alpha_2 \ge 0$, $l_3 = k - \alpha_1 \ge 0$, $l_4 = \alpha_1 + \alpha_2 + \alpha_3 - 2k \ge 0$ and we get
$$[1 - R_3(\mathbf{z})]^{-\lambda} = \sum_{\alpha \in \mathbb{N}^3, \|\alpha\|_{\infty} \le \frac{|\alpha|}{2}} \left[ \sum_{\|\alpha\|_{\infty} \le k \le \frac{|\alpha|}{2}, k \in \mathbb{N}} \frac{(\lambda)_k \tilde{b}_{1,2}^{k-\alpha_3} \tilde{b}_{1,3}^{k-\alpha_2} \tilde{b}_{2,3}^{k-\alpha_1} \tilde{b}_{1,2,3}^{\alpha_1+\alpha_2+\alpha_3-2k}}{(k-\alpha_3)!(k-\alpha_2)!(k-\alpha_1)!(\alpha_1+\alpha_2+\alpha_3-2k)!} \right] \mathbf{z}_3^{\alpha} \text{ and we get}$$
(45). Formula (39) gives (46). According to (46), we have by $k \in \mathbb{N}$, $l_1 + l_2 + l_3 + l_4 = k$; $\alpha_1 = l_1 + l_2 + l_4$, $\alpha_2 = l_1 + l_3 + l_4$, $\alpha_3 = l_2 + l_3 + l_4$, $\gamma_{(P_3,\lambda)}(\mathbf{d}\mathbf{x}) =$
$$\frac{p_{[3]}^{-\lambda}}{[\Gamma(\lambda)]^3} \exp(\theta_P, \mathbf{x}) \mathbf{x}^{(\lambda-1)\mathbf{1}_3} {}_1\mathbf{F}_3(\lambda; \tilde{b}_{1,2}x_1x_2, \tilde{b}_{1,3}x_1x_3, \tilde{b}_{2,3}x_2x_3, \tilde{b}_{1,2,3}x_1x_2x_3) \mathbf{1}_{(0,\infty)^3}(\mathbf{x})(\mathbf{d}\mathbf{x}).$$

■

**Proof of Remark 20.** The only difference with (46) is that all $c_{\alpha,\lambda}(R_3)$ defined by (49) for $\|\alpha\|_{\infty} \le \frac{|\alpha|}{2}$ are positive and the formula (40) gives (50). ■
**Proof of Remark 21.** As $\tilde{b}_{1,2,3} = 0$, $\tilde{b}_{1,2,3}^{\alpha_1+\alpha_2+\alpha_3-2k} = 1$ only for $|\frac{\alpha}{2}| = k \in \mathbb{N}$. In this case, if $\|\boldsymbol{\alpha}\|_{\infty} \le \frac{|\alpha|}{2} = k \in \mathbb{N}$, then $c_{\alpha,\lambda}(R_3) = \frac{(\lambda)_k \tilde{b}_{2,3}^{k-\alpha_1} \tilde{b}_{1,3}^{k-\alpha_2} \tilde{b}_{1,2}^{k-\alpha_3}}{(k-\alpha_1)!(k-\alpha_2)!(k-\alpha_3)!} > 0$, gives (51) and (40) gives (52). ■

**Proof of Remark 22.** If $\tilde{b}_{1,2}, \tilde{b}_{1,3}, \tilde{b}_{2,3} = 0, \tilde{b}_{1,2,3} > 0$, then

$c_{\alpha,\lambda}(R_3) = \sum_{\|\boldsymbol{\alpha}\|_{\infty} \le k \le \frac{|\alpha|}{2}, k \in \mathbb{N}} \frac{(\lambda)_k \tilde{b}_{2,3}^{k-\alpha_1} \tilde{b}_{1,3}^{k-\alpha_2} \tilde{b}_{1,2}^{k-\alpha_3} \tilde{b}_{1,2,3}^{\alpha_1+\alpha_2+\alpha_3-2k}}{(k-\alpha_1)!(k-\alpha_2)!(k-\alpha_3)!(\alpha_1+\alpha_2+\alpha_3-2k)!} \ne 0$ only for $\alpha = k\mathbf{1}_3, k \in \mathbb{N}$, and we obtain (53), and $c_{\alpha,\lambda}(R_3) = 0$ otherwise. Thus we have

$\gamma_{(P_3,\lambda)}(\mathrm{d}\mathbf{x}) = \frac{p_{[3]}^{-\lambda}}{[\Gamma(\lambda)]^3} \exp(\boldsymbol{\theta}_P, \mathbf{x}) \mathbf{x}^{(\lambda-1)} \mathbf{1}_3 \left[ \sum_{\alpha=k\mathbf{1}_3, k \in \mathbb{N}} \frac{(\lambda)_k \tilde{b}_{1,2,3}^{k}}{k!} \frac{(\mathbf{x}^{[3]})^k}{[(\lambda)_k]^3} \mathbf{1}_{(0,\infty)^3}(\mathbf{x})(\mathrm{d}\mathbf{x}) \right]$ and (7) gives (54). ■

**Proof of Theorem 23.** Let $\mathbf{1}_n = (1, \dots, 1) \in \mathbb{R}^n$, we denote by $\mathbf{1}_k = (1, \dots, 1) \in \mathbb{R}^k$ and $\mathbf{1}_{n-k} = (1, \dots, 1) \in \mathbb{R}^{n-k}$. We note that $\mathbf{X}_{[k]}$ is a random real vector such that $\mathbf{X}_{[k]} \sim \gamma_{(P_k,\lambda)}$, with $P_k(\boldsymbol{\theta}_{[k]}) = \sum_{T \in \mathfrak{P}_k} p_T \boldsymbol{\theta}_{[k]}^T = \sum_{T \in \mathfrak{P}_k} p_T \boldsymbol{\theta}_{[k]}^T$. Using (40), the Lt $L_{\mathbf{X}_{[n]\setminus[k]}}^{\mathbf{X}_{[k]}=\mathbf{X}_{[k]}}$ is given by

$$L_{\mathbf{X}_{[n]\setminus[k]}}^{\mathbf{X}_{[k]}=\mathbf{X}_{[k]}} (\boldsymbol{\theta}_{[n]\setminus[k]}) = \frac{\left[ \frac{p_{[n]}}{p_{[k]}} \right]^{-\lambda} \exp\left( (\boldsymbol{\theta}_{P_n})_{[k]} - \boldsymbol{\theta}_{P_k}, \mathbf{x}_{[k]} \right)}{\sum_{\alpha_{[k]} \in \mathbb{N}^k, c_{\alpha_{[k]}} \lambda(R_k) \neq 0} \frac{c_{\alpha_{[k]}} \lambda(R_k)}{(\lambda)_{\alpha_{[k]}}} \mathbf{x}_{[k]}^{\alpha_{[k]}}} \times \\ \sum_{\alpha = (\alpha_{[k]}, \alpha_{[n]\setminus[k]}) \in \mathbb{N}^n, c_{\alpha,\lambda}(R_n) \neq 0} \frac{\mathbf{x}_{[k]}^{\alpha_{[k]}}}{(\lambda)_{\alpha_{[k]}}} c_{\alpha,\lambda}(R_n) (\boldsymbol{\theta}_{[n]\setminus[k]} - (\boldsymbol{\theta}_{P_n})_{[n]\setminus[k]})^{-(\alpha_{[n]\setminus[k]} + \lambda \mathbf{1}_{n-k})}.$$

We obtain

$$L_{\mathbf{X}_{[n]\setminus[k]}}^{\mathbf{X}_{[k]}=\mathbf{X}_{[k]}} (\boldsymbol{\theta}_{[n]\setminus[k]}) = \frac{\exp\left( (\boldsymbol{\theta}_{P_n})_{[k]} - \boldsymbol{\theta}_{P_k}, \mathbf{x}_{[k]} \right)}{\sum_{\alpha_{[k]} \in \mathbb{N}^k, c_{\alpha_{[k]}} \lambda(R_k) \neq 0} \frac{c_{\alpha_{[k]}} \lambda(R_k)}{(\lambda)_{\alpha_{[k]}}} \mathbf{x}_{[k]}^{\alpha_{[k]}}} \left( \frac{p_{[n]}}{p_{[k]}} (\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})_{[n]\setminus[k]} \right)^{-\lambda} \times \\ \sum_{\alpha = (\alpha_{[k]}, \alpha_{[n]\setminus[k]}) \in \mathbb{N}^n, c_{\alpha,\lambda}(R_n) \neq 0} \frac{\mathbf{x}_{[k]}^{\alpha_{[k]}}}{(\lambda)_{\alpha_{[k]}}} c_{\alpha,\lambda}(R_n) (\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})_{[n]\setminus[k]}^{-\alpha_{[n]\setminus[k]}}. \quad (95)$$

Since $L_{\mathbf{X}_{[n]\setminus[k]}}^{\mathbf{X}_{[k]}=\mathbf{X}_{[k]}}$ is the Lt of a pd we have $L_{\mathbf{X}_{n-k}}^{\mathbf{X}_{k}=\mathbf{X}_{k}}(\mathbf{0}_{n-k}) = 1$ and we get

$$\frac{\exp\left( (\boldsymbol{\theta}_{P_n})_{[k]} - \boldsymbol{\theta}_{P_k}, \mathbf{x}_{[k]} \right)}{\sum_{\alpha_{[k]} \in \mathbb{N}^k, c_{\alpha_{[k]}} \lambda(R_k) \neq 0} \frac{c_{\alpha_{[k]}} \lambda(R_k)}{(\lambda)_{\alpha_{[k]}}} \mathbf{x}_{[k]}^{\alpha_{[k]}}} = \left( \frac{p_{[n]}}{p_{[k]}} (-\boldsymbol{\theta}_{P_n})_{[n]\setminus[k]} \right)^{\lambda} \left[ \sum_{\alpha = (\alpha_{[k]}, \alpha_{[n]\setminus[k]}) \in \mathbb{N}^n, c_{\alpha,\lambda}(R_n) \neq 0} \frac{\mathbf{x}_{[k]}^{\alpha_{[k]}}}{(\lambda)_{\alpha_{[k]}}} c_{\alpha,\lambda}(R_n) (-\boldsymbol{\theta}_{P_n})_{[n]\setminus[k]}^{-\alpha_{[n]\setminus[k]}} \right]^{-1}. \quad (96)$$

We carry (96) in (95) and we obtain

$$L_{\mathbf{X}_{[n]\setminus[k]}}^{\mathbf{X}_{[k]}=\mathbf{X}_{[k]}} (\boldsymbol{\theta}_{[n]\setminus[k]}) = \left[ \frac{(\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})_{[n]\setminus[k]}^{1_{n-k}}}{(-\boldsymbol{\theta}_{P_n})_{[n]\setminus[k]}^{1_{n-k}}} \right]^{-\lambda} \times \\ \frac{\sum_{\alpha = (\alpha_{[k]}, \alpha_{[n]\setminus[k]}) \in \mathbb{N}^n, c_{\alpha,\lambda}(R_n) \neq 0} \frac{\mathbf{x}_{[k]}^{\alpha_{[k]}}}{(\lambda)_{\alpha_{[k]}}} c_{\alpha,\lambda}(R_n) (\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})_{[n]\setminus[k]}^{-\alpha_{[n]\setminus[k]}}}{\sum_{\alpha = (\alpha_{[k]}, \alpha_{[n]\setminus[k]}) \in \mathbb{N}^n, c_{\alpha,\lambda}(R_n) \neq 0} \frac{\mathbf{x}_{[k]}^{\alpha_{[k]}}}{(\lambda)_{\alpha_{[k]}}} c_{\alpha,\lambda}(R_n) (-\boldsymbol{\theta}_{P_n})_{[n]\setminus[k]}^{-\alpha_{[n]\setminus[k]}}} \quad (97)$$

As we have $(\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})_{[n]\setminus[k]}^{1_{n-k}} (-\boldsymbol{\theta}_{P_n})_{[n]\setminus[k]}^{-1_{n-k}} = \prod_{i=k+1}^n [1 + \theta_i (-\boldsymbol{\theta}_{P_n})_i^{-1}]$, (97) gives (56). ■
**Proof of Remark 25.** We have $\tilde{p}_T(S_{[n]\setminus[1]}) = \frac{p_{\{1\}\cup T}}{p_1}$ and $\tilde{p}_T(S_{[n]\setminus[1]}) = -p_{([n]\setminus 1)\setminus T}(S_{[n]\setminus[1]})/\frac{p_{[n]}}{p_1} = -\frac{p_{\{1\}\cup([n]\setminus 1)\setminus T}}{p_1}/\frac{p_{[n]}}{p_1} = -\frac{p_{[n]\setminus T}}{p_1} = \tilde{p}_T(R_n)$. We also have
$$r_T(S_{[n]\setminus[1]}) = -\frac{1}{\frac{p_{[n]}}{p_1}}\left(\frac{\partial}{\partial\theta}\right)^{([n]\setminus\{1\})\setminus T}\left[\frac{1}{p_1}\frac{\partial}{\partial\theta_1}P_n(\theta_{P_n})\right] = -\frac{1}{p_{[n]}}\left(\frac{\partial}{\partial\theta}\right)^{[n]\setminus T}P_n(\theta_{P_n}) = r_T(P_n).$$

<figure><img src="image_15.png" /></figure>

**Proof of Theorem 24.** We calculate, in five steps, another expression of (57) for $k=1$ to obtain a new formula for $L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}(\theta_{[n]\setminus[1]})$ from (56).

**First step:** For $\alpha_1 \in I_1(R_n)$, let $J_{[n]\setminus[1]}(\alpha_1) = \{\alpha_{[n]\setminus[1]} \in \mathbb{N}^{n-1}, c_{\alpha_1, \alpha_{[n]\setminus[1]}, \lambda}(R_n) \neq 0\}$, then we have by (57)

$$F_1(\lambda, R_n, x_1, \theta_{[n]\setminus[1]}) = \sum_{\alpha_1 \in I_1(R_n)} \frac{x_1^{\alpha_1}}{(\lambda)^{\alpha_1}} \sum_{\alpha_{[n]\setminus[1]} \in J_{[n]\setminus[1]}(\alpha_1)} c_{\alpha_1, \alpha_{[n]\setminus[1]}, \lambda}(R_n) (\theta_{[n]} - \theta_{P_n})_{[n]\setminus[1]}^{-\alpha_{[n]\setminus[1]}}. \quad (98)$$

**Second step:** For $\alpha \in \mathbb{N}$ and $j \in \mathbb{N}$, we denote by $\langle\alpha\rangle_j$ the number $\alpha(\alpha-1)\cdots(\alpha-j+1)$, in particular $\langle\alpha\rangle_\alpha = \alpha!$ and if $j > \alpha$, $\langle\alpha\rangle_j = 0$. For $\alpha_{[k]} \in \mathbb{N}^{[k]}$, and $\alpha_i \ge \beta_i \in \mathbb{N}$, we denote by $\langle\alpha_{[k]}\rangle_{\beta_{[k]}}$ the number $\langle\alpha_1\rangle_{\beta_1} \cdots \langle\alpha_k\rangle_{\beta_k}$, in particular $\langle\alpha_{[k]}\rangle_{\alpha_{[k]}} = \alpha_{[k]}!$ and if $\exists j \in [k]$, $\beta_j > \alpha_j$, then $\langle\alpha_{[k]}\rangle_{\beta_{[k]}} = 0$. Let $\beta_1 \in \mathbb{N}$, we apply $(\partial/\partial z_1)^{\beta_1}$ to (38) with $k=1$, because $R_n$ is an affine polynomial, we obtain by [16] p. 42

$$\begin{aligned} \left(\frac{\partial}{\partial z_1}\right)^{\beta_1} [1 - R(\mathbf{z})]^{-\lambda} &= \left(\frac{\partial}{\partial z_1}\right)^{\beta_1} \sum_{\alpha \in \mathbb{N}^n} c_{\alpha, \lambda}(R) \mathbf{z}^{\alpha}(\lambda)_{\beta_1} \left(\frac{\partial}{\partial z_1} R_n(\mathbf{z})\right)^{\beta_1} [1 - R_n(\mathbf{z})]^{-(\lambda + \beta_1)} \\ &= \sum_{\alpha_1 \in I_1(R_n)} \sum_{\alpha_{[n]\setminus[1]} \in J_{[n]\setminus[1]}(\alpha_1), \alpha_1 \ge \beta_1} c_{\alpha_1, \alpha_{[n]\setminus[1]}, \lambda}(R_n) \langle\alpha_1\rangle_{\beta_1} z_1^{\alpha_1 - \beta_1} \mathbf{z}_{[n]\setminus[1]}^{\alpha_{[n]\setminus[1]}}. \end{aligned} \quad (99)$$

**Third step:** Making $z_1 = 0$ in (99), we get

$$(\lambda)_{\beta_1} [1 - R_n(0, \mathbf{z}_{[n]\setminus[1]})]^{-(\lambda + \beta_1)} \left(\frac{\partial}{\partial z_1} R_n(0, \mathbf{z}_{[n]\setminus[1]})\right)^{\beta_1} = \sum_{\alpha_{[n]\setminus[1]} \in J_{[n]\setminus[1]}(\beta_1)} c_{\beta_1, \alpha_{[n]\setminus[1]}, \lambda}(R_n) (\beta_1!) \mathbf{z}_{[n]\setminus[1]}^{\alpha_{[n]\setminus[1]}}.$$

**Fourth step:** Making $\beta_1 = \alpha_1$ and $\mathbf{z}_{[n]\setminus[1]} = (\theta_{[n]} - \theta_{P_n})_{[n]\setminus[1]}^{-1}$ in last Equality, we get

$$\sum_{\alpha_{[n]\setminus[1]} \in J_{[n]\setminus[1]}(\alpha_1)} c_{\alpha_1, \alpha_{[n]\setminus[1]}, \lambda}(R_n) (\theta_{[n]} - \theta_{P_n})_{[n]\setminus[1]}^{-\alpha_{[n]\setminus[1]}} = \frac{(\lambda)_{\alpha_1}}{\alpha_1!} [1 - R_n(0, \mathbf{z}_{[n]\setminus[1]})]^{-(\lambda + \alpha_1)} \left(\frac{\partial}{\partial z_1} R_n(0, (\theta_{[n]} - \theta_{P_n})_{[n]\setminus[1]}^{-1})\right)^{\alpha_1} \quad (100)$$

**Fifth step:** Using $\mathfrak{z}$ and $S$ in (100), we get

$$\sum_{\alpha_{[n]\setminus[1]} \in J_{[n]\setminus[1]}(\alpha_1)} c_{\alpha_1, \alpha_{[n]\setminus[1]}, \lambda}(R_n) (\theta_{[n]} - \theta_{P_n})_{[n]\setminus[1]}^{-\alpha_{[n]\setminus[1]}} = \left[ \frac{(\theta_{[n]} - \theta_{P_n})_{[n]\setminus[1]}^{-1}}{\frac{p_{[n]}}{p_1}} S(\theta_{[n]\setminus[1]}) \right]^{-\lambda} \frac{(\lambda)_{\alpha_1}}{\alpha_1!} [\mathfrak{z}(\theta_{[n]\setminus[1]})]^{\alpha_1}, \text{ hence last Equality gives}$$

$$F_1(\lambda, R_n, x_1, \theta_{[n]\setminus[1]}) = \left[ \frac{(\theta_{[n]} - \theta_{P_n})_{[n]\setminus[1]}^{-1}}{\frac{p_{[n]}}{p_1}} S(\theta_{[n]\setminus[1]}) \right]^{-\lambda} G(R_n, \mathfrak{z}(\theta_{[n]\setminus[1]}) x_1) \quad (101)$$
Substituting (101) in (56) for $k=1$ and because $S(\mathbf{0}_{[n]\setminus[1]}) = 1$ from formula (60) and (55), we get

$$L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}(\boldsymbol{\theta}_{[n]\setminus[1]}) = [Q_{[n]\setminus[1]}(\boldsymbol{\theta}_{[n]\setminus[1]})]^{-\lambda} \frac{\left[ \frac{(\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})_{[n]\setminus[1]}^{-1}}{\frac{p_{[n]}}{p_1}} S(\boldsymbol{\theta}_{[n]\setminus[1]}) \right]^{-\lambda} \mathbf{G}(R_n, \mathfrak{z}(\boldsymbol{\theta}_{[n]\setminus[1]}) x_1)}{\left[ \frac{(-\boldsymbol{\theta}_{P_n})_{[n]\setminus[1]}^{-1}}{\frac{p_{[n]}}{p_1}} S(\mathbf{0}_{[n]\setminus[1]}) \right]^{-\lambda} \mathbf{G}(R_n, \mathfrak{z}(\mathbf{0}_{[n]\setminus[1]}) x_1)} \quad (102)$$

Finally (102) and (55) gives (66). Formulas (60), (61) and (62) remain to be proven.

First, according to (24), we have $\frac{P_n(\boldsymbol{\theta}_{[n]})}{p_{[n]}(\boldsymbol{\theta}_1 - \tilde{p}_1)} = \{1 - R_n[(\boldsymbol{\theta} - \boldsymbol{\theta}_P)^{-1}]\} (\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})^{\frac{1}{n-1}}_{[n]\setminus[1]}$. We make $\theta_1 \to \infty$ in last Equality and we get

$$\lim_{\theta_1 \to \infty} \frac{P_n(\boldsymbol{\theta}_{[n]})}{p_{[n]}(\boldsymbol{\theta}_1 - \tilde{p}_1)} = \left[1 - R_n(\mathbf{0}, (\boldsymbol{\theta} - \boldsymbol{\theta}_P)^{-1}_{[n]\setminus[1]})\right] (\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})^{\frac{1}{n-1}}_{[n]\setminus[1]} \quad (103)$$

and, because $P$ is an affine polynomial, we have

$$\lim_{\theta_1 \to \infty} \frac{P_n(\boldsymbol{\theta}_{[n]})}{p_{[n]}(\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})_1} = \lim_{\theta_1 \to \infty} \frac{P(0, \boldsymbol{\theta}_{[n]\setminus[1]} + \theta_1 \frac{\partial}{\partial \theta_1} (0, \boldsymbol{\theta}_{[n]\setminus[1]})}{p_{[n]}(\boldsymbol{\theta}_1 - \tilde{p}_1)} = \frac{\frac{\partial}{\partial \theta_1} P_n(\boldsymbol{\theta}_{[n]})}{p_{[n]}}. \text{ Finally last Equality and} \\ (103) \text{ prove (60).}$$

Secondly, according to (24), with $\mathbf{z} = (\boldsymbol{\theta}_{[n]} - \boldsymbol{\theta}_{P_n})^{-1}$, that is $z_i = (\theta_i - \tilde{p}_i)^{-1}$ or $\theta_i = \tilde{p}_i + z_i^{-1}$, $i \in [n]$, we have $R_n(\mathbf{z}_{[n]}) = 1 - \frac{1}{p_{[n]}}(\prod_{j=2}^n z_j) z_1 P_n(\tilde{p}_1 + z_1^{-1}, \dots, \tilde{p}_n + z_n^{-1})$. Deriving with respect to the variable $z_1$, we obtain $\frac{\partial}{\partial z_1} R_n(z_1, z_2, \dots, z_n) = -\frac{1}{p_{[n]}}(\prod_{j=2}^n z_j) \{P_n(\tilde{p}_1 + z_1^{-1}, \dots, \tilde{p}_n + z_n^{-1}) - z_1^{-1} \frac{\partial}{\partial \theta_1} [P_n(\tilde{p}_1 + z_1^{-1}, \dots, \tilde{p}_n + z_n^{-1})]\}$.

Because $R_n$ is an affine polynomial with respect to the $n$ variables $z_1, z_2, \dots, z_n$, we know that $\frac{\partial}{\partial z_1} R_n(z_1, z_2, \dots, z_n)$ is an affine polynomial with respect to the $n-1$ variables $z_2, \dots, z_n$. Putting $z_1 = 0$ in the left-hand side of last Equality and making $z_1 \to \infty$ in the right-hand side of last Equality, we get

$$\begin{aligned} \frac{\partial}{\partial z_1} R_n (0, z_2, \dots, z_n) &= -\frac{1}{p_{[n]}} \left( \prod_{j=2}^n z_j \right) [P_n (\tilde{p}_1, \tilde{p}_2 + z_2^{-1}, \dots, \tilde{p}_n + z_n^{-1})] \text{ and} \\ \left( \prod_{j=2}^n z_j^{-1} \right) \frac{\partial}{\partial z_1} R_n (0, z_2, \dots, z_n) &= -\frac{1}{p_{[n]}} [P_n (\tilde{p}_1, \tilde{p}_2 + z_2^{-1}, \dots, \tilde{p}_n + z_n^{-1})] \text{. Substituting } z_j = (\theta_j - \tilde{p}_j)^{-1}, \\ & \quad j = 2, \dots, n, \text{ we obtain} \\ \prod_{j=2}^n (\theta_j - \tilde{p}_j) \frac{\partial}{\partial z_1} R_n (0, (\theta_2 - \tilde{p}_2)^{-1}, \dots, (\theta_n - \tilde{p}_n)^{-1}) &= -\frac{1}{p_{[n]}} [P_n (\tilde{p}_1, \theta_2, \dots, \theta_n)] \text{. Multiplying by } p_{[n]}/p_{[1]} \\ & \quad \text{we obtain (61).} \end{aligned}$$

Thirdly, applying $\frac{1}{p_{[1]}} \frac{\partial}{\partial \theta_1}$ to the equality $P_n(\boldsymbol{\theta}) = \sum_{T \subset (2, \dots, n)} p_{[1] \cup T} \boldsymbol{\theta}^{[1]} \boldsymbol{\theta}_{[n]\setminus[1]}^T + \sum_{T \in \mathfrak{P}_n, T \cap [1] \neq [1]} p_T \boldsymbol{\theta}^T$ we obtain (62). ■

**Proof of Lemma 26.** Firstly. If $\exists T \in \mathfrak{P}_n, |T| = 2, r_T \neq 0$, then without loss of generality, we can studies the case $T = [2]$, that is $r_{1,2} \neq 0$. For $\mathbf{z} = (z_1, z_2, \mathbf{0}_{n-2})$, we get $[1 - R_n(z_1, z_2, \mathbf{0}_{n-2})]^{-\lambda} = \sum_{\alpha \in \mathbb{N}^2} c_{(\alpha_1, \alpha_2, \mathbf{0}_{n-2}), \lambda}(R_n) z_1^{\alpha_1} z_2^{\alpha_2} = (1 - r_{1,2} z_1 z_2)^{-\lambda} = \sum_{l=0}^{\infty} \frac{(\lambda)_l r_{1,2}^l}{l!} z_1^l z_2^l$. Then we have $c_{(l,l,\mathbf{0}_{n-2}), \lambda} = \frac{(\lambda)_l r_{1,2}^l}{l!} \neq 0, l \in \mathbb{N}$. Finally, $\forall \alpha_1 \in \mathbb{N}, \exists \alpha_{[n]\setminus\{1\}} = (\alpha_1, \mathbf{0}_{n-2})$ such that $c_{(\alpha_1, \alpha_{[n]\setminus\{1\}}), \lambda} = \frac{(\lambda)_{\alpha_1} r_{1,2}^{\alpha_1}}{\alpha_1!} \neq 0$, and $I_1(R_n) = \mathbb{N}$.

Secondly. Let $k \in \mathbb{N}$, such that $\forall T \in \mathfrak{P}_n, |T| \leq k < n; r_T = 0$, if $\exists T \in \mathfrak{P}_n, |T| = k+1; r_T \neq 0$, then without loss of generality, we can studies the case $T = [k+1]$, that is $r_{1,...,k+1} \neq 0$.
For $\mathbf{z} = (z_1, \dots, z_{k+1}, \mathbf{0}_{n-k-1})$, we get $[1 - R_n(z_1, \dots, z_{k+1}, \mathbf{0}_{n-k-1})]^{-\lambda} = \sum_{\alpha \in \mathbb{N}^2} c_{(\alpha_1, \dots, \alpha_{k+1}, \mathbf{0}_{n-k-1}), \lambda}(R_n) z_1^{\alpha_1} \dots z_{k+1}^{\alpha_{k+1}} = (1 - r_1, \dots, r_{k+1} z_1 \dots z_{k+1})^{-\lambda} = \sum_{l=0}^{\infty} \frac{(\lambda)_l r_1^l \dots r_{k+1}^l}{l!} z_1^l \dots z_{k+1}^l$.
Then we have $c_{(\mathbf{l}_{k+1}, \mathbf{0}_{n-k+1}), \lambda} = \frac{(\lambda)_l r_1^l \dots r_{k+1}^l}{l!} \neq 0, l \in \mathbb{N}$. Finally, $\forall \alpha_1 \in \mathbb{N}, \exists \alpha_{[n]\setminus\{1\}} = (\alpha_1 \mathbf{1}_k, \mathbf{0}_{n-k-1})$ such that $c_{(\alpha_1, \alpha_{[n]\setminus\{1\}}), \lambda} = \frac{(\lambda)_l r_1^l \dots r_{k+1}^l}{l!} \neq 0$, and $I_1(R_n) = \mathbb{N}$.

**Thirdly.** Now, if $\forall T \in \mathfrak{P}_n, |T| \le n; r_T = 0$, then $R_n = 0$, and $P_n(\theta_n) = p_{[n]} \prod_{i=1}^n (\theta_i - \tilde{p}_i) = \prod_{i=1}^n (1 + (-\tilde{p}_i)^{-1} \theta_i)$, because $P_n(\mathbf{0}_n) = 1$. In this case we have $\mathbf{X} = (X_1, \dots, X_n)$, with $X_i \sim \gamma_{(-\tilde{p}_i)^{-1}, \lambda}$, $X_i, i \in [n]$ being independent. Therefore, $c_{(\alpha_1, \dots, \alpha_n), \lambda}(P_n) = 0$ unless $c_{(\mathbf{0}_n), \lambda}(P_n) = 1$, and $I_1(R_n) = \{0\}$. This completes the proof by finite induction. $\blacksquare$

**Proof of Theorem 27.** We have by (60), (61) and (63): $B(\theta_{[n]\setminus[1]}) = -\frac{1}{p_1}[P_n(0, \theta_{[n]\setminus[1]}) + \tilde{p}_1 \frac{\partial P_n}{\partial \theta_1}(0, \theta_{[n]\setminus[1]})] = -\tilde{p}_1 S_{[n]\setminus[1]}(\theta_{[n]\setminus[1]}) - \frac{1}{p_1} P_n(0, \theta_{[n]\setminus[1]})$. We deduce $\mathfrak{I}_{n-1}(\theta_{[n]\setminus[1]}) = -\tilde{p}_1 - \frac{1}{p_1} \frac{P_n(0, \theta_{[n]\setminus[1]})}{S_{[n]\setminus[1]}(\theta_{[n]\setminus[1]})}$. Then, by (66), we get

$$L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}(\theta_{[n]\setminus[1]}) = [S_{[n]\setminus[1]}(\theta_{[n]\setminus[1]})]^{-\lambda} \frac{\exp\{[-\tilde{p}_1 - \frac{1}{p_1} \frac{P_n(0, \theta_{[n]\setminus[1]})}{S_{[n]\setminus[1]}(\theta_{[n]\setminus[1]})}]|x_1\}}{\exp\{[-\tilde{p}_1 - \frac{1}{p_1}]|x_1\}}$$, which gives (69). By (21) and (68), expanding $P_n(0, \theta_{[n]\setminus[1]}) - S_{[n]\setminus[1]}(\theta_{[n]\setminus[1]})$, we obtain (70). By (21) and (68) we get
$$-(\frac{P_n(0, \theta_{[n]\setminus[1]})}{S_{[n]\setminus[1]}(\theta_{[n]\setminus[1]})} - 1) \frac{x_1}{p_1} = (\{\tilde{p}_1 - \tilde{p}_1 \sum_{T \subset [n]\setminus\{1\}, |T|>1} r_T[(\theta_{[n]} - \theta_{P_n})_{[n]\setminus\{1\}}]^{-T} + \sum_{T \subset [n]\setminus\{1\}, |T|>1} r_{\{1\} \cup T}[(\theta_{[n]} - \theta_{P_n})_{[n]\setminus\{1\}}]^{-T} + \frac{1}{p_1} - \frac{1}{p_1} \sum_{T \subset [n]\setminus\{1\}, |T|>1} r_T[(\theta_{[n]} - \theta_{P_n})_{[n]\setminus\{1\}}]^{-T})] p_{[n]}(\theta_{[n]} - \theta_{P_n})^{[n]\setminus\{1\}} \frac{x_1}{p_1} S_{[n]\setminus[1]}^{-1}(\theta_2, \dots, \theta_n),$$
then we have
$$-\left(\frac{P_n(0, \theta_2, \dots, \theta_n)}{S_{[n]\setminus[1]}(\theta_2, \dots, \theta_n)} - 1\right) \frac{x_1}{p_1} = \left\{ \begin{aligned} & (\tilde{p}_1 + \frac{1}{p_1})(\theta_{[n]} - \theta_{P_n})^{[n]\setminus\{1\}} + \sum_{k=2}^n r_{\{1,k\}} (\theta_{[n]} - \theta_{P_n})^{[n]\setminus\{1,k\}} \\ & + \sum_{T \subset [n]\setminus\{1\}, |T|>1} [r_{\{1\} \cup T} - (\tilde{p}_1 + \frac{1}{p_1}) r_T](\theta_{[n]} - \theta_{P_n})^{[n]\setminus\{1\} \cup T} \end{aligned} \right\} \frac{p_{[n]}}{p_1} x_1 S_{[n]\setminus[1]}^{-1}(\theta_2, \dots, \theta_n) \text{ and}$$
$$\text{because}$$

$$S_{[n]\setminus[1]}(\theta_2, \dots, \theta_n) = \frac{p_{[n]}}{p_1} (\theta_{[n]} - \theta_{P_n})^{[n]\setminus\{1\}} \{1 - R_{n-1}(S_{[n]\setminus[1]}^{-1})[(\theta_{[n]} - \theta_{P_n})_{[n]\setminus\{1\}}]^{-1}\} \\ = \frac{p_{[n]}}{p_1} \{(\theta_{[n]} - \theta_{P_n})^{[n]\setminus\{1\}} - \sum_{T \subset [n]\setminus\{1\}, |T|>1} r_T (\theta_{[n]} - \theta_{P_n})^{[n]\setminus\{1\} \cup T}\},$$

we get

$$(\theta_{[n]} - \theta_{P_n})^{[n]\setminus\{1\}} = \frac{p_1}{p_{[n]}} (-\tilde{p}_2)^{-1} \dots (-\tilde{p}_n)^{-1} + \sum_{T \subset [n]\setminus\{1\}, |T|>1} r_T (-\theta_{P_n})^{-T} (1 + (-\tilde{p}_1)^{-1} \theta_1, \dots, 1 + (-\tilde{p}_n)^{-1} \theta_n)^{\{[n]\setminus\{1\}\setminus T}} S_{[n]\setminus[1]}^{-1}(\theta_2, \dots, \theta_n). \quad (104)$$

Now, we have $-\left(\frac{P_n(0, \theta_2, \dots, \theta_n)}{S_{[n]\setminus[1]}(\theta_2, \dots, \theta_n)} - 1\right) \frac{x_1}{p_1} = x_1 (\tilde{p}_1 + \frac{1}{p_1}) + \sum_{T \subset [n]\setminus\{1\}, |T|>0} r_{\{1\} \cup T} \frac{p_{[n]}}{p_1} x_1 (\theta_{[n]} - \theta_{P_n})^{[n]\setminus\{1\} \cup T} S_{[n]\setminus[1]}^{-1}(\theta_2, \dots, \theta_n)$ and
$$L_{(X_2, \dots, X_n)}^{X_1=x_1}(\theta_2, \dots, \theta_n) = [S_{[n]\setminus[1]}(\theta_2, \dots, \theta_n)]^{-\lambda} \times \exp\{x_1 (\tilde{p}_1 + \frac{1}{p_1}) + \sum_{T \subset [n]\setminus\{1\}, |T|>0} r_{\{1\} \cup T} \frac{p_{[n]}}{p_1} x_1 (\theta_{[n]} - \theta_{P_n})^{[n]\setminus\{1\} \cup T} S_{[n]\setminus[1]}^{-1}(\theta_2, \dots, \theta_n)\}.$$

$$\text{Because } 1 = L_{(X_2, \dots, X_n)}^{X_1=x_1}(0, \dots, 0) =$$
$$
\begin{align*}
& \exp\left\{\left(\tilde{p}_1 + \frac{1}{p_1}\right) x_1\right\} \exp\left\{\sum_{T \subset [n] \setminus \{1\}, |T| > 0} r\{1\} \cup T \frac{p_{[n]}}{p_1} x_1 ( -\tilde{p}_2, \dots, -\tilde{p}_n )^{\{[n] \setminus \{1\}\setminus T}\} \right\}, \text{ we have} \\
& L_{(X_2, \dots, X_n)}^{X_1=x_1}(\theta_{[n]\setminus[1]}) = S_{[n]\setminus[1]}^{-\lambda}(\theta_{[n]\setminus[1]}) \times \\
& \mathbf{e}^{\left\{x_1(\tilde{p}_1 + \frac{1}{p_1}) + \sum_{T \subset [n] \setminus \{1\}, |T| > 0} r\{1\} \cup T \frac{p_{[n]}}{p_1} x_1 (\theta_{[n]} - \theta_{P_n})^{[n] \setminus \{1\} \setminus T} S_{[n]\setminus[1]}^{-1}(\theta_2, \dots, \theta_n)\right\}} \\
& = S_{[n]\setminus[1]}^{-\lambda}(\theta_{[n]\setminus[1]}) \times \\
& \mathbf{e}^{\left\{\sum_{T \subset [n] \setminus \{1\}, 0 < |T|} r\{1\} \cup T \frac{p_{[n]}}{p_1} x_1 (-\theta_{P_n})^{[n] \setminus \{1\} \setminus T} [(\mathbf{1}_n + (-\theta_{P_n})^{-1}\theta_{[n]})^{[n] \setminus \{1\} \setminus T} S_{[n]\setminus[1]}^{-1}(\theta_{[n]\setminus[1]}) - 1]\right\}}.
\end{align*}
$$

Unless $Rn = 0$, for $T \in \mathfrak{P}([n] \setminus [1])$, $\bar{T} = [n] \setminus T$, and if there is no ambiguity, for simplicity we denote
$S_T(\theta_T)$ by $S_T$, the polynomial defined by $S_T(\theta_T) = \frac{1}{p_T} (\frac{\partial}{\partial \theta})^{\bar{T}} (P_n(\theta)) = \sum_{T' \in P(T)} \frac{p_{\bar{T} \cup T'}}{p_{\bar{T}}} \theta^{T'}$

We have $S_T(\theta_T) = \sum_{T' \in \mathfrak{P}(T)} \frac{p_{\bar{T} \cup T'}}{p_{\bar{T}}} \theta^{T'} = \sum_{T' \in P(T)} q_{T'} \theta^{T'}$, with $q_{T'} = \frac{p_{\bar{T} \cup T'}}{p_{\bar{T}}}$, and for $T' \in \mathfrak{P}(T)$, we have $\tilde{q}_{T'} = - \frac{q_{T' \setminus T'}}{q_T} = \frac{p_{\bar{T} \setminus T'}}{p_{\bar{T}}} = \tilde{p}_{T'}$. Therefore, we have $\tilde{b}_{T'}(S_T) = \tilde{b}_{T'}$, and if $\gamma_{(P,\lambda)}$ is an infinitely divisible gamma distribution, $\gamma_{(S_T,\lambda)}$ is also an infinitely divisible gamma distribution.

<figure><img src="image_16.png" /></figure>

**Proof of Corollary 28.** From Proof of Corollary (17), we have $c_{\alpha,\lambda}(R_n) = 0$ unless $\alpha = k\mathbf{1}_n$, $k \in \mathbb{N}$ in which case $c_k\mathbf{1}_{n,\lambda}(R_n) = \frac{(\lambda)_k}{k!} (qp^{-n})^k$. Therefore, we have

$$
\begin{align*}
L_{\mathbf{X}_{[n]\setminus[k]}}^{\mathbf{X}_{[k]}} (\boldsymbol{\theta}_{[n]\setminus[k]}) &= \left[ \prod_{i=k+1}^{n} (1+p\theta_i)^{-\lambda} \right] \frac{\sum_{k=0}^{\infty} \frac{(x_k)^{k\mathbf{1}_k}}{\{(\lambda)_k\}^k} \frac{(\lambda)_k}{k!} (qp^{-n})^k (\boldsymbol{\theta}_{[n]} + \frac{1}{p}\mathbf{1}_n)^{-k\mathbf{1}_{n-k}}}{\sum_{k=0}^{\infty} \frac{(x_k)^{k\mathbf{1}_k}}{\{(\lambda)_k\}^k} \frac{(\lambda)_k}{k!} (qp^{-n})^k (\frac{1}{p}\mathbf{1}_n)^{-k\mathbf{1}_{n-k}}} \\
&= \left[ \prod_{i=k+1}^{n} (1+p\theta_i)^{-\lambda} \right] \frac{\sum_{k=0}^{\infty} \frac{1}{\{(\lambda)_k\}^{k-1}} \frac{1}{k!} (qp^{-k}\mathbf{x}_k^{[k]}) \prod_{i=k+1}^{n} (1+p\theta_i)^{-1}}{\sum_{k=0}^{\infty} \frac{1}{\{(\lambda)_k\}^k} \frac{(\lambda)_k}{k!} (qp^{-k}\mathbf{x}_k^{[k]})^k} \quad \text{and the definition of } F_{k-1} \text{ gives (74).}
\end{align*}
$$

<figure><img src="image_17.png" /></figure>

**Proof of Corollary 29.** Doing $k=1$ in Equality (74), we get (75). Equality (76) comes from the definition of exp. A second proof can be given by application of Theorem 24. We successively have $G(R_n, u_1) = \sum_{\alpha_1 \in \mathbb{N}} \frac{u_1^{\alpha_1}}{\alpha_1!} = \exp(u_1)$, $S_{n-1}(\theta_{[n]\setminus[1]}) = \prod_{i=2}^n (1+p\theta_i)$, $B(\theta_{[n]\setminus[1]}) = qp^{-1}$ and $\S(\theta_{[n]\setminus[1]}) = qp^{-1}[\prod_{i=2}^n (1+p\theta_i)]^{-1}$. As a result, by application of (66) we get (75). Another proof of this result is given by (72) as follows. In this case, we have $P_n(\theta) = \frac{-q}{p} + \frac{1}{p} \prod_{i=1}^n (1+p\theta_i)$, $p_T = p^{|T|-1}$, $p_{[n]} = p^{n-1}$, $\tilde{p}_T = -p^{-|T|}$, $\tilde{p}_{\{i\}} = -p^{-1}$, $\theta_{P_n} = \tilde{p} = -p^{-1}\mathbf{1}_n$, $-\tilde{p}_{[n]\setminus[1]} = p^{-(n-1)}$, $S_{[n]\setminus\{1\}}(\theta_{[n]\setminus[1]}) = \prod_{i=2}^n (1+p\theta_i)$, and

$$
P_n(\boldsymbol{\theta}) = p_{[n]} \prod_{i=1}^{n} (\theta_i + (-p^{-1})) [1 - qp^{-n} \prod_{i=1}^{n} (\theta_i + (-p^{-1}))^{-1}]. \text{ Hence we have } r_T = 0, \text{ if } 1 < |T| < n \\
\text{and } r_{[n]} = qp^{-n}. \text{ We deduce from (72): } L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1} = S_{[n]\setminus[1]}^{-\lambda} \mathbf{e}^{\{qp^{-1}x_1[S_{[n]\setminus[1]}^{-1}-1]\}} = \sum_{k=0}^{\infty} \frac{(qp^{-1}x_1)^k}{k!} \exp(-qp^{-1}x_1) S_{[n]\setminus[1]}^{-(\lambda+k)}.
$$

<figure><img src="image_18.png" /></figure>

**Proof of Theorem 30.** Indeed, $X_1 \sim \gamma_{(p,\lambda)}$, let $V_1 \sim \mathcal{P}(qp^{-1}X_1)$, we have $P(V_1 = k|X_1 = x_1) = \frac{(qp^{-1}x_1)^k}{k!} \exp(-qp^{-1}x_1)$. Clearly the variable $X_i, i = 2, \dots, n$ are conditionally independent and $X_i|(X_1 = x_1) \sim \gamma_{(p,\lambda+V_1)}$. We have

$$
L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}(\boldsymbol{\theta}_{[n]\setminus[1]}) = \sum_{v_1=0}^{\infty} P(V_1 = v_1 | X_1 = x_1) L_{\mathbf{X}_{[n]\setminus[1]}}^{X_1=x_1}(\boldsymbol{\theta}_{[n]\setminus[1]}) , \text{ and Formula (76) is verified. Finally} \\
\text{we have } \mathbf{X}_{[n]} \sim \gamma_{(P_n,\lambda)}, \text{ with } P_n(\boldsymbol{\theta}_{[n]}) = \frac{-q}{p} + \frac{1}{p} \prod_{i=1}^{n} (1+p\theta_i) . \blacksquare
$$

**Proof of Theorem 32.** From (72), we deduce the Lt of the conditional distribution of $X_2|X_1 = x_1$,
for $(X_1, X_2) \sim \gamma_{(P_2, \lambda)}$, with $P_2(\theta_1, \theta_2) = 1 + p_1\theta_1 + p_2\theta_2 + p_{1,2}\theta_1\theta_2$ with $p_1 > 0, p_2 > 0, p_{1,2} > 0$, $\tilde{b}_{1,2} = \tilde{b}_{1,2}(P_2) = p_1p_2/p_{1,2}^2 - 1/p_{1,2} > 0$. We have $S_2(\theta_2) = 1 + \frac{p_{1,2}}{p_1}\theta_2$ and

$$L_{X_2}^{X_1=x_1} = S_2^{-\lambda} \exp\{\frac{\tilde{b}_{1,2}}{(-\tilde{p}_2)} x_1(S_2^{-1} - 1)\}. \quad (105)$$

By expanding the exponential function, we obtain the following expansion:

$$L_{X_2}^{X_1=x_1} = \sum_{v_1 \in \mathbb{N}} \mathbf{P}(V_1 = v_1) S_2^{-(\lambda+v_1)}. \quad (106)$$

Equality (106) proves that $X_2|X_1$ has distribution $\gamma_{(S_2, \lambda+V_1)}$, then $\mathbf{X}_{[2]} = (X_1, X_2) \sim \gamma_{(P_2, \lambda)}$. ■

**Proof of Theorem 34.** Hence for $n=3$, we have from definition of $S_T$, $S_{2,3}(\theta_2, \theta_3) = 1 + \frac{p_{1,2}}{p_1}\theta_2 + \frac{p_{1,3}}{p_1}\theta_3 + \frac{p_{1,2,3}}{p_1}\theta_2\theta_3$, $S_2(\theta_2) = 1 + \frac{p_{1,2,3}}{p_{1,3}}\theta_2$, $S_3(\theta_3) = 1 + \frac{p_{1,2,3}}{p_{1,2}}\theta_3$, and from (73) and (27), we get

$$L_{(X_2, X_3)}^{X_1=x_1} = S_{2,3}^{-\lambda} \exp\{(-\tilde{p}_{2,3})^{-1} x_1[\tilde{b}_{\{1,2\}}(-\tilde{p}_3)S_3S_{2,3}^{-1} + \tilde{b}_{\{1,3\}}(-\tilde{p}_2)S_2S_{2,3}^{-1} + \tilde{b}_{\{1,2,3\}}S_{2,3}^{-1} - C]\} \quad (107)$$

To compute another expression for $L_{(X_2, X_3)}^{X_1=x_1}$ we use (37), then we have $(-\tilde{p}_2)(-\tilde{p}_3)S_2S_3 = (-\tilde{p}_{2,3})S_{2,3} + \tilde{b}_{2,3}$,

By respectively dividing the last equality by $S_3S_{2,3}$ and $S_2S_{2,3}$, we successively obtain $(-\tilde{p}_3)(-\tilde{p}_2)S_2S_{2,3}^{-1} = (-\tilde{p}_{2,3})S_3^{-1} + \tilde{b}_{2,3}S_3^{-1}S_{2,3}^{-1}$ and $(-\tilde{p}_2)(-\tilde{p}_3)S_3S_{2,3}^{-1} = (-\tilde{p}_{2,3})S_2^{-1} + \tilde{b}_{2,3}S_2^{-1}S_{2,3}^{-1}$. Using the two last equalities into (107), we get $L_{(X_2, X_3)}^{X_1=x_1} = S_{2,3}^{-\lambda} \exp\{x_1[\alpha_1S_2^{-1} + \alpha_2S_3^{-1} + \alpha_3S_{2,3}^{-1} + \alpha_4S_2^{-1}S_{2,3}^{-1} + \alpha_5S_3^{-1}S_{2,3}^{-1} - C]\}$, and by the condition $L_{(X_2, X_3)}^{X_1=x_1}(0, 0) = 1$, we obtain

$$L_{(X_2, X_3)}^{X_1=x_1} = S_{2,3}^{-\lambda} \exp\{x_1[\alpha_1(S_2^{-1} - 1) + \alpha_2(S_3^{-1} - 1) + \alpha_3(S_{2,3}^{-1} - 1) + \alpha_4(S_2^{-1}S_{2,3}^{-1} - 1) + \alpha_5(S_3^{-1}S_{2,3}^{-1} - 1)]\}.$$ Expanding exp in the last equality, we get (77). Equality (77) proves that $\mathbf{X}_{[3]} \sim \gamma_{(P_3, \lambda)}$. ■

**Proof of Theorem 36.** We use Theorem (32). Let $X'_2 \sim \gamma(\frac{p_{1,2}}{p_1}, \lambda+V_3+V_4+V_5)$, let $\alpha_6 = \frac{\tilde{b}_{2,3}(S_{2,3})}{-\tilde{p}_3(S_{2,3})} = \frac{\tilde{b}_{2,3}}{(-\tilde{p}_3)}$, and $V_6 \sim \mathcal{P}(\alpha_6 X'_2)$, let $X'_3 \sim \gamma_{(S_3, \lambda+V_3+V_4+V_5+V_6)}$, then $(X'_2, X'_3) \sim \gamma(S_2, \lambda+V_3+V_4+V_5)$. ■

**Proof of Theorem 39.** For $n=4$, from (73), (28) and (36), we get $(-\tilde{p}_{2,3})S_{2,3} = (-\tilde{p}_2)(-\tilde{p}_3)S_2S_3 - \tilde{b}_{2,3}$, $(-\tilde{p}_{2,4})S_{2,4} = (-\tilde{p}_2)(-\tilde{p}_4)S_2S_4 - \tilde{b}_{2,4}$, $(-\tilde{p}_{3,4})S_{3,4} = (-\tilde{p}_3)(-\tilde{p}_4)S_3S_4 - \tilde{b}_{3,4}$, and

$$L_{(X_2, X_3, X_4)}^{X_1=x_1} = S_{2,3,4}^{-\lambda} \exp\{(-\tilde{p}_{2,3,4})^{-1} x_1[\tilde{b}_{1,2}(-\tilde{p}_{3,4})S_{3,4}S_{2,3,4}^{-1} + \tilde{b}_{1,3}(-\tilde{p}_{2,4})S_{2,4}S_{2,3,4}^{-1} + \tilde{b}_{1,4}(-\tilde{p}_{2,3})S_{2,3}S_{2,3,4}^{-1} \\ + \tilde{b}_{1,2,3}(-\tilde{p}_4)S_4S_{2,3,4}^{-1} + \tilde{b}_{1,2,4}(-\tilde{p}_3)S_3S_{2,3,4}^{-1} + \tilde{b}_{1,3,4}(-\tilde{p}_2)S_2S_{2,3,4}^{-1} + \tilde{b}_{1,2,3,4}S_{2,3,4}^{-1} - C]\} \quad (108)$$

To compute $L_{(X_2, X_3, X_4)}^{X_1=x_1}$, we need to express the following expressions in terms of inverses of $S_T$, $T \in \mathfrak{P}(\{2, 3, 4\})$ $S_{3,4}S_{2,3,4}^{-1}$, $S_{2,4}S_{2,3,4}^{-1}$, $S_{2,3}S_{2,3,4}^{-1}$, $S_4S_{2,3,4}^{-1}$, $S_3S_{2,3,4}^{-1}$, $S_2S_{2,3,4}^{-1}$. From (37), we then successively have

$$(-\tilde{p}_2)(-\tilde{p}_3)S_2S_3 = (-\tilde{p}_{2,3})S_{2,3} + \tilde{b}_{2,3}, \quad (109)$$

$$(-\tilde{p}_2)(-\tilde{p}_4)S_2S_4 = (-\tilde{p}_{2,4})S_{2,4} + \tilde{b}_{2,4}, \quad (110)$$
$$(-\tilde{p}_3)(-\tilde{p}_4)S_3S_4 = (-\tilde{p}_{3,4})S_{3,4} + \tilde{b}_{3,4}, \qquad (111)$$

According to (36), we also have successively

$$S_{2,3,4} = (-\tilde{p}_{2,3,4})^{-1} \left[ (-\tilde{p}_{2,3}) S_{2,3} (-\tilde{p}_4) S_4 - \tilde{b}_{2,4} (-\tilde{p}_3) S_3 - \tilde{b}_{3,4} (-\tilde{p}_2) S_2 - \tilde{b}_{2,3,4} \right],$$

$$S_{2,3,4} = (-\tilde{p}_{2,3,4})^{-1} \left[ (-\tilde{p}_{2,4}) S_{2,4} (-\tilde{p}_3) S_3 - \tilde{b}_{2,3} (-\tilde{p}_4) S_4 - \tilde{b}_{3,4} (-\tilde{p}_2) S_2 - \tilde{b}_{2,3,4} \right],$$

$$S_{2,3,4} = (-\tilde{p}_{2,3,4})^{-1} \left[ (-\tilde{p}_{3,4}) S_{3,4} (-\tilde{p}_2) S_2 - \tilde{b}_{2,3} (-\tilde{p}_4) S_4 - \tilde{b}_{2,4} (-\tilde{p}_3) S_3 - \tilde{b}_{2,3,4} \right], \text{ and}$$

$$(-\tilde{p}_4)(-\tilde{p}_{2,3}) S_4 S_{2,3} = (-\tilde{p}_{2,3,4}) S_{2,3,4} + \tilde{b}_{2,4} (-\tilde{p}_3) S_3 + \tilde{b}_{3,4} (-\tilde{p}_2) S_2 + \tilde{b}_{2,3,4}, \qquad (112)$$

$$(-\tilde{p}_3)(-\tilde{p}_{2,4}) S_3 S_{2,4} = (-\tilde{p}_{2,3,4}) S_{2,3,4} + \tilde{b}_{2,3} (-\tilde{p}_4) S_4 + \tilde{b}_{3,4} (-\tilde{p}_2) S_2 + \tilde{b}_{2,3,4}, \qquad (113)$$

$$(-\tilde{p}_2)(-\tilde{p}_{3,4}) S_2 S_{3,4} = (-\tilde{p}_{2,3,4}) S_{2,3,4} + \tilde{b}_{2,3} (-\tilde{p}_4) S_4 + \tilde{b}_{2,4} (-\tilde{p}_3) S_3 + \tilde{b}_{2,3,4}, \qquad (114)$$

Dividing (109) by $S_3S_{2,3}$ and $S_2S_{2,3}$, (110) by $S_4S_{2,4}$ and $S_2S_{2,4}$, and (111) by $S_4S_{3,4}$ and $S_3S_{3,4}$, we obtain successively

$$(-\tilde{p}_2)(-\tilde{p}_3) \frac{S_2}{S_{2,3}} = (-\tilde{p}_{2,3}) \frac{1}{S_3} + \tilde{b}_{2,3} \frac{1}{S_3 S_{2,3}}, \qquad (115)$$

$$(-\tilde{p}_2)(-\tilde{p}_3) \frac{S_3}{S_{2,3}} = (-\tilde{p}_{2,3}) \frac{1}{S_2} + \tilde{b}_{2,3} \frac{1}{S_2 S_{2,3}}, \qquad (116)$$

$$(-\tilde{p}_2)(-\tilde{p}_4) \frac{S_2}{S_{2,4}} = (-\tilde{p}_{2,4}) \frac{1}{S_4} + \tilde{b}_{2,4} \frac{1}{S_4 S_{2,4}}, \qquad (117)$$

$$(-\tilde{p}_2)(-\tilde{p}_4) \frac{S_4}{S_{2,4}} = (-\tilde{p}_{2,4}) \frac{1}{S_2} + \tilde{b}_{2,4} \frac{1}{S_2 S_{2,4}}, \qquad (118)$$

$$(-\tilde{p}_3)(-\tilde{p}_4) \frac{S_3}{S_{3,4}} = (-\tilde{p}_{3,4}) \frac{1}{S_4} + \tilde{b}_{3,4} \frac{1}{S_4 S_{3,4}}, \qquad (119)$$

$$(-\tilde{p}_3)(-\tilde{p}_4) \frac{S_4}{S_{3,4}} = (-\tilde{p}_{3,4}) \frac{1}{S_3} + \tilde{b}_{3,4} \frac{1}{S_3 S_{3,4}}, \qquad (120)$$

Dividing (112) by $S_{2,3}S_{2,3,4}$ and $S_4S_{2,3,4}$, (113) by $S_{2,4}S_{2,3,4}$ and $S_3S_{2,3,4}$, and (114) by $S_{3,4}S_{2,3,4}$ and $S_2S_{2,3,4}$, we obtain successively

$$(-\tilde{p}_4)(-\tilde{p}_{2,3}) \frac{S_4}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_{2,3}} + \tilde{b}_{2,4} (-\tilde{p}_3) \frac{1}{S_{2,3,4}} \frac{S_3}{S_{2,3}} + \tilde{b}_{3,4} (-\tilde{p}_2) \frac{1}{S_{2,3,4}} \frac{S_2}{S_{2,3}} + \tilde{b}_{2,3,4} \frac{1}{S_{2,3,4} S_{2,3,4}}, \qquad (121)$$

$$(-\tilde{p}_4)(-\tilde{p}_{2,3}) \frac{S_{2,3}}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_4} + \tilde{b}_{2,4} (-\tilde{p}_3) \frac{1}{S_4} \frac{S_3}{S_{2,3,4}} + \tilde{b}_{3,4} (-\tilde{p}_2) \frac{1}{S_4} \frac{S_2}{S_{2,3,4}} + \tilde{b}_{2,3,4} \frac{1}{S_4 S_{2,3,4}}, \qquad (122)$$

$$(-\tilde{p}_3)(-\tilde{p}_{2,4}) \frac{S_3}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_{2,4}} + \tilde{b}_{2,3} (-\tilde{p}_4) \frac{1}{S_{2,3,4}} \frac{S_4}{S_{2,4}} + \tilde{b}_{3,4} (-\tilde{p}_2) \frac{1}{S_{2,3,4}} \frac{S_2}{S_{2,4}} + \tilde{b}_{2,3,4} \frac{1}{S_{2,4} S_{2,3,4}}, \qquad (123)$$

$$(-\tilde{p}_3)(-\tilde{p}_{2,4}) \frac{S_{2,4}}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_3} + \tilde{b}_{2,3} (-\tilde{p}_4) \frac{1}{S_3} \frac{S_4}{S_{2,3,4}} + \tilde{b}_{3,4} (-\tilde{p}_2) \frac{1}{S_3} \frac{S_2}{S_{2,3,4}} + \tilde{b}_{2,3,4} \frac{1}{S_3 S_{2,3,4}}, \qquad (124)$$

$$(-\tilde{p}_2)(-\tilde{p}_{3,4}) \frac{S_2}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_{3,4}} + \tilde{b}_{2,3} (-\tilde{p}_4) \frac{1}{S_{2,3,4}} \frac{S_4}{S_{3,4}} + \tilde{b}_{2,4} (-\tilde{p}_3) \frac{1}{S_{2,3,4}} \frac{S_3}{S_{3,4}} + \tilde{b}_{2,3,4} \frac{1}{S_{3,4} S_{2,3,4}}, \qquad (125)$$

$$(-\tilde{p}_2)(-\tilde{p}_{3,4}) \frac{S_{3,4}}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_2} + \tilde{b}_{2,3} (-\tilde{p}_4) \frac{1}{S_2} \frac{S_4}{S_{2,3,4}} + \tilde{b}_{2,4} (-\tilde{p}_3) \frac{1}{S_2} \frac{S_3}{S_{2,3,4}} + \tilde{b}_{2,3,4} \frac{1}{S_2 S_{2,3,4}}, \qquad (126)$$
Using (115) and (116) into (121), we get

$$ \begin{aligned} & (-\tilde{p}_4)(-\tilde{p}_{2,3}) \frac{S_4}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_{2,3}} + (-\tilde{p}_{2,3})(-\tilde{p}_2)^{-1} \tilde{b}_{2,4} \frac{1}{S_2 S_{2,3,4}} + (-\tilde{p}_{2,3})(-\tilde{p}_3)^{-1} \tilde{b}_{3,4} \frac{1}{S_3 S_{2,3,4}} \\ & \qquad + \tilde{b}_{2,3,4} \frac{1}{S_{2,3} S_{2,3,4}} + (-\tilde{p}_2)^{-1} \tilde{b}_{2,3} \tilde{b}_{2,4} \frac{1}{S_2 S_{2,3,4}} + (-\tilde{p}_3)^{-1} \tilde{b}_{2,3} \tilde{b}_{3,4} \frac{1}{S_3 S_{2,3,4}}. \end{aligned} \quad (127) $$

Using (117) and (118) into (123), we get

$$ \begin{aligned} & (-\tilde{p}_3)(-\tilde{p}_{2,4}) \frac{S_3}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_{2,4}} + (-\tilde{p}_{2,4})(-\tilde{p}_2)^{-1} \tilde{b}_{2,3} \frac{1}{S_2 S_{2,3,4}} + (-\tilde{p}_{2,4})(-\tilde{p}_4)^{-1} \tilde{b}_{3,4} \frac{1}{S_4 S_{2,3,4}} \\ & \qquad + \tilde{b}_{2,3,4} \frac{1}{S_{2,4} S_{2,3,4}} + (-\tilde{p}_2)^{-1} \tilde{b}_{2,3} \tilde{b}_{2,4} \frac{1}{S_2 S_{2,4} S_{2,3,4}} + (-\tilde{p}_4)^{-1} \tilde{b}_{2,4} \tilde{b}_{3,4} \frac{1}{S_4 S_{2,4} S_{2,3,4}} \end{aligned} \quad (128) $$

Using (119) and (120) into (125), we get

$$ \begin{aligned} & (-\tilde{p}_2)(-\tilde{p}_{3,4}) \frac{S_2}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_{3,4}} + (-\tilde{p}_{3,4})(-\tilde{p}_3)^{-1} \tilde{b}_{2,3} \frac{1}{S_3 S_{2,3,4}} + (-\tilde{p}_{3,4})(-\tilde{p}_4)^{-1} \tilde{b}_{2,4} \frac{1}{S_4 S_{2,3,4}} \\ & \qquad + \tilde{b}_{2,3,4} \frac{1}{S_{3,4} S_{2,3,4}} + (-\tilde{p}_3)^{-1} \tilde{b}_{2,3} \tilde{b}_{3,4} \frac{1}{S_3 S_{3,4} S_{2,3,4}} + (-\tilde{p}_4)^{-1} \tilde{b}_{2,4} \tilde{b}_{3,4} \frac{1}{S_4 S_{3,4} S_{2,3,4}} \end{aligned} \quad (129) $$

Using (129) and (128) into (122), we get

$$ \begin{aligned} & (-\tilde{p}_4)(-\tilde{p}_{2,3}) \frac{S_{2,3}}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_4} + \tilde{b}_{2,3,4} \frac{1}{S_4 S_{2,3,4}} + (-\tilde{p}_{2,3,4})(-\tilde{p}_{2,4})^{-1} \tilde{b}_{2,4} \frac{1}{S_4 S_{2,4}} + (-\tilde{p}_2)^{-1} \tilde{b}_{2,3} \tilde{b}_{2,4} \frac{1}{S_2 S_4 S_{2,3,4}} + \\ & \qquad (-\tilde{p}_4)^{-1} \tilde{b}_{3,4} \tilde{b}_{2,4} \frac{1}{S_4^2 S_{2,3,4}} + (-\tilde{p}_{2,4})^{-1} \tilde{b}_{2,3,4} \tilde{b}_{2,4} \frac{1}{S_4 S_{2,4} S_{2,3,4}} + (-\tilde{p}_2)^{-1} (-\tilde{p}_{2,4})^{-1} \tilde{b}_{2,3} \tilde{b}_{2,4} \frac{1}{S_2 S_4 S_{2,4} S_{2,3,4}} + \\ & \qquad (-\tilde{p}_4)^{-1} (-\tilde{p}_{2,4})^{-1} \tilde{b}_{2,4} \tilde{b}_{3,4} \frac{1}{S_4^2 S_{2,4} S_{2,3,4}} + (-\tilde{p}_{2,3,4})(-\tilde{p}_{3,4})^{-1} \tilde{b}_{3,4} \frac{1}{S_4 S_{3,4}} + (-\tilde{p}_3)^{-1} \tilde{b}_{2,3} \tilde{b}_{3,4} \frac{1}{S_3 S_{4,3,4}} + \\ & \qquad (-\tilde{p}_4)^{-1} \tilde{b}_{2,4} \tilde{b}_{3,4} \frac{1}{S_4^2 S_{2,3,4}} + (-\tilde{p}_{3,4})^{-1} \tilde{b}_{2,3,4} \tilde{b}_{3,4} \frac{1}{S_4 S_{3,4} S_{2,3,4}} + (-\tilde{p}_3)^{-1} (-\tilde{p}_{3,4})^{-1} \tilde{b}_{2,3} \tilde{b}_{3,4} \frac{1}{S_3 S_{4,3,4} S_{2,3,4}} + \\ & \qquad (-\tilde{p}_4)^{-1} (-\tilde{p}_{3,4})^{-1} \tilde{b}_{2,4} \tilde{b}_{3,4} \tilde{b}_{3,4} \frac{1}{S_4^2 S_{3,4} S_{2,3,4}} \end{aligned} \quad (130) $$

Using (129) and (127) into (124), we get

$$ \begin{aligned} & (-\tilde{p}_3)(-\tilde{p}_{2,4}) \frac{S_{2,4}}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_3} + \tilde{b}_{2,3,4} \frac{1}{S_3 S_{2,3,4}} + (-\tilde{p}_{2,3,4})(-\tilde{p}_{2,3})^{-1} \tilde{b}_{2,3} \frac{1}{S_3 S_{2,3}} + (-\tilde{p}_2)^{-1} \tilde{b}_{2,4} \tilde{b}_{2,3} \frac{1}{S_2 S_3 S_{2,3,4}} + \\ & \qquad (-\tilde{p}_3)^{-1} \tilde{b}_{3,4} \tilde{b}_{2,3} \frac{1}{S_3^2 S_{2,3,4}} + (-\tilde{p}_{2,3})^{-1} \tilde{b}_{2,3,4} \tilde{b}_{2,3} \frac{1}{S_3 S_{2,3} S_{2,3,4}} + (-\tilde{p}_2)^{-1} (-\tilde{p}_{2,3})^{-1} \tilde{b}_{2,3} \tilde{b}_{2,4} \frac{1}{S_2 S_3 S_{2,3} S_{2,3,4}} + \\ & \qquad (-\tilde{p}_3)^{-1} (-\tilde{p}_{2,3})^{-1} \tilde{b}_{2,3} \tilde{b}_{3,4} \frac{1}{S_3^2 S_{2,3} S_{2,3,4}} + (-\tilde{p}_{2,3,4})(-\tilde{p}_{3,4})^{-1} \tilde{b}_{3,4} \frac{1}{S_3 S_{3,4}} + (-\tilde{p}_3)^{-1} \tilde{b}_{2,3} \tilde{b}_{3,4} \frac{1}{S_3^2 S_{3,4}} + \\ & \qquad (-\tilde{p}_4)^{-1} \tilde{b}_{2,4} \tilde{b}_{3,4} \frac{1}{S_3 S_{4,2,3,4}} + (-\tilde{p}_{3,4})^{-1} \tilde{b}_{2,3,4} \tilde{b}_{3,4} \frac{1}{S_3 S_{3,4} S_{2,3,4}} + (-\tilde{p}_3)^{-1} (-\tilde{p}_{3,4})^{-1} \tilde{b}_{2,3} \tilde{b}_{3,4} \frac{1}{S_3^2 S_{3,4} S_{2,3,4}} + \\ & \qquad (-\tilde{p}_4)^{-1} (-\tilde{p}_{3,4})^{-1} \tilde{b}_{2,4} \tilde{b}_{3,4} \tilde{b}_{3,4} \frac{1}{S_3 S_{4,3,4} S_{2,3,4}} \end{aligned} \quad (131) $$

Using (128) and (127) into (126), we get

$$ \begin{aligned} & (-\tilde{p}_2)(-\tilde{p}_{3,4}) \frac{S_{3,4}}{S_{2,3,4}} = (-\tilde{p}_{2,3,4}) \frac{1}{S_2} + \tilde{b}_{2,3,4} \frac{1}{S_2 S_{2,3,4}} + [(-\tilde{p}_{2,3,4})(-\tilde{p}_{2,3})^{-1} \tilde{b}_{2,3} \frac{1}{S_2 S_{2,3}} + (-\tilde{p}_2)^{-1} \tilde{b}_{2,4} \tilde{b}_{2,3} \frac{1}{S_2^2 S_{2,3,4}} + \\ & \qquad (-\tilde{p}_3)^{-1} \tilde{b}_{3,4} \tilde{b}_{2,3} \frac{1}{S_2 S_3 S_{2,3,4}} + (-\tilde{p}_{2,3})^{-1} \tilde{b}_{2,3,4} \tilde{b}_{2,3} \frac{1}{S_2 S_{2,3} S_{2,3,4}} + (-\tilde{p}_2)^{-1} (-\tilde{p}_{2,3})^{-1} \tilde{b}_{2,3} \tilde{b}_{2,4} \frac{1}{S_2^2 S_{2,3} S_{2,3,4}} + \\ & \qquad (-\tilde{p}_3)^{-1} (-\tilde{p}_{2,3})^{-1} \tilde{b}_{2,3} \tilde{b}_{3,4} \frac{1}{S_2 S_3 S_{2,3} S_{2,3,4}}] + [(-\tilde{p}_{2,3,4})(-\tilde{p}_{2,4})^{-1} \tilde{b}_{2,4} \frac{1}{S_2 S_{2,4}} + (-\tilde{p}_2)^{-1} \tilde{b}_{2,3} \tilde{b}_{2,4} \frac{1}{S_2^2 S_{2,3,4}} + \\ & \qquad (-\tilde{p}_4)^{-1} \tilde{b}_{2,4} \tilde{b}_{3,4} \frac{1}{S_2 S_{4,2,3,4}} + (-\tilde{p}_{2,4})^{-1} \tilde{b}_{2,3,4} \tilde{b}_{2,4} \frac{1}{S_2 S_{2,4} S_{2,3,4}} + (-\tilde{p}_2)^{-1} (-\tilde{p}_{2,4})^{-1} \tilde{b}_{2,3} \tilde{b}_{2,4} \frac{1}{S_2^2 S_{2,4} S_{2,3,4}} + \\ & \qquad (-\tilde{p}_4)^{-1} (-\tilde{p}_{2,4})^{-1} \tilde{b}_{2,4} \tilde{b}_{3,4} \tilde{b}_{2,4} \frac{1}{S_2 S_4 S_{2,4} S_{2,3,4}}] \end{aligned} \quad (132) $$
Using (129), (128), (127), (130), (131) and (132) into (108), by grouping terms of the same total degree and using the condition $L_{(X_2,X_3,X_4)}^{X_1=x_1}(0,0,0)=1$, we obtain

$$
\begin{align*}
L_{(X_2,X_3,X_4)}^{X_1=x_1}(\theta_2, \theta_3, \theta_4) &= S_{2,3,4}^{-\lambda} \exp(x_1\{\alpha_1[\frac{1}{S_2}-1] + \alpha_2[\frac{1}{S_3}-1] + \alpha_3[\frac{1}{S_4}-1] + \alpha_4[\frac{1}{S_{2,3}}-1] + \alpha_5[\frac{1}{S_{2,4}}-1] + \alpha_6[\frac{1}{S_{3,4}}-1] + \alpha_7[\frac{1}{S_2S_{2,3}}-1] \\
&\quad + \alpha_8[\frac{1}{S_2S_{2,4}}-1] + \alpha_9[\frac{1}{S_3S_{2,3}}-1] + \alpha_{10}[\frac{1}{S_3S_{3,4}}-1] + \alpha_{11}[\frac{1}{S_4S_{2,4}}-1] + \alpha_{12}[\frac{1}{S_4S_{3,4}}-1] + \alpha_{13}[\frac{1}{S_{2,3,4}}-1] + \\
&\quad \alpha_{14}[\frac{1}{S_2S_{2,3,4}}-1] + \\
&\quad \alpha_{15}[\frac{1}{S_3S_{2,3,4}}-1] + \alpha_{16}[\frac{1}{S_4S_{2,3,4}}-1] + \alpha_{17}[\frac{1}{S_2S_{2,3,4}}-1] + \alpha_{18}[\frac{1}{S_3S_{2,3,4}}-1] + \\
&\quad \alpha_{19}[\frac{1}{S_4S_{2,3,4}}-1] + \alpha_{20}[\frac{1}{S_2S_3S_{2,3,4}}-1] + \alpha_{21}[\frac{1}{S_2S_4S_{2,3,4}}-1] + \alpha_{22}[\frac{1}{S_3S_4S_{2,3,4}}-1] + \alpha_{23}[\frac{1}{S_{2,3}S_{2,3,4}}-1] + \\
&\quad \alpha_{24}[\frac{1}{S_{2,4}S_{2,3,4}}-1] + \alpha_{25}[\frac{1}{S_{3,4}S_{2,3,4}}-1] + \alpha_{26}[\frac{1}{S_2S_{2,3}S_{2,3,4}}-1] + \alpha_{27}[\frac{1}{S_2S_{2,4}S_{2,3,4}}-1] + \alpha_{28}[\frac{1}{S_3S_{2,3}S_{2,3,4}}-1] + \\
&\quad \alpha_{29}[\frac{1}{S_3S_{3,4}S_{2,3,4}}-1] \\
&\quad + \alpha_{30}[\frac{1}{S_4S_{2,4}S_{2,3,4}}-1] + \alpha_{31}[\frac{1}{S_4S_{3,4}S_{2,3,4}}-1] + \alpha_{32}[\frac{1}{S_2S_{2,3}S_{2,3,4}}-1] + \\
&\quad \alpha_{33}[\frac{1}{S_2S_{2,4}S_{2,3,4}}-1] + \alpha_{34}[\frac{1}{S_3S_{2,3}S_{2,3,4}}-1] + \alpha_{35}[\frac{1}{S_3S_{3,4}S_{2,3,4}}-1] + \\
&\quad \alpha_{36}[\frac{1}{S_4S_{2,4}S_{2,3,4}}-1] + \alpha_{37}[\frac{1}{S_4S_{3,4}S_{2,3,4}}-1] + \alpha_{38}[\frac{1}{S_2S_3S_{2,3}S_{2,3,4}}-1] + \\
&\quad \alpha_{39}[\frac{1}{S_2S_4S_{2,4}S_{2,3,4}}-1] + \alpha_{40}[\frac{1}{S_3S_4S_{3,4}S_{2,3,4}}-1]]
\end{align*}
$$

Combining, we obtain (78). Definitions (79), (80), (81), (82), (83), (84), (85), (86), and (87), (88), (89) with Equality (78) give (90). This completes the proof of Theorem (39). ■

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