Markov Infinitely-Divisible Stationary Time-Reversible Integer-Valued Processes

Abstract

We prove a complete class theorem that characterizes all stationary time reversible Markov processes whose finite dimensional marginal distributions (of all orders) are infinitely divisible. Aside from two degenerate cases (iid and constant), in both discrete and continuous time every such process with full support is a branching process with Poisson or Negative Binomial marginal univariate distributions and a specific bivariate distribution at pairs of times. As a corollary, we prove that every nondegenerate stationary integer valued process constructed by the Markov thinning process fails to have infinitely divisible multivariate marginal distributions, except for the Poisson. These results offer guidance to anyone modeling integer-valued Markov data exhibiting autocorrelation.

1 Introduction

Many applications feature autocorrelated count data $X_t$ at discrete times t. A number of authors have constructed and studied stationary stochastic processes $X_t$ whose one-dimensional marginal distributions come from an arbitrary infinitely-divisible distribution family $\{\mu ^{\theta }\}$, such as the Poisson $\textsf {Po}(\theta )$ or negative binomial $\textsf {NB}(\theta ,p)$, and that are “$\textsf {AR}(1)$-like” in the sense that their autocorrelation function is ${\textsf {Corr}}[X_s,X_t] =\rho ^{|s-t|}$ for some $\rho \in (0,1)$ (Lewis, 1983; Lewis et al., 1989; McKenzie, 1988; Al-Osh and Alzaid, 1987; Joe, 1996). The most common approach is to build a time-reversible Markov process using thinning, in which the process at any two consecutive times may be written in the form

$$\begin{aligned} X_{t-1}=\xi _t+\eta _t \qquad X_t=\xi _t+\zeta _t \end{aligned}$$

with $\xi _t$, $\eta _t$, and $\zeta _t$ all independent and from the same infinitely-divisible family (see Sec. () below for details). A second construction of a stationary time-reversible process with the same one-dimensional marginal distributions and autocorrelation function, with the feature that its finite-dimensional marginal distributions of all orders are infinitely-divisible, is to set $X_t:={\mathcal {N}}(G_t)$ for a random measure ${\mathcal {N}}$ on some measure space $(E,{\mathcal {E}},m)$ that assigns independent infinitely-divisible random variables ${\mathcal {N}}(A_i)\sim \mu ^{\theta _i}$ to disjoint sets $A_i\in {\mathcal {E}}$ of measure $\theta _i=m(A_i)$, and a family of sets $\{G_t\}\subset {\mathcal {E}}$ whose intersections have measure $m\big (G_s\cap G_t\big )=\theta \rho ^{|s-t|}$ (see Sec. ()).

For the normal distribution $X_t\sim \textsf {No}(\mu ,\sigma ^2)$, these two constructions both yield the usual Gaussian $\textsf {AR}(1)$ process. The two constructions also yield identical processes for the Poisson $X_t\sim \textsf {Po}(\theta )$ distribution, but they differ for all other nonnegative integer-valued infinitely-divisible distributions. For each nonnegative integer-valued infinitely-divisible marginal distribution except the Poisson, the process constructed by thinning does not have infinitely-divisible marginal distributions of all orders (Theorem , Sec. ()), and the process constructed using random measures does not have the Markov property (Theorem , Sec. 3.5). Thus none of these is completely satisfactory for modeling autocorrelated count data with heavier tails than the Poisson distribution.

In the present manuscript we construct and characterize every process that is Markov, infinitely-divisible, stationary, and time-reversible with non-negative integer values. The formal characterization is contained in the statement of Theorem in Sec. (), which follows necessary definitions and the investigation of special cases needed to establish the general result.

1.1 Thinning Process

Any univariate infinitely-divisible (ID) distribution $\mu (dx)$ on ${\mathbb {R}}^1$ is $\mu ^1$ the for a convolution semigroup $\{\mu ^\theta :~\theta \ge 0\}$ and, for $0<\theta <\infty$ and $0<\rho <1$, determines uniquely a “thinning distribution” $\mu ^\theta _\rho (dy\mid x)$ of Y conditional on the sum $X=Y+Z$ of independent $Y\sim \mu ^{\rho \theta }$ and $Z\sim \mu ^{(1{-}\rho )\theta }$. This thinning distribution determines a unique stationary time-reversible Markov process with one-step transition probability distribution given by the convolution

$$ {\textsf {P}}[X_{t+1}\in A\mid {\mathcal {F}}_t ] = \int _{\xi +\zeta \in A} \mu ^{(1{-}\rho )\theta }(d\zeta )~\mu ^\theta _\rho (d\xi \mid X_t) $$

for Borel sets $A\subset {\mathbb {R}}$, where ${\mathcal {F}}_t=\sigma \{X_s:~s\le t\}$ is the minimal filtration. By induction the autocorrelation is ${\textsf {Corr}}(X_s,X_t) =\rho ^{|s-t|}$ for square-integrable $X_t$. The process can be constructed beginning at any $t_0\in {\mathbb {Z}}$ by setting

$$\begin{aligned} \text {X}_{t_{0}}&\sim \mu ^\theta (dx) \end{aligned}$$

(1a)

$$\begin{aligned} \xi _{t}&\sim \mu ^\theta _\rho (d\xi \mid x) \, \text {with} \, x=\left\{ \begin{array}{ll} X_{t-1} & \text {if}\,t>t_0 \\ \text {X}_{t+1} & \, \text {if}\, t<t_0\end{array}\right. \end{aligned}$$

(1b)

$$\begin{aligned} \text {X}_{t}&:= \xi _t+\zeta _t\qquad \text {for}\, \zeta _t\sim \mu ^{\theta (1{-}\rho )}(d\zeta ). \end{aligned}$$

(1c)

Time-reversibility and hence the lack of dependence of this definition on the choice of $t_0$ follows from the argument presented in the proof of Theorem in Sec. () below.

1.1.1 Thinning Example 1: Poisson

For Poisson-distributed $X_t\sim \mu ^\theta =\textsf {Po}(\theta )$ with mean $\theta >0$, for example, the thinning recursion step for $0<\rho <1$ and $t>t_0$ can be written

$$\begin{aligned} X_{t}&= \xi _t + \zeta _t \text {\quad for independent:} \\ \xi _t&\sim \textsf {Bi}\big (X_{t-1}, \rho \big ), \qquad \zeta _t \sim \textsf {Po}\big (\theta (1{-}\rho )\big ) \end{aligned}$$

and hence the joint generating function at two consecutive times is

$$\begin{aligned} \phi (s,z) \,= {\textsf {E}}\Big [ s^{X_{t-1}} z^{X_t}\Big ] \,= \text {exp}\Big \{(s+z-2)\theta (1{-}\rho )+ (s\,z-1)\theta \rho \Big \}. \end{aligned}$$

This was called the “Poisson $\textrm{AR}(1)$ Process” by McKenzie (McKenzie, 1985) and has been studied by many other authors since its introduction.

1.1.2 Thinning Example 2: Negative Binomial

In the thinning process applied to the Negative Binomial $X_t\sim \mu ^\theta =\textsf {NB}(\theta ,p)$ distribution with mean $\theta (1{-}p)/p$, recursion for $t>t_0$ takes the form

$$\begin{aligned} X_{t}&= \xi _t + \zeta _t \text {\quad for independent:}\\ \xi _t&\sim {\textsf{BB}}\big (X_{t-1};~ \theta \rho ,~\theta (1{-}\rho )\big ), \qquad \zeta _t \sim \textsf {NB}\big (\theta (1{-}\rho ),~p\big ) \end{aligned}$$

for beta-binomial distributed $\xi _t \sim {\textsf{BB}}(n;\alpha ,\beta )$ (Johnson et al., 2005, §2.2) with $n=X_{t-1}$, $\alpha =\theta \rho$, and $\beta = \theta (1{-}\rho )$, and negative binomial $\zeta _t \sim \textsf {NB}\big (\theta (1{-}\rho ),p\big )$. Thus the joint generating function is

$$\begin{aligned} \phi (s,z)&= {\textsf {E}}\Big [ s^{X_{t-1}} z^{X_t}\Big ] \nonumber \\&= p^{\theta (2-\rho )} (1-q\,s)^{-\theta (1{-}\rho )}\, (1-q\,z)^{-\theta (1{-}\rho )}\, (1-q\,s\,z)^{-\theta \rho }. \end{aligned}$$

(2)

From this one can compute the conditional generating function

$$ \phi (z\mid x)={\textsf {E}}\left[ z^{X_{t}}\mid X_{t-1}=x\right] = \left( \frac{p}{1-qz}\right) ^{\theta (1{-}\rho )}~ _2F_1(\theta \rho , -x; \theta ; 1-z) $$

where $_2F_1(a,b;c;z)$ denotes Gauss’ hypergeometric function (Abramowitz and Stegun, 1964, §15) and, from this (for comparison below),

$$\begin{aligned} {\textsf {P}}[ X_{t-1}=0, X_{t+1}=0\mid X_t=2]&= [p^{\theta (1{-}\rho )}~ _2F_1(\theta \rho , -x; \theta ; 1)]^2\nonumber \\&= \left[ p^{\theta (1{-}\rho )} (1{-}\rho )\right] ^2 \left[ \frac{1+\theta (1{-}\rho )}{1+\theta }\right] ^2. \end{aligned}$$

(3)

This process, as we will see below in Theorem , is Markov, stationary, and time-reversible, with infinitely-divisible one-dimensional marginal distributions $X_t\sim \textsf {NB}(\theta ,p)$, but the joint marginal distributions at three or more consecutive times are not ID. It appears to have been introduced by Joe (Joe, 1996, p. 665).

1.2 Random Measure Process

Another approach to the construction of processes with specified univariate marginal stationary distribution $\mu ^\theta (dx)$ is to set $X_t := {\mathcal {N}}(G_t)$ for a random measure ${\mathcal {N}}$ and a class of sets $\left\{ G_t\right\}$, as in (Wolpert and Taqqu, 2005, §3.3, 4.4). We begin with a countably additive random measure ${\mathcal {N}}(dx\,dy)$ that assigns independent random variables ${\mathcal {N}}(A_i)\sim \mu ^{|A_i|}$ to disjoint Borel sets $A_i\in {\mathcal {B}} ({\mathbb {R}}^2)$ of finite area $|A_i|$ (this is possible by the Kolmogorov consistency conditions), and a collection of sets

$$ G_t :=\left\{ (x,y):~x\in {\mathbb {R}}, ~0 \le y < \theta \lambda ~e^{-2\lambda |t-x|}\right\} $$

(shown in Fig. ()) whose intersections satisfy $|G_s\cap G_t| = \theta e^{-\lambda |s-t|}$. For $t\in {\mathbb {Z}}$, set

$$\begin{aligned} X_t := {\mathcal {N}}(G_t). \end{aligned}$$

(4)

Fig. 1

Random measure construction of process $X_t={\mathcal {N}}(G_t)$

For any n times $t_1<t_2<\dots <t_n$ the sets $\left\{ G_{t_i}\right\}$ partition ${\mathbb {R}}^2$ into $n(n+1)/2$ sets of finite area (and one with infinite area, $(\cup G_{t_i})^c$), so each $X_{t_i}$ can be written as the sum of some subset of $n(n+1)/2$ independent random variables. In particular, any $n=2$ variables $X_s$ and $X_t$ can be written as

$$ X_s={\mathcal {N}}(G_s\backslash G_t)+{\mathcal {N}}(G_s \cap G_t),\qquad X_t={\mathcal {N}}(G_t\backslash G_s)+{\mathcal {N}}(G_s \cap G_t) $$

just as in the thinning approach, so both 1-dimensional and 2-dimensional marginal distributions for the random measure process coincide with those for the thinning process of Sec. ().

Evidently the process $X_t$ constructed from this random measure is stationary, time-reversible and infinitely divisible in the strong sense that all finite-dimensional marginal distributions are ID. Although the 1- and 2-dimensional marginal distributions of this process coincide with those of the thinning process, the k-dimensional marginals may differ for $k\ge 3$, so this process cannot be Markov. We will see in Theorem below that the only nonnegative integer-valued distribution for which it is Markov is the Poisson.

1.2.1 Random Measure Example 1: Poisson

The conditional distribution of $X_{t_n}={\mathcal {N}}(G_{t_n})$ given $\{X_{t_j}: ~j<n\}$ can be written as the sum of n independent terms, $(n-1)$ of them with binomial distributions (all with the same probability parameter $p=\rho ^{|t_n-t_{n-1}|}$, and with size parameters that sum to $X_{t_{n-1}}$) and one with a Poisson distribution (with mean $\theta (1-\rho ^ {|t_n-t_{n-1}|}$). It follows by induction that the random-measure Poisson process is identical in distribution to the thinning Poisson process of Sec. ().

1.2.2 Random Measure Example 2: Negative Binomial

The random variables $X_1$, $X_2$, $X_3$ for the random measure process built on the Negative Binomial distribution $X_t\sim \textsf {NB}(\theta ,p)$ with autocorrelation $\rho \in (0,1)$ can be written as sums

$$ X_1 = \zeta _1+\zeta _{12}+\zeta _{123} \qquad X_2 = \zeta _2+\zeta _{12}+\zeta _{23}+\zeta _{123} \qquad X_3 = \zeta _3+\zeta _{23}+\zeta _{123} $$

of six independent negative binomial random variables $\zeta _s\sim \textsf {NB}(\theta _s, p)$ with shape parameters

$$ \theta _1=\theta _3=\theta (1{-}\rho ),\quad \theta _2=\theta (1{-}\rho )^2,\quad \theta _{12}=\theta _{23}=\theta \rho (1{-}\rho ),\quad \theta _{123}=\theta \rho ^2 $$

(each $\zeta _s={\mathcal {N}}\big (\cap _{t\in s} G_t\big )$ and $\theta _s=|\cap _{t\in s} G_t|$ in Fig. ()). It follows that the conditional probability

$$\begin{aligned} {\textsf {P}}[ X_1=0,~X_3=0\mid X_2=2]= & {\textsf {P}}[\zeta _1=\zeta _{12}=\zeta _{123}=\zeta _{23}=\zeta _3=0 \mid \zeta _2+\zeta _{12}\nonumber \\ & +\zeta _{23}+\zeta _{123}=2] \nonumber \\= & \frac{{\textsf {P}}[\zeta _2=2, \text { all other } \zeta _s=0]}{{\textsf {P}}[X_2=2]} \nonumber \\= & \left[ p^{\theta (1{-}\rho )}(1{-}\rho )\right] ^2{\frac{1+\theta (1{-}\rho )^2}{1+\theta }} \end{aligned}$$

(5)

differs from that of the thinning negative binomial process in Eqn. () for all $\theta >0$ and $\rho >0$. Thus this process is stationary, time-reversible, and has infinitely-divisible marginal distributions of all orders, but it cannot be Markov since its 2-dimensional marginal distributions coincide with those of the Markov thinning process but its 3-dimensional marginal distributions do not.

In Sec. () of this paper we characterize every discrete-time process that is Markov, Infinitely-divisible, Stationary, and Time-reversible with non-negative Integer values (MISTI for short). In Sec. () we first present the necessary definitions and preliminary results; in Sec. () we extend the results to continuous time, with discussion in Sec. ().

2 MISTI Processes

A real-valued stochastic process $X_t$ indexed by $t\in {\mathbb {Z}}$ is stationary if each finite-dimensional marginal distribution

$$ {\mu _{\scriptscriptstyle T}}(B):={\textsf {P}}\left[ X_T\in B\right] $$

satisfies

$$\begin{aligned} {\mu _{\scriptscriptstyle T}}(B)=\mu _{s+{\scriptscriptstyle T}}(B) \end{aligned}$$

(6a)

for each set $T\subset {\mathbb {Z}}$ of finite cardinality $|T|<\infty$, Borel set $B\in {\mathcal {B}} ({\mathbb {R}}^{|T|})$, and $s\in {\mathbb {Z}}$, where as usual “$s+T$” denotes $\{(s+t):~t\in T\}$. A stationary process is time-reversible if also

$$\begin{aligned} {\mu _{\scriptscriptstyle T}}(B)=\mu _{-{\scriptscriptstyle T}}(B) \end{aligned}$$

(6b)

(where “$-T$” is $\{-t:~t\in T\}$) and Markov if for every $t\in {\mathbb {Z}}$ and finite $T\subset \{s\in {\mathbb {Z}}:~s\ge t\}$,

$$\begin{aligned} {\textsf {P}}[ X_T\in B\mid {\mathcal {F}}_t] = {\textsf {P}}[ X_T\in B\mid X_t] \end{aligned}$$

(6c)

for all $B\in {\mathcal {B}} ({\mathbb {R}}^{|T|})$, where ${\mathcal {F}}_t:=\sigma \{X_s:~s\le t\}$. The process $X_t$ is Infinitely Divisible (ID) or, more specifically, multivariate infinitely divisible (MVID) if each ${\mu _{\scriptscriptstyle T}}$ is the n-fold convolution of some other distribution $\mu _{\scriptscriptstyle T}^{(1/n)}$ for each $n\in {\mathbb {N}}$. This is more restrictive than requiring only that the one-dimensional marginal distributions be ID and, for integer-valued processes that satisfy

$$\begin{aligned} {\mu _{\scriptscriptstyle T}}({\mathbb {Z}}^{|T|})=1, \end{aligned}$$

(6d)

it is equivalent by the Lévy-Khinchine formula (Rogers and Williams, 2000, p. 74) to the condition that each ${\mu _{\scriptscriptstyle T}}$ have characteristic function of the form

$$\begin{aligned} \int _{{\mathbb {R}}^{|T|}} e^{i\omega 'x}\,{\mu _{\scriptscriptstyle T}}(dx) = \exp \left\{ \int _{{\mathbb {Z}}^{|T|}} \big (e^{i\omega 'u}-1\big )\, \nu _{\scriptscriptstyle T}(du)\right\} ,\qquad \omega \in {\mathbb {R}}^{|T|} \end{aligned}$$

(6e)

for some finite measure $\nu _{\scriptscriptstyle T}$ on ${\mathcal {B}} ({\mathbb {Z}}^{|T|})$. Call a process $X_t$ or its distributions ${\mu _{\scriptscriptstyle T}}(du)$ MISTI if it is Markov, nonnegative Integer-valued, Stationary, Time-reversible, and Infinitely divisible, i.e., satisfies Eqns. (–). We now turn to the problem of characterizing all MISTI distributions.

2.1 Three-dimensional Marginals

By stationarity and the Markov property all MISTI finite-dimensional distributions ${\mu _{\scriptscriptstyle T}}(du)$ are determined completely by the marginal distribution for $X_t$ at two consecutive times; to exploit the MVID property we will study the three-dimensional marginal distribution for $X_t$ at any set T of ${|T|}=3$ consecutive times— say, $T=\{1,2,3\}$. By Eqn. () we can represent $X_{\{1,2,3\}}$ in the form

$$ X_1 = \sum i {N_{i++}}\qquad X_2 = \sum j {N_{+j+}}\qquad X_3 = \sum k {N_{++k}} $$

for independent Poisson-distributed random variables

$$\begin{aligned} {N_{ijk}} \mathrel {\mathop {\sim }\limits ^{\textrm{ind}}}\textsf {Po}({\lambda _{\,ijk}}) \end{aligned}$$

with means ${\lambda _{\,ijk}} := \nu (\{(i,j,k)\})$; here and hereafter, a subscript “$+$” indicates summation over the entire range of that index— ${\mathbb {N}}_0=\left\{ 0,2\dots \right\}$ for $\left\{ {N_{ijk}}\right\}$ and $\left\{ {\lambda _{\,ijk}}\right\}$, ${\mathbb {N}}=\left\{ 1,2,\dots \right\}$ for $\left\{ {\theta _{j}}\right\}$. The sums ${\theta _{j}}:={\lambda _{\,+j+}}$ for $j\ge 1$ characterize the univariate marginal distribution of each $X_t$— for example, through the probability generating function (pgf)

$$ \varphi ({z}) := {\textsf {E}}[ {z}^{X_t} ] = \exp \left[ \sum \nolimits _{j\ge 1}\big ({z}^j-1\big ){\theta _{j}}\right] . $$

To avoid trivial technicalities we will assume that $0 < {\textsf {P}}[X_t=1]=\varphi '(0) ={\theta _{1}}e^{-{\theta _{+}}}$, i.e., ${\theta _{1}}>0$. Now set ${r_{i}}:={\lambda _{\,i1+}}/{\theta _{1}}$, and for later use define functions:

$$\begin{aligned} \psi _j({s},{t}) := \sum _{i,k\ge 0}{s}^i{t}^k{\lambda _{\,ijk}}\quad p({s}) := \psi _1(s,1)/{\theta _{1}}=\sum _{i\ge 0}{s}^i \,{r_{i}}\quad P({z}) := \sum _{j\ge 1}{z}^j \,{\theta _{j}}. \end{aligned}$$

(7)

Since ${r_{i}}$ and ${\theta _{j}}$ are nonnegative and summable (by Eqns. (, )), $p({s})$ and $P({z})$ are analytic on the open unit ball ${\mathbb {U}}\subset {\mathbb {C}}$ and continuous on its closure. Similarly, since ${\lambda _{\,ijk}}$ is summable, each $\psi _j({s},{t})$ is analytic on ${\mathbb {U}}^2$ and continuous on its closure. Note $\psi _j(1,1)={\theta _{j}}$, $p(0)={r_{0}}$ and $p(1)=1$, while $P(0)=0$ and $P(1)={\theta _{+}}$; also $\varphi ({z}) = \exp \left\{ P({z})-{\theta _{+}}\right\}$. Each $\psi _j(s,t)=\psi _j(t,s)$ is symmetric by Eqn. (), as are the conditional probability generating functions:

$$\begin{aligned} {\varphi ({s},{t}\mid z)} := {\textsf {E}}\big [{s}^{X_1}{t}^{X_3}\mid X_2=z\big ]. \end{aligned}$$

2.1.1 Conditioning on $X_2=0$

By the Markov property Eqn. (), $X_1$ and $X_3$ must be conditionally independent given $X_2$, so the conditional probability generating function must factor:

$$\begin{aligned} {\varphi ({s},{t}\mid 0)}&:= {\textsf {E}}\big [{s}^{X_1}{t}^{X_3}\mid X_2=0\big ] = {\textsf {E}}\big [{s}^{\sum _{i\ge 0}i{N_{i0+}}}~ {t}^{\sum _{k\ge 0}k{N_{+0k}}}\big ]\nonumber \\&= \text {exp}\Big \{\sum _{i,k\ge 0}\big ({s}^i{t}^k-1\big ){\lambda _{\,i0k}}\Big \}\nonumber \\&\equiv \varphi ({s},1\mid 0)~ \varphi (1,{t}\mid 0). \end{aligned}$$

(8)

Taking logarithms,

$$\begin{aligned} {\sum \big ({s}^i{t}^k-1\big ){\lambda _{\,i0k}}}&\equiv {\sum \big ({s}^i-1\big ){\lambda _{\,i0k}}} +{\sum \big ({t}^k-1\big ){\lambda _{\,i0k}}}\nonumber \end{aligned}$$

or, for all ${s}$ and ${t}$ in the unit ball in ${\mathbb {C}}$,

$$\begin{aligned} 0&\equiv \sum ({s}^i-1)({t}^k-1\big ){\lambda _{\,i0k}}. \end{aligned}$$

(9)

Thus ${\lambda _{\,i0k}}=0$ whenever both $i>0$ and $k>0$ and, by symmetry,

$$\begin{aligned} \varphi (1,z\mid 0)=\varphi (z,1\mid 0)&=\exp \left\{ \sum \nolimits _{i\ge 0}(z^i-1){\lambda _{\,i00}}\right\} .\nonumber \end{aligned}$$

2.1.2 Conditioning on $X_2=1$

Similarly

$$\begin{aligned} {\varphi ({s},{t}\mid 1)} := {\textsf {E}}\big [{s}^{X_1}{t}^{X_3}\mid X_2=1\big ]&= {\textsf {E}}\big [{s}^{\sum _{i\ge 0}i({N_{i0+}} + {N_{i1+}})}\quad {t}^{\sum _{k\ge 0}k({N_{+0k}} + {N_{+1k}})} \mid {N_{+1+}}=1\big ]\\&= {\varphi ({s},{t}\mid 0)} {\textsf {E}}\big [{s}^{\sum _{i\ge 0}i{N_{i1+}}}\quad {t}^{\sum _{k\ge 0}k{N_{+1k}}} \mid {N_{+1+}}=1\big ]\\&= {\varphi ({s},{t}\mid 0)} \left\{ \sum _{i,k\ge 0}{s}^i{t}^k\,\big [{\lambda _{\,i1k}}/{\lambda _{\,+1+}}\big ]\right\} \end{aligned}$$

since $\left\{ {N_{i1k}}\right\}$ is conditionally multinomial given ${N_{+1+}}$ and independent of $\left\{ {N_{i0k}}\right\}$. By the Markov property this too must factor, as $\varphi ({s},{t}\mid 1)=\varphi ({s},1\mid 1) \,\varphi (1,{t}\mid 1)$, so by Eqn. ()

$$\begin{aligned} {\theta _{1}} \left\{ \sum \nolimits _{i,k\ge 0}{s}^i{t}^k{\lambda _{\,i1k}}\right\}&= \left\{ \sum \nolimits _{i\ge 0}{s}^i {\lambda _{\,i1+}}\right\} \left\{ \sum \nolimits _{k\ge 0}{t}^k {\lambda _{\,+1k}}\right\} \end{aligned}$$

or, since ${\lambda _{\,i1k}}={\lambda _{\,k1i}}$ by Eqns. (, ),

$$\begin{aligned} \psi _1({s},{t})&:= \sum \nolimits _{i,k\ge 0}{s}^i{t}^k {\lambda _{\,i1k}} = {\theta _{1}} p({s})\,p({t}),\\ {\varphi ({s},{t}\mid 1)}&= {\varphi ({s},{t}\mid 0)}\, p({s})\, p({t}). \end{aligned}$$

2.1.3 Conditioning on $X_2=2$

The event $\left\{ X_2=2\right\}$ for $X_2:=\sum _{j\ge 1}j{N_{+j+}}$ can happen in two ways: either ${N_{+1+}} =2$ and each ${N_{+j+}}=0$ for $j\ge 2$, or ${N_{+2+}}=1$ and ${N_{+j+}}=0$ for $j=1$ and $j\ge 3$, with ${N_{+0+}}$ unrestricted in each case. These two events have probabilities $({\theta _{1}}^2/2)e^{-{\theta _{+}}}$ and $({\theta _{2}})e^{-{\theta _{+}}}$, respectively, so the joint generating function for $\{X_1,X_3\}$ given $X_2=2$ is

$$\begin{aligned} {\varphi ({s},{t}\mid 2)}:= & {\textsf {E}}\big [{s}^{X_1}{t}^{X_3}\mid X_2=2\big ]\nonumber \\= & {\textsf {E}}\big [{s}^{\sum _{i\ge 0}i({N_{i0+}} + {N_{i1+}} + {N_{i2+}})}\quad {t}^{\sum _{k\ge 0}k({N_{+0k}} + {N_{+1k}} + {N_{+2k}})} \mid {N_{+1+}}+2{N_{+2+}}=2\big ]\nonumber \\= & {\varphi ({s},{t}\mid 0)}\left\{ \frac{{\theta _{1}}^2/2}{{\theta _{1}}^2/2+{\theta _{2}}} \left[ \sum _{i,k\ge 0}{s}^i{t}^k{\lambda _{\,i1k}}/{\lambda _{\,+1+}}\right] ^2 \right. \nonumber \\ & \left. + \frac{{\theta _{2}}}{{\theta _{1}}^2/2+{\theta _{2}}} \left[ \sum _{i,k\ge 0}{s}^i{t}^k{\lambda _{\,i2k}}/{\lambda _{\,+2+}}\right] \right\} \nonumber \\= & \frac{{\varphi ({s},{t}\mid 0)}}{{\theta _{1}}^2/2+{\theta _{2}}}\left\{ \frac{{\theta _{1}}^2}{2} \left[ \sum _{i,k\ge 0}{s}^i{t}^k{\lambda _{\,i1k}}/{\theta _{1}}\right] ^2 + {\theta _{2}} \left[ \sum _{i,k\ge 0}{s}^i{t}^k{\lambda _{\,i2k}}/{\theta _{2}}\right] \right\} \nonumber \\= & \frac{{\varphi ({s},{t}\mid 0)}}{{\theta _{1}}^2/2+{\theta _{2}}}~ \left\{ \frac{{\theta _{1}}^2}{2} p({s})^2 p({t})^2 + \psi _2(s,t)\right\} . \end{aligned}$$

(10)

In view of Eqn. (), this will factor in the form $\varphi ({s},{t}\mid 2)= \varphi ({s},1\mid 2)\, \varphi (1,{t}\mid 2)$ as required by Markov property Eqn. () if and only if for all s, t in the unit ball:

$$\begin{aligned} \left[ \frac{{\theta _{1}}^2}{2}+{\theta _{2}}\right] \left[ \frac{{\theta _{1}}^2}{2} p({s})^2 p({t})^2 + \psi _2(s,t)\right]&= \left[ \frac{{\theta _{1}}^2}{2} p({s})^2 + \psi _2(s,1)\right] \left[ \frac{{\theta _{1}}^2}{2} p({t})^2 + \psi _2(1,t)\right] \end{aligned}$$

or

$$\begin{aligned} \frac{{\theta _{1}}^2}{2} \Big [{\theta _{2}} p(s)^2 p(t)^2 - p(s)^2 \psi _2(1,t) - \psi _2(s,1) p(t)^2 + \psi _2(s,t)\Big ] \\ = \Big [ \psi _2(s,1)\,\psi _2(1,t)-{\theta _{2}}\psi _2(s,t) \Big ]. \end{aligned}$$

To satisfy the ID requirement of Eqn. (), this must hold with each ${\theta _{j}}$ replaced by ${\theta _{j}}/n$ for each integer $n\in {\mathbb {N}}$. Since the left and right sides are homogeneous in $\theta$ of degrees 3 and 2 respectively, this will only happen if each square-bracketed term vanishes identically, i.e., if

$$\begin{aligned} {\theta _{2}} \psi _2(s,t) \equiv \psi _2(s,1)\psi _2(1,t) \end{aligned}$$

and

$$\begin{aligned} 0&= {\theta _{2}} \Big [{\theta _{2}} p(s)^2 p(t)^2 - p(s)^2 \psi _2(1,t) - \psi _2(s,1) p(t)^2\Big ] + \psi _2(s,1) \psi _2(1,t)\nonumber \\&= \left[ {\theta _{2}} p(s)^2-\psi _2(s,1)\right] \left[ {\theta _{2}} p(t)^2-\psi _2(1,t)\right] , \end{aligned}$$

so

$$\begin{aligned} \psi _2(s,t)&:= \sum _{i,k\ge 0}{s}^i{t}^k{\lambda _{\,i2k}} = {\theta _{2}} p(s)^2\,p(t)^2,\nonumber \\ {\varphi ({s},{t}\mid 2)}&= {\varphi ({s},{t}\mid 0)}\, p({s})^2 p({t})^2. \end{aligned}$$

2.1.4 Conditioning on $X_2=j$

The same argument applied recursively, using the Markov property for each $j\ge 1$ in succession, leads to:

$$\begin{aligned} \Big [\frac{{\theta _{1}}^j}{j!}+\dots +{\theta _{1}}{\theta _{j-1}}\Big ] \Big [{\theta _{j}} p(s)^j p(t)^j - p(s)^j\psi _j(1,t) -\psi _j(s,1)p(t)^j + \psi _j(s,t)\Big ] \\ = \Big [\psi _j(s,1)\psi _j(1,t)- {\theta _{j}} \psi _j(s,t)\Big ] \end{aligned}$$

so

$$\begin{aligned} \psi _j(s,t) := \sum _{i,k\ge 0}{s}^i{t}^k{\lambda _{\,ijk}} = {\theta _{j}}\, p({s})^j p({t})^j,\qquad j\ge 1 \end{aligned}$$

(11)

and consequently

$$\begin{aligned} {\varphi ({s},{t}\mid j)} = {\textsf {E}}\big [{s}^{X_1}{t}^{X_3}\mid X_2=j\big ] = \left[ \varphi ({s},1\mid 0)\, p({s})^j\right] ~ \left[ \varphi (1,{t}\mid 0)\, p({t})^j\right] . \end{aligned}$$

Conditionally on $\left\{ X_2=j\right\}$, $X_1$ and $X_3$ are distributed independently, each as the sum of j independent random variables with generating function p(s), plus one with generating function $\varphi (s,1\mid 0)$— so $X_t$ is a branching process (Harris, 1963) whose unconditional three-dimensional marginal distributions have generating function:

$$\begin{aligned} \varphi ({s},{z},{t})&:= {\textsf {E}}\big [{s}^{X_1}{z}^{X_2}{t}^{X_3}\big ]\nonumber \\&= {\varphi ({s},{t}\mid 0)} \sum _{j\ge 0} {z}^j p({s})^j p({t})^j {\textsf {P}}[X_2=j]\nonumber \\&= {\varphi ({s},{t}\mid 0)} {\textsf {E}}\left[ z p({s}) p({t})\right] ^{X_2}\nonumber \\&= {\varphi ({s},{t}\mid 0)} \varphi \big ( z p({s}) p({t})\big )\nonumber \\&= {\varphi ({s},{t}\mid 0)} \exp \big [P\big ( z p({s}) p({t})\big )-{\theta _{+}}\big ]. \end{aligned}$$

(12)

See Secs. and for further development of this branching process representation.

2.2 Stationarity

Without loss of generality we may take ${\lambda _{\,000}} = 0$. By Eqn. () with ${s}=0$ and ${t}=1$ we have ${\lambda _{\,0j+}}={\theta _{j}}{r_{0}}^j$; by Eqn. () we have ${\lambda _{\,i00}} = {\lambda _{\,i0+}}$. By time-reversibility we conclude that ${\lambda _{\,i00}} = 0$ for $i=0$ and, for $i\ge 1$,

$$\begin{aligned} {\lambda _{\,i00}} = {\theta _{i}} {r_{0}}^i . \end{aligned}$$

(13)

Now we can evaluate

$$ \varphi ({s},{t}\mid 0) = \exp \left\{ P({s}\,{r_{0}})+P({t}\,{r_{0}})-2P({r_{0}})\right\} $$

and, from this and Eqn. (), evaluate the joint generating function for $X_{\{1,2,3\}}$ as:

$$\begin{aligned} \varphi ({s},{z},{t}) =\text {exp}\left\{ P\big ( z\, p({s}) p({t})\big )-{\theta _{+}} +P({s}\,{r_{0}})+P({t}\,{r_{0}})-2P({r_{0}})\right\} ,\quad j\ge 1 \end{aligned}$$

(14)

and so that for $X_{\{1,2\}}$ as:

$$\begin{aligned} \varphi ({s},{z},1) =\text {exp}\left\{ P\big ( z\, p({s})\big )-{\theta _{+}} + P({s}\,{r_{0}})-P({r_{0}})\right\} . \end{aligned}$$

(15)

Now consider Eqn. () with ${t}=1$,

$$\begin{aligned} \sum _{i\ge 0}{s}^i{\lambda _{\,ij+}} = {\theta _{j}}\, p({s})^j. \end{aligned}$$

(16)

It follows first for $j=1$ and then for $i=1$ that

$$\begin{aligned} {\lambda _{\,i1+}}&= {\theta _{1}} {r_{i}}&i&\ge 1 \\ {\lambda _{\,1j+}}&= {\theta _{j}} [ j {r_{0}}^{j-1} {r_{1}} ]&j&\ge 1 \end{aligned}$$

so again by time reversibility with $i=j$, since ${\theta _{1}}>0$, we have

$$\begin{aligned} {r_{j}} = {\theta _{j}} [j\, {r_{0}}^{j-1} {r_{1}}]/ {\theta _{1}}\qquad j\ge 1. \end{aligned}$$

(17)

Thus ${r_{0}}$, ${r_{1}}$, and $\{{\theta _{j}}\}$ determine all the $\{{r_{j}}\}$ and so all the $\{{\lambda _{\,ijk}}\}$ by Eqns. (, ) and hence the joint distribution of $\{X_t\}$.

Now consider Eqn. () first for $j=2$ and then $i=2$:

$$\begin{aligned} \sum _{i\ge 0}{s}^i {\lambda _{\,ij+}}&= {\theta _{j}}\,\left[ \sum \nolimits _{i\ge 0}{s}^i {r_{i}}\right] ^j \\ {\lambda _{\,i2+}}&= {\theta _{2}} \sum _{k=0}^i {r_{k}} {r_{i-k}}&i&\ge 2 \\ {\lambda _{\,2j+}}&= {\theta _{j}} \left[ j {r_{0}}^{j-1} {r_{2}} + \left( {\begin{array}{c}j\\ 2\end{array}}\right) {r_{0}}^{j-2} {r_{1}}^2 \right]&j&\ge 2. \end{aligned}$$

Equating these for $i=j\ge 2$ (by time-reversibility) and applying Eqn. () for $0< k<i$ (the cases $k=0$ and $k=i$ need to be handled separately),

$$\begin{aligned} {r_{0}}^{i-2} {r_{1}}^2 \left[ {\theta _{2}} \sum _{0<k<i}{\theta _{k}}{\theta _{i-k}} k(i-k)-{\theta _{i}}\frac{i(i-1)}{2} {\theta _{1}}^2\right] =0. \end{aligned}$$

(18)

3 The Solutions

Eqn. () holds for all $i\ge 2$ if ${r_{0}}=0$ or ${r_{1}}=0$, leaving ${r_{j}}=0$ by Eqn. () for all $j\ge 2$, hence ${r_{0}}+{r_{1}}=1$ and $\left\{ {\theta _{j}}\right\}$ is restricted only by the conditions ${\theta _{1}}>0$ and ${\theta _{+}}<\infty$.

3.1 The Constant Case

The case ${r_{0}}=0$ leads to ${r_{1}}=1$ and ${r_{j}}=0$ for all $j\ne 1$, so $p(z)\equiv z$. By Eqn. () the joint pgf is

$$ \varphi ({s},{z},{t}) =\exp \left\{ P( {s}\,{z}\,{t})-{\theta _{+}}\right\} , $$

so $X_1=X_2=X_3$ and all $\{X_t\}$ are identical, with an arbitrary ID distribution.

3.2 The IID Case

The case ${r_{1}}=0$ leads to ${r_{0}}=1$ and ${r_{j}}=0$ for all $j\ne 0$ so $p(z)\equiv 1$ and

$$\begin{aligned} \varphi ({s},{z},{t}) =\exp \left\{ P({s})+P({z})+P({t})-3{\theta _{+}}\right\} \end{aligned}$$

by Eqn. (), making all $\{X_t\}$ independent, with identical but arbitrary ID distributions.

3.3 The Poisson Case

Aside from these two degenerate cases, we may assume ${r_{0}}>0$ and ${r_{1}}>0$, and (by Eqn. ()) rewrite Eqn. () in the form:

$$\begin{aligned} {r_{i}} = \frac{{r_{2}}}{{r_{1}}^2(i-1)}\sum _{k=1}^{i-1} {r_{k}} {r_{i-k}},\quad i\ge 2, \nonumber \end{aligned}$$

whose unique solution for all integers $i\ge 1$ (by induction) is

$$\begin{aligned} {r_{i}} = {r_{1}}({r_{2}}/{r_{1}})^{i-1}. \end{aligned}$$

(19)

If ${r_{2}}=0$, then again ${r_{i}}=0$ for all $i\ge 2$ but, by Eqn. (), ${\theta _{j}}=0$ for all $j\ge 2$; thus $P(z)={\theta _{1}} z$ so each $X_t\sim \textsf {Po}({\theta _{1}})$ has a Poisson marginal distribution with mean ${\theta _{1}}={\theta _{+}}$. In this case ${r_{0}}+{r_{1}}=1$, $p(z)={r_{0}}+{r_{1}}z$, and the two-dimensional marginals (by Eqn. ()) of $X_1$, $X_2$ have joint pgf

$$\begin{aligned} \varphi ({s},{z})&= \text {exp}\left\{ P\big ( z\, p({s})\big )-{\theta _{+}} +P({s}\,{r_{0}})-P({r_{0}})\right\} \\&= \text {exp}\left\{ {\theta _{1}}{r_{0}}(s+z-2)+{\theta _{1}}{r_{1}}(sz-1)\right\} , \nonumber \end{aligned}$$

(20)

the bivariate Poisson distribution (Johnson et al., 1997, § 37.2), so $X_t$ is the familiar “Poisson $\textrm{AR}(1)$ Process” of McKenzie (McKenzie, 1985; McKenzie, 1988) (with autocorrelation $\rho ={r_{1}}$) considered in Sec. (). Its connection with Markov branching processes was recognized earlier (Steutel et al., 1983). By Eqn. () the conditional distribution of $X_{t+1}$, given ${\mathcal {F}}_t :=\sigma \left\{ X_s:~s\le t\right\}$, is that of the sum of $X_t$ independent Bernoulli random variables with pgf p(s) and a Poisson innovation term with pgf $\exp \{P({r_{0}} s)-P({r_{0}})\}$, so the Markov process $X_t$ may be written recursively starting at any $t_0$ as

$$\begin{aligned} X_{t_0}&\sim \textsf {Po}({\theta _{+}}) \\ X_{t}&= \xi _t+\zeta _t,\text {\quad where } \xi _t \sim \textsf {Bi}(X_{t-1}, {r_{1}}) \text { and } \zeta _t \sim \textsf {Po}({\theta _{t}}{r_{0}}) \end{aligned}$$

(all independent) for $t> t_0$, the thinning construction of Sec. ().

3.4 The Negative Binomial case

Finally if ${r_{0}}>0$, ${r_{1}}>0$, and ${r_{2}}>0$, then (by Eqn. ()) ${r_{i}}= {r_{1}}(q{r_{0}})^{i-1}$ for $i\ge 1$ and hence (by Eqn. ()) ${\theta _{j}}=\alpha q^j/j$ for $j\ge 1$ with $q:=(1-{r_{0}}-{r_{1}})/{r_{0}}(1-{r_{0}})$ and $\alpha := {\theta _{1}}/q$. The condition ${\theta _{+}}<\infty$ entails $q<1$ and ${\theta _{+}}=-\alpha \log (1{-}q)$. The 1-marginal distribution is $X_t\sim \textsf {NB}(\alpha ,p)$ with $p:=(1{-}q)$, and the functions $P(\cdot )$ and $p(\cdot )$ are $P(z)=-\alpha \log (1-qz)$, $p(s)={r_{0}}+{r_{1}} s/(1-q{r_{0}} s)$, so the joint pgf for the 2-marginal distribution of $X_1,X_2$ is

$$\begin{aligned} \varphi ({s},{z})&= \text {exp}\left\{ P\big ( z\, p({s})\big )-{\theta _{+}} +P({s}\,{r_{0}})-P({r_{0}})\right\} \nonumber \\&= p^{2\alpha }[(1-q\rho )-q(1{-}\rho )({s}+{z}) +q(q-\rho ){s}{z}]^{-\alpha } \end{aligned}$$

(21)

with one-step autocorrelation $\rho :=(1{-}{r_{0}})^2/{r_{1}}$. This bivariate distribution was introduced as the “compound correlated bivariate Poisson”(Edwards and Gurland, 1961), but we prefer to call it the Branching Negative Binomial distribution. In the branching formulation $X_{t}$ may be viewed as the sum of $X_{t-1}$ iid random variables with pgf $p(s)={r_{0}} +{r_{1}} s/(1-q{r_{0}} s)$ and one with pgf $\exp \left\{ P(s{r_{0}})-P({r_{0}})\right\} =(1-q{r_{0}})^\alpha (1-q{r_{0}}\,s)^{-\alpha }$. The first of these may be viewed as $Y_t$ plus a random variable with the $\textsf {NB}(Y_t,1{-}q{r_{0}})$ distribution, for $Y_t\sim \textsf {Bi}(X_{t-1},1-{r_{0}})$, and the second has the $\textsf {NB}(\alpha ,1{-}q{r_{0}})$ distribution, so a recursive updating scheme beginning with $X_{t_0} \sim \textsf {NB}(\alpha ,p)$ is:

$$ X_t = Y_t+\zeta _t,\text {\quad where } Y_t \sim \textsf {Bi}(X_{t-1}, ~1{-}{r_{0}})\text { and } \zeta _t\sim \textsf {NB}(\alpha +Y_t,~1{-}q{r_{0}}). $$

In the special case of $\rho =q$ the joint pgf simplifies to $\varphi ({s},{z}) = p^\alpha [1+q(1-{s}-{z})]^{-\alpha }$ and the joint distribution of $X_1,X_2$ reduces to the negative trinomial distribution (Johnson et al., 1997, Ch. 36) with pmf

$$\begin{aligned} {\textsf {P}}[X_1=i,X_2=j]&= \frac{\Gamma (\alpha +i+j)}{\Gamma (\alpha )~i!~j!} \left( \frac{1-q}{1+q}\right) ^\alpha \,\left( \frac{q}{1+q}\right) ^{i+j} \end{aligned}$$

and simple recursion $X_t\mid X_{t-1}\sim \textsf {NB}\big (\alpha +X_{t-1}, ~\frac{1}{1+q}\big )$.

3.5 Results

We have just proved:

Theorem 1

Let $\left\{ X_t\right\}$ be a Markov process indexed by $t\in {\mathbb {Z}}$ taking values in the non-negative integers ${\mathbb {N}}_0$ that is stationary, time-reversible, has infinitely-divisible marginal distributions of all finite orders, and satisfies ${\textsf {P}}[X_t=1]>0$. Then $\left\{ X_t\right\}$ is one of four processes:

• 1. $X_t\equiv X_0\sim \mu _0(dx)$ for an arbitrary ID distribution $\mu _0$ on ${\mathbb {N}}_0$ with $\mu _0(\{1\})>0$;

• 2. $X_t\mathrel {\mathop {\sim }\limits ^{\textrm{iid}}}\mu _0(dx)$ for an arbitrary ID distribution $\mu _0$ on ${\mathbb {N}}_0$ with $\mu _0(\{1\})>0$;

• 3. For some $\theta >0$ and $0<\rho <1$, $X_t\sim \textsf {Po}(\theta )$ with bivariate joint generating function

  $$ {\textsf {E}}\left[ {s}^{X_1}~{z}^{X_2}\right] = \exp \left\{ \theta (1{-}\rho )({s}-1) + \theta (1{-}\rho )({z}-1) + \theta \rho ({s}{z}-1) \right\} $$

  and hence autocorrelation ${\textsf {Corr}}(X_s,X_t)= \rho ^{|s-t|}$ and recursive update

  $$ X_t = \xi _t+\zeta _t,\text {\quad where } \xi _t \sim \textsf {Bi}(X_{t-1}, ~\rho )\text { and } \zeta _t\sim \textsf {Po}\big (\theta (1{-}\rho )); $$

• 4. For some $\alpha >0$, $0<p<1$, and $0<\rho <1$, $X_t\sim \textsf {NB}(\alpha ,p)$, with bivariate joint generating function

  $$ {\textsf {E}}\left[ {s}^{X_1}~{z}^{X_2}\right] = p^{2\alpha }[(1-q\rho )-q(1{-}\rho )(s+z)+q(q-\rho )sz]^{-\alpha } $$

  where $q=1{-}p$, and hence autocorrelation ${\textsf {Corr}}(X_s,X_t)= \rho ^{|s-t|}$ and recursive update

  $$ X_t = Y_t+\zeta _t,\text {\quad where } Y_t \sim \textsf {Bi}\big (X_{t-1}, ~\rho \,p/(1-\rho q)\big )\text { and } \zeta _t\sim \textsf {NB}\big (\alpha +Y_t,~p/(1-\rho q)\big ). $$

Note the limiting cases of autocorrelation $\rho =1$ and $\rho =0$ in cases 3., 4. are subsumed by the degenerate cases 1. and 2., respectively. The theorem follows from this.

From this theorem follows:

Theorem 2

Let $\left\{ \mu ^\theta :~\theta \ge 0\right\}$ be an ID semigroup of probability distributions on the nonnegative integers ${\mathbb {N}}_0$ with $\mu ^\theta (\{1\})>0$. Fix $\theta >0$ and $0<\rho <1$ and let $\left\{ X_t\right\}$ be the “thinning process” of Eqn. (1) in Sec. () with the representation

$$\begin{aligned} X_{t-1}=\xi _t+\eta _t \qquad X_t=\xi _t+\zeta _t \end{aligned}$$

(22)

for each $t\in {\mathbb {Z}}$ with independent

$$ \xi _t\sim \mu ^{\rho \theta }(d\xi )\qquad \eta _t\sim \mu ^{(1{-}\rho )\theta }(d\eta )\qquad \zeta _t\sim \mu ^{(1{-}\rho )\theta }(d\zeta ).$$

Then $X_t$ is Markov, stationary, time-reversible, and nonnegative integer valued, but it does not have infinitely-divisible marginal distributions of all orders unless $\{\mu ^\theta \}$ is the Poisson family.

Proof

By construction $X_t$ is obviously Markov and stationary. The joint distribution of the process at consecutive times is symmetric (see Eqn. ()) since the marginal and conditional pmfs

$$ p(x) := \mu ^\theta (\{x\}),\quad q(y\mid x) := \frac{\sum _z \mu ^{\rho \theta }(\{z\}) ~\mu ^{(1{-}\rho )\theta }(\{x-z\}) ~\mu ^{(1{-}\rho )\theta }(\{y-z\})}{\mu ^{\theta }(\{x\})} $$

of $X_t$ and $X_t\mid X_{t-1}$ satisfy the symmetric relation

$$\begin{aligned} p(x)~q(y\mid x) = q(x\mid y)~ p(y). \end{aligned}$$

Applying this inductively, for any $s<t$ and any $\{x_s,\cdots ,x_t\}\subset {\mathbb {N}}_0$ we find

$$\begin{aligned} {\textsf {P}}[X_s=x_s,\cdots ,X_t=x_t]&= p(x_s)\hspace{4mm} q(x_{s+1}\mid x_s) \hspace{1mm} q(x_{s+2}\mid x_{s+1})\cdots q(x_t\mid x_{t-1}) \\&= q(x_s\mid x_{s+1}) p(x_{s+1}) \hspace{1mm} q(x_{s+2}\mid x_{s+1}) \cdots q(x_t\mid x_{t-1}) \\&=\cdots \\&= q(x_s\mid x_{s+1}) q(x_{s+1}\mid x_{s+2})\cdots q(x_{t-1}\mid x_t)\hspace{1mm} p(x_t), \end{aligned}$$

and so the distribution of $X_t$ is time-reversible. Now suppose that it is also ID. Then by Theorem it must be one of the four specified processes: constant, iid, branching Poisson, or branching negative binomial.

Since $\rho <1$ it cannot be the constant $\left\{ X_t\equiv X_0\right\}$ process; since $\rho >0$ it cannot be the independent $\left\{ X_t\mathrel {\mathop {\sim }\limits ^{\textrm{iid}}}\mu ^\theta (dx)\right\}$ process. The joint generating function $\phi (s,z)$ at two consecutive times for the negative binomial thinning process, given in Eqn. (), differs from that for the negative binomial branching process, given in Eqn. (). The only remaining option is the Poisson branching process of Sec. (). $\square$

Theorem 3

Let $\left\{ \mu ^\theta :~\theta \ge 0\right\}$ be an ID semigroup of probability distributions on the nonnegative integers ${\mathbb {N}}_0$ with $\mu ^\theta (\{1\})>0$. Fix $\theta >0$ and $0<\rho <1$ and let $\left\{ X_t\right\}$ be the “random measure process” of Eqn. () in Sec. (). Then $X_t$ is ID, stationary, time-reversible, and nonnegative integer valued, but it is not a Markov process unless $\{\mu ^\theta \}$ is the Poisson family.

Proof

By construction $X_t$ is ID, stationary, and time-reversible; suppose that it is also Markov. Then by Theorem it must be one of the four specified processes: constant, iid, branching Poisson, or branching negative binomial.

Since $\rho <1$ it cannot be the constant $\left\{ X_t\equiv X_0\right\}$ process; since $\rho >0$ it cannot be the independent $\left\{ X_t\mathrel {\mathop {\sim }\limits ^{\textrm{iid}}}\mu ^\theta (dx)\right\}$ process. The joint generating function $\phi (s,z)$ at two consecutive times for the negative binomial random measure process coincides with that for the negative binomial thinning process, given in Eqn. (), and differs from that for the negative binomial branching process, given in Eqn. (). The only remaining option is the Poisson branching process of Sec. (). $\square$

4 Continuous Time

Now consider a nontrivial ${\mathbb {N}}_0$-valued time-reversible stationary Markov processes $X_t$ indexed by continuous time $t\in {\mathbb {R}}$. The restriction of any such process to integer times $t\in {\mathbb {Z}}$ will still be Markov, hence MISTI, so there are but two possibilities for the restriction of $X_t$ to $t\in {\mathbb {Z}}$: one with univariate Poisson marginal distributions, and one with univariate Negative Binomial distributions. Both do in fact exist, as shown below in Secs. and . For any positive integer $q\in {\mathbb {N}}$ the restriction of such a process $X_t$ to $t\in {\mathbb {Z}}/q:=\{p/q:~p\in {\mathbb {Z}}\}$ will also still be Markov, hence MISTI, as one can see by considering the related MISTI process $Y_p:=X_{p/q}$ indexed by $p\in {\mathbb {Z}}$, so the restriction of $X_t$ to ${\mathbb {Z}}/q$ must also be either the Poisson or the Negative Binomial MISTI process. Since the union of $\{{\mathbb {Z}}/q_k\}$ for finitely-many $\{q_k \in \mathbb {Q}\}$ is in ${\mathbb {Z}}/q_*$ for some $q_*\in \mathbb {Q}$, it follows that the restriction of $X_t$ to $t\in \mathbb {Q}$ is also either the Poisson or the Negative Binomial MISTI process. Finally, one can show that $X_t$ for each $t\in {\mathbb {R}}$ is almost-surely equal to the limit from above of $\{X_s:~s\in \mathbb {Q}, s>t\}$, and conclude that every ${\mathbb {N}}_0$-valued time-reversible stationary Markov processes $X_t$ indexed by continuous time $t\in {\mathbb {R}}$ is also one of the four types described in Theorem . Now we construct the two nontrivial ones explicitly.

4.1 Continuous-Time Poisson Branching Process

Fix $\theta >0$ and $\lambda >0$ and construct a nonnegative integer-valued Markov process with generator

$$\begin{aligned} {\mathfrak {A}}f(x)&= \frac{\partial }{\partial s} {\textsf {E}}[ f(X_s)-f(X_t)\mid X_t=x ] \Big |_{s=t} \nonumber \\&= \lambda \theta \big [f(x+1)-f(x)]\big ] + \lambda x \big [f(x-1)-f(x)\big ] \end{aligned}$$

(23a)

or, less precisely but more intuitively, for all $i,j\in {\mathbb {N}}_0$ and $\epsilon >0$,

$$\begin{aligned} {\textsf {P}}\big [X_{t+\epsilon }=i\mid X_t=j\big ] = o(\epsilon ) + \left\{ \begin{array}{lll} \epsilon \lambda \theta & i=j+1 \\ 1-\epsilon \lambda (\theta +j) & i=j \\ \epsilon \lambda j & i=j-1. \\ \end{array}\right. \end{aligned}$$

(23b)

$X_t$ could be described as a linear death process with immigration. In Sec. () we verify that its univariate marginal distribution and autocorrelation are

$$\begin{aligned} X_t&\sim \textsf {Po}(\theta )\\ {\textsf {Corr}}(X_s,X_t)&= e^{-\lambda |s-t|}, \end{aligned}$$

and its restriction to integer times $t\in {\mathbb {Z}}$ is precisely the process described in Sec. () item 3, with one-step autocorrelation $\rho =e^{-\lambda }$.

4.2 Continuous-Time Negative Binomial Branching Process

Now fix $\theta >0$, $\lambda >0$, and $0<p<1$ and construct a nonnegative integer-valued Markov process with generator

$$\begin{aligned} {\mathfrak {A}}f(x)&= \frac{\partial }{\partial s} {\textsf {E}}[ f(X_s)-f(X_t)\mid X_t=x ] \big |_{s=t} \nonumber \\&= \frac{\lambda (\alpha +x)(1{-}p)}{p} \big [f(x+1)-f(x)]\big ] + \frac{\lambda x}{p} \big [f(x-1)-f(x)\big ] \end{aligned}$$

(24a)

or, for all $i,j\in {\mathbb {N}}_0$ and $\epsilon >0$,

$$\begin{aligned} {\textsf {P}}\big [X_{t+\epsilon }=i\mid X_t=j\big ] = o(\epsilon ) + \left\{ \begin{array}{lll} \epsilon \lambda (\alpha +j)(1{-}p)/p & i=j+1 \\ 1-\epsilon \lambda [(\alpha +j)(1{-}p)+j]/p & i=j \\ \epsilon \lambda j/p & i=j-1, \end{array}\right. \end{aligned}$$

(24b)

so $X_t$ is a linear birth-death process with immigration. The univariate marginal distribution and autocorrelation (see Sec. ()) are now

$$\begin{aligned} X_t&\sim \textsf {NB}(\alpha ,p)\\ {\textsf {Corr}}(X_s,X_t)&= e^{-\lambda |s-t|}, \end{aligned}$$

and its restriction to integer times $t\in {\mathbb {Z}}$ is precisely the process described in Sec. () item 4, with autocorrelation $\rho =e^{-\lambda }$.

4.3 Markov Branching (Linear Birth/Death) Processes

The process $X_t$ of Sec. () can also be described as the size of a population at time t if individuals arrive in a Poisson stream with rate $\lambda \theta$ and die or depart independently after exponential holding times with rate $\lambda$; as such, it is a continuous-time Markov branching process.

Similarly, that of Sec. () can be described as the size of a population at time t if individuals arrive in a Poisson stream with rate $\lambda \alpha (1{-}p)/p$, give birth (introducing one new individual) independently at rate $\lambda (1{-}p)/p$, and die or depart at rate $\lambda /p$. In the limit as $p\rightarrow 1$ and $\alpha \rightarrow \infty$ with $\alpha (1{-}p)\rightarrow \theta$ this will converge in distribution to the Poisson example of Sec. ().

4.4 Marginal Distributions

Here we verify that the Poisson and Negative Binomial distributions are the univariate marginal stationary distributions for the Markov chains with generators ${\mathfrak {A}}$ given in Eqn. (23) and Eqn. (24), respectively.

Denote by $\pi ^0_i={\textsf {P}}[X_t=i]$ the pmf for $X_t$ and by $\pi ^\epsilon _i={\textsf {P}}[X_{t+\epsilon }=i]$ that for $X_{t+\epsilon }$, and by $\varphi _0(s)={\textsf {E}}[s^{X_t}]$ and $\varphi _\epsilon (s)={\textsf {E}}[s^{X_{t+\epsilon }}]$ their generating functions. The stationarity requirement that $\varphi _0(s)\equiv \varphi _\epsilon (s)$ will determine $\varphi (s)$ and hence $\{\pi _i\}$ uniquely.

4.4.1 Poisson

From Eqn. () for $\epsilon >0$ we have

$$\begin{aligned} \pi ^\epsilon _i&= \epsilon \lambda \theta \pi ^0_{i-1} & + [1-\epsilon \lambda (\theta +i)] \pi ^0_{i} & +\epsilon \lambda (i+1) \pi ^0_{i+1} & + o(\epsilon ). \nonumber \end{aligned}$$

Multiplying by $s^i$ and summing, we get:

$$\begin{aligned} \varphi _\epsilon (s)&= \epsilon \lambda \theta s \sum _{i\ge 1} s^{i-1} \pi ^0_{i-1} & + [1-\epsilon \lambda \theta ] \varphi _0(s) - \epsilon \lambda s \sum _{i\ge 0}i s^{i-1} \pi ^0_{i} & + \epsilon \lambda \sum _{i\ge 0}(i+1)s^i \pi ^0_{i+1} & + o(\epsilon ) \nonumber \\&= \epsilon \lambda \theta s \varphi _0(s) & + [1-\epsilon \lambda \theta ] \varphi _0(s) - \epsilon \lambda s \varphi _0'(s) & + \epsilon \lambda \varphi _0'(s) & + o(\epsilon ) \nonumber \end{aligned}$$

so

$$\begin{aligned} \varphi _\epsilon (s)-\varphi _0(s)&= \epsilon \lambda (s-1)\left[ \theta \varphi _0(s)-\varphi _0'(s)\right] + o(\epsilon ) \end{aligned}$$

and stationarity ($\varphi _0(s)\equiv \varphi _\epsilon (s)$) entails $\lambda =0$ or $\varphi _0'(s)/\varphi _0(s) \equiv \theta$, so $\log \varphi _0(s) \equiv (s-1)\theta$ and:

$$\begin{aligned} \varphi _0(s)&= \exp \left\{ (s-1)\theta \right\} \end{aligned}$$

so $X_t\sim \textsf {Po}(\theta )$ is the unique stationary distribution.

4.4.2 Negative Binomial

From Eqn. () for $\epsilon >0$ we have

$$\begin{aligned} \pi ^\epsilon _i&= (\epsilon \lambda (1{-}p)/p)(\alpha +i-1)~\pi ^0_{i-1} + \{1-(\epsilon \lambda /p)[(\alpha +i)(1{-}p)+i]\}~ \pi ^0_{i}\\&+ (\epsilon \lambda /p)(i+1)~ \pi ^0_{i+1}+ o(\epsilon ) \\ \varphi _\epsilon (s)&= (\epsilon \lambda (1{-}p)/p)\alpha ~ s\varphi _0(s) + (\epsilon \lambda (1{-}p)/p)~ s^2\varphi _0'(s)\\&+ \varphi _0(s) - (\epsilon \lambda (1{-}p)/p) \alpha \varphi _0(s) - (\epsilon \lambda /p)((1{-}p)+1)~s\varphi _0'(s)\\&+ (\epsilon \lambda /p)~\varphi _0'(s)+ o(\epsilon ) \\ \varphi _\epsilon (s)-\varphi _0(s)&=(\epsilon \lambda /p)~\left\{ \varphi _0(s)~\alpha (1{-}p)(s-1) + \varphi _0'(s)~[(1{-}p)s^2 -((1{-}p)+1)s +1]\right\} + o(\epsilon ) \\&=(\epsilon \lambda /p)(s-1)~\left\{ \varphi _0(s)~\alpha (1{-}p)+ \varphi _0'(s)~((1{-}p)s-1)\right\} + o(\epsilon ) \nonumber \end{aligned}$$

so either $\lambda =0$ (the trivial case where $X_t\equiv X$) or $\lambda >0$ and:

$$\begin{aligned} \varphi _0'(s)/\varphi _0(s)&= \alpha (1{-}p)(1-(1{-}p)s)^{-1} \\ \log \varphi _0(s)&= -\alpha \log (1-(1{-}p)s)+\alpha \log (p) \\ \varphi _0(s)&= p^\alpha (1-(1{-}p)s)^{-\alpha } \end{aligned}$$

and $X_t\sim \textsf {NB}(\alpha ,p)$ is the unique stationary distribution.

4.4.3 Alternate Proof

A detailed-balance argument (Hoel et al., 1972, p. 105) shows that the stationary distribution $\pi _i:={\textsf {P}}[X_t=i]$ for linear birth/death chains is proportional to

$$\begin{aligned} \pi _i \,\propto \prod _{0\le j<i} \frac{\beta _{j}}{\delta _{j+1}} \end{aligned}$$

where $\beta _j$ and $\delta _j$ are the birth and death rates when $X_t=j$, respectively. For the Poisson case, from Eqn. () this is

$$\begin{aligned} \pi _i \, \propto \prod _{0\le j<i} \frac{\lambda \theta }{\lambda (j+1)} = \theta ^i/i!, \end{aligned}$$

so $X_t\sim \textsf {Po}(\theta )$, while for the Negative Binomial case from Eqn. () we have

$$\begin{aligned} \pi _i \, \propto \prod _{0\le j<i} \frac{\lambda (\alpha +j)(1{-}p)/p}{\lambda (j+1)/p} = \frac{\Gamma (\alpha +i)}{\Gamma (\alpha )\,i!}~ (1{-}p)^i, \end{aligned}$$

so $X_t\sim \textsf {NB}(\alpha ,p)$. In each case the proportionality constant is $\pi _0=P[X_t=0]$: $\pi _0=e^{-\theta }$ for the Poisson case, and $\pi _0=p^\alpha$ for the negative binomial.

4.4.4 Autocorrelation

Aside from the two trivial (iid and constant) cases, MISTI processes have finite pth moments for all $p<\infty$ and, in particular, have finite variance and well-defined autocorrelation. It follows by the Markov property and induction that the autocorrelation must be of the form

$$\begin{aligned} {\textsf {Corr}}[X_s,X_t] = \rho ^{-|t-s|} \end{aligned}$$

for some $\rho \in [-1,1]$. In both the Poisson and negative binomial cases the one-step autocorrelation $\rho$ is nonnegative; without loss of generality we may take $0<\rho <1$.

5 Discussion

The condition $\mu ^\theta (\{1\})>0$ introduced in Sec. () to avoid trivial technicalities is equivalent to a requirement that the support $\mathop {\textrm{spt}}(\mu ^\theta ) ={\mathbb {N}}_0$ be all of the nonnegative integers. Without this condition, for any MISTI process $X_t$ and any integer $k\in {\mathbb {N}}$ the process $Y_t=k\,X_t$ would also be MISTI, leading to a wide range of essentially equivalent processes.

The branching approach of Sec. () could be used to generate a wider class of continuous-time stationary Markov processes with ID marginal distributions (Vervaat, 1979; Steutel et al., 1983). If families of size $k\ge 1$ immigrate independently in Poisson streams at rate $\lambda _k$, with $\sum _{k\ge 1} \lambda _k\log k<\infty$, and if individuals (after independent exponential waiting times) either die (at rate $\delta >0$) or give birth to some number $j\ge 1$ of progeny (at rate $\beta _j\ge 0$), respectively, with $\delta > \sum _{j\ge 1} j\,\beta _j$, then the population size $X_t$ at time t will be a Markov, infinitely-divisible, stationary processes with nonnegative integer values. Unlike the MISTI processes, these may have infinite pth moments if $\sum _{k\ge 1} \lambda _k k^p=\infty$ for some $p>0$ and, in particular, may not have finite means, variances, or autocorrelations.

Unless $\lambda _k =0$ and $\beta _j=0$ for all $k,j>1$, however, these will not be time-reversible, and hence not MISTI. Decreases in population size are always of unit size (necessary for the Markov property to hold), while increases might be of size $k>1$ (if immigrating family sizes exceed one) or $j>1$ (if multiple births occur).