Negative Binomial as Gamma-Poisson Mixture or Compound Logarithmic Poisson: can this correspondence be generalized to other distributions?

Question

Preamble

A random variable $X$ with a negative binomial distribution can be characterized in three ways:

1. [Negative Binomial] $X\sim\operatorname{NegBin}(r,p)$ for some $r$ and $p$;

2. [Gamma-Poisson Mixture] $X\mid\theta\sim\operatorname{Poisson}(\theta)$ and $\theta\sim\operatorname{Gamma}(\alpha,\beta)$ for some $\alpha$ and $\beta$;

3. [Compound Poisson-Logarithmic] $X=\sum_{i=1}^NY_i$ where $N\sim\operatorname{Poisson}(\lambda)$ and the $Y_i$ are i.i.d. with $Y_i\sim\operatorname{Logarithmic}(p)$ for some $\lambda$ and $p$.

Moreover there are explicit relationships between the parameter pairs $(r,p)$, $(\alpha,\beta)$, and $(\lambda,p)$.

Question

Is there a more general correspondence between Poisson mixtures and compund poisson distributions? In other words is there a theorem similar to the one below?

Maybe Theorem. The following are equivalent:

1. [Poisson Mixture] $X\mid\theta\sim\operatorname{Poisson}(\theta)$ and $\theta\sim{F}$ for some distribution $F$ satisfying some nice properties;

2. [Compound Poisson] $X=\sum_{i=1}^NY_i$ where $N\sim\operatorname{Poisson}(\lambda)$ and the $Y_i$ are i.i.d. with $Y_i\sim G$ for some distribution $G$ satisfying some nice properties.

Moreover it is possible to derive $F$ from $G$ and vice versa.

Additional question

Can the result be further generalized by replacing Poisson with some other distribution?

Answer 1

Let $$ M_\theta(z)=E e^{z\theta} $$ denote the moment generating function of $\theta$. Since $X|\theta\sim \operatorname{Poisson}(\theta)$, the probability generating function (pgf) of $X|\theta$ is
$$ G_{X|\theta}(z)=Ez^X|\theta = e^{\theta(z-1)}. $$ Hence, using the law of total expectation, the pgf of $X$ is $$ G_X(z)=Ez^X=E(Ez^X|\theta)=Ee^{\theta(z-1)} =M_\theta(z-1). \tag{1} $$

If, at the same time, $$ X = \sum_{i=1}^N Y_i $$ where the $Y_i$'s are iid with pgf $G_Y(z)$ and $N \sim \operatorname{Poisson}(\lambda)$ such that the pgf of $N$ is $ G_N(z)=e^{\lambda(t-1)}$, then the pgf of $X$ must also be equal to \begin{align} G_X(z) &=Ez^X \\&=E E(z^X|N) \\&=E E(z^{Y_1+Y_2+\dots+Y_N}|N) \\&=E (G_Y(z)^N|N) \\&=G_N(G_Y(z)). \tag{2} \end{align} Hence, the pgf of the iid $Y_i$'s given $M_\theta(z)$ is $$ G_Y(z)=G_N^{-1}(G_X(z))=\frac1\lambda(\ln G_X(z) + 1)=\frac1\lambda(\ln M_\theta(z-1) + 1). \tag{3} $$ from which the probability mass function $P(Y=k)=G_Y^{(k)}(0)/k!$, $k=0,1,2,\dots$ can be computed.

Of course, if any of the point masses are negative, $X$ can not be represented simultaneously both as a Poisson mixture and as a compound Poisson distribution.

Example 1: Suppose that $\theta \sim \operatorname{Bern}(1/2)$ such that $M_\theta(z)=\frac{1+e^z}2$. Note that this means that $X$ has a zero-inflated Poisson distribution. Using (3) and the fact that $G_Y(1^-)=\sum_{k=0}^\infty P(Y_i=k)=1$ we must have $\lambda=1$ and \begin{align} G_Y(z)&=\ln\frac{1+e^{z-1}}2+1 \\&=\ln\frac{1 + e}2 + \frac z{1 + e} + \frac{e z^2}{2 (1 + e)^2} + \frac{(e - 1) e z^3}{6 (1 + e)^3} + \frac{e (1 - 4 e + e^2) z^4}{24 (1 + e)^4} + ... \end{align} which would imply that $P(Y_i=4)=-0.00147<0$.

Example 2: There are other cases than the negative binomial case described in the original post, however. Suppose that the $Y_i$'s have pgf $G_Y(z)=e^{z-1}$, that is, each $Y_i \sim \operatorname{Poisson}(1)$.

Using (2), the pgf of $X$ is then $$ G_X(z)=G_N(G_Y(z))=e^{\lambda(e^{z-1}-1)} $$ and using (1), the mgf of $\theta$ would need to be $$ M_{\theta}(z)=G_X(z+1)=e^{\lambda(e^z-1)} $$ implying that $\theta\sim \operatorname{Poisson}(\lambda)$. So in this case $X$ can be seen both as a Poisson-Poisson mixture and as a sum of a Poisson distributed number of iid Poisson(1) random variables.

Answer 2

At the moment this is just a partial answer. I'm going to start by deriving the requirement for these two mixture forms to be equal. This will get you some of the way by at least showing the equation that your functions must solve. At the moment I can't see a full solution.

We can see from the first mixture that the support of $X$ is the non-negative integers, so let's work in that space. Suppose we let $p_X$ be the mass function and $m_X$ be the moment-generating function for $X$. I will take $F$ and $G$ to be the density functions for the distributions of $\theta$ and $Y_i$. The first mixture form can be characterised by the requirement:

$$\text{Mixture 1} \quad \quad \quad p_X(x) = \int \limits_\Theta \text{Pois}(x|\theta) \cdot F(\theta) \ d\theta,$$

and the second mixture form can be characterised by the requirement:

$$\text{Mixture 2} \quad \quad \quad m_X(t) = \sum_{n=0}^\infty m_Y(t)^n \cdot \text{Pois}(n|\lambda).$$

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Equating the mixture forms: To solve your problem, let's work in terms of moment-generating functions, and simplify things as far as we can. From the first mixture form we must have:

$$\begin{align} m_X(t) &= \mathbb{E}(e^{tX}) \\[12pt] &= \sum_{x=0}^\infty e^{tx} p_X(x) \\[8pt] &= \sum_{x=0}^\infty e^{tx} \int \limits_\Theta \text{Pois}(x|\theta) \cdot F(\theta) \ d\theta. \\[6pt] \end{align}$$

Expanding the moment-generating function $m_Y$ in the second mixture form gives:

$$\begin{align} m_X(t) &= \sum_{n=0}^\infty m_Y(t)^n \cdot \text{Pois}(n|\lambda) \\[6pt] &= \sum_{n=0}^\infty \mathbb{E}(e^{tY})^n \cdot \text{Pois}(n|\lambda) \\[6pt] &= \sum_{n=0}^\infty \Bigg( \sum_{y=0}^\infty e^{ty} \cdot G(y) \Bigg)^n \cdot \text{Pois}(n|\lambda). \\[6pt] \end{align}$$

Equating the two forms for the moment-generating function then gives the equivalence requirement:

$$\boxed{\quad \sum_{n=0}^\infty \Bigg( \sum_{y=0}^\infty e^{ty} \cdot G(y) \Bigg)^n \cdot \text{Pois}(n|\lambda) = \sum_{x=0}^\infty e^{tx} \int \limits_\Theta \text{Pois}(x|\theta) \cdot F(\theta) \ d\theta \quad}$$

Consequently, what we require to solve your mixture problem, is some forms for the densities $F$ and $G$ and the parameter $\lambda$ that satisfy the equivalence requirement. If you can obtain conditions on the densities that ensure that the above equation is satisfied then you will have established the equivalence of the two mixtures.