Given an integer observation $v$, I consider it comes from a Poisson distribution centered at $\boldsymbol{\theta}^T \boldsymbol{\lambda}$ where $\boldsymbol{\theta, \lambda}$ are K-dimensional integer vectors.

\begin{align} v \sim Poisson(\sum_k\theta_k\lambda_k) & \quad\text{ or }\quad v = \sum_kPoisson(\theta_k\lambda_k)\\ \lambda &\sim Gamma (\alpha, \beta) \end{align}

I we marginalize over $\lambda$, we get the following generative model:

\begin{align} v &= \sum_k NegativeBinomial(\alpha, p_k = \frac{\theta_k}{\beta + \theta_{k}}) \end{align}

and the probability distribution gives: \begin{align} p(v | \boldsymbol{P})=& \sum_{c_1 + ... c_K = v} \prod_k \left[ \frac{\Gamma(\alpha_k + c_{k})}{\Gamma(\alpha)c_{k}!} (1 - p_{k})^{\alpha} p_{k}^{c_{k}} \right] \end{align}

where the sum is over all integer partitions (or compositions, since $c_i = 0$ is allowed) of $v$: $c_1+...+c_k = v$

Looking for some similar distribution, I found the "Negative Binomial of order $k$". However, the partitions of the Negative Binomial of order $k$ are such that \begin{align} \sum_k kc_k = v \end{align}

Does my probability distribution correspond (or is similar) to some known distribution?


1 Answer 1


The distribution of $\nu$ can be derived in a slightly different way. We can write $\nu = \sum_k \nu_k$, $\nu_k \sim \text{Poisson}(\theta_k \lambda_k)$. Each product $\theta_k \lambda_k$ has (let us say; it depends on the parameterization of the Gamma) distribution $\text{Gamma}(\alpha, \beta \theta_k)$. Integrating out $\lambda_k$ gives us a Negative Binomial distribution for each of the $\nu_k$ with parameters $\alpha, \frac{\beta\theta_k}{1 + \beta\theta_k}$. Thus, the distribution of $\nu$ is that of the sum of $K$ negative binomial variates with shape parameter $\alpha$ and probability parameters $\frac{\beta\theta_k}{1 + \beta\theta_k}$.

Unfortunately, this distribution does not have a tractable form (unless all the $\theta_k$ are equal,) nor does it have a name, as far as I know.

  • $\begingroup$ Actually I use this decomposition of a sum of Poissons, too :) Thank you so much for your feedback @jbowman, I'm desperately trying to prove it has only one mode (ignoring the k! symmetries) and I'd like to be sure it has not been studied before in some obscure paper from the 60's ! (I'll let the question open some more days in case someone has something else to say) $\endgroup$
    – alberto
    Apr 26, 2017 at 21:04
  • $\begingroup$ Btw, are you sure it gives you a sum of NB and not exactly the equation I wrote? (A sum, for all possible combinations of $v_k$, of the product of K Negative Binomials) I'd be very surprised... Note that you don't know the "true" combination of $v_k$ and that's why it appears the sum over the partitions in my derivation. $\endgroup$
    – alberto
    Apr 26, 2017 at 21:10
  • 1
    $\begingroup$ It looks the same, I was being more conceptual than mathematical in my explanation. I believe, though, that what you would like is a description of the conditions under which the sum of unimodal variates is itself unimodal. (not so in general: math.stackexchange.com/questions/70651/…). Log-concavity does the job for independent variates, which holds for negative binomial distributions with shape parameter $\geq 1$. (See arxiv.org/pdf/math/0502548.pdf). $\endgroup$
    – jbowman
    Apr 27, 2017 at 3:01
  • $\begingroup$ I'd seen this paper but it looks like its time to red it thoroughly. Hoewever, I forgot to mention I look for unimodality... in the loglikelihood (I want to estimate the $\mathbf{p}$) $\endgroup$
    – alberto
    Apr 27, 2017 at 9:18
  • $\begingroup$ One last question @jbowman, if you don't mind. Had you already seen this distribution before? $\endgroup$
    – alberto
    Apr 28, 2017 at 10:09

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