I have a Negative Multinomial distribution, defined as:

\begin{align} P(k_1,...k_D | k_0, \mathbf{p}) = \frac{\Gamma(k_0 + \sum_i k_i) }{\Gamma(\alpha)\prod_{i} k_{i}!} p_0^{k_0} \prod_{i=1}^D p_i^{k_{i}} \end{align}

I want to compute the sum over some pre-computed subset of count vectors $\mathcal{K}$:

\begin{align} \sum_{\mathbf{k} \in \mathcal{K}} P(k_1,...k_D | k_0, \mathbf{p}) = \sum_{\mathbf{k} \in \mathcal{K}} \frac{\Gamma(k_0 + \sum_i k_i) }{\Gamma(\alpha)\prod_{i} k_{i}!} p_0^{k_0} \prod_{i=1}^D p_i^{k_{i}} \end{align}

How can I do it?

Or maybe it can't be done analytically?

My final goal is to find the MLE estimate of $\mathbf{p}$ for that subset $\mathcal{K}$.

Any answer or reference is very welcome.


1 Answer 1


I think using MLE here may give you misleading conclusions. You know that some outcome in $\mathcal{K}$ has occured and you want to compute and maximise the likelihood $L(\mathbf{p})=P(k\in\mathcal{K})$. For example, suppose the experiment is stopped when $k_0=1$ events in the category 0 has occured and the number of events in the other categories were any $$ k\in \mathcal{K}=\{(10,0,0,0),(0,0,0,10)\} $$ This tells you that $\mathbf{p}$ must be either quite close to $\mathbf{p}=(1/11,10/11,0,0,0\}$ or $\mathbf{p}=(1/11,0,0,0,10/11\}$ and this will appear as two optima in $L(\mathbf p)$ close to these values.

You may be better off doing Bayesian inference, perhaps with a Dirichlet prior on $\mathbf{p}$. The resulting posterior would then be a Dirichlet mixture with components associated with each element in $\mathcal{K}$.

  • $\begingroup$ Thanks Jarle. Yes, symmetric values will give symmetric optima, I see. Yet I can't (or I'm not interested in) use Bayesian priors here since I want to understand the properties of this model, that arose as by-product of another one. $\endgroup$
    – alberto
    Mar 13, 2017 at 12:45
  • $\begingroup$ Ok, I guess the likelihood may be unimodal if all elements of $\mathcal{K}$ are clustered together, in which case MLE may give reasonable results. If so, I don't really see where the difficulty with computing the MLEs of $\mathbf{p}$ (numerically by brute force), but perhaps an analytic expression is available if $\mathcal{K}$ has a specific form? $\endgroup$ Mar 13, 2017 at 13:00
  • $\begingroup$ (sorry @Jarle, I just saw your comment) Yeap, exactly, I'm hoping that in some cases (e.g.: when K contains all integer partitions of a number) it has an analytic solution. $\endgroup$
    – alberto
    Mar 27, 2017 at 9:31

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