Suppose we have that the distribution of a count $x$ is given by a binomial distribution: $x\sim \mathcal{B}(n,\theta)$. Further, suppose that the sample size $n\sim NB(\lambda,\kappa)$, where this represents a negative binomial distribution with mean $\lambda$ and variance $\lambda^2/\kappa$. I can write a Mathematica function that implements this parameterisation of the negative binomial as follows:

fCustomNB[λ_, κ_] := NegativeBinomialDistribution[κ, κ/(κ + λ)]

which has desired mean (variance also but omitting to keep question short):

FullSimplify@Mean[fCustomNB[λ, κ]]


I can also use Mathematica to find the distribution of $x$ where I have marginalised $n$ out of the distribution (i.e. the distribution of $x|\theta, \lambda, \kappa$) as follows:

aDist = ParameterMixtureDistribution[
          BinomialDistribution[n, θ],
             n \[Distributed] fCustomNB[λ, κ]];

I can then evaluate the PDF of this using:

Assuming[{λ > 0, κ > 0, 0 < θ < 1, x >= 0}, PDF[aDist, x]]

which gives me this, somewhat heathen, function:

(1 - θ)^(-x + Ceiling[x]) θ^x κ^κ λ^ Ceiling[x] (κ + λ)^(-κ - Ceiling[x]) Binomial[Ceiling[x], x] Binomial[-1 + κ + Ceiling[x], -1 + κ] Hypergeometric2F1[ 1, κ + Ceiling[x], 1 - x + Ceiling[ x], (λ - θ λ)/(κ + λ)]

The part of this that is particularly difficult to translate into other languages (this distribution represents part of a larger model already coded up in Stan) is the Hypergeometric bit.

The bottom line is, can I simplify this expression in terms of more vanilla-like functions?

Edit: as finishing writing this question, I realised we can -- so I include my answer below in case it helps others!


1 Answer 1


Since $x$ is a count variable, just state it in the assumptions. Then everything is hunky dory (whatever hunky dory is):

Assuming[{λ > 0, κ > 0, 0 < θ < 1, x >= 0, x ∈ Integers},
         PDF[aDist, x]]

which evaluates to:

κ^κ (θ λ)^x (κ + θ λ)^(-x - κ) Binomial[-1 + x + κ, -1 + κ]

Less heathen, past self, I think you'll agree.

Bottom line: always state all of your assumptions to guarantee best chance of simple expressions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.