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I was wondering if someone could clarify the following for me. In the paper "Inference for the binomial $N$ parameter" by Adrian Raftery, his first example outlines the posterior of $N$ given $x$ as

$$ p(N|x) \propto (N!)^{-1}\left\{\prod_{i=1}^{n}\binom{N}{x_i}\right\}\int_{0}^{1}\int_{0}^{\infty}\theta^{-N+S}(1-\theta)^{nN-S}\lambda^N\text{exp}(-\lambda/\theta)p(\lambda,\theta)d\lambda d\theta $$

He assumes a Poisson distribution as the prior distribution for $N$, so I understand where the

$$ (N!)^{-1}\lambda^N\exp\left(\lambda/\theta\right) $$

comes from. However, why is the exponent of $e$ $\lambda/\theta$? Thanks!

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  • $\begingroup$ What are $\lambda$ and $\theta$? $\endgroup$
    – Sycorax ♦
    Commented Nov 30, 2014 at 14:27
  • $\begingroup$ Related: stats.stackexchange.com/questions/123367/… $\endgroup$
    – kjetil b halvorsen ♦
    Commented Nov 30, 2014 at 18:14
  • $\begingroup$ In the paper the author assumes a Poisson prior distribution for N with mean $\mu$. The author says then that marginally, each $x_i$ has a Poisson distribution with a mean $\lambda = \mu\theta$. The author decides to specify the prior distribution in terms of $(\lambda, \theta)$. So I'm assuming that $\lambda = \mu\theta$ and $\theta$ is the success probability. Hope this helps. Thanks! $\endgroup$
    – TSP
    Commented Nov 30, 2014 at 20:53

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A Poisson distribution for $N$ with mean $\mu$ has density $$ (N!)^{-1} \mu^N \exp(-\mu) $$ Since $\mu = \lambda/\theta$, this explains the exponent.

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