The expected number of events for a Poisson distribution under a Gamma prior with parameters $\alpha$ and $\beta$ (with mean $\alpha/\beta$) is:

$$ \newcommand{\paren}[1]{\left(#1\right)} \begin{aligned} \sum_{k=1}^\infty k & \cdot P(k \mid \alpha, \beta) \\ = \sum_{k=1}^\infty k & \int_0^\infty P(k \mid \lambda) p(\lambda \mid \alpha, \beta) \, d\lambda \\ = \sum_{k=1}^\infty k & \int_0^\infty \paren{\frac{\lambda^k e^{-\lambda}}{k!}} \paren{\frac{\beta^\alpha}{\Gamma(\alpha)} \lambda^{\alpha-1} e^{-\beta \lambda}} \> d\lambda \\ = \sum_{k=1}^\infty k & \> \frac{\beta^\alpha}{k! \> \Gamma(\alpha)} \int_0^\infty \lambda^{k+\alpha-1} e^{-(\beta+1)\lambda} \> d\lambda \\ = \sum_{k=1}^\infty k & \> \frac{\beta^\alpha}{(\beta+1)^{\alpha+k}} \frac{\Gamma(k+\alpha)}{\Gamma(\alpha) \Gamma(k+1)} \\ = \sum_{k=1}^\infty & \frac{\beta^\alpha}{(\beta+1)^{\alpha+k}} \frac{\Gamma(k+\alpha)}{\Gamma(\alpha) \Gamma(k)} \\ = \sum_{k=1}^\infty & \frac{\beta^\alpha}{(\beta+1)^{\alpha+k}} \frac{1}{B(\alpha, k)} \\ = \frac{\beta^\alpha}{(\beta+1)^\alpha} & \sum_{k=1}^\infty \frac{(\beta+1)^{-k}}{B(\alpha, k)} \\ = \paren{\frac{\beta}{\beta+1}}^\alpha & \sum_{k=1}^\infty \frac{(\beta+1)^{-k}}{B(\alpha, k)} \end{aligned} $$ where $\Gamma$ is the Gamma function and $B$ is the Beta function.

Is there any way to further simplify this expression?

Intuitively, I would expect it to be at or near the expected number of events for a Poission distribution with mean equal to the expected $\lambda$ from the prior,

$$ \begin{aligned} & \mathbb{E}_k \text{Poisson}\left(\lambda = \mathbb{E}_\lambda(\text{Gamma}(\alpha, \beta))\right) \\ = {} & \> \mathbb{E}_k \text{Poisson}\left(\lambda = \alpha/\beta\right) \\ = {} & \> \alpha/\beta \end{aligned} $$


I found a description of the Gamma-Poisson distribution, which confirms my intuition that the expected value is in fact $\alpha/\beta$.

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    $\begingroup$ Hint: what is the posterior predictive distribution of the Poisson-Gamma model? I think you're making this a bit more difficult by cranking directly at the definition of expectation. $\endgroup$ – Reinstate Monica Mar 13 '17 at 18:27
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    $\begingroup$ Try to use double expectation formula $\DeclareMathOperator{\E}{\mathbb{E}} \E \E (X \mid Y) = \E X$ $\endgroup$ – kjetil b halvorsen Mar 13 '17 at 18:28
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    $\begingroup$ stats.stackexchange.com/questions/163236/… $\endgroup$ – Reinstate Monica Mar 13 '17 at 19:57

From the link given by Sycorax above it was pointed out that

$$ \frac{d^k \beta^\alpha}{(d+\beta)^{k+\alpha}} \frac{\Gamma(k+\alpha)}{\Gamma(k+1)\Gamma(\alpha)} = \binom{k+\alpha-1}{\alpha-1} (1-p)^\alpha p^k $$

is the PDF of a Negative Binomial Distribution with parameters $(\alpha,p=d/(d+\beta))$.

We can insert this halfway through the derivation in my original question (with $d = 1$) to get:

$$ \sum_{k=0}^\infty k \cdot P(k \mid \alpha, 1/(1+\beta)) = \mathbb{E}\left[k\right] = \alpha/\beta $$

Very cool!

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    $\begingroup$ This answer would be more clear if you cleaned up the notation to match that of your question. For example, you don't have $d$ in the question. $\endgroup$ – Reinstate Monica Mar 14 '17 at 21:34
  • $\begingroup$ edited to make more clear $\endgroup$ – Mageek Mar 14 '17 at 21:54

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