Discussion
Suppose we have 10 observations \(\underset{\sim}{y} = (y_{1}, y_{2}, \ldots, y_{10})\) from Poisson \((\lambda)\) and we assume a Gamma \((a, b)\) prior distribution for \(\lambda\).
\[ \begin{aligned} \Big(p(\lambda) &= \{\Gamma(a)\}^{-1} b^{a} e^{-b\lambda} \lambda^{a-1}, \\ &\quad a \gt 0,\ b \gt 0\Big) \end{aligned} \]
Then the posterior predictive distribution \(p(y_{\text{new}} / \underset{\sim}{y})\) of a new observation \(y_{\text{new}}\) is given by :
Answer: B
The posterior predictive distribution is Negative Binomial \(\left(a + \sum y_{i},\ b + 10\right)\).
With a Gamma \((a, b)\) prior (rate \(b\)) and 10 Poisson observations, the posterior is Gamma \((a + \sum y_{i},\ b + 10)\), by conjugacy. A new observation is Poisson \((\lambda)\) given \(\lambda\), and mixing a Poisson over a gamma distribution gives a negative binomial:
\[ \begin{aligned} &p(y_{\text{new}} \mid \underset{\sim}{y}) \\ &= \binom{\alpha + y - 1}{y} \left(\frac{\beta}{\beta + 1}\right)^{\alpha} \\ &\quad \times \left(\frac{1}{\beta + 1}\right)^{y} \end{aligned} \]
with \(\alpha = a + \sum y_{i}\) and \(\beta = b + 10\), the parameters named in the option.
The two gamma options describe the posterior of \(\lambda\) (one with its parameters swapped), not the distribution of a new count, which must be discrete. The Poisson option ignores the uncertainty in \(\lambda\), which is exactly what makes the predictive distribution overdispersed.
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