BDA3 Chapter 2 Exercise 6

Here’s my solution to exercise 6, chapter 2, of Gelman’s Bayesian Data Analysis (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

Considering the negative binomial variable \(y\) as a gamma-Poisson variable, we derive expressions for the mean and variance.

From equation 1.6, \(\mathbb E (y) = \mathbb E (\mathbb E(y \mid \theta))\). Since \(y \mid \theta \sim \dpois(10n\theta)\), it follows that \(\mathbb E (y \mid \theta) = 10n\theta\). The rate \(\theta \sim \dgamma(\alpha, \beta)\) so \(\mathbb E(\theta) = \frac{\alpha}{\beta}\). Thus, \(\mathbb E(y) = 10n\mathbb E(\theta) = 10n \frac{\alpha}{\beta}\).

We also have \(\mathbb V(\theta) = \frac{\alpha}{\beta^2}\) since \(\theta \sim \dgamma(\alpha, \beta)\), and \(\mathbb V(y \mid \theta) = 10n\theta\) since \(y \mid \theta \sim \dpois(10n\theta)\). Thus,

\[ \begin{align} \mathbb V (y) &= \mathbb E(\mathbb V(y \mid \theta)) + \mathbb V (\mathbb E (y \mid \theta)) \\ &= \mathbb E(10n\theta) + \mathbb V (10n\theta) \\ &= 10n\frac{\alpha}{\beta} + (10n)^2\frac{\alpha}{\beta^2} \qquad \square \end{align} \]