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.

\(\DeclareMathOperator{\dbinomial}{binomial} \DeclareMathOperator{\dbern}{Bernoulli} \DeclareMathOperator{\dgamma}{gamma} \DeclareMathOperator{\dpois}{Poisson} \DeclareMathOperator{\dbeta}{beta}\)

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,

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