BDA3 Chapter 5 Exercise 7

Here’s my solution to exercise 7, chapter 5, of Gelman’s Bayesian Data Analysis (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

\(\DeclareMathOperator{\dbinomial}{Binomial} \DeclareMathOperator{\dbern}{Bernoulli} \DeclareMathOperator{\dpois}{Poisson} \DeclareMathOperator{\dnorm}{Normal} \DeclareMathOperator{\dt}{t} \DeclareMathOperator{\dcauchy}{Cauchy} \DeclareMathOperator{\dexponential}{Exp} \DeclareMathOperator{\duniform}{Uniform} \DeclareMathOperator{\dgamma}{Gamma} \DeclareMathOperator{\dinvgamma}{InvGamma} \DeclareMathOperator{\invlogit}{InvLogit} \DeclareMathOperator{\logit}{Logit} \DeclareMathOperator{\ddirichlet}{Dirichlet} \DeclareMathOperator{\dbeta}{Beta}\)

Part a

Suppose \(y \mid \theta \sim \dpois(\theta)\) with prior \(\theta \sim \dgamma(\alpha, \beta)\). Let’s derive the expectation and variance of \(y\). Using equation 1.8 (page 21), the expectation is

\[ \begin{align} \mathbb E (y) &= \mathbb E \left( \mathbb E(y \mid \theta) \right) \\ &= \mathbb E \left( \theta \right) \\ &= \frac{\alpha}{\beta} . \end{align} \]

Using equation 1.9 (page 21), the variance is

\[ \begin{align} \mathbb V (y) &= \mathbb E \left( \mathbb V (y \mid \theta) \right) + \mathbb V \left( \mathbb E (y \mid \theta) \right) \\ &= \mathbb E \left( \theta \right) + \mathbb V (\theta) \\ &= \frac{\alpha}{\beta} + \frac{\alpha}{\beta^2} \\ &= \alpha \frac{1 + \beta}{\beta^2} . \end{align} \]

Part b

Suppose \(y \mid \mu, \sigma \sim \dnorm(\mu, \sigma)\) with prior \(p(\mu, \sigma^2) \propto \sigma^{-2}\). Then the expectation of \(\mu \mid y\) is

\[ \begin{align} \mathbb E \left( \mu \mid y \right) &= \mathbb E \left( \mathbb E (\mu \mid \sigma^2, y) \mid y \right) \\ &= \mathbb E \left( \bar y \mid y \right) \\ &= \bar y . \end{align} \]

For posterior expectations, we condition on the data, which allows us to treat \(y\), \(n\), and \(s\) as constants. Since \(\theta := \sqrt{n} (\mu - \bar y) / s\) is a linear function of \(\mu\), its posterior expectation is zero. For this to hold, it is necessary that \(n \ge 2\) for \(s\) to be well-defined (to avoid division by zero). Moreover, the first identity implicitly assumes that the expectation \(\mathbb E (u \mid y)\) is well-defined. Combining the calculation of \(p(\mu \mid y)\) with this proof shows that it is well-defined only when \(n \ge 3\).

The posterior variance is

\[ \begin{align} \mathbb V \left( \frac{\sqrt{n}}{s} (\mu - \bar y) \mid y \right) &= \mathbb E \left( \mathbb V \left(\frac{\sqrt{n}}{s} (\mu - \bar y) \mid \sigma^2, y \right) \mid y \right) + \mathbb V \left( \mathbb E \left(\frac{\sqrt{n}}{s} (\mu - \bar y) \mid \sigma^2, y \right) \mid y \right) \\ &= \mathbb E \left( \frac{n}{s^2} \mathbb V \left( \mu \mid \sigma^2, y \right) \mid y \right) + \mathbb V \left( 0 \mid y \right) \\ &= \mathbb E \left(\frac{n}{s^2} \frac{\sigma^2}{n} \mid y \right) + 0 \\ &= \frac{n - 1}{n - 3} . \end{align} \]

Since \(n - 3\) appears in the denominator, it is necessary that \(n \ge 4\). Again, for the first identity to hold, it is implicitly assumed that the variance on the left hand side is finite. It can be verified that the variance is finite for \(n \ge 4\).