BDA3 Chapter 2 Exercise 17

Here’s my solution to exercise 17, chapter 2, of Gelman’s Bayesian Data Analysis (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

\(\DeclareMathOperator{\dbinomial}{binomial} \DeclareMathOperator{\dbern}{Bernoulli} \DeclareMathOperator{\dpois}{Poisson} \DeclareMathOperator{\dnorm}{normal} \DeclareMathOperator{\dcauchy}{Cauchy} \DeclareMathOperator{\dgamma}{gamma} \DeclareMathOperator{\invlogit}{invlogit} \DeclareMathOperator{\logit}{logit} \DeclareMathOperator{\dbeta}{beta}\)

We’ll show that highest posterior invervals are not invariant under parameter transformations.

Suppose \(\frac{nv}{\sigma^2} \mid \sigma^2 \sim \chi_n^2\) with (improper) prior \(\sigma \propto \sigma^{-1}\). From equation (2.19), page 52, the prior density for \(\sigma^2\) is

\[ p(\sigma^2) = p(\sigma) (2\sigma)^{-1} \propto \frac{1}{\sigma^2} . \]

Thus the posteriors are

\[ \begin{align} p(\sigma^2 \mid y) &= \left( \frac{1}{\sigma} \right)^n e^{\frac{-nv}{2 \sigma^2}} \\ p(\sigma \mid y) &= \left( \frac{1}{\sigma} \right)^{n - 1} e^{\frac{-nv}{2 \sigma^2}} , \end{align} \]

where we have dropped any multiplicative constants that don’t depend on \(\sigma\).

Since the posteriors are continuous everywhere, we can assume that the highest posterior regions are collections of closed intervals. Let \(a, b\) be two boundary points on the highest posterior density region of \(p(\sigma^2 \mid y)\). Using continuity and the defining property of highest posterior regions, the density at \(a\) is equal to the density at \(b\), i.e.

\[ \left( \frac{1}{a} \right)^n e^{\frac{-nv}{2 a^2}} = \left( \frac{1}{b} \right)^n e^{\frac{-nv}{2 b^2}} . \]

Assume for contradiction that the highest posterior region for \(p(\sigma \mid y)\) is the square root of the region for \(p(\sigma^2 \mid y)\). Then by continuity, \(\sqrt{a}, \sqrt{b}\) are endpoints on the highest posterior region for \(p(\sigma \mid y)\). Thus

\[ \left( \frac{1}{a} \right)^{n - 1} e^{\frac{-nv}{2 a^2}} = \left( \frac{1}{b} \right)^{n - 1} e^{\frac{-nv}{2 b^2}} . \]

The two equalities above are equivalent to \(\frac{1}{a} = \frac{1}{b}\), which implies

\[ a = b \qquad ↯ \]

This is true of any two boundary points, so the highest posterior region is a point. This contradicts the fact that the highest posterior region contains 95% probability mass. Therefore, the highest posterior regions are not invariant under reparameterisation.