BDA3 Chapter 3 Exercise 10

Posted on 21 October, 2018 by Brian
Tags: bda chapter 3, solutions, bayes, f, scaled inverse chi2, inverse chi2, chi2
Category: bda3

Here’s my solution to exercise 10, chapter 3, of Gelman’s Bayesian Data Analysis (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

\(\DeclareMathOperator{\dbinomial}{Binomial} \DeclareMathOperator{\dbern}{Bernoulli} \DeclareMathOperator{\dpois}{Poisson} \DeclareMathOperator{\dnorm}{Normal} \DeclareMathOperator{\dt}{t} \DeclareMathOperator{\dcauchy}{Cauchy} \DeclareMathOperator{\dexponential}{Exp} \DeclareMathOperator{\duniform}{Uniform} \DeclareMathOperator{\dgamma}{Gamma} \DeclareMathOperator{\dinvgamma}{InvGamma} \DeclareMathOperator{\invlogit}{InvLogit} \DeclareMathOperator{\dinvchi}{InvChi2} \DeclareMathOperator{\dsinvchi}{SInvChi2} \DeclareMathOperator{\dchi}{Chi2} \DeclareMathOperator{\dnorminvchi}{NormInvChi2} \DeclareMathOperator{\logit}{Logit} \DeclareMathOperator{\ddirichlet}{Dirichlet} \DeclareMathOperator{\dbeta}{Beta}\)

For \(j = 1, 2\), let

\[ \begin{align} y_j \mid \mu_j \sigma_j^2 &\sim \dnorm(\mu_j, \sigma_j^2) \\ p(\mu_j, \log \sigma_j^2) &\propto 1. \end{align} \]

We show that

\[ \frac{s_1^2 \sigma_2^2}{s_2^2 \sigma_1^2} \sim F(n_1 - 1, n_2 - 1) . \]

Equation 3.5 in the book shows that \(\sigma_j^2 \mid y \sim \dinvchi(n_j - 1, s_j^2)\). It follows that \(\frac{\sigma_j^2}{(n_j - 1) s_j^2} \sim \dinvChi(n_j - 1)\). Thus, \(\frac{(n_j - 1) s_j^2}{\sigma_j^2} \sim \dchi(n_j - 1)\). The result follows from the fact that the ratio of two \(\chi^2\) random variables (divided by the ratio of their degrees of freedom) has an \(F\)-distribution.