BDA3 Chapter 2 Exercise 8

Here’s my solution to exercise 8, chapter 2, of Gelman’s Bayesian Data Analysis (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

With prior \(\theta \sim \dnorm(180, 40)\), sampling distribution \(y \mid \theta \sim \dnorm(\theta, 20)\), and \(n\) sampled students with average weight \(\bar y = 150\), it follows from 2.11 that the posterior mean is

\[ \begin{align} \mu := \mathbb E(\theta \mid \bar y) &= \frac{\frac{180}{1600} + \frac{150n}{400}}{\frac{1}{1600} + \frac{n}{400}} \\ &= \frac{60(3 + 10n)}{1600} \cdot \frac{1600}{1 + 4n} \\ &= \frac{60(3 + 10n)}{1 + 4n} \\ 1 / \sigma^2 := 1 / \mathbb V (\theta \mid \bar y) &= \frac{1}{1600} + \frac{n}{400} \\ &= \frac{1 + 4n}{1600} . \end{align} \]

So \(\theta \mid \bar y ~ \dnorm \left( \frac{60(3 + 10n)}{1 + 4n}, \frac{40}{\sqrt{1 + 4n}} \right)\).

It follows from the calculations shown in the book that the posterior predictive distribution is \(\tilde y \mid y \sim \dnorm(\mu, \sigma + 20)\).

We can obtain 95% posterior intervals as follows.

mu <- function(n) 60 * (3 + 10 * n) / (1 + 4 * n)
sigma <- function(n) 40 / sqrt(1 + 4 * n)

percentiles <- c(0.05, 0.95)

theta_posterior_interval <- qnorm(percentiles, mu(10), sigma(10))
y_posterior_interval <- qnorm(percentiles, mu(10), sigma(10) + 20)

With a sample of size of 10, we get θ ϵ [140.5, 161] and \(\tilde y\) ϵ [107.6, 193.9].

theta_posterior_interval <- qnorm(percentiles, mu(100), sigma(100))
y_posterior_interval <- qnorm(percentiles, mu(100), sigma(100) + 20)

With a sample of size of 100, we get θ ϵ [146.8, 153.4] and \(\tilde y\) ϵ [113.9, 186.3].