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 θ∼normal(180,40), sampling distribution y∣θ∼normal(θ,20), and n sampled students with average weight ˉy=150, it follows from 2.11 that the posterior mean is

μ:=E(θ∣ˉy)=1801600+150n40011600+n400=60(3+10n)1600⋅16001+4n=60(3+10n)1+4n1/σ2:=1/V(θ∣ˉy)=11600+n400=1+4n1600.

So θ∣ˉy∼normal(60(3+10n)1+4n,40√1+4n). When n=0 this is exactly the prior, and when n=∞ this is 150 (the observed mean) with zero variance.

It follows from the calculations shown in the book that the posterior predictive distribution is ˜y∣y∼normal(μ,√σ2+400).

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), sqrt(sigma(10)^2 + 400))

With a sample of size of 10, we get θ ϵ [140.5, 161] and ˜y ϵ [116.3, 185.2].

theta_posterior_interval <- qnorm(percentiles, mu(100), sigma(100))
y_posterior_interval <- qnorm(percentiles, mu(100), sqrt(sigma(100)^2 + 400))

With a sample of size of 100, we get θ ϵ [146.8, 153.4] and ˜y ϵ [117, 183.1].

Both of these posterior intervals for θ are very similar to the frequentist confidence intervals, especially in the case n=100.

qnorm(percentiles, 150, 20 / sqrt(10))

## [1] 139.597 160.403

qnorm(percentiles, 150, 20 / sqrt(100))

## [1] 146.7103 153.2897

We would expect them to become more similar as n increases, because both means and standard deviations converge to the same values for large n.