# BDA3 Chapter 2 Exercise 9

Here’s my solution to exercise 9, chapter 2, of Gelman’s Bayesian Data Analysis (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

$$\DeclareMathOperator{\dbinomial}{binomial} \DeclareMathOperator{\dbern}{Bernoulli} \DeclareMathOperator{\dnorm}{normal} \DeclareMathOperator{\dgamma}{gamma} \DeclareMathOperator{\invlogit}{invlogit} \DeclareMathOperator{\logit}{logit} \DeclareMathOperator{\dbeta}{beta}$$

The data show 650 people in support of the death penalty and 350 against. We explore the effect of different priors on the posterior.

First let’s find the prior with a mean of 0.6 and standard deviation 0.3. The mean of the $$\dbeta(\alpha, \beta)$$ distribution is

$\frac{3}{5} = \frac{\alpha}{\alpha + \beta}$

which implies that $$\alpha = 1.5 \beta$$. The variance is

\begin{align} \frac{9}{100} &= \frac{\alpha \beta}{(\alpha + \beta)^2 (\alpha + \beta + 1)} \\ &= \frac{3}{2} \frac{\beta^2}{\frac{25}{4}\beta^2 \frac{5\beta + 2}{2}} \\ &= \frac{3}{2}\frac{4}{25}\frac{2}{5\beta + 2} \\ &= \frac{12}{25(5\beta + 2)} \\ &\Leftrightarrow \\ 5\beta + 2 &= \frac{12}{25}\frac{100}{9} \\ &= \frac{16}{9}, \end{align}

which implies that $$\beta = \frac{2}{3}$$. Thus $$\alpha = 1$$. Let’s check we’ve done the maths correctly.

α <- 1
β <- 2 / 3

list(
'mean_diff' = 3 / 5 - α / (α + β),
'variance_diff' = 9 / 100 - α * β / ((α + β)^2 * (α + β + 1))
)
## $mean_diff ## [1] -1.110223e-16 ## ##$variance_diff
## [1] -1.387779e-17

The value 1e-16 is computer-speak for 0.

Since $$\beta < 1 <= \alpha$$, we see one maximum at 1.

tibble(x = seq(0, 1, 0.001), y = dbeta(x, α, β)) %>%
ggplot() +
aes(x, y) +
geom_area(fill = 'skyblue') +
labs(
x = 'x',
y = 'beta(x | α, β)',
title = 'Beta prior with mean 0.3 and standard deviation 0.6',
subtitle = str_glue('α = {α}, β = {signif(β, 3)}')
)

The beta distribution is self-conjugate so the posterior is $$\dbeta(0.6 + 650, 0.4 + 350)$$.

Let’s plot the posterior with priors of different strength. We can increase the strength of the prior whilst keeping the mean constant by multiplying $$\alpha$$ and $$\beta$$ by the same constant c. We will use $$c \in \{ 1, 10, 100, 1000\}$$. In the plot below, we have restricted the x-axis to focus on the differences in the shape of the posteriors.

support <- 650
against <- 350

expand.grid(magnitude = 0:3, x = seq(0, 1, 0.001)) %>%
as_tibble() %>%
mutate(
c = 10^magnitude,
a_prior = α * c,
b_prior = β * c,
y = dbeta(x, support + a_prior, against + b_prior),
prior_magnitude = factor(as.character(10^magnitude))
) %>%
ggplot() +
aes(x, y, colour = prior_magnitude) +
geom_line() +
scale_x_continuous(limits = c(0.55, 0.75)) +
labs(
x = 'x',
y = 'beta(x | support + a, against + b)',
title = 'Beta posterior with different priors',
subtitle = str_glue(paste(
'a = {α} * 10^magnitude, b = {β} * 10^magnitude',
'support = 650, against = 350',
sep = '\n'
)),
colour = 'Magnitude of the prior'
)

Magnitudes 1 and 10 give very similar results close to the maximum likelihood estimate of 65%. The higher magnitudes pull the mean towards the prior mean of 60%.