BDA3 Chapter 2 Exercise 16

Here’s my solution to exercise 16, chapter 2, of Gelman’s Bayesian Data Analysis (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

Suppose \(y \mid \theta \sim \dbinomial(n, \theta)\) and \(\theta \sim \dbeta(\alpha, \beta)\). Then

\[ \begin{align} p(y) &= \int_0^1 p(y \mid \theta) p(\theta) d\theta \\ &= \int_0^1 \binom{n}{y} \theta^y (1 - \theta)^{n - y} \theta^{\alpha - 1} (1 - \theta)^{\beta - 1} d\theta \\ &= \binom{n}{y} \frac{\Gamma (y + \alpha) \Gamma (n - y + \beta)}{\Gamma (n + \alpha + \beta)} . \end{align} \]

If this density is constant in \(y\) for any \(n\), then

\[ \binom{n}{y} \Gamma (y + \alpha) \Gamma (n - y + \beta) \]

is also constant in \(y\) for any \(n\). In particular, for \(n = 1\), we can evaulate this at \(y = 0\) and \(y = n\) to give

\[ \Gamma (\alpha) \Gamma (1 + \beta) = \Gamma (\beta) \Gamma (1 + \alpha) . \]

Since \(\Gamma (1 + \alpha) = \alpha \Gamma(\alpha)\), it follows that

\[ \beta \Gamma (\alpha) \Gamma (\beta) = \alpha \Gamma (\alpha) \Gamma (\beta) . \]

Using the fact that the gamma function is always positive on the reals, we can conclude \(\alpha = \beta\).

Now, for \(n = 2\), we can evaluate at \(y = 0\) and \(y = 1\) to get

\[ \begin{align} \Gamma(\alpha) \Gamma(\alpha + 2) &= 2 \Gamma(\alpha + 1)^2 \\ \Leftrightarrow (\alpha + 1) \alpha \Gamma (\alpha)^2 &= 2 \alpha^2 \Gamma (\alpha)^2 \\ \Leftrightarrow \alpha + 1 &= 2\alpha \\ \Leftrightarrow \alpha &= 1 \end{align} , \]

where we have used the facts that \(\alpha > 0\), \(\Gamma(\alpha + 1) = \alpha \Gamma(\alpha)\), and that the gamma function is positive on the reals. Therefore, \(\alpha = \beta = 1\). In other words, if the beta-binomial distribution is uniform, then \(\alpha = \beta = 1\).