BDA3 Chapter 2 Exercise 15

Here’s my solution to exercise 15, 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 \(Z \sim \dbeta(\alpha, \beta)\). Then for \(m, n \in \mathbb N\) we have

\[ \begin{align} \mathbb E \left[ Z^m (1 - Z)^n \right] &= \frac{1}{B(\alpha, \beta)} \int_0^1 z^m (1 - z)^n z^{\alpha - 1} (1 - z)^{\beta - 1} dz \\ &= \frac{1}{B(\alpha, \beta)} \int_0^1 z^{m + \alpha - 1} (1 - z)^{n + \beta - 1} dz \\ &= \frac{1}{B(\alpha, \beta)} \frac{\Gamma (m + \alpha) \Gamma( n + \beta )}{\Gamma (m + n + \alpha + \beta)} \\ &= \frac{(\alpha + \beta -1)!}{(\alpha - 1)! (\beta - 1)!} \frac{(m + \alpha - 1)! (n + \beta - 1)!}{(m + n + \alpha + \beta - 1)!} , \end{align} \]

where \(B\) is the beta function and \(\Gamma\) is the gamma function.

I’m not sure how the above integral is so useful for this exercise since calculating the mean and variance only require \(n = 0\) and \(m = 1, 2\).

The mean of \(Z\) is the above expectation when \(m = 1\) and \(n = 0\), giving

\[ \mathbb E (Z) = \frac{(\alpha + \beta - 1)!}{(\alpha - 1)! (\beta - 1)!} \frac{\alpha! (\beta - 1)!}{(\alpha + \beta)!} = \frac{\alpha}{\alpha + \beta} . \]

The second moment is given by \(m = 2\) and \(n = 0\), which is

\[ \mathbb E(Z^2) = \frac{(\alpha + \beta - 1)!}{(\alpha - 1)! (\beta - 1)!} \frac{(\alpha +1)! (\beta - 1)!}{(\alpha + \beta + 1)!} = \frac{(\alpha + 1) \alpha}{(\alpha + \beta + 1) (\alpha + \beta)} . \]

It follows that the variance is

\[\begin{align} \mathbb E (Z^2) - \mathbb E (Z)^2 &= \frac{(\alpha + 1) \alpha}{(\alpha + \beta + 1) (\alpha + \beta)} - \frac{\alpha^2}{(\alpha + \beta)^2} \\ &= \frac{(\alpha + 1) \alpha (\alpha + \beta)}{(\alpha + \beta + 1) (\alpha + \beta)^2} - \frac{\alpha^2(\alpha + \beta + 1)}{(\alpha + \beta + 1) (\alpha + \beta)^2} \\ &= \frac{ \alpha^2 (\alpha + \beta) + \alpha (\alpha + \beta) - \alpha^2(\alpha + \beta) - \alpha^2 }{(\alpha + \beta + 1) (\alpha + \beta)^2} \\ &= \frac{ \alpha (\alpha + \beta) - \alpha^2 }{(\alpha + \beta + 1) (\alpha + \beta)^2} \\ &= \frac{ \alpha \beta }{(\alpha + \beta + 1) (\alpha + \beta)^2} . \end{align}\]