# BDA3 Chapter 2 Exercise 19

Here’s my solution to exercise 19, chapter 2, of Gelman’s *Bayesian Data Analysis* (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

Let’s show that the gamma distribution is conjugate to the exponential distribution. That is, we suppose \(y \mid \theta \sim \dexponential(\theta)\) with prior \(\theta \sim \dgamma(\alpha, \beta)\), and show that the posterior is also gamma distributed.

The posterior is

\[ \begin{align} p(\theta \mid y) &\propto \theta^n e^{-\theta \sum_1^n y_i} \cdot \theta^{\alpha - 1} e^{-\beta \theta} \\ &= \theta^{n + \alpha - 1} e^{-\theta \left(\beta + \sum_1^n y_i \right)} \end{align} \]

which implies \(\theta \mid y \sim \dgamma(\alpha + n, \beta + \sum_1^n y_i)\).

Suppose now that we wish to do inference on \(\phi := \theta^{-1}\). We will show that \(\phi\) has an inverse gamma distribution if \(\theta\) has a gamma distribution. Indeed,

\[ \begin{align} p(\phi) &\propto p(\theta) \left\vert {\frac{d\phi}{d\theta}} \right\vert^{-1} \\ &= \theta^{\alpha - 1} e^{-\beta \theta} \cdot\theta^2 \\ &= \phi^{-\alpha - 1} e^{-\frac{\beta}{\phi}} , \end{align} \]

which corresponds to an \(\dinvgamma(\alpha, \beta)\) distribution.

Suppose that the lifetime of a light bulb can be modelled as an exponential distribution with rate \(\theta\). Let’s compare inferences using the two different parameterisations above. For \(\theta \sim \dgamma(\alpha, \beta)\), the prior variance is \(\frac{\alpha}{\beta^2}\) and the mean is \(\frac{\alpha}{\beta}\), so the prior coefficient of variation is \(\alpha^{-\frac{1}{2}}\). We are given that the prior coefficient of variation is 0.5, so \(\alpha = 4\). The posterior coefficient of variation is \((4 + n)^{-\frac{1}{2}}\). If we wish this to be at most 0.1, then we would need to test at least \(n = 96\) light bulbs.

For \(\phi \sim \dinvgamma(\alpha, \beta)\), the prior variance is

\[ \frac{\beta^2}{(\alpha - 1)^2 (\alpha - 2)} \]

and the mean is \(\frac{\beta}{\alpha - 1}\), so the prior coefficient of variation is \((\alpha - 2)^{-\frac{1}{2}}\). With a prior coefficient of variation is 0.5, we have \(\alpha = 6\). The posterior coefficient of variation is \((6 + n)^{-\frac{1}{2}}\). If we wish this to be at most 0.1, then we would need to test at least \(n = 94\) light bulbs.