BDA3 Chapter 3 Exercise 7

Here’s my solution to exercise 7, chapter 3, of Gelman’s Bayesian Data Analysis (BDA), 3rd edition. There are solutions to some of the exercises on the book’s webpage.

Suppose we observe \(b\) bikes and \(v\) other vehicles passing a section of road within an hour. We can model the counts as Poisson distributed

\[ \begin{align} b \mid \theta_b &\sim \dpois(\theta_b) \\ v \mid \theta_v &\sim \dpois(\theta_v) \end{align} \]

or as binomial distributed

\[ \begin{align} b \mid n, p &\sim \dbinomial(n, p) \end{align} \]

where \(n\) is the number of trials and \(p\) is the probability of observing a bike. Let

\[ p := \frac{\theta_b}{\theta_b + \theta_v} . \]

We are supposed to show that this definition of \(p\) gives the two models the same likelihood, but I’m stuck. At best I can show that the expectations are different

\[ \mathbb E (b \mid \theta_b) = \theta_b \\ \mathbb E (b \mid n, p) = np = n\frac{\theta_b}{\theta_b + \theta_v} \]

which suggests the conditioning should be done differently.