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.

\(\DeclareMathOperator{\dbinomial}{Binomial} \DeclareMathOperator{\dbern}{Bernoulli} \DeclareMathOperator{\dpois}{Poisson} \DeclareMathOperator{\dnorm}{Normal} \DeclareMathOperator{\dt}{t} \DeclareMathOperator{\dcauchy}{Cauchy} \DeclareMathOperator{\dexponential}{Exp} \DeclareMathOperator{\duniform}{Uniform} \DeclareMathOperator{\dgamma}{Gamma} \DeclareMathOperator{\dinvgamma}{InvGamma} \DeclareMathOperator{\invlogit}{InvLogit} \DeclareMathOperator{\logit}{Logit} \DeclareMathOperator{\ddirichlet}{Dirichlet} \DeclareMathOperator{\dbeta}{Beta}\)

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.