Suppose I am interested in learning about the proportion $p$ of the population with a certain property (e.g. the proportion who are over 6ft tall). I observe $n$ binary data points, $X_1$, ..., $X_n$ (so $\sum X_i/n$ is the proportion of individuals in my sample with the property). Let us further assume that I sampled the individuals randomly from the population.
If I want to conduct a Bayesian analysis, it seems tempting to
- Quantify my initial uncertainty about $p$ using a normal prior, i.e. $p \sim N(p_0, \sigma_0^2)$ where $p_0$ is my initial 'best guess' about $p$ and $\sigma_0^2$ quantifies my initial uncertainty.
- Assume that, whatever the value of $p$ turns out to be, the data drawn follows a normal distribution centered around $p$. In other words, I assume that every data point $X_i \sim N(p, \sigma^2)$, i.e. we have not just a normal prior but also a normal likelihood.
Assuming the double normal model (i.e. normal prior + normal likelihood) greatly simplifies calculations: e.g. my posterior mean is then a weighted average of my prior mean $p_0$ and the sample average $\sum X_i/n$. However, is this a sensible way to proceed? I have some reservations:
- Obviously, the proportion $p$ is restricted to $[0, 1]$, but the normal distribution's support is the entire real line. So it is perhaps a bit odd to assume a normal prior. This point applies quite generally; but I guess it might not matter in practice since practically all the probability mass will lie in [0, 1] if $p_0$ and $\sigma^2_0$ are chosen appropriately.
- Given the actual proportion $p$, each variable follows a Bernouilli distribution (not normal!) But maybe this is a not such a problem in light of the central limit theorem (since the joint distribution of the data is binomial, which then approximates the normal)?
In light of these points, it is ever sensible to estimate a proportion using the double normal model? Does anyone ever do this? Or it is better to use, e.g. a beta distributed prior and a binomial likelihood?
