I'm currently reading Kruschke's excellent "Doing Bayesian Data Analysis" book. However, the chapter on hierarchical logistic regression (Chapter 20) is somewhat confusing.

Figure 20.2 describes a hierarchical logistic regression where the Bernoulli parameter is defined as a the linear function on the coefficients transformed through a sigmoid function. This seems to be the way hierarchical logistic regression is posed in most of the examples I've seen in other sources online as well. For instance - http://polisci2.ucsd.edu/cfariss/code/SIMlogit02.bug

However, when the predictors are nominal, he adds a layer in the hierarchy - the Bernoulli parameter is now drawn from a beta distribution (Figure 20.5) with parameters determined by mu and kappa, where mu is the sigmoid transformation of the linear function of coefficients, and kappa uses a gamma prior.

This seems reasonable and analogous to the coin-flipping example from chapter 9, but I don't see what having nominal predictors has to do with adding a beta distribution. Why wouldn't one do this in the case of metric predictors and why was the beta distribution added for the nominal predictors?

EDIT: Clarification on the models I'm referring to. First, a logistic regression model with metric predictors (no beta prior). This is similar to other examples of hierarchical logistic regression, such as the bugs example above:

$$ y_i \sim \operatorname{Bernoulli}(\mu_i) \\ \mu_i = \operatorname{sig}(\beta_0 + \sum_j \beta_j x_{ji} ) \\ \beta_0 \sim N(M_0, T_0) \\ \beta_j \sim N(M_\beta, T_\beta) \\ $$

Then the example with nominal predictors. Here's where I don't quite understand the role of the "lower" level of the hierarchy (incorporating the logistic outcome into a beta prior for a binomial) and why it should be different than the metric example.

$$ z_i \sim \operatorname{Bin}(\theta_i, N) \\ \theta_i \sim \operatorname{Beta}(a_j, b_j) \\ a_j = \mu_j \kappa \\ b_j = (1- \mu_j) \kappa \\ \kappa \sim \Gamma(S_\kappa, R_\kappa) \\ \mu_j = \operatorname{sig}(\beta_0 + \sum_j \beta_j x_{ji} ) \\ \beta_0 \sim N(M_0, T_0) \\ \beta_j \sim N(0, \tau_\beta) \\ \tau_\beta = 1/\sigma_{\beta}^2 \\ \sigma_{\beta}^2 \sim \operatorname{folded t} (T_t, DF) $$


The two models you compare have many extraneous features, and I think you can restate your question more clearly in the context of the following two simplified models:

Model 1:

\begin{align} y_i | \mu_i &\sim \operatorname{Bern}( \mu_i ) \\ \mu_i &\sim \pi(\mu_i) \end{align}

Model 2:

\begin{align} y_i | \theta_i & \sim \operatorname{Bern}( \theta_i ) \\ \theta_i | \mu_i,\kappa &\sim \operatorname{Beta}\big( \mu_i\kappa, (1-\mu_i)\kappa \big) \\ \mu_i&\sim \pi(\mu_i) \end{align}

Your questions are: (1) what role is played by the beta distribution; and related, (2) how (if at all) is Model 2 different from Model 1?

On the surface these appear to be pretty different models, but in fact, the marginal distributions of $\mu_i$ in both models are identical. The posterior distribution of $\mu_i$ in Model 1 is \begin{gather} p(\mu_i|y_i) \propto \mu_i^{y_i}(1-\mu_i)^{1-y_i}\pi(\mu_i) \end{gather} whereas the marginal posterior distribution of $\mu_i$ in Model 2 is: \begin{align} p(\mu_i|y_i,\kappa) &\propto \int^1_0 \frac{\theta_i^{y_i + \mu_i\kappa - 1}(1-\theta_i)^{\kappa(1-\mu_i)-y_i}}{B\big(\kappa\mu_i,\kappa(1-\mu_i)\big)} d\theta \,\pi(\mu_i) \\ &\propto \frac{B\big(y_i+\mu_i\kappa,1-y_i+\kappa(1-\mu_i)\big)\pi(\mu_i) }{B\big(\kappa\mu_i,\kappa(1-\mu_i)\big)} \\ &\propto \mu_i^{y_i}(1-\mu_i)^{1-y_i} \pi(\mu_i) \end{align}

Thus any advantage gained from using Model 2 is computational. Overparameterizing hierarchical models, such as the addition of $\theta_i$ in Model 2, can sometimes improve the efficiency of the sampling procedure; for example, by introducing conditionally conjugate relationships between groups of parameters (see Jack Tanner's answer), or by breaking correlation among parameters of interest (google "Parameter Expansion").


The reason for drawing the Bernoulli parameter from a beta distribution is that the beta is conjugate to the binomial. Using a conjugate prior distribution enables a closed-form solution to finding the posterior.

EDIT: clarifying. Either model will work. Even with MCMC, it's useful to have conjugate priors because that permits the use of specialized samplers for various types of distributions that are more efficient than generic samplers. For example, see the JAGS user manual sec. 4.1.1 and sec 4.2.

  • $\begingroup$ There may not be enough context from the book in my question, but these analyses are performed with Gibbs sampling, so a closed form representation of the posterior isn't necessary. In the example I linked, the bernoulli parameter isn't fixed as a beta distribution, but arises from a sigmoid transformation of the linear predictors, which have normally distributed coefficients. This is also how Kruschke presents an earlier example (with metric predictors) in the chapter as well (bernoulli parameter is just the sigmoid transformation of the linear function with normally distributed coefficients) $\endgroup$
    – user4733
    May 17 '12 at 21:28
  • $\begingroup$ @user4733 Jack Tanner is right about beta being the conjugate prior to bernoulli samples. it seems like more than a coincidence that it was chosen. Yes you may be doing Gibbs sampling to get the posterior distribution but in a hierarchical model there is more than one prior involved and it could be that you are putting a prior on a hyperparameter (a parameter for a family of prior distributions. it is a prior on the prior if you will. In that context it may be convenient to use a conjugate prior. Some of your description of the book is confusing to us. $\endgroup$ May 17 '12 at 21:55
  • 1
    $\begingroup$ You are taking little excerpts that create gaps in our ability to understand what is going on. You need to describe the model and the hierarchy of priors better for us to help (at least for me)> $\endgroup$ May 17 '12 at 21:57
  • $\begingroup$ Added some descriptions to the hierarchical models I'm referring. Hopefully it helps. $\endgroup$
    – user4733
    May 17 '12 at 22:24

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