Consider the following model $$\left \{ \begin{array}{l} y_i = x_i\beta + z_ib + \varepsilon_i,\\\\ b_i \sim \mathcal N(0, \Sigma), \quad \varepsilon_i \sim \mathcal N(0, \sigma^2), \end{array} \right.$$


\begin{equation} \Sigma = \begin{bmatrix} \sigma_1^2 & \rho\sigma_1\sigma_2 \\ \rho\sigma_1\sigma_2 & \sigma_2^2 \end{bmatrix}. \end{equation}

The log-likelihood function for this model is given by $$\begin{eqnarray} \ell(\theta) & = & \sum_{i = 1}^n p(y_i; \theta)\\ & = & \sum_{i = 1}^n \int p(y_i, b_i; \theta), \\ & = & \sum_{i = 1}^n \int p(y_i \mid b_i; \theta) \, p(b_i; \theta) \; db_i, \end{eqnarray}$$

where $\theta = (\beta, \sigma, \sigma_1, \sigma_2, \rho)$ denotes the parameters of the model.

Suppose I fix $\theta = \hat{\theta}$, and want to calculate the empirical Bayes estimates $\hat{b}_i = \text{argmax}_b \{\log p(y_i, b_i; \hat{\theta})\}$.

Is $\log p(y_i, b_i; \hat{\theta})$ considered a log-likelihood? I understand that the $b_i$ are random effects and not parameters, but they are unknown quantities. Would it be wrong to term this quantity a log-likelihood?


1 Answer 1


It's a little bit semantics. Namely, to do empirical Bayes you need to write down the posterior distribution of the random effects $b_i$ given the data $y_i$ and (the maximum likelihood) estimates of the parameters $\hat \theta$, i.e.,

$$p(b_i \mid y_i, \hat \theta) \propto p(y_i \mid b_i, \hat \theta) p(b_i \mid \hat \theta).$$

Now in this expression, the likelihood is the first term $p(y_i \mid b_i, \hat \theta)$ and the prior the second term $p(b_i \mid \hat \theta)$. Hence, to find the modes $\hat b_i$ of this posterior, you need to find the mode of the combined likelihood and prior terms, which is equivalent to finding the mode of $\log p(y_i, b; \hat \theta)$ wrt $b$.

With that being said though, these two combined terms have been called a likelihood. An example of this is the H-likelihood approach for fitting mixed models.

  • $\begingroup$ This makes sense. Although, from a Bayesian perspective, I find it somewhat confusing that the terms "posterior", and "prior" are used. $\endgroup$
    – JLee
    Nov 29, 2019 at 12:54

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