Hi, I am trying to understand how elpd fits into the model comparison picture and specifically how it relates to (traditional) log-likelihood ratios.

I sometimes compare models by the ratio of their log-likelihoods obtained by leave-on-out cross-validation (or k-fold cross-validation as an approximation):

\operatorname{LLR}_{1,2} = \log\left(\frac{\operatorname{L_1}}{\operatorname{L_2}}\right) = \log(\operatorname{L_1}) - \log(\operatorname{L_2})

Likelihoods are usually calculated using a point estimate for \theta (often obtained by MLE/MAP estimation through optimization). Assuming that \theta_{-i} is the parameter estimate obtained by fitting the model using all data points except y_i, the likelihood can be written as follows (for discrete outcomes):

\operatorname{L} = \sum^{N}_{i=1}{} \operatorname{Pr}(y_{i}|\hat \theta_{-i})

This seems to be awfully similar to how the elpd is calculated, assuming we draw S samples from the posterior obtained by using all data points except y_i (again, for discrete outcomes):

\widehat{\mathrm{elpd}}_{i}=\log \left(\frac{1}{S} \sum_{s=1}^{S} Pr\left(y_{i} \mid \theta_{-i,s}\right)\right)

\widehat{\mathrm{elpd}}_{\mathrm{full}}=\sum_{i=1}^{N} \widehat{\mathrm{elpd}_{i}}

\widehat{\operatorname{elpd-diff}}_{1,2} = \widehat{\mathrm{elpd}}_{\mathrm{full,2}} - \widehat{\mathrm{elpd}}_{\mathrm{full,2}}.

So could we say that elpd differences are a â€śfull Bayesian versionâ€ť of a log-likelihood ratio that accounts for (a) the prior distribution and (b) uncertainty about \theta? As far as I can tell, elpd differences and log-likelihood ratios should be identical for flat priors and symmetric posteriors, right?