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if ("sigma"%in% names(formula$pforms)) {
# Case with a model on sigmasigma_prior<-brms::prior(
"normal(-0.7, 0.4)", class="Intercept", dpar="sigma"
)
} elseif ("sigma"%in% names(formula$pfix)) {
# Case with sigma fixed to a constantsigma_prior<-NULL
} else {
# Case with no model on sigmasigma_prior<-brms::prior(
"lognormal(-0.7, 0.4)", class="sigma", lb=0, ub="NA"
)
}
Within epidist_family_prior.lognormal. This is to deal with 3 cases:
Model on the dpar
No model on the dpar
The dpar set to a constant
Ideally we should be automating parts of this. Specifically here, setting the prior in the case that there is no model. Can this be obtained as a tranformation of the case where there is a model? And then it can be done at a higher level e.g. in epidist_prior.
This also flags issues with the priors being connected to link functions, which can be changed by the user, and we don't currently account properly for.
Edit: we have also already discussed another issue which would provide a solution to this #266. Summary: just enforce there is always a model ~ 1. This is mentioned in #307.
The text was updated successfully, but these errors were encountered:
one of this and #306 seems like a duplicate. What do you mean by no model? Do you mean no linear model but still a parameter (i.e sigma vs no sigma)? if so can you clarify and add in the already discussed option of enforcing that we always have a linear model (i.e the default is sigma ~ 1.
As of PR #282 we have code like:
Within
epidist_family_prior.lognormal
. This is to deal with 3 cases:dpar
dpar
dpar
set to a constantIdeally we should be automating parts of this. Specifically here, setting the prior in the case that there is no model. Can this be obtained as a tranformation of the case where there is a model? And then it can be done at a higher level e.g. in
epidist_prior
.This also flags issues with the priors being connected to link functions, which can be changed by the user, and we don't currently account properly for.
Edit: we have also already discussed another issue which would provide a solution to this #266. Summary: just enforce there is always a model
~ 1
. This is mentioned in #307.The text was updated successfully, but these errors were encountered: