Thanks! Out of curiosity, will the newer version of the plot make it to cran anytime soon?
I did. If I make this change and I run 4 chains, things don’t work:
some chains had errors; consider specifying chains = 1 to debughere are whatever error messages were returned
[[1]]
Stan model 'stan-72e478f44dfb' does not contain samples.
[[2]]
Stan model 'stan-72e478f44dfb' does not contain samples.
My guess is that this is the same problem we are discussing on this thread. @bbbales2 thoughts?
Things get even weirder. I tried to fit this with only one chain, and everything runs smoothly. My guess is that this has something to do with the starting values when you do 4 chains vs only 1 chain. So, given that this works with only one chain, I thought I should share some output:
Computed from 1000 by 1795 log-likelihood matrix
Estimate SE
elpd_loo -8716.7 46.2
p_loo 553.4 21.2
looic 17433.4 92.3
Pareto k diagnostic values:
Count Pct
(-Inf, 0.5] (good) 1290 71.9%
(0.5, 0.7] (ok) 388 21.6%
(0.7, 1] (bad) 106 5.9%
(1, Inf) (very bad) 11 0.6%
See help('pareto-k-diagnostic') for details.
@avehtari if I understood you correctly, this figure shows that the negative binomial is doing a much better job than the Poisson model. Is that correct?
Could you show me the more elegant solution?