glmmTMB syntax to brm() syntax

   See below.  The two models (glmmTMB and brms) give sufficiently
similar estimates that I'm confident that the specifications match.

set.seed(101)
library(glmmTMB)
library(brms)
library(broom.mixed)
library(tidyverse)

dd <- data.frame(ID = rep(1:100, each = 10),
                  TRIAL_INDEX = rep(1:10, 100),
                  con = rnorm(1000))
dd$pic_percent <- simulate_new(
     ~ con + (0+con | ID) +
         (0+con | TRIAL_INDEX),
     ziformula = ~1,
     family = beta_family(),
     newdata = dd,
     newparams = list(beta = c(0, 0.5), theta = rep(-1,2),
                      betadisp = 1, betazi = -2))[[1]]


m1 <- glmmTMB(pic_percent ~ con + (0+con | ID) +
     (0+con | TRIAL_INDEX),
         data=dd,
         family = beta_family(),
         ziformula = ~1)

##
https://mvuorre.github.io/posts/2019-02-18-analyze-analog-scale-ratings-with-zero-one-inflated-beta-models/
m2 <- brm(
     bf(pic_percent ~ con + (0+con | ID) +
            (0+con | TRIAL_INDEX),
     zi = ~ 1),
     data=dd,
     family = zero_inflated_beta()
)


(purrr::map_dfr(list(glmmTMB = m1, brms = m2), tidy, .id = "model")
     |> select(model, effect, component, group, term, estimate)
     |> pivot_wider(names_from = model, values_from = estimate)
)



On 10/23/24 19:13, Simon Harmel wrote:

Hello all,

I was wondering what is the closest equivalent of my glmmTMB syntax below
in brms::brm() syntax?

glmmTMBglmmTMB(pic_percent ~ con +
                       (0+con | ID) +
                       (0+con | TRIAL_INDEX),
                     data=DATA,
         family = beta_family(),
         ziformula = ~1)


Thank you,

Simon

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Dr. Benjamin Bolker
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