rbayesianstanbrms

BRMS: on what scale are the estimated coefficients of a model with a log link function?


I am using the brms package to fit a model (brms). However, I am unsure on how to interpret the resulting coefficients/effects. Specifically, on what scale are the coefficients if the link function is a log?

Example from the documentation:

fit1 <- brm(count ~ zAge + zBase * Trt + (1|patient),
            data = epilepsy, family = poisson())
#>  Family: poisson 
#>   Links: mu = log 
#> Formula: count ~ zAge + zBase * Trt + (1 | patient) 
#>    Data: epilepsy (Number of observations: 236) 
#>   Draws: 4 chains, each with iter = 2000; warmup = 1000; thin = 1;
#>          total post-warmup draws = 4000
#> 
#> Group-Level Effects: 
#> ~patient (Number of levels: 59) 
#>               Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> sd(Intercept)     0.58      0.07     0.46     0.73 1.01      768     1579
#> 
#> Population-Level Effects: 
#>            Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> Intercept      1.77      0.11     1.54     1.99 1.00      753     1511
#> zAge           0.09      0.08    -0.07     0.26 1.00      830     1429
#> zBase          0.70      0.12     0.47     0.95 1.00      678     1389
#> Trt1          -0.26      0.16    -0.59     0.05 1.01      709     1356
#> zBase:Trt1     0.05      0.17    -0.29     0.37 1.01      721     1404
#> 
#> Draws were sampled using sampling(NUTS). For each parameter, Bulk_ESS
#> and Tail_ESS are effective sample size measures, and Rhat is the potential
#> scale reduction factor on split chains (at convergence, Rhat = 1).

If I want to represent zAge on the original scale of the data, should I take exp(zAge), or can I directly use zAge?


Solution

  • The coefficients are on the log scale. Assuming that zAge is a standardized/Z-scored version of age, zAge = 0.09 means that an increase of 1 standard deviation in age would lead to an increase of 0.09 log-counts, or approximately a 9% increase in counts, or more precisely an multiplicative increase of exp(0.09) = 1.094 = a 9.4% increase in counts.