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Computes log marginal likelihood via bridge sampling, which can be used in the computation of bayes factors and posterior model probabilities. The brmsfit method is just a thin wrapper around the corresponding method for stanfit objects.

Usage

# S3 method for class 'brmsfit'
bridge_sampler(samples, recompile = FALSE, ...)

Arguments

samples

A brmsfit object.

recompile

Logical, indicating whether the Stan model should be recompiled. This may be necessary if you are running bridge sampling on another machine than the one used to fit the model. No recompilation is done by default.

...

Additional arguments passed to bridge_sampler.stanfit.

Details

Computing the marginal likelihood requires samples of all variables defined in Stan's parameters block to be saved. Otherwise bridge_sampler cannot be computed. Thus, please set save_pars = save_pars(all = TRUE) in the call to brm, if you are planning to apply bridge_sampler to your models.

The computation of marginal likelihoods based on bridge sampling requires a lot more posterior draws than usual. A good conservative rule of thump is perhaps 10-fold more draws (read: the default of 4000 draws may not be enough in many cases). If not enough posterior draws are provided, the bridge sampling algorithm tends to be unstable leading to considerably different results each time it is run. We thus recommend running bridge_sampler multiple times to check the stability of the results.

More details are provided under bridgesampling::bridge_sampler.

Examples

# \dontrun{
# model with the treatment effect
fit1 <- brm(
  count ~ zAge + zBase + Trt,
  data = epilepsy, family = negbinomial(),
  prior = prior(normal(0, 1), class = b),
  save_pars = save_pars(all = TRUE)
)
#> Compiling Stan program...
#> Start sampling
#> 
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 1).
#> Chain 1: 
#> Chain 1: Gradient evaluation took 2.9e-05 seconds
#> Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.29 seconds.
#> Chain 1: Adjust your expectations accordingly!
#> Chain 1: 
#> Chain 1: 
#> Chain 1: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 1: Iteration:  200 / 2000 [ 10%]  (Warmup)
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#> Chain 1: Iteration: 1000 / 2000 [ 50%]  (Warmup)
#> Chain 1: Iteration: 1001 / 2000 [ 50%]  (Sampling)
#> Chain 1: Iteration: 1200 / 2000 [ 60%]  (Sampling)
#> Chain 1: Iteration: 1400 / 2000 [ 70%]  (Sampling)
#> Chain 1: Iteration: 1600 / 2000 [ 80%]  (Sampling)
#> Chain 1: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 1: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 1: 
#> Chain 1:  Elapsed Time: 0.114 seconds (Warm-up)
#> Chain 1:                0.115 seconds (Sampling)
#> Chain 1:                0.229 seconds (Total)
#> Chain 1: 
#> 
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 2).
#> Chain 2: 
#> Chain 2: Gradient evaluation took 2.2e-05 seconds
#> Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.22 seconds.
#> Chain 2: Adjust your expectations accordingly!
#> Chain 2: 
#> Chain 2: 
#> Chain 2: Iteration:    1 / 2000 [  0%]  (Warmup)
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#> Chain 2: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 2: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 2: 
#> Chain 2:  Elapsed Time: 0.114 seconds (Warm-up)
#> Chain 2:                0.125 seconds (Sampling)
#> Chain 2:                0.239 seconds (Total)
#> Chain 2: 
#> 
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 3).
#> Chain 3: 
#> Chain 3: Gradient evaluation took 2.3e-05 seconds
#> Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.23 seconds.
#> Chain 3: Adjust your expectations accordingly!
#> Chain 3: 
#> Chain 3: 
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#> Chain 3: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 3: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 3: 
#> Chain 3:  Elapsed Time: 0.112 seconds (Warm-up)
#> Chain 3:                0.109 seconds (Sampling)
#> Chain 3:                0.221 seconds (Total)
#> Chain 3: 
#> 
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 4).
#> Chain 4: 
#> Chain 4: Gradient evaluation took 2.1e-05 seconds
#> Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.21 seconds.
#> Chain 4: Adjust your expectations accordingly!
#> Chain 4: 
#> Chain 4: 
#> Chain 4: Iteration:    1 / 2000 [  0%]  (Warmup)
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#> Chain 4: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 4: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 4: 
#> Chain 4:  Elapsed Time: 0.115 seconds (Warm-up)
#> Chain 4:                0.117 seconds (Sampling)
#> Chain 4:                0.232 seconds (Total)
#> Chain 4: 
summary(fit1)
#>  Family: negbinomial 
#>   Links: mu = log 
#> Formula: count ~ zAge + zBase + Trt 
#>    Data: epilepsy (Number of observations: 236) 
#>   Draws: 4 chains, each with iter = 2000; warmup = 1000; thin = 1;
#>          total post-warmup draws = 4000
#> 
#> Regression Coefficients:
#>           Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> Intercept     1.89      0.07     1.75     2.04 1.00     4394     3210
#> zAge          0.11      0.05     0.01     0.21 1.00     4859     3241
#> zBase         0.72      0.05     0.61     0.83 1.00     4399     3019
#> Trt1         -0.19      0.10    -0.39     0.01 1.00     4295     3013
#> 
#> Further Distributional Parameters:
#>       Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> shape     2.35      0.32     1.80     3.06 1.00     4066     3210
#> 
#> 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).
bridge_sampler(fit1)
#> Iteration: 1
#> Iteration: 2
#> Iteration: 3
#> Iteration: 4
#> Bridge sampling estimate of the log marginal likelihood: -672.2953
#> Estimate obtained in 4 iteration(s) via method "normal".

# model without the treatment effect
fit2 <- brm(
  count ~ zAge + zBase,
  data = epilepsy, family = negbinomial(),
  prior = prior(normal(0, 1), class = b),
  save_pars = save_pars(all = TRUE)
)
#> Compiling Stan program...
#> Start sampling
#> 
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 1).
#> Chain 1: 
#> Chain 1: Gradient evaluation took 3.2e-05 seconds
#> Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.32 seconds.
#> Chain 1: Adjust your expectations accordingly!
#> Chain 1: 
#> Chain 1: 
#> Chain 1: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 1: Iteration:  200 / 2000 [ 10%]  (Warmup)
#> Chain 1: Iteration:  400 / 2000 [ 20%]  (Warmup)
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#> Chain 1: Iteration:  800 / 2000 [ 40%]  (Warmup)
#> Chain 1: Iteration: 1000 / 2000 [ 50%]  (Warmup)
#> Chain 1: Iteration: 1001 / 2000 [ 50%]  (Sampling)
#> Chain 1: Iteration: 1200 / 2000 [ 60%]  (Sampling)
#> Chain 1: Iteration: 1400 / 2000 [ 70%]  (Sampling)
#> Chain 1: Iteration: 1600 / 2000 [ 80%]  (Sampling)
#> Chain 1: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 1: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 1: 
#> Chain 1:  Elapsed Time: 0.108 seconds (Warm-up)
#> Chain 1:                0.096 seconds (Sampling)
#> Chain 1:                0.204 seconds (Total)
#> Chain 1: 
#> 
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 2).
#> Chain 2: 
#> Chain 2: Gradient evaluation took 2.3e-05 seconds
#> Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.23 seconds.
#> Chain 2: Adjust your expectations accordingly!
#> Chain 2: 
#> Chain 2: 
#> Chain 2: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 2: Iteration:  200 / 2000 [ 10%]  (Warmup)
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#> Chain 2: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 2: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 2: 
#> Chain 2:  Elapsed Time: 0.105 seconds (Warm-up)
#> Chain 2:                0.108 seconds (Sampling)
#> Chain 2:                0.213 seconds (Total)
#> Chain 2: 
#> 
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 3).
#> Chain 3: 
#> Chain 3: Gradient evaluation took 2e-05 seconds
#> Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.2 seconds.
#> Chain 3: Adjust your expectations accordingly!
#> Chain 3: 
#> Chain 3: 
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#> Chain 3: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 3: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 3: 
#> Chain 3:  Elapsed Time: 0.101 seconds (Warm-up)
#> Chain 3:                0.12 seconds (Sampling)
#> Chain 3:                0.221 seconds (Total)
#> Chain 3: 
#> 
#> SAMPLING FOR MODEL 'anon_model' NOW (CHAIN 4).
#> Chain 4: 
#> Chain 4: Gradient evaluation took 2.3e-05 seconds
#> Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.23 seconds.
#> Chain 4: Adjust your expectations accordingly!
#> Chain 4: 
#> Chain 4: 
#> Chain 4: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 4: Iteration:  200 / 2000 [ 10%]  (Warmup)
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#> Chain 4: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 4: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 4: 
#> Chain 4:  Elapsed Time: 0.102 seconds (Warm-up)
#> Chain 4:                0.104 seconds (Sampling)
#> Chain 4:                0.206 seconds (Total)
#> Chain 4: 
summary(fit2)
#>  Family: negbinomial 
#>   Links: mu = log 
#> Formula: count ~ zAge + zBase 
#>    Data: epilepsy (Number of observations: 236) 
#>   Draws: 4 chains, each with iter = 2000; warmup = 1000; thin = 1;
#>          total post-warmup draws = 4000
#> 
#> Regression Coefficients:
#>           Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> Intercept     1.80      0.05     1.70     1.90 1.00     4665     3241
#> zAge          0.13      0.05     0.02     0.23 1.00     4317     3255
#> zBase         0.72      0.06     0.62     0.84 1.00     4529     3103
#> 
#> Further Distributional Parameters:
#>       Estimate Est.Error l-95% CI u-95% CI Rhat Bulk_ESS Tail_ESS
#> shape     2.33      0.32     1.78     3.02 1.00     3962     3009
#> 
#> 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).
bridge_sampler(fit2)
#> Iteration: 1
#> Iteration: 2
#> Iteration: 3
#> Iteration: 4
#> Iteration: 5
#> Bridge sampling estimate of the log marginal likelihood: -671.6805
#> Estimate obtained in 5 iteration(s) via method "normal".
# }