# Understanding output of multinomial multilevel model using brms package

Dear stan community,
I am trying to fit a multinomial multilevel model with 6 discrete categories and 2 levels of variance using the brms-package with family = categorical( link = “logit”).

First a few words about the data and my general aim:

We conducted an ambulatory assessment study and captured data from 119 participants over the course of 10 days. In each situation each participant could choose from 6 discrete categories (musical styles). In addition several variables measuring the situation (e.g. current mood) were measured. The data is not balanced, meaning that the amount of situations reported per participant varies. We also measured variables characterizing the participants (e.g. personality).
Our goal is to predict the outcome variable (musical styles, 6 categories) and we are interested in associations of variables on different variance levels (2 levels) with the outcome variable.
In detail, we are interested in:

1. Associations between variables on level one (within situation) such as current mood and the outcome
2. Associations between variables on level two (between subject) such as personality and the outcome
3. Spliting the variance (ICC) of the two levels to see which percentage of the variance is attributable to each level.

What I did so far:
I started by fitting an intercept-only model using family = categorical(link = “logit”).

m1 <- brm(Style.Self.Code ~ 1 + (1|PersID), data = DF_brms, family = categorical(link = "logit"))

Brms fits the model to the data without any problems but I have some problems understanding the summary output that I could not find any information in the vignette about:

Family: categorical
Links: muOtherculturesLatin = identity; muPop = identity; muRockMetal = identity; muTechnoEDM = identity; muVolksmusikSchlager = identity
Formula: Style.Self.Code ~ 1 + (1 | PersID)
Data: DF_brms (Number of observations: 1561)
Samples: 4 chains, each with iter = 2000; warmup = 1000; thin = 1;
total post-warmup samples = 4000

Group-Level Effects:
~PersID (Number of levels: 104)
Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
sd(muOtherculturesLatin_Intercept)     1.89      0.31     1.35     2.55       1163 1.01
sd(muPop_Intercept)                    1.60      0.24     1.18     2.11       1148 1.00
sd(muRockMetal_Intercept)              1.77      0.28     1.29     2.40       1010 1.00
sd(muTechnoEDM_Intercept)              1.92      0.26     1.46     2.49       1091 1.01
sd(muVolksmusikSchlager_Intercept)     1.43      0.47     0.53     2.42        593 1.01

Population-Level Effects:
Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
muOtherculturesLatin_Intercept     0.04      0.31    -0.62     0.58       1211 1.00
muPop_Intercept                    1.00      0.24     0.52     1.46       1511 1.00
muRockMetal_Intercept              0.84      0.26     0.31     1.33       1113 1.00
muTechnoEDM_Intercept              1.31      0.26     0.78     1.80       1205 1.00
muVolksmusikSchlager_Intercept    -1.54      0.44    -2.54    -0.82        936 1.00

Samples were drawn using sampling(NUTS). For each parameter, Eff.Sample
is a crude measure of effective sample size, and Rhat is the potential
scale reduction factor on split chains (at convergence, Rhat = 1).
1. Why the identity function is shown in the second row starting with Links? Shouldn’t it be the logit function linked here?
2. How to interpret the intercept coefficients? Are these exponentiated odds-ratios?

If I look at the predicted values using predict(m1) I have the following question:

P(Y = Blues & Jazz) P(Y = Other cultures & Latin) P(Y = Pop) P(Y = Rock & Metal) P(Y = Techno & EDM) P(Y = Volksmusik & Schlager)
[1,]             0.05950                       0.01350    0.09875             0.53800             0.27950                      0.01075
[2,]             0.05575                       0.01325    0.09475             0.54500             0.28175                      0.00950
[3,]             0.05150                       0.01725    0.09700             0.54075             0.28225                      0.01125
[4,]             0.05875                       0.01125    0.10550             0.55000             0.26700                      0.00750
[5,]             0.05825                       0.01525    0.10275             0.53300             0.28000                      0.01075
[6,]             0.05475                       0.01200    0.10100             0.53400             0.28875                      0.00950
[7,]             0.05175                       0.01175    0.09800             0.54550             0.28350                      0.00950
[8,]             0.05700                       0.01425    0.10375             0.52100             0.29350                      0.01050
[9,]             0.04750                       0.01325    0.10400             0.55450             0.27000                      0.01075
[10,]             0.05875                       0.01150    0.09950             0.54125             0.27650                      0.01250
...
1. Why do the predicted probabilities vary between situations (i.e., from line to line)? The first 41 lines belong to one person (level-2 unit). As far as I understand it, there should be one probability for each person and category.

My last question addresses the variance split:

1. Is it possible to calculate an intraclass-correlation coefficient for such a model?

I will be very happy about any suggestions regarding one of these four questions.

1 Like

Sorry, I can provide only partial answer,

1. I don’t know, hopefully @paul.buerkner can give more hints what Links: muOtherculturesLatin = identity means in a categorical mode .
2. The intercept coefficients are AFAIK log odds ratios. But in general I find it best to avoid interpreting coefficients and interpret predictions instead - much more intuitive and less prone to error
3. No, there are multiple probabilities for each person, representing individual samples from the posterior distribution. I.e. We are not only uncertain about the outcome, given the model coefficeints, but we are also uncertain about the model coefficients themselves.
4. This should IMHO be possible directly from the posterior samples - just calculate the correlation for each sample separately, which will give you the posterior distribution of the correlations.

Best of luck!

It should indeed say logit not identity as the link function in the summary output. I will fix that.

I’d encourage you to model the random effects as correlated. Paul shows the brms syntax in this series of posts: Correlated random effects in categorical model?

1 Like

AFAIK predict() generates “future” multinomial outcomes based on your model, then (at the default setting) calculates summary statistics based on that. If you want to merely see the model-fitted probabilities (which are indeed fixed for each subject in an intercept-only model), you would have to use fitted() instead.

Hey everyone,
I have a similar question to FabianG’s #1. I am fitting a model using the gamma family…
chl.C3 <- brm(formula = chlorophyll ~ NAO2yrlag + SpringDate + CCBStrat + (1|Year),
family = “Gamma”, prior = chl.C2.prior,
warmup = 1000, inits = 0,
iter = 4000, chains = 2,
control = list(stepsize = 0.01, max_treedepth = 15,

The BRMS vignette indicates that the default link for the gamma family is inverse, not log, so I’m confused as to why the second line of the output indicates mu = log. Any thoughts?

Family: gamma
Links: mu = log; shape = identity
Formula: chlorophyll ~ NAO2yrlag + SpringDate + CCBStrat + (1 | Year)
Data: all.data (Number of observations: 100)
Samples: 2 chains, each with iter = 4000; warmup = 1000; thin = 1;
total post-warmup samples = 6000

Group-Level Effects:
~Year (Number of levels: 20)
Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
sd(Intercept) 0.16 0.08 0.01 0.32 1023 1.00

Population-Level Effects:
Estimate Est.Error l-95% CI u-95% CI Eff.Sample Rhat
Intercept 0.86 0.94 -0.92 2.80 4874 1.00
NAO2yrlag -0.05 0.02 -0.10 0.00 3821 1.00
SpringDate -0.01 0.01 -0.02 0.00 4818 1.00
CCBStrat 0.20 0.06 0.08 0.32 6000 1.00

Thanks for any help!