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In conditional logit models, global intercepts cannot be estimated as they do not influence the conditional probability of a positive outcome within groups. I understand the intercept term gets canceled out in the derivation of the equation, just as any coefficients for which predictor values are constant within groups; however, would it make sense to estimate the variance among group-intercepts when group is specified as a random term?

Example: I'm looking at consumer choice between two alternative products, one is always chosen. The products differ in their attributes and I'm modeling the conditional probability that one is chosen based on the combination of those attributes. The outcome (being chosen) is grouped by test (strata). Some consumers appear more than once in the data, and some products appear in more than one test. I expect certain products to be more likely to be chosen due to inherent, unobserved attributes. I also expect certain consumers to have inherent preferences (e.g. value attribute1 more than other consumers).

| Test | Consumer | Product | Chosen | Attr1 | Attr2|
| ---- | -------- | ------- | ------ | ----- | ---- | 
| 1    | A        | a       | 0      |  0    | 0.5  | 
| 1    | A        | b       | 1      |  1    | 0.8  | 
| 2    | B        | c       | 1      |  1    | 0.0  | 
| 2    | B        | a       | 0      |  1    | 0.1  | 
| 3    | A        | d       | 1      |  0    | 0.0  | 
| 3    | A        | e       | 0      |  1    | 0.8  |
  1. Can I include a random intercept for Product?
  2. Does it make sense to include random terms for Consumer given there are no between-test estimations?

I'm working in R, rstanarm (but my question is about model specification so feel free to chime in with examples from frequentist approach). stan_clogit() does give me estimates for both randomgroup level intercepts (i.e. Product and Consumer), but I'm not sure if these are meaningful.

Estimates:                              mean   sd   10%   50%   90%
b[(Intercept) Product:a]                0.0    1.1 -1.3   0.0   1.4 
b[(Intercept) Product:b]               -0.4    1.3 -2.1  -0.4   1.1 
...
b[(Intercept) Consumer:P]               -0.1    1.4 -1.3   0.0   1.1 
b[(Intercept) Consumer:Y]               0.0    1.4 -1.2   0.0   1.3 
Sigma[Product:(Intercept),(Intercept)]  2.0    1.6  0.6   1.6   4.0 
Sigma[Consumer:(Intercept),(Intercept)] 2.0    4.9  0.0   0.5   5.1

MCMC diagnostics                           mcse Rhat n_eff
b[(Intercept) Product:a]                   0.0  1.0  5589 
...
b[(Intercept) Consumer:Y]                  0.0  1.0  3509 
Sigma[Product:(Intercept),(Intercept)]     0.0  1.0  1510 
Sigma[Consumer:(Intercept),(Intercept)]    0.1  1.0  4280 
log-posterior                              0.3  1.0  1126 
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1 Answer 1

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If you were doing this by maximising the conditional likelihood there wouldn't be any information about the consumer intercepts. With a Bayesian analysis you can always get estimates out, because you put priors in, but the posterior may not be very different from the prior for those parameters.

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  • $\begingroup$ "With a Bayesian analysis you can always get estimates out, because you put priors in". Could you say a little more about this? I can't fully understand how going Bayesian allows you to estimate these parameters. Is this fundamentally a different model? $\endgroup$
    – awhug
    Nov 16, 2022 at 7:37
  • $\begingroup$ Thank you, @Thomas. Any idea how come random intercept variance (for Product in this case) is estimable when intercepts are not identified? $\endgroup$
    – Olifa
    Nov 16, 2022 at 15:07
  • $\begingroup$ If you have a proper prior distribution over a parameter and the data give you no information about the parameter, the posterior is just the same as the prior. So you can always get estimates, even if the parameters are non-identifiable. There might also be some leakage of information from other parameters that aren't exactly independent, so your posterior might also not be exactly your prior. But your posterior is definitely going to be sensitive to the prior $\endgroup$ Nov 17, 2022 at 2:14

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