We use the underlying continuous latent response variable approach, also discussed in McKelvey & Zavoina (1975) in J of Math'l Soc. You also find this discussed in the Snijders & Bosker multilevel book.
Jinhua Zhao posted on Tuesday, March 31, 2009 - 9:07 am
Does Mplus calculate a McFadden's R-square for nominal dependent variables?
If not, shall I run a model without any predictors and compare the H0 value of the loglikelihoods of the null model and the full model? How do I specify a null model?
Thank you, Jinhua
Jinhua Zhao posted on Tuesday, March 31, 2009 - 9:11 am
To get a null model, I tried to regress the nominal dependent variable on constant ONE=1; But mplus does not allow it because it ONE has a variance of zero.
How do I specify the null model then please?
Jinhua Zhao posted on Tuesday, March 31, 2009 - 9:31 am
I tried a null model with MODEL section left empty. And mplus gives a H0 value of the loglikelihood. Is this the correct way?
Then when I introduce the latent variables to the multinomial logit model, the H0 value of the loglikelihood seems to be on a different order of magnitude. For example in my case: Ho for Null Model: -3502.791 Ho for Model with observed covariates: -2200.72 Ho for Model with latent variable: -21436.896
Then how do I make a meaningful comparison and calculate the R-square?
Jinhua Zhao posted on Tuesday, March 31, 2009 - 3:47 pm
Thank you, Prof. Muthen! This works with your suggestion.
Then after I introduced LV to the MNL model, the H0 value of the loglikelihood becomes an order of magnitude bigger. Large portion of H0 value is from the LV measurement and SEM model part, so no matter how well we do in the discrete choice model part, the R2 will be always small. For example, in my case: Before I introduce LV, H0 for the null model is -2720.4; H0 for the full model is -1510.4; R2 = 0.445 After introducing LV, H0 for the null model (constant only for the choice model part) is -27155.679; H0 for the full model is -26718.724. Now, even the h0 increases quite a bit but R2 is only 0.017. So this seems not a fair comparison between models, and R2 is arbitrarily low for models with LV. Is there a way we can measure the model fit just for the MNL choice model part?
Hi, I am trying to compare two non-nested models with binary outcomes, one criteria mentioned in literature is the explanatory power of the models in prediction (R-square), from the output and below R-square title, i got R-square value for u1 as .98. As i know that R-square in social science hardly exceeds .6, and i perceive R-square of .98 as a very high value...any comments? Thanks,
Yes, that sounds atypical. But also note that R2 for binary outcomes is not that informative given that it refers to a latent response variable - not the observed binary outcome - and no residual variance is estimated.
Right, there is a residual variance reported, but it is not an estimated free parameter. Rather, it is a remainder, computed as 1-(the explained variance). This is in line with probit/logit regression where there is no estimated, free residual variance. See our Topic 2 video for further discussion of this.
If you can do the model in ML, you will have access to BIC.