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I run an ordinal regression model and I wanted to check the proportional odds assumption. In order to do that I used the VGAM package and I run olr twice, the first under the assumption and the second without the assumption. Below is the code and the results

> fit1 <- vglm(stage ~Ki67+Cyclin_E,family=cumulative(parallel=T))
> summary(fit1)

vglm(formula = stage ~ Ki67 + Cyclin_E, family = cumulative(parallel = T))

Pearson Residuals:
                     Min       1Q  Median      3Q    Max
logit(P[Y< = 1]) -3.1177 -0.43593 0.37246 0.53111 1.4927
logit(P[Y< = 2]) -3.8479  0.14119 0.18785 0.28679 1.9903

                Estimate Std. Error  z value
(Intercept):1  2.2414705   1.091225  2.05409
(Intercept):2  3.2164214   1.178916  2.72829
Ki67          -0.1157273   0.039889 -2.90124
Cyclin_E       0.0085266   0.028626  0.29786

Number of linear predictors:  2 

Names of linear predictors: logit(P[Y< = 1]), logit(P[Y< = 2])

Dispersion Parameter for cumulative family:   1

Residual deviance: 50.82946 on 62 degrees of freedom

Log-likelihood: -25.41473 on 62 degrees of freedom

Number of iterations: 5 

> fit2 <- vglm(grade ~Ki67+Cyclin_E,family=cumulative(parallel=F),maxit=50)
> summary(fit2)

vglm(formula = grade ~ Ki67 + Cyclin_E, family = cumulative(parallel = F), 
    maxit = 50)

Pearson Residuals:
                     Min       1Q   Median      3Q    Max
logit(P[Y< = 1]) -1.1870 -0.65271 -0.23199 0.44910 3.4798
logit(P[Y< = 2]) -2.6235 -0.70599  0.27305 0.72691 2.8544

               Estimate Std. Error   z value
(Intercept):1 -0.059702   0.928078 -0.064328
(Intercept):2  2.687277   1.050909  2.557097
Ki67:1        -0.100832   0.047754 -2.111483
Ki67:2        -0.101817   0.036567 -2.784377
Cyclin_E:1     0.018768   0.022708  0.826474
Cyclin_E:2    -0.015416   0.022927 -0.672390

Number of linear predictors:  2 

Names of linear predictors: logit(P[Y< = 1]), logit(P[Y< = 2])

Dispersion Parameter for cumulative family:   1

Residual deviance: 78.93318 on 78 degrees of freedom

Log-likelihood: -39.46659 on 78 degrees of freedom

Number of iterations: 34 

In order to check if the difference of the two models is significant I run the next command

[1] 0.03072927

As you see the result is that the 2 models differ and so the proportional odds assumption isn't true. But if you see the coefficients about the Ki67 (Cyclin is not significantly important so i guess i can skip it) they are almost the same. In that case what should I do? I believe that I could stick with the model under the po assumption but I'd like to know what others think

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If you don't think the proportional odds assumption is reasonable, then you could fit a multinomial logistic model instead of an ordinal logistic model. There are more parameters but fewer assumptions... –  Macro Apr 10 '12 at 12:54
On the contrary, I believe that the proportional odds holds despite the results of the test since the coefficients don't differ almost at all.Regarding the use of a multinomial logistic model isn't it wrong to use it when you know the response variable is ordered? –  Nick Apr 10 '12 at 15:03
It's not "wrong" to use the multinomial logistic model. True, it doesn't take advantage of the ordinal structure in the data but, as I said, the ordinal model is a submodel of the multinomial model. Therefore, any fit achievable with the ordinal model is achievable with the multinomial model. If the proportional odds assumption does hold, you're sacrificing parsimony by using the multinomial model. –  Macro Apr 10 '12 at 15:23
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1 Answer

Rather than test the significance of the differences in the deviances, why not plot the predicted probabilities from each model against each other in a scatterplot? You might also plot the differences between the two predicted probabilities against the predicted probability from the first model. Finally you might look at a boxplot of the differences.

If the predicted values are similar, then the models are, for most purposes anyway, giving similar results.

Another idea is to compare the parameter estimates on the independent variables in the two models, but this might require some rescaling.

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Can you be more precise about the word "similar"? Since there is a significant reduction in deviance, there is some explanatory power in the additional parameters, so we know a priori that there will be some differences in the predictions. Similar could mean anything from seeing that the predictor effects have the same sign as before, to seeing that the percentage of the time the predicted classes are different is "small" (what is small?) –  Macro Apr 10 '12 at 16:16
I think "similar" depends on context - are the differences big enough that people in the field in question will care about them? Certainly there will be some difference in the predictions, but is it worth the complexity? Not so much in the sense of a test of statistical sig., but in the sense of "are these differences worth writing about?" The size of difference that will matter will vary a lot, depending on the nature of the study, how precise theory is, how much is known, etc. –  Peter Flom Apr 11 '12 at 10:35
@Peter Flom- I'll check your idea about the scatterplot. In the meantime I did the olr is SPSS and it didn't reject the assumption. Moreover I found on the net a pdf by Agresti where he faces a similar problem and his answer is the following The improvement in fit is statistically significant, but perhaps not substantively significant; effect of dose is moderately negative for each cumulative probability which I think is the case in my problem since in both models the Ki67 coef. are similar. Also could you explain to me what do you mean by saying rescaling of the the parameter estimates? –  Nick Apr 11 '12 at 13:36
Always glad when a top statistician like Agresti has said something similar to what I said! :-) The parameter estimates might not be directly comparable (I am not sure on this) because they are estimating different things. When the DV changes (even if only in scale) the meaning of the parameters changes. I was thinking this might be like changes inches to feet in the DV. (Again, I'm not sure on this). –  Peter Flom Apr 11 '12 at 14:07
Sorry I forgot to tell you the name of the pdf in case you or someone else is interested. Well, it's called Examples of Using R for Modeling Ordinal Data. –  Nick Apr 12 '12 at 10:43
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