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Suppose $Y$ is an ordinal response variable with 4 levels (i.e. $Y$ can take on the values $1,2,3$ or $4$). We want to model $Y$ with $X_1, X_2$ using a proportional odds model. If $Y$ takes on the values of $1,2,3$ and $4$ an equal number of times (e.g. $Y = 1$ 4 times, $Y = 2$ 4 times, $Y = 3$ 4 times, and $Y= 4$ 4 times), will the estimates of some of the coefficients be 0?

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    $\begingroup$ Not necessarily - it depends on the values of $X_1, X_2$. $\endgroup$ – Macro Feb 12 '12 at 21:37
  • $\begingroup$ @Macro: Suppose $X_1$ and $X_2$ are binary and take on 1 and 0 an equal number of times. $\endgroup$ – Tony J Feb 12 '12 at 22:13
  • $\begingroup$ Whether or not you will get 0 coefficients depends on how those outcomes correspond to the values of $Y$. If all possible combinations of $X_1,X_2$ occur equally often for each value of $Y$, then the coefficients for $X_1,X_2$ will be 0. $\endgroup$ – Macro Feb 12 '12 at 22:53
  • $\begingroup$ @Macro: How would you interpret the regression coefficients if they are 0? What if the y intercept is 0? $\endgroup$ – Tony J Feb 12 '12 at 23:08
  • $\begingroup$ If you're fitting a 4-level ordinal response using a proportional odds model then there should be three "threshold" values that are returned, not just a single intercept (ordinal logistic regression is equivalent to a regression model of a logistic random variable that is thresholded to produce ordinal levels - the thresholds returned denote the changeover points on that latent scale). In a binary logistic regression model, a '0' intercept and '0' values for all coefficients tells me half of all samples a 0s and half are 1s, and that predictors are distributed exactly the same in both halves. $\endgroup$ – Macro Feb 13 '12 at 1:01
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Macro's answered this pretty well in the comments but to illustrate, and in case there were any doubt remaining that the answer is "not necessarily, indeed it is unlikely":

> library(MASS)
> x1 <- sample(0:1, 1000, replace=T)
> x2 <- sample(0:1, 1000, replace=T)
> y <- ordered(sample(1:4, 1000, replace=T))
> polr(y~x1+x2)
Call:
polr(formula = y ~ x1 + x2)

Coefficients:
      x1       x2 
 0.16223 -0.05669 

Intercepts:
     1|2      2|3      3|4 
-1.04553  0.05007  1.19741 

Residual Deviance: 2769.64 
AIC: 2779.64 
> polr(y~x1*x2)
Call:
polr(formula = y ~ x1 * x2)

Coefficients:
     x1      x2   x1:x2 
 0.2466  0.0277 -0.1638 

Intercepts:
     1|2      2|3      3|4 
-1.00048  0.09521  1.24313 

Residual Deviance: 2769.12 
AIC: 2781.12 
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