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I have read from multiple sources (like this pdf) that in a linear regression model, an ordinal variable can be obtained by imposing linear restrictions on its dummy coding.

e.g. $x$ is an ordinal variable ranging from 1 to 5, so

$$y_1= b_0+ b_1x+e$$

is just a nested model of $$y_2=b_0+b_1(x=2)+b_2(x=3)+b_3(x=4)+b_4(x=5)+e$$

Which linear restrictions should be imposed on $y_2$ in order to obtain $y_1$?

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  • $\begingroup$ If x is constant, then the regression returns the mean of y and the results of [1] and [2] are equivalent although the intercept won't be identical unless x = 1. Otherwise I don't see that the statement is ever correct. [1] in general yields a single sloping line, [2] a series of horizontal lines, depending on how many distinct values of x are observed. $\endgroup$
    – Nick Cox
    Commented Jul 28, 2013 at 9:55
  • $\begingroup$ The original example from Verbeek's "Guide to modern econometrics" is: [1] y = b0 + b1*ln(x) + e, with x ranging from 1 to 5; and [2] y = b0 + b1*(x=2) + b2*(x=3) + b3*(x=4) + b4*(x=5) + e. "the previous model is nested within the current model and imposes three restrictions" (p. 74) $\endgroup$
    – Stefano Tombolini
    Commented Jul 28, 2013 at 10:17
  • $\begingroup$ That is even more puzzling. Why do you think you can ignore the logarithm here? $\endgroup$
    – Nick Cox
    Commented Jul 28, 2013 at 10:22
  • $\begingroup$ The mean is already returned by b0, adding a constant regressor only creates collinearity. $\endgroup$
    – Stefano Tombolini
    Commented Jul 28, 2013 at 10:25
  • $\begingroup$ I figured it out by myself. The three restrictions for the Verbeek's problem are: b1 - ln(2)/ln(3) * b2 = 0; b3 - ln(4)/ln(5) * b4 = 0; b2 - ln(3)/ln(4) * b3 = 0. Similar solutions exist also when dealing directly with x rather than ln(x). $\endgroup$
    – Stefano Tombolini
    Commented Jul 28, 2013 at 14:31

1 Answer 1

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The three restrictions for the problem are: b1 - 1/2 * b2 = 0; b3 - 3/4 * b4 = 0; b2 - 2/3 * b3 = 0.

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