I have fitted a general linear model $$y=\beta_0+\beta_1x_1+\beta_2x_2+\beta_3x_3,$$ whose log likelihood is $L_u$.

Now I wish to test if the coefficients are the same.

  • First, overall test: the log likelihood of the reduced model $y=\beta_0+\beta_1\cdot(x_1+x_2+x_3)$ is $L_r$. By likelihood ratio test, the full model is significantly better than the reduced one with $p=0.02$.
  • Next, $\beta_1=\beta_2$? The reduced model is $y=\beta_0+\beta_1\cdot(x_1+x_2)+\beta_2x_3$. The result is, $\beta_1$ is NOT different from $\beta_2$ with $p=0.15$.
  • Similarly, $\beta_1=\beta_3$? They are different with $p=0.007$.
  • Finally, $\beta_2=\beta_3$? They are NOT different with $p=0.12$.

This is quite confusing to me, because I expect the overall $p$ to be smaller than $0.007$, since obviously $\beta_1=\beta_2=\beta_3$ is a much stricter criterion than $\beta_1=\beta_3$ (who generates $p=0.007$).

That is, since I am already "$0.007$ confident" that $\beta_1=\beta_3$ does not hold, I should be "more confident" that $\beta_1=\beta_2=\beta_3$ does not hold. So my $p$ should go down.

Am I testing them wrongly? Otherwise, where am I wrong in the reasoning above?

  • $\begingroup$ I assume x1, x2 and x3 are different levels of a similar factor, dummy coded. Then, I think, such surprising results could arise from differing number of independent replicates (= experimental units) in each levels. $\endgroup$
    – Rodolphe
    Jun 16, 2015 at 11:05

1 Answer 1


That is, since I am already "0.007 confident" that $\beta_1=\beta_3$ does not hold, I should be "more confident" that $\beta_1=\beta_2=\beta_3$ does not hold. So my p should go down

Short answer : Your likelihood should go down. But here, the p-values do not measure the likelihood, but whether the release of some constraints provides a significant improvement on the likelihood. That's why it's not necessarily easier to reject $\beta_1=\beta_2=\beta_3$ than to reject $\beta_1=\beta_3$ because you need to show much better likelihood improvements in the most constrained model to prove that the release of 2 degrees of freedom to reach the full model was "worth it".

Elaboration : Let's draw a graph of likelihood improvements. likelihood graph
The only constraint to avoid a contradiction is that the likelihood improvements must be equal with the sum of likelihood improvement from the indirect path. That's how I found the p-value from the step 1 of the indirect path : $$\frac{L_3}{L_1}=\frac{L_3}{L_2}\times\frac{L_2}{L_1}$$ By likelihood improvements, I mean the log likelihood ratio represented by the $\Delta$Chi-squared, that's why they are summed in the graph. With this schema, one can discard the apparent contradiction because much of the likelihood improvement of the direct path comes from the release of only one degree of freedom ($\beta_1=\beta_3$).
I would suggest two factors that can contribute to this pattern.

  • $\beta_2$ has a large confidence interval in the full model
  • $\beta_2$ is around the mean of $\beta_3$ and $\beta_1$ in the full model

Under these conditions, there is not a big likelihood improvement by releasing one degree of freedom from $\beta_3=\beta_1=\beta_2$ model to the $\beta_3=\beta_1$ model because in the later model the estimation of $\beta_2$ can be close from the two other coefficients.

From this analysis and the two other p-values you gave one could suggest that maybe $\frac{\beta_3+\beta_1}{2}=\beta_2$ can provide a good fit.


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