I have a question about what defines “nested models” that can be compared using the likelihood ratio test (LRT). For example, given the 3 linear models…

M1: Y = Intercept + B1 * X1 + B2 * X2
M2: Y = Intercept + B1 * X1
M3: Y = Intercept + B3 * (0.5 * X1 + 0.5 * X2)

M1 and M2 are nested models because M2 is a special case of M1 with B2 = 0. M1 and M2 can be compared using the LRT with 1 degree of freedom.

The model M3 might arise if you decided to model Y as a function of a pre-specified linear combination of X1 and X2 (in this example I’ve used the mean of X1 and X2). Are M1 and M3 nested models? My thinking is that M3 is a special case of M1 with B1 = B3 * 0.5 and B2 = B3 * 0.5. Can M1 and M3 be compared using the LRT? With 1 degree of freedom?


Yes, M1 is nested in M3.

The models satisfying M3 are the precisely the subspace of models satisfying M1 that also satisfy B1=B2. That's a one-dimensional restriction, so you can test the null hypothesis B1=B2 by comparing M1 and M3 with 1 df.


By specifying the linear combination, you break the nesting.

Say that $M1$ turns out to be $\hat{y}=3-x_1+2x_2$.

$M3$ can’t ever be that equation, as $M3$ requires the coefficients on $x_1$ and $x_2$ to be equal.

You can adjust the linear combination of $x_1$ and $x_2$ but will always be able to find fitted values in $M1$ that cannot happen in $M3$.

  • $\begingroup$ Thanks for your answer. My understanding is that nested models are models that can be obtained by restricting a parameter in a more complex model to be 0. Suppose we rewrite your M1 example as Y = 3 + (0.5 * X1 + 0.5 * X2) - 1.5 * X1 + 1.5 * X2. From this perspective, is my M3 example Y = 3 + (0.5 * X1 + 0.5 * X2) nested? (because the second X1 coefficient (-1.5) and the second X2 coefficient (1.5) are 0). My comment looks kinda wrong to me, but I'm not sure why.. $\endgroup$
    – Scarper
    Mar 25 '20 at 15:39
  • $\begingroup$ I suppose so, but then you have perfect collinearity by using the $x_1$ variable twice and putting different parameters on each instance. I don’t know for sure what perfect collinearity does to a likelihood ratio test, but I suspect that it isn’t good. $\endgroup$
    – Dave
    Mar 25 '20 at 16:06

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