Question

Show that under the normality assumption, the F test is equivalent to the likelihood ratio test.

Answer 1

The ratio s 2 Y /σˆ 2 is, of course, the F-statistic, up to constants not depending on the data. Since, for this problem, the likelihood ratio test and the F test use equivalent test statistics, if we fix the same size or level α for the two tests, they will have exactly the same power. In fact, even for more complicated linear models — the “general linear tests” of the textbook — the F test is always equivalent to a likelihood ratio test, at least when the presumptions of the former are met. The likelihood ratio test, however, applies to problems which do not involve Gaussian-noise linear models, while the F test is basically only good for them. If you can only remember one of the two tests, remember the likelihood ratio test.

Other constraints Setting p − q parameters to zero is really a special case of imposing p − q linearly independent constraints on the p parameters. For instance, requiring θ2 = θ1 while θ3 = −2θ1 is just as much a two-parameter restriction as fixing θ2 = θ3 = 0. This is because we could transform to a new set of parameters, say ψ1 = θ1, ψ2 = θ2 − θ1, ψ3 = θ3 + 2θ1, where the restrictions are ψ2 = ψ3 = 0, and we can transform back to the θ parameters without loss of information. So the theory of the likelihood ratio test applies whenever we have linearly independent constraints. More generally, that theory applies under the following (admittedly rather complicated) conditions: • Under the null model, θ must obey equations f1(θ) = 0, f2(θ) = 0, . . . fp−q(θ) = 0. • Any θ which obeys those equations is in the null model. • There is an invertible function g where, writing ψ = g(θ), in the null model, ψ always has ψq+1, . . . ψp = 0, and under the alternative, ψ is unrestricted.

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