Chapter 10.11, Problem 111E

Suppose that we are interested in testing the simple null hypothesis H₀:θ = θ₀ versus the simple alternative hypothesis H_a: θ = θ_a. According to the Neyman–Pearson lemma, the test that maximizes the power at θ_a has a rejection region determined by

$\frac{L\left({\theta }_0\right)}{L\left({\theta }_a\right)}<k.$

In the context of a likelihood ratio test, if we are interested in the simple H₀ and H_a, as stated, then ${\Omega }_0=\left\{{\theta }_0\right\}$, ${\Omega }_a=\left\{{\theta }_a\right\}$, and $\Omega =\left\{{\theta }_0,{\theta }_a\right\}$.

1. a Show that the likelihood ratio λ is given by

   $\lambda =\frac{L\left({\theta }_0\right)}{\max \left\{L\left({\theta }_0\right),L\left({\theta }_a\right)\right\}}=\frac{1}{\max \left\{1,\frac{L\left({\theta }_a\right)}{L\left({\theta }_0\right)}\right\}}.$

2. b Argue that λ < k if and only if, for some constant $k^{\prime }$

$\frac{L\left({\theta }_0\right)}{L\left({\theta }_a\right)}<k^{\prime }.$

1. c What do the results in parts (a) and (b) imply about likelihood ratio tests when both the null and alternative hypotheses are simple?