# Does Neyman-Pearson Lemma consider the case when the likelihood ratio equals the critical value?

Here are three different versions of Neyman-Pearson lemma. They differ in that the first two (books) ignore the case when the likelihood ratio equals the critical value, while the last one (Wikipedia) doesn't. I feel not ignoring is natural, because ignoring will leave the set where the likelihood ratio equals the critical value to be undecided. So I was wondering why some ignore while some don't? Thanks and regards!

1. In Casella and Berger's Statistical Inference 2. In Bickel and Doksum's Mathematical Statistics Vol 1  Especially in (c), the event excludes the case when the likelihood ratio $$L(x, \theta_0, \theta_1)$$ equals the critical value $$k$$. So how can we say the $$\phi_k$$ ahd $$\phi$$ are equivalent up to a set of probability zero?

3. In Wikipedia:

when performing a hypothesis test between two point hypotheses $$H_0: θ = θ_0$$ and $$H_1: θ = θ_1$$, then the likelihood-ratio test which rejects $$H_0$$ in favour of $$H_1$$ when $$\Lambda(x)=\frac{ L( \theta _0 \mid x)}{ L (\theta _1 \mid x)} \leq \eta$$ where $$P(\Lambda(X)\leq \eta\mid H_0)=\alpha$$ is the most powerful test of size $$α$$ for a threshold $$η$$.

• Formally, the critical value will be in the rejection region; to see that it must be so, see the definition of p-value. May 2 '13 at 5:52
• @Glen_b: Yes, I also thought so, until I saw the two books. why do they ignore the case of likelihood ratio equal the critical value?
– Tim
May 2 '13 at 8:34
• If they're dealing with continuous cases, there'd be no need to. May 2 '13 at 12:44
• @Glen_b: But in both books, the distributions are not necessarily continuous.
– Tim
May 2 '13 at 12:51
• Wikipedia is incorrect in its use of $\le$ instead of $\lt$. Most likely both C&B and B&D state, perhaps in footnotes, that in any case where the boundary of the rejection region can be observed with nonzero probability, an auxiliary randomization device can be used to achieve the intended test size. Thus I speculate that Wikipedia is the source that ignores this issue and the other two either discuss it elsewhere or bury it in assumptions or notes.
– whuber
Jun 28 '13 at 18:05

• Well, the acceptance region and the rejection region should form a partition of the range of $X$, don't they? If you ignore $\{x: \varphi_k(x)=k\}$, then the union of the acceptance region and the rejection region will be a proper subset of the range of $X$, not forming a partition.