# 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, 2013 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, 2013 at 8:34
• If they're dealing with continuous cases, there'd be no need to. May 2, 2013 at 12:44
• @Glen_b: But in both books, the distributions are not necessarily continuous.
– Tim
May 2, 2013 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, 2013 at 18:05

Perusal of the derivation of the existence part of the Neyman-Pearson lemma in $$\rm [I], ~[II]$$ will show that the test function $$\varphi(\mathbf x)$$ takes $$\gamma$$ when $$f(\mathbf x|\theta_1) =cf(\mathbf x|\theta_0).$$ This warrants an elaboration: let $$f(\mathbf x|\theta_1)/f(\mathbf x|\theta_0)\equiv \mathrm Y; ~\mathbb P_{\theta_0}(\mathrm Y> c) =: \alpha(c).~\alpha(\cdot)$$ is non-increasing, right continuous and $$\mathbb P_{\theta_0} (\mathrm Y=c) = \alpha(c^{-})-\alpha(c).\tag 1\label{a}$$

Now, for $$\alpha \in (0, 1),$$

\begin{align}\mathbb E_{\theta_0}\varphi(\mathbf x) &= \mathbb P_{\theta_0}(\mathrm Y> c) +\gamma\mathbb P_{\theta_0} (\mathrm Y=c) \\&=\alpha.\tag{2.I}\label{b}\end{align}

Also, \begin{align}1-\mathbb E_{\theta_0}\varphi(\mathbf x) &= \mathbb P_{\theta_0}(\mathrm Y\leq c) -\gamma\mathbb P_{\theta_0} (\mathrm Y=c) \\&=1-\alpha.\tag{2.II}\label{c}\end{align}

$$\bullet$$ If in $$\eqref{b}, ~\mathbb P_{\theta_0}\left(\mathrm Y> c^\prime\right)=\alpha$$ (i.e. in $$\eqref{c}, \mathbb P_{\theta_0}\left(\mathrm Y\leq c^\prime\right) =1-\alpha$$), then $$k=c^\prime$$ and $$\gamma$$ is assigned $$0.$$

$$\bullet$$ Otherwise, there exists $$c_0$$ such that $$\mathbb P_{\theta_0}(\mathrm Y< c_0)\leq 1-\alpha< \mathbb P_{\theta_0}(\mathrm Y\leq c_0)$$ i.e. $$\alpha(c_0) <\alpha\leq\alpha(c_0^{-});$$ take $$k=c_0$$ and \begin{align}\gamma &=\frac{\mathbb P_{\theta_0}(\mathrm Y\leq c_0)-(1-\alpha)}{ \mathbb P_{\theta_0}(\mathrm Y=c_0)}\\ &= \frac{\alpha-\mathbb P_{\theta_0}(\mathrm Y> c_0)}{ \mathbb P_{\theta_0}(\mathrm Y=c_0)}\\ &= \frac{\alpha -\alpha(c_0)}{\alpha(c^{-})-\alpha(c_0)}.\tag 3\label d\end{align}

$$\bullet$$ Note in $$\eqref{d},$$ in case $$\alpha(c_0) =\alpha(c_0^{-}) ,$$ then, as $$\mathbb P_{\theta_0}(\mathrm Y= c_0) =0, ~\varphi(\cdot)$$ is defined almost everywhere.

Authors Casella and Berger in their book is restricting their discussion to what is a non-randomized test. $$(8.3.2)$$ specifies that the test function $$\varphi(\cdot)$$ has to be of size $$\alpha$$ for the set $$R:=\{\mathbf x: \mathrm Y > k\}.$$ Hence, there is no need to assign any $$\gamma$$ to the set $$\{\mathbf x: \mathrm Y = k\}.$$

In fact, in $$\rm [III],$$ the author, in stating (and deriving) the Neyman-Pearson lemma, defined the critical region $$G$$ of size $$\alpha$$ for a certain $$\delta> 0$$ as

$$G = G(\delta) :=\left\{\mathbf x: \frac{\mathcal L_{\theta_0}(\mathbf x) }{\mathcal L_{\theta_1}(\mathbf x) }\leq \delta\right\}.\tag 4\label e$$

Moral of the story is the test function has to be of size $$\alpha:$$ if randomization is needed to attain it, then so be it. Otherwise not. Also, as in $$\eqref{e},$$ care must be taken as to what is constituting the critical region to make it of the desired size.

## References:

$$\rm [I]$$ Testing Statistical Hypotheses, E. L. Lehmann, Joseph P. Romano, Springer Science$$+$$Business Media, $$2005,$$ sec. $$3.2,$$ p. $$61.$$

$$\rm [II]$$ Mathematical Statistics: A Decision Theoretic Approach, Thomas S. Ferguson, Academic Press, $$1967,$$ sec. $$5.1,$$ pp. $$202-203.$$

$$\rm [III]$$ Mathematical Statistics, Wiebe R. Pestman, Walter de Gruyter GmbH & Co., $$2009,$$ sec. $$\rm III. 1,$$ pp. $$161-162.$$

There is no boundary cases to my understanding, because Neyman-Pearson Lemma (in 1, 2, and 3) are testing hypothesis at two points, not regions.

• 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.
– Tim
May 2, 2013 at 2:25