# Testing a Null Hypothesis Nested by the Alternative Hypothesis

Consider a parameter of interest $$\beta \in \mathcal{B}$$, and two hypotheses $$H_0:\;\beta\in\mathcal{B_0}\quad versus \quad H_1:\;\beta\in\mathcal{B}_1$$ where $$\mathcal{B}_0\cup\mathcal{B_1}=\mathcal{B}$$ and $$\mathcal{B}_0\subset\mathcal{B_1}$$.

That is, the null hypothesis is a subset of the alternative hypothesis.

To be specific, I am dealing with hypotheses $$H_0:\;\beta\geq0\quad versus \quad H_1:\;\beta\geq c$$ where $$c$$ is an unknown negative constant (i.e. $$c<0$$ and $$c$$ is unknown).

Is there any method or approach related to this situation?

(this post is closely related to my past posting Testing a special inequality hypothesis)

• Conventionally, the class of distributions $\mathcal P$ is classified into two mutually exclusive subclasses $\mathcal H_0\uplus \mathcal H_1= \mathcal P.$ That is the basis of hypothesis testing. The current situation is pretty unorthodox and it seems evident there would be cases where both would be true. Commented Sep 22, 2022 at 4:45
• @User1865345 Not only mutually exclusive, but complementary. Commented Sep 22, 2022 at 5:39

If you are interested in the probability of $$H_0$$ compared to the probability of $$H_1$$, you might want to approximate the density and then integrate it over both $$H_0$$ and $$H_1$$.
• @Glen_b the simple-vs-simple hypotheses consider population families with just two members, so in this sense, all "possibilities" (i.e. all populations in the considered family) are contained in either $H_0$ or $H_1$. See also the first paragraph of chapter 6 in Jun Shao's "Mathematical Statistics". Commented Sep 22, 2022 at 8:00
• @Glen_b I agree that in general we only need disjointness. E.g. Young and Smith (2005) write on page 65 "Throughout we have a parameter space $\Theta$, and consider hypotheses of the form $H_0:\theta \in \Theta_0$ vs. $H_1:\theta \in \Theta_1$, where $\Theta_0$ and $\Theta_1$ are two disjoint subsets of $\Theta$, possibly, but not necessarily, satisfying $\Theta_0 \cup\Theta_1=\Theta$." Commented Sep 22, 2022 at 10:22
• @statmerkur But what is then $\Theta \setminus (\Theta_0 \cup \Theta_1)$ used for? Why not setting $\Theta = \Theta_0 \cup \Theta_1$ ? Commented Sep 22, 2022 at 14:13