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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)

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    $\begingroup$ 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. $\endgroup$ Sep 22, 2022 at 4:45
  • $\begingroup$ @User1865345 Not only mutually exclusive, but complementary. $\endgroup$
    – Alexis
    Sep 22, 2022 at 5:39

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By definition (see e.g. Jun Shao "Mathematical Statistics", 2nd edition, chapter 6, first paragraph) of statistical hypothesis tests, the null and the alternative hypothesis must be disjoint. Moreover, both hypotheses together must contain all possibilities.

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 $H_1$.

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    $\begingroup$ I'm not sure that the requirement to contain all possibilities is necessarily absolute. e.g. This would appear to typically preclude the testing of simple-vs-simple hypotheses as in the Neyman-Pearson lemma. (That's not to say I don't think it's generally both an important concept and a good strategy.) $\endgroup$
    – Glen_b
    Sep 22, 2022 at 7:52
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    $\begingroup$ @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". $\endgroup$
    – frank
    Sep 22, 2022 at 8:00
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    $\begingroup$ @User1865345 Yes, they also agree with Shao's definition. $\endgroup$
    – frank
    Sep 22, 2022 at 9:28
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    $\begingroup$ @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 $\Theta_0 \cup\Theta_1=\Theta$." $\endgroup$
    – statmerkur
    Sep 22, 2022 at 10:22
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    $\begingroup$ @statmerkur But what is then $\Theta \setminus (\Theta_0 \cup \Theta_1)$ used for? Why not setting $\Theta = \Theta_0 \cup \Theta_1$ ? $\endgroup$
    – frank
    Sep 22, 2022 at 14:13

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