# Hypothesis test of union of dependent hypothesis

I have dependent correlation coefficients $$\rho_{AB},\,\rho_{AC}$$, and $$\rho_{AD}$$ (with equal sample sizes for $$\small{A,B,C,D}$$).

I want to test whether $$\rho_{AB}$$ is neither equal to $$\rho_{AC}$$ nor to $$\rho_{AD}$$.

In other words, how can I test the "combined" null hypothesis $$H_{0,combined} : \, \big\{ H_0: \rho_{AB} = \rho_{AC} \, \bigcup \, H_0: \rho_{AB} = \rho_{AD} \big\}$$, i.e. where the alternate hypothesis is $$H_{1,combined} : \, \big\{ H_1: \rho_{AB} \neq \rho_{AC} \, \bigcap \, H_1: \rho_{AB} \neq \rho_{AD}\big\}$$?

PS I could test the individual hypotheses $$H_0: \rho_{AB} = \rho_{AC}$$ and $$H_0: \rho_{AB} = \rho_{AD}$$ using the T-statistic $$T_{xy,xz} = (\hat\rho_{xy}-\hat\rho_{xz})/σ[\hat\rho_{xy}-\hat\rho_{xz}]$$ on the raw coefficients (or their Fisher-z transform) with $$σ[.]$$ estimated via randomizing or bootstrapping.

PPS Can I use the approach Wolfgang discussed at How to test for a significant difference between several dependent correlation coefficients? (in his example he tests whether $$\rho_{AB}$$ "is different from the rest" using $$y=[\rho_{AB},\rho_{AC},\rho_{AD}]′=Xβ+e$$ where $$X = \begin{pmatrix}1 & 1 \\ 1 & 0 \\ 1 & 0 \end{pmatrix}$$)?

This is an intersection-union test. Therefore, the size of the test is bounded above by the maximum size of all of the individual tests. Therefore, you may test each individual hypothesis at level $$\alpha$$, and the overall procedure is level $$\alpha$$. This is because the joint rejection region for all of the hypotheses is a subset of the largest rejection region among all of the hypotheses. Therefore, the probability of incorrectly rejecting the null cannot exceed the probability for the largest rejection region. Note that this is not necessarily a size $$\alpha$$ test. In other words, we only have that $$P(\text{type I error}) \leq \alpha$$. Thus, this approach can be conservative.
• Yes, simply put the rejection region is the intersection of the rejection regions of each individual test. We only reject $H_0$ if all of the individual null hypotheses are rejected. Since the rejection region is an intersection, it is a subset of each individual rejection region. Therefore, it must have smaller probability than all of the individual rejection regions. So, if each individual test is conducted at level $\alpha$, then the intersection cannot have probability larger than $\alpha$ since it is a subset. So this procedure is level $\alpha$, but not necessarily size $\alpha$. Dec 20, 2019 at 17:35