# Optimal Concordant Tests
## 2 Methods
### Concordant Tests Based on Gamma Distribution
: _(Theorem 5.1 of [PERSON], [PERSON], and [PERSON] 1967 [13]) Let \(X_{1},\ \cdots,X_{n}\) be associated random variables, \(X=(X_{1},\cdots,X_{n}),S_{i}=f_{i}(X),i=1,\cdots,k\), then \(P[S_{1}\leq s_{1},\cdots,S_{k}\leq s_{k}]\geq\prod_{i=1}^{k}[S_{i}\leq s_{i}]\) and \(P[S_{1}>s_{1},\cdots,S_{k}>s_{k}]\geq\prod_{i=1}^{k}[S_{i}>s_{i}]\) for all \(s_{1},s_{2},\cdots,s_{k}\)._  
**Lemma 3**. : _For integer \(n\geq 1\), let \(X=\left(X_{1},\cdots,X_{n}\right)\in\left(0,1\right)^{n}\) have independent components. Set \(T_{\mathrm{G}\left(\alpha\right)}^{L}=T_{\mathrm{G}\left(\alpha\right)}\left(X_ {1},\cdots,X_{n}\right)\), and \(T_{\mathrm{G}\left(\alpha\right)}^{R}=T_{\mathrm{G}\left(\alpha\right)}\left(1 -X_{1},\cdots,1-X_{n}\right)\). Then, for any \(A_{L}>0\), and \(A_{R}>0\), \(P\Big{[}T_{\mathrm{G}\left(\alpha\right)}^{L}>A_{L},T_{\mathrm{G}\left(\alpha \right)}^{R}>A_{R}\Big{]}\leq P\Big{[}T_{\mathrm{G}\left(\alpha\right)}^{L}>A_ {L}\Big{]}P\Big{[}T_{\mathrm{G}\left(\alpha\right)}^{R}>A_{R}\Big{]}\)._  
Proof. : Denote \(f(X)=T_{\mathrm{G}\left(\alpha\right)}^{L}=-\sum_{i=1}^{n}F_{\mathrm{G}\left( \alpha\right)}^{-1}(1-P_{i})\) and \(g(X)=T_{\mathrm{G}\left(\alpha\right)}^{R}=\sum_{i=1}^{n}F_{\mathrm{G}\left( \alpha\right)}^{-1}(P_{i})\). It is easy to see that both \(f\) and \(g\) are nondecreasing functions of \(X\). And the components of \(X\) are independent and hence associated. Therefore, we have  
\[ \begin{array}{c}P\Big{[}T_{\mathrm{G}\left(\alpha\right)}^{L}>A_{L},T_{ \mathrm{G}\left(\alpha\right)}^{R}>A_{R}\Big{]}=P[-f(X)>A_{L},\,g(X)>A_{R}]=P[f (X)<\\ -A_{L},\,g(X)>A_{R}]=P[g(X)>A_{R}]-P[f(X)>-A_{L},\,g(X)>A_{R}]\leq\\ P[g(X)>A_{R}]-P[f(X)>-A_{L}]P[g(X)>A_{R}]=P[g(X)>A_{R}]P[f(X)<\\ -A_{L}]=P\Big{[}T_{\mathrm{G}\left(\alpha\right)}^{R}>A_{R}\Big{]}P\Big{[}T_{ \mathrm{G}\left(\alpha\right)}^{L}>A_{L}\Big{]}.\end{array} \]  
The concordant test, \(T_{\mathrm{G}\left(\alpha\right)}^{C}\), defined in (4) has the following property. **Theorem 1**. : _Let \(X=\left(X_{1},\,\cdots,X_{n}\right)\in\left(0,1\right)^{n}\) be a random vector with independent components, and denote \(T_{\mathrm{G}\left(\alpha\right)}^{L}=T_{\mathrm{G}\left(\alpha\right)}\left(X _{1},\cdots,X_{n}\right)\), \(T_{\mathrm{G}\left(\alpha\right)}^{R}=T_{\mathrm{G}\left(\alpha\right)}\left( 1-X_{1},\cdots,1-X_{n}\right)\), and \(T_{\mathrm{G}\left(\alpha\right)}^{C}=max\Big{(}T_{\mathrm{G}\left(\alpha \right)}^{L},T_{\mathrm{G}\left(\alpha\right)}^{R}\Big{)}\). For any \(A\in\mathbb{R}\), let \(P_{L}=P\Big{[}T_{\mathrm{G}\left(\alpha\right)}^{L}>A\Big{]}\), \(P_{R}=P\Big{[}\,T_{\mathrm{G}\left(\alpha\right)}^{R}>A\Big{]}\), and \(P_{\mathrm{C}}=P\Big{[}T_{\mathrm{G}\left(\alpha\right)}^{C}>A\Big{]}\). Then \(P_{L}+P_{R}-P_{L}P_{R}\leq\ P_{\mathrm{C}}\leq P_{L}+P_{R}\)_  
Proof. : From Lemmas 2 and 3, we have \(P_{\mathrm{C}}=P\Big{[}T_{\mathrm{G}\left(\alpha\right)}^{C}>A\Big{]}=P\Big{[} T_{\mathrm{G}\left(\alpha\right)}^{L}>A\) or \(T_{\mathrm{G}\left(\alpha\right)}^{R}>A\Big{]}\)  
\[\begin{array}{c}=\,P\Big{[}\,T_{\mathrm{G}\left(\alpha\right)}^{L}>A\Big{]} +P\Big{[}\,T_{\mathrm{G}\left(\alpha\right)}^{R}>A\Big{]}-P\,\Big{[}\,T_{ \mathrm{G}\left(\alpha\right)}^{L}>A,T_{\mathrm{G}\left(\alpha\right)}^{R}>A \Big{]}\geq\,P\Big{[}\,T_{\mathrm{G}\left(\alpha\right)}^{L}>A\Big{]}\\ +\,P\Big{[}\,T_{\mathrm{G}\left(\alpha\right)}^{R}>A\Big{]}-P\Big{[}\,T_{ \mathrm{G}\left(\alpha\right)}^{L}>A\Big{]}P\Big{[}\,T_{\mathrm{G}\left( \alpha\right)}^{R}>A\Big{]}=P_{L}+P_{R}-P_{L}P_{R}\). On the other hand, from Bonferroni inequality, we have \(P_{\mathrm{C}}\leq P\Big{[}\,T_{\mathrm{G}\left(\alpha\right)}^{L}>A\Big{]}+P \Big{[}\,T_{\mathrm{G}\left(\alpha\right)}^{R}>A\Big{]}=P_{L}+P_{R}\). Under the condition specified in Theorem 1 (i.e., the global null hypothesis), it is easy to see that \(T_{\mathrm{G}\left(\alpha\right)}^{L}\) and \(T_{\mathrm{G}\left(\alpha\right)}^{R}\) have the same distribution; hence, \(P\Big{[}T_{\mathrm{G}\left(\alpha\right)}^{L}>A\Big{]}=P\Big{[}T_{\mathrm{G} \left(\alpha\right)}^{R}>A\Big{]}\).