# Optimal Tests for Combining \(p\)-Values
## Appendix A Proof of Theorems
If we define \(T^{\prime}_{G(\alpha)}=\left(T_{G(\alpha)}-na\right)/\sqrt{na}=\sum_{i=1}^{n} \left[F_{G(\alpha)}^{-1}(1-P_{i})-a\right]/\sqrt{na}\), then \(T^{\prime}_{G(\alpha)}\to N(0,1)\) (\(\alpha\to\infty\)). On the other hand, since \(T^{\prime}_{G(\alpha)}\) is a linear transformation of \(T_{G(\alpha)}\), it is easy to show that \(T_{G(\alpha)}\equiv T^{\prime}_{G(\alpha)}\) for any \(\alpha>0\). Hence, \(T_{G(\infty)}\equiv\underset{\alpha\to\infty}{lim}T^{\prime}_{G(\alpha)}=Z_{p}\). \(\square\)  
Proof of Theorem 2.: First, we show that \(f_{\alpha,c}(p)=(1-c)^{\alpha}\exp\Bigl{[}cF_{G(\alpha)}^{-1}(1-p)\Bigr{]}\) is a PDF. Let \(y=F_{G(\alpha)}^{-1}(1-x)\), then \(x=1-F_{G(\alpha)}(y)=\int_{y}^{\infty}t^{\alpha-1}\exp(-t)/\Gamma(\alpha)dt\), and \(dx=-y^{\alpha-1}\exp(-y)/\Gamma(\alpha)dy\). Hence, \(\int_{0}^{1}f_{X}(x)dx=\int_{0}^{1}(1-c)^{\alpha}\exp\Bigl{[}cF_{G(\alpha)}^{ -1}(1-x)\Bigr{]}dx=\int_{0}^{\infty}(1-c)^{\alpha}\exp(cy)y^{\alpha-1}\exp(-y)\)\(/\Gamma(\alpha)dy=1\) as\((1-c)^{\alpha}\exp(cy)y^{\alpha-1}\exp(-y)/\Gamma(\alpha)=(1-c)^{\alpha}y^{\alpha-1}\exp [-(1-c)y]/\Gamma(\alpha)\), the PDF of \(Gamma(\alpha,1-c)\). However, under the global null hypothesis, \(P_{i}\sim\text{U}(0,1)\), the log-likelihood ratio under the global null and alternative hypotheses is \(\sum_{i=1}^{n}\ln\bigl{(}f_{\alpha,c}(P_{i})\bigr{)}=a(\alpha)+c\sum_{i=1}^{n}F_{ G(\alpha)}^{-1}(1-P_{i})\), where \(a(\alpha)=-n\alpha\ln(1-c)\), a constant. Therefore, by the Neyman-Pearson lemma [43], \(T_{G(\alpha)}\) is UMP under the specified condition. \(\square\)  
Proof of Theorem 3.: Since the unconstrained MLE for \(c\) is \(\xi_{a_{0}}=1-n\alpha_{0}/\sum_{i=1}^{n}F_{G(\alpha_{0})}^{-1}(1-p_{i})\), when \(\hat{\epsilon}_{\alpha_{0}}\leq 0\), \(T_{CLRT,A_{0}}=0\). On the other hand, when \(\hat{\epsilon}_{\alpha_{0}}>0\), i.e., \(T_{G(\alpha_{0})}=\sum_{i=1}^{n}F_{G(\alpha_{0})}^{-1}(1-p_{i})\)\(>n\alpha_{0}\), \(\hat{\epsilon}_{CLRT,A_{0}}=\hat{\epsilon}_{\alpha_{0}}\), and \(T_{CLRT,A_{0}}=2n\alpha_{0}ln\bigl{(}n\alpha_{0}/\sum_{i=1}^{n}F_{G(\alpha_{0})}^{ -1}(1-p_{i})\bigr{)}+2\sum_{i=1}^{n}F_{G(\alpha_{0})}^{-1}(1-p_{i})\)\(p_{i})-2n\alpha_{0}=2n\alpha_{0}ln\Big{(}n\alpha_{0}/T_{G(\alpha_{0})}\Big{)}+2T_{G (\alpha_{0})}-2n\alpha_{0}=2n\alpha_{0}ln(n\alpha_{0})-2n\alpha_{0}ln\Big{(}T_{G( \alpha_{0})}\Big{)}+2T_{G(\alpha_{0})}-2n\alpha_{0}\). For any \(t>0\), let \(A=\big{\{}t\big{|}t<T_{CLRT,\alpha_{0}}\big{\}}\); it is easy to show that \(A=\big{\{}t\big{|}t<T_{CLRT,\alpha_{0}}\big{\}}=\Big{\{}t\big{|}2n\alpha_{0}ln (n\alpha_{0})-2n\alpha_{0}ln\Big{(}T_{G(\alpha_{0})}\Big{)}+2T_{G(\alpha_{0}) }-2n\alpha_{0}>t\big{\}}=\\ \Big{\{}t\big{|}2\Big{(}T_{G(\alpha_{0})}-n\alpha_{0}\Big{)}>t+2n\alpha_{0}ln \Big{(}T_{G(\alpha_{0})}/n\alpha_{0}\Big{)}\Big{\}}\). Let \(f(x)=x-n\alpha_{0}ln(x)-n\alpha_{0}-t/2+n\alpha_{0}\ln(n\alpha_{0})\), then for \(x>n\alpha_{0}\), \(f^{\prime}(x)=1-n\alpha_{0}/x>0\), and \(f(x)\) is an increasing function of \(x\), but \(\lim\limits_{x\rightarrow(n\alpha_{0})+}f(x)=-t/2<0\), and \(f(kn\alpha_{0})=(k-1-\ln(k))n\alpha_{0}-t/2>0\) for large \(k\). Hence, there must exist a unique \(x_{0}\in(n\alpha_{0},\infty)\) such that \(f(x_{0})=0\). Accordingly, \(Pr\big{[}T_{CLRT,\alpha_{0}}>t\big{]}=Pr\big{[}T_{CLRT,\alpha_{0}}>t,\,\delta _{CLRT,\alpha_{0}}>0\big{]}=Pr\big{[}2n\alpha_{0}ln(n\alpha_{0})-2n\alpha_{0} ln\Big{(}T_{G(\alpha_{0})}\Big{)}+\\ 2T_{G(\alpha_{0})}-2n\alpha_{0}>t\) and \(T_{G(\alpha_{0})}>n\alpha_{0}\big{]}=Pr\big{[}T_{G(\alpha_{0})}-n\alpha_{0}ln \Big{(}T_{G(\alpha_{0})}\Big{)}-n\alpha_{0}-t/2+\\ n\alpha_{0}ln(n\alpha_{0})>0,\,T_{G(\alpha_{0})}>n\alpha_{0}\big{]}=Pr\big{[} T_{G(\alpha_{0})}>t_{0}\big{]}\), where \(t_{0}\) is the root of \(f(x)\), i.e., \(f(t_{0})=0\). This shows that when \(T_{G(\alpha_{0})}>n\alpha_{0}\),\(T_{CLRT,\alpha_{0}}\) and \(T_{G(\alpha_{0})}\) have the same \(p\)-value. On the other hand, when \(T_{G(\alpha_{0})}\leq n\alpha_{0}\), \(T_{CLRT,\alpha_{0}}=0\); hence, \(Pr\big{[}T_{CLRT,\alpha_{0}}=0\big{]}=Pr\big{[}T_{G(\alpha_{0})}<n\alpha_{0} \big{]}\).