# Optimal Tests for Combining \(p\)-Values
## Appendix A Proof of Theorems  
To prove Theorem 1, we need the following results:  
**Lemma A1** (Theorem 1 of [PERSON], [PERSON], and [PERSON] 2017). : _If \(Y_{\alpha}=Gamma(\alpha,1)\), then \(\underset{\alpha\to 0+}{lim}-\alpha ln(Y_{\alpha})\,\sim Exp(1)\) in distribution [42]._  
**Corollary A1**. : \(Pr\Big{(}\underset{\alpha\to 0+}{lim}Y_{\alpha}>1\Big{)}=0\)_._  
Proof of Corollary A1. : From Lemma A1, \(\Pr\Big{(}\underset{\alpha\to 0+}{lim}Y_{\alpha}>1\Big{)}=\Pr(\underset{ \alpha\to 0+}{lim}\ln(Y_{\alpha})>0)=\Pr(\lim\limits_{\alpha\to 0+}-\alpha\ln(Y_{ \alpha})<0)=\Pr(Exp(1)<0)=0\). \(\square\)  
**Corollary A2**. : _Let \(Y=\underset{\alpha\to 0+}{lim}Y_{\alpha}^{-\alpha}\), where \(Y_{\alpha}=Gamma(\alpha,1)\), then the PDF of \(Y\) is \(f_{Y}(y)=1/y^{2}\) for \(y\in(1,\infty)\)._  
Proof of Corollary A2. : Notice that \(Y=\underset{\alpha\to 0+}{lim}Y_{\alpha}^{-\alpha}=\exp\big{[}\ln( \underset{\alpha\to 0+}{lim}Y_{\alpha}^{-\alpha})\big{]}\!=\exp\big{[} \underset{\alpha\to 0+}{lim}-\alpha\ln(Y_{\alpha})\big{]}\). But from Lemma A1, \(\underset{\alpha\to 0+}{lim}-\alpha\ln(Y_{\alpha})\sim Exp(1)\). Hence, the PDF of \(Y\) is \(f_{Y}(y)=\exp(-\ln(y))/y=1/y^{2}\) for \(y\in[1,\infty)\). \(\square\)  
**Lemma A2**. : _Let \(0<p_{1}<p_{2}<1\) and \(q_{i}^{(a)}=F_{G(a)}^{-1}(p_{i})\)\((i=1,2)\). Denote \(r_{a}=q_{2}^{(a)}/q_{1}^{(a)}\), then we have \(\underset{\alpha\to 0+}{lim}r_{\alpha}=\infty\). \(\square\)_  
Proof of Lemma A2. : Suppose \(\underset{\alpha\to 0+}{lim}r_{\alpha}=\infty\) does not hold; then, there exists a constant \(R\) such that \(r_{\alpha}<R\) for any \(\alpha>0\). However, \(0<p_{2}-p_{1}=\Pr\Bigl{(}q_{1}^{(a)}<Y_{\alpha}<q_{2}^{(a)}\Bigr{)}=\Pr\Bigl{[} -\alpha\ln\Bigl{(}q_{2}^{(a)}\Bigr{)}<-\alpha\ln(Y_{\alpha})<-\alpha\ln\Bigl{(} q_{1}^{(a)}\Bigr{)}\Bigr{]}=\Pr[-\alpha\ln\Bigl{(}q_{1}^{(a)}\Bigr{)}-\alpha\ln(r_{ \alpha})<-\alpha\ln(Y_{\alpha})<-\alpha\ln\Bigl{(}q_{1}^{(a)}\Bigr{)}\]  
**Corollary A3**. : \(\underset{\alpha\to 0+}{lim}T_{G(\alpha)}(P_{1},P_{2},\cdots,P_{n})= \underset{\alpha\to 0+}{lim}F_{G(\alpha)}^{-1}\Bigl{(}1-P_{(1)}\Bigr{)}\)_, where \(P_{(1)}\) is the smallest value of \(P_{1},P_{2},\cdots,P_{n}\)._  
Proof of Corollary A3. : This is a direct consequence of Lemma A2. \(\square\)  
Proof of Theorem 1.: Now, we prove Theorem 1:  
1. Denote \(Q_{1}=\underset{\alpha\to 0+}{lim}F_{G(\alpha)}^{-1}\Bigl{(}1-P_{(1)} \Bigr{)}\). From Lemma A1 and Corollary A3, \(P_{T_{G(0)}}=\underset{\alpha\to 0+}{lim}\Pr(\text{Gamma}(n\alpha,1)>Q_{1})= \underset{\alpha\to 0+}{lim}\Pr(Y_{n\alpha}>Q_{1})=\underset{\alpha\to 0+}{lim}\Pr(-n \alpha\ln(Y_{n\alpha})<-n\alpha\ln(Q_{1}))=1-\exp[n\alpha\ln(Q_{1})]\). But, \(\exp[n\alpha\ln(Q_{1})]=\exp[\ln\bigl{(}Q_{1}^{n\alpha}\bigr{)}]=Q_{1}^{n \alpha}=\left(Q_{1}^{\alpha}\right)^{n}=\left\{\exp[\alpha\ln(Q_{1})]\right\} ^{n}=\left[\underset{\alpha\to 0+}{lim}\Pr(-\alpha\ln(Y_{\alpha})>\alpha\ln(Q_{1})) \right]^{n}=\left[\underset{\alpha\to 0+}{lim}\Pr(Y_{\alpha}<Q_{1})]^{n}=\left(1-P_{(1)} \right)^{n}\). Hence, \(P_{T_{G(0)}}=1-\left(1-P_{(1)}\right)^{n}=P_{T_{p}}\), and \(T_{G(0)}\equiv T_{p}\). 2. From the property of gamma distribution, we know that \(Gamma(v/2,2)=\chi_{v}^{2}\), a chi-square distribution with \(v\) df. However, the sum of \(n\)_id_\(\chi_{v}^{2}\) is \(\chi_{nv}^{2}\). Hence, let \(v=1\) or \(\alpha=0.5\), \(T_{G(0.52)}\equiv\chi_{n}^{2}\). However, from Proposition 2, \(T_{G(0.52)}\equiv T_{G(0.5)}\); therefore, \(T_{G(0.5)}\equiv\chi_{n}^{2}\). 3. As in (ii), when \(v=2\) and \(\alpha=1,\ Gamma(1,2)=\chi_{2}^{2}\); therefore, \(T_{G(1)}\equiv T_{G(1,2)}\equiv F_{p}\). 4. From the property of gamma distribution, we know that \(Gamma(\alpha,\beta)\to N(\alpha/\beta,\alpha/\beta^{2})\) (\(\alpha\to\infty\)). Hence, for \(\beta=1,\ Gamma(\alpha,1)\to N(\alpha,\alpha)\), and \(Gamma(n\alpha,1)\to N(n\alpha,n\alpha)\).