the triplet $(\gamma, U,
V)$ forms a set of orthogonal parameters in the sense of @CoxReid1987, that is, the off-diagonal elements of the expected Fisher information matrix are all zero. Propositions \[prop1sub2sec4\] and \[prop2sub2sec4\] give Jeffreys’ priors for $(\gamma, U, V)$ and $(\gamma, \lambda_0, \lambda_1)$, which are derived in Appendix 1.

\[prop1sub2sec4\] Jeffreys’ prior under the parameterization $(\gamma,U,V)$ is $$\label{eq1sub2sec4}
\pi_J(\gamma, u, v) \propto \dfrac{\gamma^{1/2-1}(1-\gamma
)^{1/2-1}}{(1+\gamma)} (u + rv)^{1/2}
u^{1/2-1}(1-u)^{1/2-1} v^{1/2-1}(1-v)^{1/2-1}$$ for $0<\gamma, u, v<1$ and it is proper, where $r=\dfrac
{m_{+1}}{m_{+0}}$ is the ratio of the sample sizes in the two treatment groups.

Under Jeffreys’ prior, the nuisance parameter, $\gamma$, is independent of both $U$ and $V$ and its marginal prior distribution is given by $$\begin{aligned}
\label{eq2sub2sec4}
\pi_J(\gamma) = \dfrac{\sqrt{2}}{\pi} \,\frac{\gamma^{1/2-1}
\;(1-\gamma)^{1/2-1}}{(1+\gamma)}, \qquad0<\gamma<1.\end{aligned}$$ However, Jeffreys’ prior depends indirectly on the sample sizes in both groups through their ratio, $r$.

\[prop2sub2sec4\] In the original space, Jeffreys’ prior reduces to $$\begin{aligned}
\label{eq3sub2sec4}
\pi_J(\gamma, \lambda_0, \lambda_1) & \propto&
\sqrt{\frac{(1+\gamma) (\lambda_0+r\lambda_1)}{\gamma
(1-\gamma) \lambda_0 \lambda_1
\big[1-(1+\gamma)\lambda_0\big] \big[1-(1+\gamma)\lambda_1\big]}},\\
& & (\gamma, \lambda_0, \lambda_1) \in\Omega.\nonumber\end{aligned}$$

The posterior distribution resulting from the use of Jeffreys’ prior is $$\begin{aligned}
\label{eq4sub2sec4}
\pi_J\big(\gamma, \lambda_0, \lambda_1|D\big) \propto&
\dfrac{2^{m_{1+}+1/2}}{{\mbox{{\bf B}}}(m_{1+}+1/2,m_{2+}+1/2)} \gamma^{m_{1+}+1/2-1}
(1-\gamma)^{m_{2+}+1/2-1}(1+\gamma)^{1/2}\nonumber\\
& \times(\lambda_0 + r\lambda_1)^{1/2}\,\lambda_0^{m_{10}+m_{20}+1/2-1}
\big[1-(1+\gamma)\lambda_0\big]^{m_{00}+1/2-1} \nonumber\\
& \times\lambda_1^{m_{11}+m_{21}+1/2-1}\big[1-(1+\gamma)\lambda_1\big
]^{m_{01}+1/2-1},
\qquad(\gamma,\lambda_0,\lambda_1)\in\Omega.\end{aligned}$$

