Proof. Each stratum has \(n_k = 2\), with one case and one control. A concordant pair has \(a_k d_k = b_k c_k = 0\). A pair in which only the case is exposed has \(a_k = d_k = 1\), \(b_k = c_k = 0\), contributing \(\tfrac 12\) to \(R_+\); one in which only the control is exposed contributes \(\tfrac 12\) to \(S_+\). So \(\widehat {\OR }_{\text {MH}} = (n_{10}/2)/(n_{01}/2)\).

For the test, a concordant pair has \(n_{1k} = 0\) or \(n_{0k} = 0\), so \(\Var (a_k) = 0\) and \(a_k = E(a_k)\); it contributes nothing. A discordant pair has all margins \(1\) and \(n_k = 2\), so \(E(a_k) = \tfrac 12\) and \(\Var (a_k) = 1/(4 \times 1) = \tfrac 14\). Summing over discordant pairs, \(\sum a_k - \sum E(a_k) = n_{10} - (n_{10} + n_{01})/2 = (n_{10} - n_{01})/2\) and \(\sum \Var (a_k) = (n_{10} + n_{01})/4\), giving \((n_{10} - n_{01})^2/(n_{10} + n_{01})\). □