Proof. \(\sum _k a_k d_k/n_k = \sum _k (b_k c_k/n_k)\,\widehat {\OR }_k\); divide by \(\sum _k b_k c_k/n_k\). □

Proof. Write \(R_k = a_k d_k/n_k\) and \(S_k = b_k c_k/n_k\). The key fact is that \(E(R_k - \psi S_k) = 0\) in every stratum, whatever its size: under the product-binomial model of a case-control stratum, with exposure probabilities \(q_{1k}\) among \(m_{1k}\) cases and \(q_{0k}\) among \(m_{0k}\) controls, \(a_k\) and \(d_k\) are independent, so \(E(a_k d_k) = m_{1k}q_{1k}\,m_{0k}(1 - q_{0k})\) and \(E(b_k c_k) = m_{1k}(1 - q_{1k})\,m_{0k}q_{0k}\). Their ratio is the stratum odds ratio \(\psi \). (In this model \(n_k = m_{1k} + m_{0k}\) is fixed, so dividing by it changes nothing.)

With \(K\) fixed and the strata growing, \(R_k/n_k\) and \(S_k/n_k\) converge in probability to their means, and \(\sum _k R_k/\sum _k S_k\) converges to \(\sum _k E(R_k)/\sum _k E(S_k) = \psi \). With the sizes bounded and \(K \to \infty \), \(\sum _k R_k/K\) and \(\sum _k S_k/K\) are averages of independent bounded variables, so by the law of large numbers their difference \(\frac 1K\sum _k (R_k - \psi S_k)\) tends to \(0\), while \(\frac 1K\sum _k S_k\) stays bounded away from \(0\) under mild conditions. Hence the ratio tends to \(\psi \). □