Power of a two-sample rare-event comparison

For two independent samples of size $n$ with event probabilities $p_0,p_1$, put $\delta =p_0-p_1$, $\bar{p}=(p_0+p_1)/2$, $s_0^2=2\bar{p}(1-\bar{p})/n$ and $s_1^2=[p_0(1-p_0)+p_1(1-p_1)]/n$. The normal approximation to a two-sided level-$\alpha$ test has powerFor rare events, independent Poisson distributions approximate the counts. Conditional on their total, the null count allocation is binomial distribution with probability one half; under the alternative the allocation probability is $p_0/(p_0+p_1)$. Averaging its exact rejection probability over the alternative total supplies a discrete power calculation.

// Target: probability-theory.bigb