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For observed pairings between two labels $A,B$, let $N_{\mathrm{same}}=N_{AA}+N_{BB}$ and $N_{\mathrm{cross}}=N_{AB}+N_{BA}$, where the first label denotes the female's group and the second the male's. A mating assortment contrast isIt lies between $-1$ and $1$. Positive $I$ describes an excess of same-label pairings over cross-label pairings; whether that is evidence of assortative mating depends on the null mathematical model and the assay. The four counts form a contingency table.

Here is an original scope check. Suppose independently paired females and males have probabilities $p_f,p_m$ of carrying label $A$. ThenWith $p_f=p_m=1/2$, the expected value is zero. With $p_f=p_m=4/5$, it is $9/25=0.36$, despite pairing being independent of label. This counterexample concerns unequal pool composition; it does not replace the group introductions and contingency table tests in the dietary divergence and mating isolation in Drosophila pseudoobscura experiment. Even with balanced pools, finite observed samples fluctuate: a positive $I$ alone is not a hypothesis test, much less a universal species boundary.

Panmixia is random mating within a breeding population, without systematic choice by genotype or phenotype. In a homogeneous genetic model it makes the two parental gamete draws independent, yielding the Hardy-Weinberg equilibrium genotype probabilities $p_i^2$ and $2p_i{p}_j$. It is a modelling assumption; assortative mating and population stratification are distinct ways a population can depart from it.

// Target: causal-inference.bigb

Suppose assortative mating raises the dizygotic genetic correlation from $1/2$ to $\rho >1/2$. Thenso the formula returnsThus using the nominal one-half genetic correlation produces downward bias in estimated heritability, assuming the other ACE assumptions remain valid.