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A case–control study selects subjects according to an outcome: cases have the outcome and controls represent an appropriate comparison population. It compares their prior exposures or genetic variants to study association. Its retrospective sampling must be accounted for when interpreting risks; selection bias and confounding can distort an unadjusted association. In genetics, matched family pseudo-control genotypes can replace unrelated controls for a conditional transmission disequilibrium test.

Label each parent's two allele copies by its realized transmission: let $(A,B)$ be the transmitted and untransmitted alleles of the first parent and $(C,D^{\prime })$ those of the second. Under random mating and Hardy-Weinberg equilibrium, the four parental allele draws are independent with population frequencies $\pi$. Random Mendelian segregation merely swaps the two independent copies within each parent, so the transmitted/untransmitted labelled pair still has probability ${\pi }_a{\pi }_b$; the second pair has probability ${\pi }_c{\pi }_d$.

The child receives $A,C$. If $\mathcal{D}$ is its disease event, its penetrance is $k{\psi }_A{\psi }_C$ and its overall disease probability is $k{Z}^2$. ThereforeThis proves the transmitted and untransmitted alleles under multiplicative penetrance factorization. Under these assumptions, the transmitted pair $a/c$ has the affected-case distribution, while the complementary pair $b/d$ has the population distribution and is independent of the transmitted pair. It is the family pseudo-control genotype.

As in part (a), the formula records labelled transmissions. When observed genotypes are unordered, sum over every compatible parental-origin transmission. This includes multiplicities for overlapping parental alleles or two heterozygous parents producing a heterozygous child; it avoids interpreting one product as the total probability of all observationally identical configurations. The homogeneous random-mating model is also essential to the unconditional independence: mixing ancestry strata can correlate the case and pseudo-control through their shared family background.

For each family combine the two untransmitted parental alleles into one family pseudo-control genotype. A heterozygous child of two heterozygous parents has a heterozygous pseudo-control, whatever its parent-specific transmission phase. Combining the rows givesFor example, pseudo-control genotype $1/1$ receives contributions $1,2,5,5$ from the relevant rows; genotype $2/2$ receives $3,1,2,6$, and the remaining 25 are heterozygous. These counts also give the untransmitted allele totals 51 and 49.

The estimated allele frequencies are ${\hat{p}}_1=0.51$, ${\hat{p}}_2=0.49$. The Hardy-Weinberg equilibrium expected genotype counts areThey are almost identical to the observations, so these pseudo-control counts conform extremely closely to HWE proportions. The usual Pearson discrepancy is only about $8.0\times 10^{-6}$. This numerical agreement is not a general guarantee of HWE for family pseudo-controls, whose construction involves mating structure and case ascertainment in a genetic study.

Under random mating and Hardy-Weinberg equilibrium, label each parental allele by whether it is transmitted. Let $(A,B)$ and $(C,D)$ be the transmitted/untransmitted pairs of two parents. Before ascertainment all four draws are independent with population frequencies $\pi$. If offspring disease weights the transmitted pair by ${\psi }_A{\psi }_C$, thenThe transmitted pair is an affected-case genotype and the untransmitted pair a population family pseudo-control genotype under this homogeneous model. For unordered genotypes, compatible labelled transmissions must be summed. Conditioning on parental genotypes instead removes nuisance population frequencies and gives the usual transmission disequilibrium test.

// Target: statistical-modelling.bigb