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First cousins are children of full siblings. An otherwise outbred first-cousin pair shares two noninbred grandparents and has kinship coefficient $1/16$.

// Target: biology.bigb

For first cousins in an otherwise outbred pedigree, compare the copies transmitted from their related parents, who are full siblings. These transmissions trace to the same shared grandparent with probability $1/2$ and then to the same homologous copy of that grandparent with probability $1/2$, giving $k_1=1/4$. The unrelated parental contributions cannot match, so two shared copies have probability zero. Selecting one random copy from each cousin chooses those two potentially shared copies with probability $1/4$, giving kinship coefficient $k_1/4=1/16$.

Let $J$ be the largest number of chromosome copies that can be paired across the two full siblings as identical by descent, counting repeated copies when a child is autozygous. For offspring of otherwise outbred first cousins, with probability $3/4$ the parental copies have no common ancestral origin, giving ordinary sibling sharing $(1/4,1/2,1/4)$. With probability $1/4$, write the parental copies as $(A,B)$ and $(A,C)$ with three distinct origins. Each child independently has one of $AA,AB,AC,BC$, equally likely. Among sixteen ordered child pairs, two share no copy, ten share one, and four share two. The mixture gives the displayed probabilities. These three numbers do not determine all Jacquard identity coefficients; the formula $\varphi =\mathbb{E}[J]/4$ is restricted to noninbred individuals and fails here.

In an outbred pedigree, the two full siblings receive the same paternal copy with probability $1/2$, and independently the same maternal copy with probability $1/2$. Thus the number $J$ of shared copies has a binomial distribution with probabilities $(1/4,1/2,1/4)$, and $E[J/2]=1/2$.

// Target: biology.bigb

For full siblings in an outbred pedigree, condition on both being affected under multiplicative diallelic penetrance and on residual disease outcomes being independent given their genotypes. The usual Mendelian IBD sharing of full siblings probabilities $(1/4,1/2,1/4)$ are weighted by $p^2{\alpha }^4,p^2{\alpha }^2\beta ,p^2{\beta }^2$, respectively, giving the displayed conditional sharing probabilities. Since $\beta -{\alpha }^2=\varphi (1-\varphi )(\theta -1{)}^2\ge 0$, $z_1\le 1/2$, strictly for a polymorphic locus with a nontrivial effect.