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For full siblings in an outbred pedigree, Mendelian IBD sharing of full siblings gives prior sharing probabilities $(1/4,1/2,1/4)$ for $J=0,1,2$. Let $A$ be the event that both siblings are affected. Bayes' rule weights those probabilities by the pair penetrances, giving immediatelyTo obtain the one-shared-copy probability, assign each allele a risk multiplier $R$, equal to $1$ with probability $1-\varphi$ and $\theta$ with probability $\varphi$. Then $\mathbb{E}R=\alpha$ and $\mathbb{E}{R}^2=\beta$. With $J=1$, write the siblings' penetrances as $p{R}_s{R}_1$ and $p{R}_s{R}_2$, where the shared multiplier $R_s$ and the two unshared multipliers are independent. Conditional independence of the disease outcomes givesThe total pair probability is therefore $p^2(\beta +{\alpha }^2{)}^2/4$, and the three ascertained sharing probabilities areIn particular,The one-IBD proportion is below one half for a polymorphic locus with a genuine penetrance effect. Equality holds when $\beta ={\alpha }^2$, namely when $\varphi \in \{0,1\}$ or $\theta =1$; there is then no variation in the allele risk multiplier. The inequality follows from the different ascertainment weights for zero, one and two shared copies, rather than from applying the unascertained sibling probabilities to affected pairs. This is sibling IBD ascertainment under multiplicative penetrance. All affected-pair conditional probabilities presume $\mathrm{Pr}(A)>0$; if disease never occurs, conditioning on affected pairs is undefined.

For the usual no-linkage reference model with unrelated, noninbred parents, define $J$ as the number of homologous allele copies shared identical by descent by two full siblings at the marker. The two siblings receive the same paternal copy with probability $1/2$, and independently receive the same maternal copy with probability $1/2$. Therefore Mendelian IBD sharing of full siblings givesHenceFor affected siblings, absence of genetic linkage makes the marker transmissions independent of selection by disease, so these are the relevant null moments. This is a fresh outbred reference calculation, not an application of these sharing probabilities to the inbred siblings $E,F$ in part (b). Without the outbred-parent assumption the three ordinary sharing probabilities require modification.

Use identity-by-descent sharing of inbred full siblings, with $J$ the maximum number of copies that can be paired across the children as identical by descent. If two copies within one child have the same ancestral origin, they still count as two copies when both can be paired.

With probability $3/4$ the cousin parents share no ancestral copy. Conditional on this event, their children have ordinary Mendelian IBD sharing of full siblings, namely $(1/4,1/2,1/4)$. With probability $1/4$, the parental copies can be labelled $(A,B)$ and $(A,C)$, with $A,B,C$ distinct ancestral origins. Each child independently receives one of $AA,AB,AC,BC$, with probability $1/4$ each. Among the sixteen ordered child pairs, $AA$ paired with $BC$ in either order gives $J=0$; four equal genotype-origin pairs give $J=2$; the remaining ten give $J=1$. The mixture isThese are inbred children, so the three sharing probabilities do not retain all within-child identities. The full Jacquard identity coefficients are $(1,0,2,1,2,1,15,30,12)/64$. In particular, their kinship coefficient is $9/32$; applying the outbred formula $(P(J=1)+2P(J=2))/4$ here would be incorrect.

Label the two paternal ancestral marker copies $P_1,P_2$ and the two maternal copies $M_1,M_2$. Treat these as four distinct ancestral copies in an outbred pedigree; equal observed alleles need not imply identity by descent. By Mendelian segregation, each child independently receives either paternal copy with probability $1/2$. Thus the indicator $I_P$ that the two full siblings inherit the same paternal copy satisfiesThe same argument gives $P(I_M=1)=1/2$ for the maternal copy. Paternal and maternal meioses are independent, so $I_P$ and $I_M$ are independent. Consequently the number $J=I_P+I_M$ of copies shared identical by descent has a binomial distribution, givingThis is Mendelian IBD sharing of full siblings. It describes an ordinary, unselected sib pair: conditioning on a shared disease phenotype can change the distribution if there is genetic linkage to its causal locus.

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.