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For $N$ individuals in monogamous families of exactly two offspring, let ${\Phi }_G$ be mean pairwise kinship coefficient, $F_G$ mean inbreeding and ${\Phi }_{m,G}$ mean kinship of mates. A random distinct offspring pair are siblings with probability $1/(N-1)$. Accounting separately for within-family and cross-family pairs gives the displayed recurrence, while $F_{G+1}={\Phi }_{m,G}$. Random mating makes ${\Phi }_{m,G}={\Phi }_G$. Sibling avoidance changes the mating-pair mean; it does not create unrelated founders or stop eventual drift to ancestral identity in a closed finite population.

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 $k_1/4=1/16$.

For an otherwise outbred pedigree, each of the two shared grandparents contributes $(1/2{)}^{2+2+1}=1/32$ to the kinship coefficient of first cousins. Thus their child has inbreeding coefficient $1/16$. Equivalently, cousin kinship is one quarter of the $1/4$ kinship of their sibling parents.

// Target: biology.bigb

The inbreeding coefficient is the probability that the two allele copies at a locus within an individual are identical by descent. It equals the kinship coefficient of the individual's parents. It concerns ancestral identity, rather than merely homozygosity of observed allelic states.

// Target: biology.bigb

For two individuals carrying labelled copies $(a,b)$ and $(c,d)$, condensed identity coefficients give the probabilities of nine identity by descent patterns, identifying states that differ only by exchanging the two homologues within either individual. They retain within-individual autozygosity, which three ordinary outbred sharing coefficients omit. One convention isEach block consists of copies identical by descent, and different blocks have distinct ancestral origins. For siblings whose parents are otherwise outbred first cousins, these coefficients are $(1,0,2,1,2,1,15,30,12)/64$. This follows by conditioning on whether the parents share one ancestral copy and enumerating the four independent Mendelian segregation choices to their two children. The kinship coefficient is ${\Delta }_1+({\Delta }_3+{\Delta }_5+{\Delta }_7)/2+{\Delta }_8/4$, here $9/32$. Weir's discussion of identity states uses this convention.

Take “cousins” to mean first cousins with unrelated, outbred founders and unrelated spouses in the intervening generation. Interpret a homozygous chromosome as two copies identical by descent, and first use an idealization of independent, nonrecombining chromosome blocks. Ordinary sequence homozygosity is a different quantity: unrelated copies can have the same allele, so its probability also requires allele frequencies. Similarly, identity of an entire recombining chromosome requires a genetic map, not just a pedigree.

At a single autosomal locus, the probability of autozygosity in their child is the parents' kinship coefficient. Each of the two common grandparents contributes a path of length two from each parent, with probability $(1/2{)}^{2+2+1}$. Thus the first-cousin inbreeding coefficient isIf there are $L$ independent autosomal blocks, the probability that none is autozygous is $(1-F{)}^L$. Hence the usual 23-block idealization givesThe exponent describes the stated independent-block approximation; it is not a claim that all 23 human chromosome pairs inherit like autosomes.

Humans have 22 autosomal pairs. A son has one X and one Y and cannot have two homologous X copies. A daughter can have an autozygous X, with probability $F_X$ determined by the sex-linked cousin kinship calculation in part (b). Thus a sex-aware version iswhere $p_D$ is the probability of a daughter, or is one or zero for a specified daughter's or son's sex. The independent assortment approximation applies between the autosomes and the X. Opposite-sex parentage alone does not determine $F_X$: the parents may be cousins through their mothers, their fathers or one of each. Without these pedigree and chromosome-block assumptions, the wording does not specify a unique probability of an entire homozygous human chromosome.

Use the conventional additive pedigree relatedness $r_{UV}=2{\varphi }_{UV}$, where ${\varphi }_{UV}$ is the kinship coefficient, the probability that randomly selected gene copies from $U$ and $V$ are identical by descent. For noninbred individuals this convention gives ordinary full siblings relatedness $1/2$, as required. It is important that this is not the probability of sharing at least one copy: that probability is $3/4$ for ordinary full siblings.

Let the parents be $M,F$. They are otherwise outbred first cousins, so ${\varphi }_{MF}=1/16$ and ${\varphi }_{MM}={\varphi }_{FF}=1/2$. For two children $C,D$, independent Mendelian segregation givesTherefore the autosomal additive relatedness isThe additional $1/16$ beyond $1/2$ comes from maternal-paternal cross-origin matches. Each child has inbreeding coefficient $1/16$. The additive coefficient is a sum of descent probabilities rather than a literal bounded probability in all inbred pedigrees; a relationship normalized to self-relatedness would be $r_{CD}/(1+F)=9/17$. Likewise, identity-by-descent sharing of inbred full siblings gives a different matched-copy statistic. The exam's calibration to $1/2$ supports the additive convention, but the quoted probability wording needs this qualification.

For sex chromosomes the route of cousinship matters, not merely the fact that the parents have opposite sex. Let $P_M$ be the mother-cousin's parent in the shared sibling pair, and $P_F$ the father-cousin's parent in that pair. Let $F_X$ be the probability that the mother's randomly transmitted X matches the father's single X by descent. With no founder inbreeding, the possibilities areIn the first two rows the father-cousin is related through his father, whose X he does not inherit. In the third row his mother is a sister of the mother-cousin's father: the relevant brother and sister match their maternal X copies with probability $1/4$, and the mother-cousin transmits her paternal X with probability $1/2$, giving $1/8$. In the last row the two sisters' random X gametes match with probability $3/8$: their common father's X contributes $1/4$, and their common mother's two X copies contribute $1/8$. Transmission by the mother-cousin halves this to $3/16$. This derives the sex-linked cousin kinship values.

For an unweighted comparison by chromosome pair, two daughters have sex-pair additive relatedness $3/4+F_X$: their paternal X always matches, maternal X matches with probability $1/2$, and the two cross-origin terms each contribute $F_X$. Two sons have mean sex-pair relatedness $3/4$, from an X match of probability $1/2$ and a common paternal Y with probability one. A daughter-son pair has sex-pair value $1/4+F_X/2$, with the Y contribution zero. These calculations treat chromosomes as unrecombined blocks and weight X and Y equally; chromosome-length-weighted or locus-specific coefficients are different.

For two children with independent equally likely sexes, these three cases have weights $1/4,1/4,1/2$. Their mean sex-pair relatedness is $1/2+F_X/2$, so averaging the 22 autosomal pairs and the sex pair givesIf the cousins' route has $F_X=0$, this is $103/184$, slightly below $9/16$. If, as an additional sampling assumption, the four parent-sex routes in the table are equally likely, $\mathbb{E}{F}_X=5/64$ and the answer is $1653/2944$. These are different legitimate numerical corrections under different specified pedigrees. There is no unique sex-chromosome correction from opposite-sex cousinhood alone. The table and formula give it once the missing pedigree and offspring-sex information is supplied.

Assume a closed population with even size $N$, unrelated founders, monogamous mating, nonoverlapping generations and exactly two offspring per couple. Sex labels are exchangeable for the autosomal calculation, with enough individuals of each sex to form the stipulated pairs. A randomly chosen distinct offspring pair are siblings with probability $1/(N-1)$: among the $\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ pairs, exactly $N/2$ are within-family pairs. This fixed-family-size fact, rather than independent sampling of all parents, determines the recurrence.

First give the elementary chromosome-sharing idealization consistent with taking self-relatedness as one and ordinary sibling relatedness as $1/2$. In this model each individual carries one ancestral label, passed through one randomly chosen parent. Let $r_G$ be mean sharing between distinct individuals, and let $m_G$ be the mean sharing between mates. Siblings share their parental label with probability $(1+m_G)/2$. The average relatedness between parents from different couples isTo see the latter count, the ordered pairs from different couples are all ordered distinct pairs minus the $N$ ordered mating pairs. Each pair of parents from different couples contributes equally to the four possible cross-family offspring relationships. Combining sibling and nonsibling offspring givesWith random mating, $m_G=r_G$ in expectation, so $r_0=0$ givesThe mean sharing tends to one, on a generation scale of order $2(N-1)$. It is not linear indefinitely: repeated ancestry saturates the probability. This is an exact result for the stated one-label approximation, not an exact diploid formula after substantial inbreeding has developed.

Now prohibit full-sibling matings, with $N\ge 4$ and representative sampling of allowed nonsibling mating pairs. The following closed recurrences assume that opposite-sex mating pairs have the same mean autosomal sharing as all nonsibling pairs; a sex-specific or assortative scheme requires separate mating means. In the exchangeable one-label model write $u_G$ for mean relatedness of a nonsibling pair, so $m_G=u_G$. The two required recurrences areThe second relation includes sharing through grandparents and more distant ancestors; setting $u_G=0$ forever would be wrong. For $v_G=1-r_G$, $w_G=1-u_G$, the linear recurrence matrix isIts eigenvalues areboth of magnitude less than one for $N\ge 4$. Thus $r_G\to 1$ even when siblings never mate. Explicitly, $1-r_G=A{\lambda }_{+}^G+(1-A){\lambda }_{-}^G$, where $A=[1-1/(2(N-1))-{\lambda }_{-}]/({\lambda }_{+}-{\lambda }_{-})$. Sibling avoidance prevents immediate close-relative matings, but it does not make the remaining population unrelated. In a fixed-two-offspring model it need not reduce average pairwise sharing at every generation: a smaller mating mean actually increases the displayed increment in $r_G$.

For a fully diploid formulation, self-kinship also changes with inbreeding and cannot remain artificially fixed. Let ${\Phi }_G$ be mean kinship coefficient between distinct individuals, $F_G$ mean inbreeding coefficient, and ${\Phi }_{m,G}$ mean kinship of mates. A sibling pair has kinship $(1+F_G)/4+{\Phi }_{m,G}/2$, because parental self-kinship is $(1+F_G)/2$. Repeating the pair count yields finite-population chromosomal coancestry:Initially ${\Phi }_0=F_0=0$. For random mating take ${\Phi }_{m,G}={\Phi }_G$. An explicit solution is the affine matrix recurrenceIts characteristic polynomial is ${\lambda }^2-[1-1/(2(N-1))]\lambda -1/[4(N-1)]$; its two roots have magnitude less than one. Hence ${\Phi }_G,F_G\to 1$. The normalized relationship $2{\Phi }_G/(1+F_G)$ tends to one, while unnormalized additive relationship $2{\Phi }_G$ tends to two, demonstrating why that coefficient cannot always be read literally as a probability.

With sibling avoidance replace ${\Phi }_{m,G}$ by the nonsibling mean $U_G$ and addThese equations compute every generation's kinship and inbreeding, including delayed consanguinity through cousins. Under random allowed mating in a connected closed population, backward ancestral gene-copy lineages have a positive chance to meet in each sufficiently long block of generations; repeated blocks force eventual coalescence. Thus sibling avoidance alone does not prevent long-term identity by descent and genetic drift. Maintaining distinct isolated breeding groups, introducing immigrants, or changing offspring-number variance gives a different model. The elementary exponential formula answers the fixed-self-sharing approximation; the diploid recurrences give the consistent inbreeding-aware answer.

The kinship coefficient ${\varphi }_{XY}$ is the probability that one allele copy sampled uniformly from each of individuals $X,Y$, independently conditional on their copies, is identical by descent. The inbreeding coefficient $F_X$ is the probability that the two homologous allele copies within $X$ are identical by descent. For distinct offspring and older individual, Mendelian segregation givesFor the last identity, two independent selections choose the same copy with probability $1/2$, and choose distinct copies with probability $1/2$, in which case identity has probability $F_X$. An unrelated, noninbred founder has $F=0$ and self-kinship $1/2$, while distinct such founders have zero mutual kinship coefficient. The source's statement of zero founder kinship must be read as zero kinship between distinct founders; zero self-kinship would contradict the definition.

The self-kinship coefficient of the inbred individual $D$ is ${\varphi }_{DD}=(1+F_D)/2=5/8$. Since $E$ receives a uniformly chosen parental allele from each of $D,G$,The ${\varphi }_{DD}$ term is essential: self-kinship coefficient is not zero, even for a noninbred individual.

The founder $C$ is unrelated to the descendants of $A,B$, while the kinship coefficient between $D$ and its parents' sibling $U$ isAs $G$ has parents $C,U$, the kinship coefficient between $D,G$ isThe parents of $E$ are $D,G$, soThe same calculation gives $F_F=1/8$.

To identify the unlabelled individuals in the original pedigree, call the three children of $A,B$ from left to right $U,V,W$. Then $G$ has parents $C,U$, $D$ has parents $V,W$, and $E,F$ have parents $G,D$. The three full siblings $U,V,W$ are noninbred because $A,B$ are unrelated noninbred founders in a pedigree. Their pairwise kinship coefficients areSince $D$ is the child of the sibling pair $V,W$, its inbreeding coefficient is

The full siblings $E,F$ have parents $G,D$. Since $G$ is noninbred, ${\varphi }_{GG}=1/2$, whereas ${\varphi }_{DD}=5/8$ and ${\varphi }_{GD}=1/8$. Applying the parental kinship coefficient recursion to both siblings givesThe familiar outbred-sibling value $1/4$ does not apply to this inbred pedigree.

Only the copies inherited through the two related parents can share recent identity by descent; the other parental contributions come from unrelated founders in a pedigree. The related parents are full siblings. Their transmitted copies trace to the same shared grandparent with probability $1/2$, and then to the same copy of that grandparent with probability $1/2$. Thus the first-cousin identity-by-descent probabilities areTo compute the kinship coefficient, one random copy from each cousin selects those potentially shared contributions with probability $1/4$. Hence

An offspring is autozygous precisely when its two transmitted parental copies are identical by descent. Therefore its inbreeding coefficient equals the parental kinship coefficient, giving the first-cousin inbreeding coefficient

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.

The kinship coefficient ${\varphi }_{XY}$ is the probability that a randomly selected allele copy from individual $X$ and an independently selected copy from individual $Y$ at the same genetic locus are identical by descent. The sampling is of gene copies, not of whole genotypes. For a noninbred individual $X$, sampling its two copies with replacement gives ${\varphi }_{XX}=1/2$; in general ${\varphi }_{XX}=(1+F_X)/2$, where $F_X$ is its inbreeding coefficient.

For two full siblings in an outbred pedigree, the two sampled copies are both paternal with probability $1/4$, both maternal with probability $1/4$, and of different parental origin with probability $1/2$. Given that both are paternal, their independent Mendelian segregation transmissions select the same paternal ancestral copy with probability $1/2$; likewise for both maternal. The different-parent case cannot be identical by descent when the parents are unrelated. ThereforeEquivalently, conditional on an IBD-sharing count $J$, the chance that the two random copies match is $J/4$, so ${\varphi }_{\mathrm{sib}}=E[J]/4=1/4$. The expected shared proportion $E[J/2]=1/2$ and the usual additive relationship coefficient $2\varphi =1/2$ are twice this kinship coefficient; confusing these conventions would double the subsequent inbreeding coefficient.

Let the first cousins be $X,Y$, with parents $(P,U)$ and $(Q,V)$ respectively, where $P,Q$ are full siblings and $U,V$ have no ancestral relationship to one another or to that sibling family. To compute ${\varphi }_{XY}$, choose one allele copy from each cousin. They originate from $P$ and $Q$ with probability $1/4$. All other parental combinations have zero kinship coefficient under the otherwise outbred pedigree assumption. HenceThe two gene copies within a child of $X,Y$ are random transmissions from these parents. They are identical by descent precisely in the event defining parental kinship, so the child's inbreeding coefficient isFor a direct ancestral-path check, there are two shared noninbred grandparents. Through either grandparent, the two cousin-to-grandparent paths have two meioses each, and matching the ancestral grandparental copy supplies a further factor $1/2$. Each contributes $2^{-2-2-1}=1/32$, giving the same first-cousin inbreeding coefficient. Additional ancestral relationships or inbred grandparents would change the result.