Why Use Pedigree Recursions?
A pedigree describes who is related to whom through parentage. Kinship coefficients translate that structure into a probability: how likely is it that alleles sampled from two individuals trace to the same ancestral copy? A recursion is useful because it builds this probability from parental relationships, rather than requiring every ancestral path to be enumerated separately.
The calculations below concern a single autosomal locus, with diploid individuals and ordinary Mendelian transmission. They give pedigree-based expectations under the recorded relationships; they do not describe the exact inheritance realized across every position in a genome.
IBD means that the sampled alleles are copies of the same ancestral allele through descent, rather than merely carrying the same nucleotide by chance. Kinship is symmetric: \(\phi(i,j)=\phi(j,i)\). The diagonal convention matters: when two independent draws are made from one individual, there is a one-half chance of drawing the same homolog twice. The other one-half of the time, the two distinct homologs are drawn, and they are IBD with probability \(F_i\).
Deriving the Recursion
Suppose \(k\) is a nonfounder with parents \(s\) and \(d\). At the locus under consideration, an allele sampled from \(k\) came from \(s\) with probability one-half and from \(d\) with probability one-half. Condition on which parent supplied that allele. For any other individual \(i\), the probability of IBD is the corresponding parental kinship probability averaged over those two equally likely sources.
The inbreeding relation follows because the two distinct homologs of \(k\) are one allele inherited from each parent. The probability those two inherited alleles are IBD is precisely the probability that an allele drawn from \(s\) and one drawn from \(d\) are IBD, which is \(\phi(s,d)\). For founders, no parental recursion applies: their mutual kinship and inbreeding must be specified as assumptions or supplied from known ancestry.
A short derivation for two offspring
Let \(a\) and \(b\) be distinct children of the same parents \(s\) and \(d\). A sampled allele from each child can come from either parent. The four source combinations each have probability one-quarter. Averaging their kinship probabilities gives
This derivation uses symmetry, \(\phi(s,d)=\phi(d,s)\), and the diagonal formula for each parent. It shows why the familiar full-sibling value depends on assumptions about the parents: it is \(1/4\) when both are noninbred and unrelated, but it can be larger if they are inbred or related.
Worked Example: Parent–Offspring Kinship
Suppose founders \(A\) and \(B\) are unrelated and noninbred, and their child is \(C\). Thus \(\phi(A,B)=0\) and \(\phi(A,A)=1/2\). Calculate the kinship of \(A\) and \(C\).
The allele sampled from \(C\) came from \(A\) with probability one-half and from \(B\) with probability one-half. Therefore,
The pedigree kinship is \(0.2500\). The calculation depends on the stated founder assumptions; if \(A\) and \(B\) were related, the \(\phi(A,B)\) term would not be zero.
Worked Example: Full Siblings
Let \(C\) and \(D\) be full siblings whose parents \(A\) and \(B\) are unrelated, noninbred founders. The parental diagonal kinships are \(\phi(A,A)=\phi(B,B)=1/2\), and the cross-parent kinships are zero. Apply the four-source recursion.
The full-sibling kinship is \(0.2500\) under these assumptions. Each child inherits one allele from each parent; the calculation averages the IBD probabilities over all four possible parental-source pairings.
Worked Example: Half Siblings
Suppose \(C\) is a child of \(A\) and \(B\), and \(D\) is a child of \(A\) and \(E\). Assume all three founders \(A,B,E\) are unrelated and noninbred. The shared parent is \(A\); the other parent differs between the children.
The sampled alleles can have four parental-source combinations: \(A\) with \(A\), \(A\) with \(E\), \(B\) with \(A\), or \(B\) with \(E\). Only the first has nonzero kinship under the assumptions, and \(\phi(A,A)=1/2\). Thus,
The half-sibling kinship is \(0.1250\). If the other parents \(B\) and \(E\), or either of them and \(A\), were related, the additional cross-kinship terms would need to be included.
Worked Example: A Child of Related Parents and the Diagonal
Suppose \(S\) and \(T\) are full siblings whose parents are unrelated, noninbred founders. The full-sibling calculation gives \(\phi(S,T)=1/4\). Let \(Z\) be their child. Find \(F_Z\) and \(\phi(Z,Z)\).
By the offspring inbreeding recursion, the probability that \(Z\)’s distinct homologs are IBD is \(F_Z=\phi(S,T)=1/4=0.2500\). For two independent allele draws from \(Z\), the same homolog is selected with probability one-half; otherwise the distinct homologs are selected, and those are IBD with probability \(1/4\). Hence
The value \(0.6250\) is the diagonal kinship under the independent-draw definition. It is not the probability that \(Z\)’s two distinct homologs are IBD; that probability is \(F_Z=0.2500\).
Applying Recursions Reliably
A pedigree calculation proceeds from founders toward their descendants. Specify the founders’ kinships and inbreeding coefficients first. Then, for each nonfounder, calculate its kinship with previously defined individuals using the parental average, and calculate its inbreeding coefficient from its parents’ kinship. The diagonal is obtained from \(F_i\), not by applying the off-diagonal recursion to the same individual.
Record which founders are assumed unrelated and noninbred, or enter their known kinship and inbreeding values. Do not silently treat all founders as unrelated when the pedigree says otherwise.
For a nonfounder \(k\) with parents \(s,d\), use \(\phi(i,k)=\{\phi(i,s)+\phi(i,d)\}/2\) for each \(i\ne k\).
Set \(F_k=\phi(s,d)\), then use \(\phi(k,k)=(1+F_k)/2\), preserving the distinction between kinship and IBD of distinct homologs.
Verify that pairwise values are symmetric and that every zero cross-kinship comes from a stated founder or pedigree assumption.
Assumptions, Diagnostics, and Alternatives
These recursions assume an accurately recorded pedigree, diploid inheritance, and Mendelian segregation at the locus. They ordinarily omit mutation and require a clear convention for founders: “unrelated” means their alleles are treated as not IBD with one another under the chosen pedigree base. If founders have shared ancestry not represented in the pedigree, setting their cross-kinship to zero can understate descendant kinship. For a genome-wide interpretation, the values are pedigree expectations; realized sharing can vary among relatives and across the genome.
Several checks expose common errors. Kinship is symmetric, and under the usual probability definition it lies between zero and one. For an outbred individual, \(\phi(i,i)=1/2\); with inbreeding it is greater than one-half according to \((1+F_i)/2\). A parent–offspring pair from unrelated, noninbred parents has kinship \(1/4\), while full siblings have \(1/4\) and half siblings \(1/8\) under the same baseline assumptions. These familiar values are useful checks, not universal constants independent of pedigree context.
A frequent mistake is to say that \(\phi(i,i)\) is the probability that the two distinct homologs of \(i\) are IBD, and then use \((1+F_i)/2\). That mixes two definitions: the probability for distinct homologs is \(F_i\), whereas \(\phi(i,i)\) uses two independent draws, which may select the same homolog. The corrected calculation is \(1/2+(1/2)F_i\). Another error is to set the kinship of two parents to zero merely because they are from different branches of a drawn pedigree; this is justified only if the founder assumptions support it.
One can calculate simple relationships by memorized constants or enumerate parental-source paths, but both approaches become error-prone as pedigrees grow or parents are related. Recursion applies the same conditioning rule systematically and makes assumptions visible in the terms used. Genotype-based estimates provide a different kind of evidence: they use observed genomic data rather than only recorded descent, and are subject to their own estimation choices. They should not be confused with the pedigree expectation derived here.
Check Your Understanding
Use the stated assumptions in each question. Brief solutions follow the questions.
- For an outbred individual \(Q\), what are \(F_Q\) and \(\phi(Q,Q)\)? Explain why they differ.
- Two unrelated, noninbred founders have a child. What is the kinship between the child and either parent?
- Two full siblings have parents with \(F_s=F_d=0\) and \(\phi(s,d)=0.10\). Use the derived formula to calculate their kinship.
- Two half siblings share a parent \(A\) with \(F_A=0.20\). Their other parents are unrelated to one another and to \(A\), and are noninbred. Calculate the half-sibling kinship.
- A student reports that an individual’s distinct homologs have IBD probability \(0.30\), so the diagonal kinship is \(0.30\). Correct the calculation.
Solutions. (1) \(F_Q=0\), while \(\phi(Q,Q)=(1+0)/2=0.5\); the independent draws select the same homolog half the time. (2) The recursion gives \(\{1/2+0\}/2=0.25\). (3) \(\phi(a,b)=1/4+(0+0)/8+0.10/2=0.30\). (4) Only the shared-parent term contributes, so \(\phi=\phi(A,A)/4=\{(1+0.20)/2\}/4=0.15\). (5) The stated \(0.30\) is \(F_i\); the diagonal kinship is \((1+0.30)/2=0.65\).