Combining Pairwise Sharing Within Multiplex Families

Why Summarize Sharing by Family?

A multiplex family has several affected relatives, so it contributes more than one affected-relative pair to a linkage analysis. Those pairwise observations can be useful summaries of sharing, but they are not separate, unrelated families: two pairs may contain the same person. The key is to summarize the sharing at the family level while retaining the family as the unit that supplies independent information.

For example, three affected siblings contribute three distinct pairs. If one sibling shares a marker allele identically by descent (IBD) with each of the other two, those two pair scores both involve the same person. Treating them as independent observations ignores their shared family structure. This tutorial defines a simple family-level summary and shows how that dependence affects standard errors.

Definition: In a family with $m_f$ affected relatives, the number of distinct unordered relative pairs is $M_f=\binom{m_f}{2}=m_f(m_f-1)/2$. Let $S_{fij}$ be the pairwise sharing score for affected relatives $i$ and $j$ in family $f$. The family-level mean sharing score is $$ T_f=\frac{1}{M_f}\sum_{i<j}S_{fij}. $$ It gives each pair within the family equal weight, then represents the family by one summary.

For a full-sib pair, a simple IBD sharing score is the proportion of the pair's two alleles that are shared IBD: $S_{fij}$ is 0, $1/2$, or 1 when the pair shares zero, one, or two alleles IBD. Under the usual Mendelian null of no linkage, the expected sharing score for a full-sib pair is $1/2$. In practice, a pair score may instead be an estimate based on marker information; the summary formula is the same.

Calculating a Family Summary

The pair count increases quickly with family size. A family of two affected relatives contributes one pair; a family of three contributes three; and a family of four contributes six. The family mean averages the available pair scores, so families with more affected relatives do not automatically receive more weight when family means are later averaged.

Worked Example: Three Affected Siblings

Suppose a hypothetical family has three affected full siblings, A, B, and C. At a marker, their pairwise IBD sharing scores are $S_{AB}=1$, $S_{AC}=0.5$, and $S_{BC}=0.5$. Calculate the number of pairs and the family mean.

Count the pairs. There are $$ M_f=\binom{3}{2}=\frac{3(2)}{2}=3 $$ distinct pairs: AB, AC, and BC.

Average their scores. $$ T_f=\frac{1+0.5+0.5}{3}=\frac{2}{3}\approx0.667. $$

The family's mean pairwise score is about 0.667, above the full-sib null expectation of 0.5. This is a descriptive summary of the observed sharing in this family; by itself, it does not establish statistical evidence for linkage. The three scores also cannot be treated as three independent family observations.

The interpretation depends on what is averaged. An equal-family average answers, “What is the mean family summary for a family drawn from the target population?” A pooled pair average answers a different question: “What is the mean score for a pair drawn from all observed pairs?” In the pooled average, a family with six pairs contributes six times as much as a family with one pair. Neither weighting is universally correct; the choice should match the target summary.

Worked Example: Equal-Family and Pooled-Pair Averages

Consider two hypothetical families. Family A has one affected pair with score 1. Family B has three affected pairs with scores 0, 0.5, and 0.5. Calculate the family means, the equal-family average, and the pooled pair average.

Family summaries. Family A has $T_A=1$. Family B has $$ T_B=\frac{0+0.5+0.5}{3}=\frac{1}{3}\approx0.333. $$

Equal-family average. Giving each family equal weight gives $$ \frac{T_A+T_B}{2}=\frac{1+1/3}{2}=\frac{2}{3}\approx0.667. $$

Pooled pair average. There are four pairs in total, so the pair-weighted average is $$ \frac{1+0+0.5+0.5}{4}=\frac{2}{4}=0.500. $$

The difference is not an arithmetic error. The equal-family average weights the two families equally, whereas the pooled average gives Family B three quarters of the pair weight. Any report should make clear which summary is being used.

Why Pair Scores Within a Family Are Dependent

Pairs that share a relative are connected through the same pedigree and genetic transmissions. For example, $S_{AB}$ and $S_{AC}$ both involve sibling A. A change in A's inherited alleles can affect both scores. Consequently, knowing one pair's sharing may provide information about another pair's sharing. The pair scores therefore need not be independent, even when the families themselves were sampled independently.

This dependence matters for uncertainty. For $M_f$ pair scores in a family, the variance of their mean is $$ \operatorname{Var}(T_f)=\frac{1}{M_f^2}\left\{\sum_{i<j}\operatorname{Var}(S_{fij})+ 2\sum_{(i<j)<(k<l)}\operatorname{Cov}(S_{fij},S_{fkl})\right\}. $$ The second sum includes covariances between distinct pairs. If the pairs are independent, these covariances are zero. In a multiplex family, that condition generally cannot be assumed. The covariances depend on the pedigree, the pair relationships, and the score being analyzed; they need not all have the same value or even the same sign.

A standard error calculated as though all pairs were independent omits those covariance terms. It can therefore misstate uncertainty. The common remedy for a summary that gives equal weight to families is to calculate one $T_f$ per independent family and estimate variability across those family summaries.

Key result: If $F$ independent families are sampled from a population of families and the target is their mean family summary, estimate it by $\bar{T}=\sum_{f=1}^{F}T_f/F$. Its estimated standard error is $s_T/\sqrt{F}$, where $s_T$ is the sample standard deviation of the $F$ family summaries. This calculation treats families—not their constituent pairs—as the independent sampling units.

Estimating Uncertainty Across Families

The family-level standard error is appropriate under the stated sampling structure: the families are independent observations from the target family population, and the summaries $T_f$ are defined consistently. The sample standard deviation across families is $$ s_T=\sqrt{\frac{\sum_{f=1}^{F}(T_f-\bar{T})^2}{F-1}}, \qquad SE(\bar{T})=\frac{s_T}{\sqrt{F}}. $$ This does not require the pair scores within a family to be independent. Their dependence is already contained in the observed variation among the family summaries. With few families, the standard error itself may be imprecise, and any interval or test should use an appropriate small-sample method rather than treating the number of pairs as the sample size.

Worked Example: Standard Error From Five Families

Suppose five hypothetical independent families have family mean scores $0.75$, $0.50$, $0.625$, $0.375$, and $0.75$. Calculate the mean, the sample standard deviation, and the family-level standard error.

Mean across families. $$ \bar{T}=\frac{0.75+0.50+0.625+0.375+0.75}{5} =\frac{3.00}{5}=0.600. $$

Sample variance across families. The deviations from 0.600 are 0.150, $-0.100$, 0.025, $-0.225$, and 0.150. Thus $$ s_T^2= \frac{0.150^2+(-0.100)^2+0.025^2+(-0.225)^2+0.150^2}{5-1} =\frac{0.10625}{4}=0.0265625. $$ Therefore $s_T=\sqrt{0.0265625}\approx0.1630$.

Standard error. $$ SE(\bar{T})=\frac{0.1630}{\sqrt{5}}\approx0.0729. $$ The estimated mean family-level score is 0.600, with a family-level standard error of about 0.0729 score units. This standard error describes sampling variation across families under the stated independent-family sampling assumption; it is not a standard error based on the number of pairs.

Comparing Pair-Level and Family-Level Calculations

A small example makes the consequence of the independent-pair assumption visible. Suppose three independent hypothetical families each contain three affected siblings. Each family contributes three pair scores, but the three pairs within a family are related through their shared siblings. The pair scores below are hypothetical IBD proportions.

Worked Example: A Naive Pair Standard Error

Let Family A have scores $1, 0.5, 0.5$, Family B have scores $0, 0.5, 0.5$, and Family C have scores $1, 0.5, 0.5$. Compare the standard error obtained by treating all nine pairs as independent with the standard error based on the three family means.

Pair-level calculation. The nine scores have sum 5 and mean $5/9\approx0.5556$. Their sum of squares is $2(1^2)+6(0.5^2)+0^2=3.5$. The sum of squared deviations is $$ \sum(S-\bar{S})^2=3.5-9\left(\frac{5}{9}\right)^2 =\frac{13}{18}. $$ Thus the sample variance across the nine pairs is $$ s_{\text{pair}}^2=\frac{13/18}{9-1}=\frac{13}{144}\approx0.09028. $$ If the nine pair scores were independent, the estimated standard error would be $$ SE_{\text{naive}}=\sqrt{\frac{s_{\text{pair}}^2}{9}} =\sqrt{\frac{13}{1296}}\approx0.1002. $$

Family-level calculation. The family means are $2/3$, $1/3$, and $2/3$, whose mean is $5/9$. Their sample variance is $$ s_T^2= \frac{(2/3-5/9)^2+(1/3-5/9)^2+(2/3-5/9)^2}{3-1} =\frac{1}{27}. $$ The standard error across the three independent families is $$ SE(\bar{T})=\sqrt{\frac{1/27}{3}}=\frac{1}{9}\approx0.1111. $$

In this example the naive pair calculation gives a smaller standard error than the family-level calculation. The example does not imply that the naive standard error is always too small by a fixed amount: the direction and size of the error depend on the covariances. The general problem is that the independent-pair calculation assumes those covariances are zero without justification.

Practical Choices and Common Pitfalls

A family-level analysis should state both how pair scores are combined within each family and how the family summaries are combined across families. It should also identify the sampling unit supporting the standard error. When families have different numbers of affected relatives, equal weighting of family summaries and pooling all pair scores answer different questions, as shown above.

Wrong reasoning: “There are 30 affected-relative pairs, so the standard error should use $n=30$.”
Correction: If those 30 pairs come from only six independent families, the pair scores are clustered within six families. For an equal-family target, calculate six family summaries and estimate the standard error across those six values.
Wrong reasoning: “A family with more affected relatives is automatically more informative, so it should always receive more weight.”
Correction: More pairs do not necessarily mean proportionally more independent information. Choose equal-family weighting or pair-weighting according to the estimand, then account for the family clustering in uncertainty calculations.

These summaries are useful because they reduce a complex set of within-family pair comparisons to a clearly interpretable family-level quantity. They do not remove the need to account for pedigree structure, nor do they by themselves provide a complete linkage test. Their main contribution here is to make the unit of summary and the unit of independent replication explicit.

Key takeaway: First calculate a sharing summary within each multiplex family. Then base uncertainty for an equal-family mean on variation across independent families. Multiple pairs from one family add detail about that family, but they do not become independent families merely because they are listed separately.

Check Your Understanding

Use the definitions and examples above to answer these questions.

How many unordered pairs are contributed by a family with five affected relatives?
A family has four pair scores of 0.5, 0.5, 1, and 0. What is its family-level mean sharing score?
Explain why two pair scores involving the same affected relative may be dependent.
When does an equal-family average differ from a pooled pair average, and what does each summarize?
Six independent families contribute 24 total pairs. For an equal-family mean, which sample size determines the standard error, and why?