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Nonparametric Linkage Methods · Tutorial 255 of 989

Valid Permutations Under Family Ascertainment

Learn to match each NPL null simulation to the way families were recruited and to the exchangeability assumptions the simulation requires.

Intermediate 12 min read

What You'll Learn

  • Define exchangeability for a family-based NPL randomization
  • Explain why family ascertainment must be included in the conditioning scheme
  • Distinguish constrained affection-label permutations from marker gene-dropping
  • Identify which pedigree, phenotype, marker, and missingness features to retain
  • Calculate an exact or Monte Carlo permutation p-value under a stated null scheme
  • Diagnose invalid permutations that break family structure or selection rules

Why Ascertainment Changes the Randomization Question

Nonparametric linkage (NPL) statistics assess whether affected relatives share marker alleles or chromosomal segments more often than expected under a null of no linkage. A permutation calibration estimates that null distribution by repeatedly randomizing some part of the data and recalculating the statistic. The crucial question is not simply which values can be shuffled. It is which randomizations have the stated null distribution after accounting for how the families entered the study?

Family ascertainment is the rule used to recruit or select pedigrees. For example, a study may recruit a pedigree only if at least two siblings are affected, or may require a particular affected relative to be the proband. Those rules shape the observed affection patterns. A null replicate that ignores them may create families that could not have been sampled, or give too much probability to patterns that the study design excluded. In either case, the simulated tail probability need not calibrate the NPL statistic.

Definition: A randomization is exchangeable under a specified null if, conditional on the design features held fixed and the ascertainment rule, the observed arrangement and the arrangements generated by the randomization have the same probability under that null. A valid permutation scheme must reproduce the conditional null distribution relevant to the analysis; shuffling values is not by itself evidence of exchangeability.

Here, the design features can include the pedigree relationships, covariates, affection labels, observed or simulated marker data, and the recruitment rule. Which of these are held fixed depends on the randomization method. There is no single universally valid operation called “the NPL permutation”: a conditional affection-label permutation and a Mendelian gene-drop answer different conditional questions.

Two Conditional Routes to a Null Distribution

Let \(T\) denote the NPL statistic, \(Y\) the affection labels, \(M\) the marker information, \(D\) the pedigree and relevant covariates, and \(A\) the ascertainment event. A valid null replicate must reflect the distribution of the statistic under the null and under the conditions used by the analysis. In particular, if recruitment was based on \(Y\), then the selected data are conditional on \(A\), which is a property of \(Y\).

One route is to hold the marker data and pedigree fixed and randomize affection labels. This is appropriate only when the labels are exchangeable among the individuals being permuted after conditioning on the design and ascertainment rule. The permissible assignments may need to preserve the number affected, a fixed affected proband, covariate strata, or other recruitment features. If the phenotype model gives different probabilities to individuals, an unrestricted shuffle is not justified merely because everyone belongs to one family.

A second route is to hold the pedigree and observed affection pattern fixed and simulate marker inheritance under the no-linkage null, often by gene-dropping. Founders’ marker alleles or haplotypes, the marker map and allele model, the pedigree relationships, and the study’s missingness pattern must be treated consistently with the chosen design. Descendant marker genotypes are then generated through Mendelian transmissions under the null. If founders are untyped, their marker alleles must be handled by a stated model rather than silently treated as known.

These routes should not be conflated. In a label permutation, observed marker information is retained and phenotype labels move only among exchangeable positions. In a gene-drop, the affection labels remain fixed while marker inheritance is simulated. A method may also use a carefully defined conditional distribution of inheritance given marker observations, but it must specify what is being conditioned on and how the statistic is recalculated. In particular, one cannot claim to retain all observed child genotypes exactly while freely replacing the inheritance pattern that those genotypes constrain.

Ascertainment and Exchangeability Conditions

The key issue is whether the selection rule is invariant over the proposed randomizations. If a study enrolls families because a designated proband is affected, a permutation that makes that proband unaffected does not preserve the sampled design. If recruitment requires at least two affected siblings, an assignment with fewer than two affected siblings is not a possible selected family. Sometimes a constraint such as a fixed affection count makes every proposed assignment satisfy the rule, but that alone does not establish exchangeability: individuals may still differ in age, sex, covariates, or probability of selection.

Under the no-linkage null, marker inheritance is independent of affection status at the trait locus, conditional on the relevant pedigree and design, when the marker region was not itself used to select families and the null model correctly represents the transmissions. This provides the rationale for holding selected affection labels fixed while gene-dropping marker inheritance. It does not make arbitrary permutations of affection labels valid. Nor does it protect a simulation that ignores marker-dependent recruitment, genotyping-based exclusions, or other selection features.

A practical audit asks four questions before any randomization is run:

  • What was selected? State the recruitment or exclusion rule, including any designated proband or affected-relative requirement.
  • What is held fixed? Specify pedigree structure, affection labels, covariates, founder information, marker model, and missingness as appropriate.
  • What is randomized? Name the precise labels, transmissions, or other quantities that change in each replicate.
  • Why are those outcomes exchangeable? State the null and the conditions under which every allowed replicate has the required probability.

Worked Example: Exact Null Distribution for Two Ascertainment-Selected Families

Suppose a hypothetical study recruits two unrelated pedigrees, each because its two siblings are affected. The analysis uses one affected sib pair from each family. Let \(X_j\) be the number of alleles identical by descent (IBD) shared by pair \(j\) at a marker, and let \(S=X_1+X_2\). Under the no-linkage null for a sib pair, \(X_j\) is 0, 1, or 2 with probabilities \(1/4\), \(1/2\), and \(1/4\), respectively. Assume the marker transmissions in the two pedigrees are independent and the marker was not part of the recruitment rule.

The affection labels and the fact that both pairs qualified for recruitment are held fixed. Under the stated null, the marker transmissions are independent of those affection-based selection events. Therefore, for each pair, the null probability-generating expression is

$$ P(X_j=0)+P(X_j=1)z+P(X_j=2)z^2 =\frac{1+2z+z^2}{4}. $$

For two independent pairs, multiply the expressions:

$$ \left(\frac{1+2z+z^2}{4}\right)^2 =\frac{1+4z+6z^2+4z^3+z^4}{16}. $$

Thus \(S\) has probabilities \(1/16,4/16,6/16,4/16,1/16\) for values 0 through 4. If the observed total is \(S=4\), the one-sided null probability of sharing at least that much is \(P(S\geq 4)=1/16=0.0625\). This calculation is valid for the stated design because the affected-pair ascertainment is fixed and does not alter marker transmission probabilities under the null. It would need reconsideration if marker results had influenced recruitment.

Constraining Affection-Label Permutations

When an analysis does permute affection labels, the allowed assignments must reflect both exchangeability and ascertainment. Conditioning on the number affected is a common constraint, but it is not automatically sufficient. A fixed proband status, covariate strata, or family-specific recruitment requirement may impose further restrictions. Even a constrained set of assignments is valid only if the remaining individuals are exchangeable under the null, conditional on the variables used to define the constraints.

Worked Example: A Fixed Affected Proband

Consider a hypothetical pedigree with four siblings. Sibling 1 is the designated proband, and the study enrolled the family only if the proband was affected and exactly two of the four siblings were affected. Suppose, for illustration, that the other three siblings are exchangeable under the conditional null after accounting for measured covariates. The observed affection pattern is \((1,1,0,0)\), where 1 denotes affected and the first position is the proband.

An unrestricted permutation of the two affected labels among four siblings would allow \(\binom{4}{2}=6\) assignments. Three of these have an unaffected proband, so they do not satisfy the recruitment condition. The valid conditional assignments instead keep sibling 1 affected and choose one additional affected sibling from siblings 2, 3, and 4:

$$ \binom{3}{1}=3. $$

The three allowed patterns are \((1,1,0,0)\), \((1,0,1,0)\), and \((1,0,0,1)\). Under the stated exchangeability assumption, each has probability \(1/3\) conditional on recruitment and on exactly two affected siblings. If age or another covariate makes siblings 2, 3, and 4 nonexchangeable, these three patterns cannot simply be assigned equal probability; the null model must account for those differences. The calculation identifies the permissible assignments but does not replace the exchangeability assumption.

Gene-Dropping While Preserving Family and Marker Structure

In gene-dropping replicates, pedigree relationships are not shuffled: parents remain parents and siblings remain siblings. A marker allele is passed from each parent according to the specified Mendelian transmission model. For a multipoint analysis, transmissions across markers must also respect the map and recombination model; independently drawing a fresh inheritance state at every marker would generally erase the dependence along a chromosome. The statistic is recomputed from the simulated marker data using the same analysis rule as for the observed data.

“Preserve marker information” therefore means preserving the relevant marker model and design, not necessarily copying every observed descendant genotype into every replicate. A simulation might condition on typed founder haplotypes, use allele frequencies to model untyped founders, retain the marker map, and keep the locations of missing genotypes fixed while generating the values that would be observed. Alternatively, a specifically conditional procedure may condition on observed marker genotypes and sample compatible inheritance states. Those are different null distributions and must not be described as if they were the same procedure.

Likewise, missingness should not be handled by silently changing which relatives contribute from replicate to replicate unless that change is part of the specified model. A common design-preserving choice is to hold the observed typing pattern fixed and apply it to each replicate. This keeps the statistic’s information pattern comparable while allowing simulated marker inheritance to vary. If missingness depends on genotype or affection status, simply fixing its pattern may not suffice; that dependence needs to be represented in the null mechanism.

Worked Example: Monte Carlo Calibration With a Fixed Pedigree Design

Suppose a hypothetical NPL analysis has 12 selected pedigrees, each recruited under a documented affected-relative rule. The analysis uses a one-sided statistic \(T\), with larger values indicating greater affected-relative sharing. A gene-drop procedure fixes the pedigree structures and affection labels, uses the specified founder marker information and map, simulates marker transmissions under the no-linkage null, and retains the observed typing pattern. Assume 999 null replicates are generated and 37 have \(T^{(b)}\geq T_{\mathrm{obs}}\).

For a Monte Carlo test, use the observed statistic together with the simulated statistics in the exceedance calculation:

$$ \hat{p}_{\mathrm{MC}} =\frac{1+\sum_{b=1}^{999} I\!\left(T^{(b)}\geq T_{\mathrm{obs}}\right)}{999+1} =\frac{1+37}{1000} =0.0380. $$

Here \(I(\cdot)\) equals 1 when the condition is true and 0 otherwise. The estimated one-sided null tail probability is 0.0380 under the specified gene-drop model and conditioning scheme. The added 1 in numerator and denominator prevents a zero Monte Carlo p-value and treats the observed arrangement as part of the randomized comparison. This result is only as reliable as the stated null mechanism: changing the recruitment rule, founder model, or marker simulation can change the calibration.

Why the Conditioning Must Be Explicit

A permutation p-value is a tail probability under a particular conditional null, not a general probability that the null hypothesis is true. If the randomization fails to preserve the selection process, its reference distribution may not describe the NPL statistic for the families actually observed. This is especially consequential in linkage studies because family recruitment often deliberately enriches the sample for affected relatives.

The short justification for holding affection labels fixed in a gene-drop is a conditional-independence argument. Under the no-linkage null, and assuming marker data did not determine recruitment, marker transmissions have the same null distribution among pedigrees selected by affection status as among otherwise comparable pedigrees with those affection labels. Symbolically, the intended implication is

$$ P(\text{marker transmissions}\mid D,Y,A,H_0) = P(\text{marker transmissions}\mid D,H_0), $$

for the specified transmission model and ascertainment event \(A\). This equality explains why a gene-drop can keep \(Y\) and \(A\) fixed. It is not automatic: it can fail if the marker region affects the trait, if the marker was used in recruitment, or if the simulation omits relevant selection or dependence. Those possibilities require a different null model or additional conditioning.

Key takeaway: A valid NPL randomization preserves the sampled design. Keep pedigree relationships intact; retain or model affection labels and ascertainment explicitly; and handle marker transmissions, marker information, and missingness according to a stated null mechanism. Use a shuffle only when its exchangeability conditions are justified.

Common Mistake

Wrong reasoning: “All siblings are relatives in the same family, so their affection labels can be permuted freely; the permutation automatically accounts for family ascertainment.”

Correction: Relatedness alone does not imply exchangeability. A designated affected proband, a minimum affected count, differences among siblings, or a marker-based selection rule can restrict or change the probabilities of assignments. State the recruitment event, condition on the appropriate features, and justify equal probabilities only for assignments that are genuinely exchangeable under the null. If that justification is unavailable, use a null simulation that represents the ascertainment and transmission process rather than an unrestricted shuffle.

Check Your Understanding

For each question, identify the conditioning or exchangeability issue that determines whether the proposed randomization is valid.

  1. A family is recruited because its designated proband is affected. What must an affection-label permutation preserve?
  2. Why does fixing the number of affected siblings not, by itself, prove that their labels are exchangeable?
  3. In a gene-drop, which parts of the pedigree and marker design should remain fixed or be explicitly modeled?
  4. A marker was used to decide which families were invited. Why might a gene-drop that conditions only on affection labels fail to represent the selected sample?
  5. In a one-sided Monte Carlo test, 24 of 499 null replicates equal or exceed the observed statistic. Calculate the corrected exceedance estimate using the observed statistic in the comparison.
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