Small-Sample Null Distributions For NPL Tests

Why Small-Sample Calibration Needs Attention

A nonparametric linkage (NPL) statistic measures whether affected relatives share marker alleles identical by descent (IBD) more often than expected under no linkage. The central calibration question is: how unusual would this statistic be if the marker and disease locus were not linked? With many independent pedigrees, a standardized NPL statistic is often compared with a normal distribution, or a related quadratic statistic with a chi-square distribution. With only a few pedigrees, those reference distributions may describe the tail poorly.

The reason is not simply that the total number of observed relative pairs is small. Relatives within a pedigree share inheritance information, so their sharing contributions are generally dependent. The effective number of independent units is closer to the number of independent pedigrees. Moreover, family-level sharing scores may be discrete, skewed, or different in spread across pedigrees. Standardizing a statistic to have mean zero and variance one does not by itself make its distribution normal.

Definition: Let $S_j$ be the sharing score for pedigree $j$, and let $E_0(S_j)$ and $\operatorname{Var}_0(S_j)$ denote its mean and variance under the no-linkage model. A general standardized NPL score is $$ Z_{\mathrm{NPL}}= \frac{\sum_{j=1}^{J}\{S_j-E_0(S_j)\}} {\sqrt{\operatorname{Var}_0\left(\sum_{j=1}^{J}S_j\right)}}. $$ If pedigrees are independent under the null, the denominator is $\sqrt{\sum_{j=1}^{J}\operatorname{Var}_0(S_j)}$. The particular definition of $S_j$ depends on the NPL method; the calibration problem is to find the null distribution of the resulting statistic.

The no-linkage distribution must reflect the statistic actually used and the information available in each pedigree. For an exact calculation, this means deriving the distribution of each family score under the relevant Mendelian no-linkage model and combining those distributions. If marker genotypes leave IBD status uncertain, the calculation must account for that uncertainty rather than acting as though estimated IBD states were observed without error.

These calculations rely on a stated null model. In broad terms, the no-linkage model specifies Mendelian transmission at the marker and the conditioning used for the family data, including the pedigree structure and ascertainment. Its validity requires an appropriate inheritance model, correctly handled marker information, and independence between pedigrees. If the calculation conditions on observed phenotype information, that conditioning must also be appropriate to the ascertainment scheme. A precise tail probability under a misspecified model is not a reliable test.

Exact and Asymptotic Reference Distributions

An exact null distribution gives the probabilities of all possible values of the chosen statistic under the specified no-linkage model. It can be obtained by enumerating possible inheritance states when enumeration is manageable. Under independent pedigrees with discrete family scores, the distribution of the sum can also be found by convolution: combine the possible scores and probabilities for one pedigree with those for the next, and continue until all pedigrees are included.

For a prespecified one-sided test of excess sharing, the exact p-value is the null probability of a statistic at least as large as the observed value. Because an exact distribution may be discrete, this probability can be noticeably larger than a normal-approximation p-value. That difference is not a computational error: the exact probability respects the attainable values and their probabilities.

The normal approximation is motivated by the central limit theorem: sums of independent contributions can approach a normal distribution as the number of contributions grows, under suitable conditions. It is an approximation, not a guarantee for a handful of pedigrees. Likewise, chi-square calibration for a quadratic or likelihood-based statistic is generally an asymptotic result. A few pedigrees, a highly discrete statistic, or markedly unequal family information can make either approximation inaccurate, particularly in the tail where significance is judged.

Worked Example: Exact Tail Probability for Four Sib-Pair Pedigrees

Suppose four independent, hypothetical pedigrees each contribute one affected sib pair. Let $K_j$ be the number of alleles shared IBD by the pair in pedigree $j$, so $K_j$ can be 0, 1, or 2. Under the Mendelian no-linkage null for a sib pair, $P_0(K_j=0)=1/4$, $P_0(K_j=1)=1/2$, and $P_0(K_j=2)=1/4$. Assume the pedigrees are independent and the observed IBD states are known for this illustration.

The sharing total is $T=\sum_{j=1}^{4}K_j$. Its null mean is $4(1)=4$, and its variance is $4(1/2)=2$, since one sib-pair IBD count has mean 1 and variance $1/2$. Suppose the four hypothetical observed counts are 2, 2, 1, and 2, giving $T_{\mathrm{obs}}=7$.

For the normal approximation, the standardized value is $$ Z_{\mathrm{NPL}}=\frac{7-4}{\sqrt{2}}=2.121. $$ The one-sided standard-normal tail probability is approximately $P(Z\geq2.121)=0.0169$.

For the exact calculation, define $D_j=K_j-1$. Each $D_j$ is $-1,0,$ or $1$, with probabilities $1/4,1/2,$ and $1/4$. The event $T\geq7$ is the event $\sum D_j\geq3$. It occurs if exactly three of the four $D_j$'s equal 1 and the remaining one equals 0, or if all four equal 1. Therefore, $$ P_0(T\geq7)= \binom{4}{3}\left(\frac14\right)^3\left(\frac12\right) +\left(\frac14\right)^4 =\frac{8}{256}+\frac{1}{256} =\frac{9}{256}=0.03516. $$

Under the stated null model, the exact one-sided p-value is 0.0352, rather than the normal approximation of 0.0169. The exact result provides less evidence against no linkage in this example. The discrepancy arises because only a small, discrete set of totals is possible.

Convolving Family-Specific Null Distributions

Pedigrees need not have identical information or identical score distributions. For independent pedigrees, an exact distribution can still be constructed by combining their individual distributions, provided each has been correctly specified under the null. The following example uses a simple indicator score to make the calculation transparent; it illustrates the calibration principle rather than defining every NPL statistic.

Worked Example: Different Sharing Probabilities Across Pedigrees

Consider three independent, hypothetical pedigrees. Let $X_j=1$ if pedigree $j$ has a prespecified high-sharing outcome and $X_j=0$ otherwise. Suppose the no-linkage probabilities are $P_0(X_1=1)=0.50$, $P_0(X_2=1)=0.25$, and $P_0(X_3=1)=0.75$, reflecting different family information. The observed outcomes are $1,1,1$, so $T=X_1+X_2+X_3=3$.

The exact upper-tail probability is the probability that all three outcomes are 1: $$ P_0(T\geq3)=P_0(X_1=1)P_0(X_2=1)P_0(X_3=1) =(0.50)(0.25)(0.75)=0.09375. $$ The product is justified by the assumed independence of pedigrees. No common success probability is required.

For comparison, the null mean of $T$ is $0.50+0.25+0.75=1.50$. Its variance is $$ (0.50)(0.50)+(0.25)(0.75)+(0.75)(0.25)=0.625. $$ Thus the observed standardized value is $(3-1.50)/\sqrt{0.625}=1.897$, whose standard-normal one-sided tail is about 0.0289. That approximation is much smaller than the exact value 0.0938. With only three contributions, the standardized sum is still discrete and the normal tail does not represent its attainable probabilities well.

More generally, if the null probabilities for family $j$'s score are known, the distribution of the aggregate score follows from summing over all combinations of family scores. For example, if scores are integer-valued, a probability-generating function can encode each family's distribution, and multiplying the functions gives the distribution of their sum. Enumeration is often straightforward for a few small pedigrees; it may become expensive when pedigrees have many possible inheritance states or the statistic is more complicated.

Simulated No-Linkage Distributions

When exact enumeration is impractical, a no-linkage distribution can be approximated by repeatedly generating data under the specified null model and recalculating the same NPL statistic. In a linkage setting, such simulations may generate marker inheritance under Mendelian transmission while maintaining the conditioning and family structure required by the analysis. This is a model-based simulation of the no-linkage reference distribution; it is not a license to ignore ascertainment, observed marker information, or uncertainty in IBD inference.

Let $T_{\mathrm{obs}}$ be the observed statistic and let $T_1^*,\ldots,T_B^*$ be statistics from $B$ null simulations. If $R$ simulated statistics are at least as large as $T_{\mathrm{obs}}$, an upper-tail Monte Carlo estimate is $(R+1)/(B+1)$ when the observed result and null replicates are exchangeable under the null procedure. The added one avoids reporting a zero p-value merely because a finite set of simulations did not exceed the observed statistic. The estimate remains subject to Monte Carlo error: it is not the exact model-based tail probability.

Worked Example: Estimating a Tail From Null Simulations

Suppose an observed, hypothetical NPL statistic is $T_{\mathrm{obs}}=2.40$. A specified no-linkage simulation procedure generates $B=999$ replicate statistics using the same pedigree structure and analysis. Assume 42 replicates are at least 2.40. The estimated one-sided Monte Carlo p-value is $$ \widehat{p}_{\mathrm{MC}}=\frac{R+1}{B+1} =\frac{42+1}{999+1}=0.0430. $$

A rough measure of simulation uncertainty is the binomial standard error $\sqrt{\widehat{p}_{\mathrm{MC}}(1-\widehat{p}_{\mathrm{MC}})/B}$, which here is approximately $$ \sqrt{\frac{(0.0430)(0.9570)}{999}}=0.00642. $$

The simulation therefore estimates a one-sided null tail probability of about 0.043, with non-negligible Monte Carlo variability at this simulation size. This is evidence against the specified no-linkage model at a 0.05 threshold, but the estimate should not be reported as though it were an exact probability. More simulations can reduce simulation error; they cannot correct a misspecified null model.

Choosing and Checking the Calibration

A practical calibration decision begins with the statistic and the independent units, not with a software default. Check how many independent pedigrees contribute, whether one or a few pedigrees dominate the score, and whether the null statistic has a coarse or skewed distribution. Then determine whether the exact distribution is tractable or whether simulation is needed. The tail used for calibration must match the prespecified alternative: an upper tail for excess sharing, for example, is not interchangeable with a two-sided tail.

Specify the statistic and direction.
State how family information contributes to the NPL statistic and whether evidence is assessed in an upper tail or another prespecified region.

Specify the no-linkage model.
Identify the inheritance assumptions, family conditioning, marker information, and any treatment of uncertain IBD states that determine the null distribution.

Assess whether asymptotic calibration is credible.
Consider the number of independent pedigrees, discreteness, skewness, and differences in information across families; a large pair count alone is not enough.

Calculate the reference tail.
Use exact enumeration or convolution when feasible, or simulate the same statistic under the specified no-linkage model and report the simulation precision.

Exact or simulated calibration matters because NPL evidence is judged in the tail, where small errors in a reference distribution can alter a conclusion. It is especially useful when there are few independent pedigrees, when the statistic takes only a limited number of values, or when family contributions vary considerably. It also makes clear what a reported p-value depends on: not only the observed sharing, but also the inheritance model and the conditioning used to define “no linkage.”

Common Mistake

Wrong reasoning: “The standardized NPL score is 2.1, so its p-value must be the standard-normal tail, even though only four pedigrees contributed.”
Correction: Standardization sets the null mean and variance; it does not establish a normal shape. With few pedigrees, calculate or simulate the statistic’s no-linkage distribution when feasible, and use its tail probability for the stated test.
Wrong reasoning: “There are many relative pairs, so the large-sample approximation must be reliable.”
Correction: Pairs within the same pedigree can be dependent. Assess the number and influence of independent pedigrees, and use the family-level null distribution rather than treating every pair as independent.

Key takeaway: An exact or simulated small-sample NPL p-value is meaningful only relative to a clearly specified no-linkage model. When the number of independent pedigrees is small, the statistic’s actual discrete null distribution can differ substantially from its normal or chi-square approximation.

Check Your Understanding

Try these questions without looking back.

Why can a large number of relative pairs still provide only a small effective number of independent contributions to an NPL statistic?
In the four-pedigree example, why was the exact upper-tail probability larger than the normal-approximation tail?
What assumptions justify multiplying the family-specific probabilities in the three-pedigree example?
What does an exact null distribution account for that variance standardization alone does not?
If 0 of 500 simulated null statistics exceed the observed statistic, why should the estimated p-value not be reported as zero?