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Nonparametric Linkage Methods · Tutorial 250 of 989

Kong-Cox Allele-Sharing LOD Scores

Build the Kong–Cox likelihood from a family sharing statistic and its no-linkage distribution, then interpret the maximized LOD as evidence for excess sharing.

Intermediate 10 min read

What You'll Learn

  • Define the Kong–Cox exponential-tilt likelihood for a family sharing statistic.
  • Construct the likelihood ratio using family-specific no-linkage distributions.
  • Maximize the log likelihood over a nonnegative excess-sharing parameter.
  • Calculate likelihood ratios and LOD scores for hypothetical affected-relative data.
  • Distinguish a maximized LOD from a p-value and a biological recombination parameter.
  • Recognize assumptions and pitfalls in combining information across pedigrees.

Why Use a Likelihood Score for Allele Sharing?

Nonparametric linkage analysis asks whether affected relatives share alleles identical by descent (IBD) at a marker more often than expected without linkage. A sharing statistic summarizes the evidence within a pedigree; its no-linkage distribution describes the values expected under the null model. The Kong–Cox approach uses that distribution to construct a likelihood ratio favoring excess sharing, without specifying a disease-allele frequency and penetrance model.

The key idea is to tilt the null distribution toward larger sharing-statistic values. The tilt is controlled by a parameter that is estimated from the observed sharing. The maximum likelihood ratio, reported on a base-10 logarithmic scale, is the Kong–Cox allele-sharing LOD. It summarizes the strength of evidence against no excess sharing under this particular likelihood construction; it is not itself a significance probability.

Definition: Let \(S_i\) be a prespecified allele-sharing statistic for pedigree \(i\), and let \(P_0\) denote its distribution under no linkage. For a nonnegative parameter \(\delta\), define the exponentially tilted distribution $$ P_\delta(S_i=s)=\frac{P_0(S_i=s)e^{\delta s}}{M_i(\delta)}, \qquad M_i(\delta)=E_0\!\left[e^{\delta S_i}\right]. $$ For a continuous statistic, the same definition applies to its density, with the expectation taken under its null distribution. The Kong–Cox LOD is the maximum, over \(\delta\geq 0\), of the base-10 log likelihood ratio of the tilted model to the null model.

The normalizing term \(M_i(\delta)\) makes the tilted probabilities sum or integrate to one. At \(\delta=0\), the tilt disappears, so the distribution is exactly the no-linkage distribution. For \(\delta>0\), higher values of \(S_i\) receive relatively more probability. The parameter \(\delta\) measures the strength of this mathematical tilt; it is not a recombination fraction or a direct estimate of a biological effect.

If pedigrees are independent under the model and their observed statistics are \(s_1,\ldots,s_m\), the joint likelihood ratio is

$$ \frac{L(\delta)}{L(0)} =\prod_{i=1}^{m}\frac{e^{\delta s_i}}{M_i(\delta)} =\exp\!\left\{\delta\sum_{i=1}^{m}s_i-\sum_{i=1}^{m}\log M_i(\delta)\right\}. $$

Thus, the log likelihood ratio on the natural logarithm scale is \(\ell(\delta)=\delta\sum_i s_i-\sum_i\log M_i(\delta)\). The LOD at a specified \(\delta\) is \(\ell(\delta)/\ln(10)\); the maximized LOD is \(\max_{\delta\geq0}\ell(\delta)/\ln(10)\). If each pedigree has the same null distribution, the denominators are identical. If their null distributions differ, each pedigree must retain its own normalizing term.

Worked Examples

Worked Example: One Affected Sib Pair at a Fixed Tilt

Suppose one hypothetical affected sib pair has \(K\), the number of alleles shared IBD, as its sharing statistic. Under no linkage, \(P_0(K=0)=1/4\), \(P_0(K=1)=1/2\), and \(P_0(K=2)=1/4\). The pair is observed to share \(K=2\). Calculate the likelihood ratio and LOD at \(\delta=\ln(2)\).

Here \(e^\delta=2\), so the normalizing constant is $$ M(\delta)=\frac14+\frac12(2)+\frac14(2^2)=\frac94. $$

The observed value is \(s=2\). Its likelihood ratio against the null at this specified tilt is $$ \frac{L(\delta)}{L(0)}=\frac{e^{2\delta}}{M(\delta)} =\frac{2^2}{9/4}=\frac{16}{9}=1.778. $$

Therefore, the LOD at this fixed tilt is \(\log_{10}(16/9)=0.2499\). Under the tilted model, the probability of observing \(K=2\) is \((1/4)(4)/(9/4)=4/9\), compared with \(1/4\) under the null. This example evaluates one specified \(\delta\); it has not yet maximized over possible tilts.

Worked Example: Maximizing Across Independent Sib Pairs

Consider three hypothetical affected sib pairs from independent pedigrees, each with the same null distribution for \(K\) as above. Their observed sharing counts are \(2,2,0\). Find the maximized Kong–Cox LOD.

Write \(q=e^\delta\), so \(q\geq1\). For each pair, \(M(\delta)=(1+q)^2/4\). The sum of the observed sharing counts is \(4\), so the joint likelihood ratio is $$ R(q)=\frac{q^4}{\{(1+q)^2/4\}^3} =\frac{64q^4}{(1+q)^6}. $$

Differentiate the natural log likelihood ratio with respect to \(\delta\). Since \(q=e^\delta\), the derivative is $$ \frac{d\log R}{d\delta}=4-\frac{6q}{1+q}. $$ Setting it equal to zero gives \(4(1+q)=6q\), hence \(q=2\), or \(\widehat{\delta}=\ln(2)\). The derivative at \(\delta=0\) is \(4-3=1>0\), so the boundary \(\delta=0\) is not the maximum; the stationary point is the maximum.

Substitute \(q=2\): $$ R(2)=\frac{64(2^4)}{(1+2)^6} =\frac{1024}{729}=1.405, \qquad \mathrm{LOD}=\log_{10}(1024/729)=0.1475. $$

The observed sharing gives a likelihood about 1.405 times as large at the fitted tilt as under no linkage. The maximized LOD is modest. It is not the probability that linkage is true, nor does it by itself determine a p-value.

Worked Example: Why Family-Specific Null Distributions Matter

Suppose two hypothetical independent pedigrees have different sharing-statistic distributions. For pedigree A, \(S_A\) takes values 0 and 2 with null probabilities \(1/2\) each; its observed value is 2. For pedigree B, \(S_B\) takes values 0, 1, and 2 with null probabilities \(1/4,1/2,1/4\); its observed value is 1. Calculate the joint likelihood ratio and LOD at \(\delta=\ln(2)\).

With \(q=e^\delta=2\), the two normalizing terms are $$ M_A(\delta)=\frac12+\frac12(2^2)=\frac52, \qquad M_B(\delta)=\frac14+\frac12(2)+\frac14(2^2)=\frac94. $$

The observed values contribute \(e^{\delta(2+1)}=2^3=8\) to the numerator. Using both family-specific denominators gives $$ R(\delta)=\frac{8}{(5/2)(9/4)} =\frac{64}{45}=1.422, \qquad \mathrm{LOD}=\log_{10}(64/45)=0.1530. $$

The null distributions are not interchangeable: pedigree A has no probability of a statistic value of 1, while pedigree B does. The example is an evaluation at a fixed tilt, not a maximization. Replacing the two normalizing terms by a single pooled term would generally define a different likelihood.

How Maximization Works

The maximization is conveniently understood through the derivative of the log likelihood ratio. For each pedigree, the tilted expectation of \(S_i\) is $$ E_\delta(S_i)=\frac{E_0[S_i e^{\delta S_i}]}{M_i(\delta)}. $$ Differentiating the joint log likelihood ratio gives $$ \ell'(\delta)=\sum_{i=1}^{m}s_i-\sum_{i=1}^{m}E_\delta(S_i). $$

At a finite interior maximum, the observed total sharing equals the total expected sharing under the fitted tilted distributions. The curvature is $$ \ell''(\delta)=-\sum_{i=1}^{m}\operatorname{Var}_\delta(S_i)\leq0. $$ The variance identity follows by differentiating the tilted expectation: \(dE_\delta(S_i)/d\delta=\operatorname{Var}_\delta(S_i)\). Therefore the log likelihood ratio is concave in \(\delta\), provided the relevant expectations exist. Concavity means a stationary point is a global maximum, although a maximum may instead occur at the boundary or be approached as \(\delta\) tends to infinity.

1
Specify the statistic and its null distribution.
Define each pedigree's allele-sharing summary and determine its distribution under no linkage, using the pedigree and the chosen way of accounting for marker information.
2
Form the tilted likelihood.
For each pedigree, calculate \(M_i(\delta)=E_0[e^{\delta S_i}]\), then combine the likelihood contributions if the pedigrees are independent.
3
Maximize over excess-sharing tilts.
Maximize \(\ell(\delta)\) subject to \(\delta\geq0\). A nonpositive derivative at zero places the maximum at zero; otherwise, solve for an interior maximum or check the limiting boundary.
4
Report the base-10 score in context.
Convert the maximized natural log likelihood ratio by dividing by \(\ln(10)\), and describe it as relative likelihood evidence under the specified sharing model.

Assumptions and Interpretation

The construction depends on a valid null distribution for the chosen sharing statistic. In linkage analysis, this distribution must reflect the pedigree structure and the no-linkage model for marker inheritance; uncertainty in marker-based IBD information must be handled consistently with the statistic being used. The exponential tilt is a specified alternative family of distributions, not a complete model of disease inheritance. It rewards larger sharing-statistic values while retaining the null distribution as its reference.

The product likelihood also assumes that the pedigree contributions being multiplied are independent under the model. This is a reasonable construction for distinct, non-overlapping pedigrees with independent inheritance data; related or overlapping units cannot simply be treated as independent. Any weights or transformations used to define the sharing statistics should be fixed as part of the analysis rather than chosen after examining which version gives the largest score.

A LOD is a logarithm of a likelihood ratio: a LOD of 1 corresponds to a likelihood ratio of 10, while a LOD of 0 corresponds to a ratio of 1. Maximizing over \(\delta\) gives the best fit within the specified tilted family, which is why the maximized value summarizes evidence for excess sharing. It does not establish that a marker causes disease, identify a causal variant, or express the probability that the null hypothesis is true. Calibration of a test statistic and genome-wide interpretation require additional inferential steps.

Common Mistakes

  • Calling the LOD a p-value. A likelihood ratio compares how well two models describe the data; it is not the probability of data this extreme under the null. A LOD of 0.15 does not mean a 15% p-value.
  • Dropping the normalizing constant. Using only \(e^{\delta\sum_i s_i}\) rewards larger sharing without accounting for how the alternative distribution changes. The terms \(M_i(\delta)\) are essential to a valid likelihood.
  • Using one null distribution for every pedigree. If the statistic has different null distributions across pedigrees, its normalizing constants generally differ too. Preserve the family-specific terms.
  • Interpreting \(\widehat{\delta}\) as a recombination fraction. It is the fitted strength of an exponential tilt, not a map distance or a direct biological parameter.
  • Assuming the fitted parameter must be positive. The allowed range includes \(\delta=0\). If the likelihood does not improve for a positive tilt, the maximum is at the no-excess-sharing boundary.
Key takeaway: The Kong–Cox allele-sharing LOD compares the observed sharing statistic with its no-linkage distribution after tilting that distribution toward excess sharing. Its maximized value is a log-base-10 likelihood ratio under this model, not a p-value or a biological parameter estimate.

Check Your Understanding

Use the likelihood construction and interpretation rules to answer these questions.

  1. Why must \(M_i(\delta)\) appear in the denominator of the tilted distribution?
  2. For independent pedigrees with the same null distribution, what determines the numerator of their joint likelihood ratio?
  3. What does a fitted value \(\widehat{\delta}=0\) mean in this model?
  4. Why is a maximized LOD not a p-value?
  5. What assumption justifies multiplying the likelihood contributions from separate pedigrees?
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