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Variance Components and Family QTLs · Tutorial 295 of 989

A Worked Ascertainment Correction

Work through an extreme-proband ascertainment correction and see how conditioning changes the likelihood ratio for a family QTL.

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What You'll Learn

  • Define the ascertainment event and the conditional likelihood for a recruited family
  • Calculate recruitment probabilities under competing Gaussian family models
  • Divide each phenotype likelihood by its model-specific recruitment probability
  • Compare unconditioned and ascertainment-corrected likelihood ratios
  • Explain what a change in the likelihood ratio does—and does not—say about QTL evidence
  • Identify when the ascertainment correction depends on the correct recruitment mechanism

Why Recruitment Changes the Likelihood

A family study may recruit families because one member has an unusually high or low trait value. The observed families are then not a random sample of all families: they were selected using phenotype information. If the likelihood treats them as though they had been sampled at random, it can give misleading comparisons between QTL models. The correction is to use the likelihood conditional on the event that caused recruitment.

This tutorial works through a two-member family example. The first member is the proband, whose phenotype triggers recruitment; the second member is also phenotyped. We compare a null model \(H_0\) with no QTL contribution to the family covariance against an alternative model \(H_1\) with a QTL contribution. The point is not that an extreme proband always strengthens or weakens QTL evidence. Rather, the change depends on how likely that recruitment event is under each model.

Definition: Let \(Y\) denote the vector of phenotypes observed in a family, and let \(A\) be the event that the family meets the recruitment rule. Under hypothesis \(H_j\), the ascertainment-corrected likelihood for a recruited family with observed phenotype vector \(y\) is $$ L_{\mathrm{cond},j}(y)=f_j(y\mid A)=\frac{f_j(y)}{P_j(A)}, $$ provided that \(y\) satisfies \(A\) and \(P_j(A)>0\). Here \(f_j(y)\) is the joint phenotype density under \(H_j\), and \(P_j(A)\) is the probability of recruitment under that same model.

The numerator is a joint density for all phenotypes being analyzed. The denominator is the probability of the selection event, calculated under the model being evaluated. It is generally not enough to divide by the recruitment probability under just one hypothesis: each model implies its own probability of producing a recruited family.

Worked Example: The Recruitment Probabilities

Worked Example: The Recruitment Probabilities

Suppose the proband phenotype \(Y_1\) has mean zero under both models. Under \(H_0\), \(Y_1\sim N(0,1)\); under \(H_1\), \(Y_1\sim N(0,1.918)\). A family is recruited if the proband exceeds \(c=1.1095\) trait units, so \(A=\{Y_1>1.1095\}\). These are hypothetical model values.

For a normal variable, standardizing the threshold gives \(P(Y_1>c)=1-\Phi(c/\sigma)\), where \(\Phi\) is the standard normal cumulative distribution function. Under \(H_0\),

$$ P_0(A)=1-\Phi\left(\frac{1.1095}{\sqrt{1}}\right) =1-\Phi(1.1095)\approx 0.1336. $$

Thus about 13.36% of families meet the recruitment rule under \(H_0\). Under \(H_1\), the proband standard deviation is \(\sqrt{1.918}\approx1.3849\) trait units, and

$$ P_1(A)=1-\Phi\left(\frac{1.1095}{\sqrt{1.918}}\right) =1-\Phi(0.8011)\approx 0.2115. $$

About 21.15% of families meet the same rule under \(H_1\). The larger proband variance makes an extreme value more likely, even though the proband mean is zero under both models. This calculation uses the marginal distribution of the proband, because recruitment depends only on \(Y_1\).

From the Unconditioned to the Conditional Likelihood

For the family likelihood calculation, suppose the observed phenotype vector is \(y=(1.15,2.03)\) trait units. The first component exceeds the threshold, so this is a recruited family. Under each hypothesis, assume the full phenotype vector follows a bivariate normal distribution with mean zero and a positive-definite covariance matrix. The covariance matrices may differ between hypotheses, as required for the two variance-component models.

The bivariate normal density used here is

$$ f_j(y)=\frac{1}{2\pi|\Sigma_j|^{1/2}} \exp\left(-\frac{1}{2}y^\mathsf{T}\Sigma_j^{-1}y\right), $$

where \(\Sigma_j\) is the covariance matrix under \(H_j\), and \(|\Sigma_j|\) is its determinant. For this hypothetical example, evaluating that density at the observed family vector gives \(f_0(y)=0.010484\) and \(f_1(y)=0.031239\) per squared trait unit. These are joint densities, not probabilities of observing the exact vector. The covariance matrices are assumed known for this worked comparison; estimating them is a separate issue.

Worked Example: Correcting Both Family Likelihoods

Under \(H_0\), divide the joint density by the probability of recruitment under \(H_0\):

$$ L_{\mathrm{cond},0}(y) =\frac{0.010484}{0.1336} \approx 0.07847 \quad\text{per squared trait unit}. $$

Under \(H_1\), use its own recruitment probability:

$$ L_{\mathrm{cond},1}(y) =\frac{0.031239}{0.2115} \approx 0.1477 \quad\text{per squared trait unit}. $$

The conditional likelihood ratio in favor of \(H_1\) over \(H_0\) is therefore

$$ \frac{L_{\mathrm{cond},1}(y)}{L_{\mathrm{cond},0}(y)} =\frac{0.1477}{0.07847} \approx 1.882. $$

For comparison, the unconditioned likelihood ratio is \(0.031239/0.010484\approx2.980\). In this example, conditioning reduces the relative support for \(H_1\): the alternative model predicts recruitment more often, so observing a family known to have been recruited is less surprising under \(H_1\) than an unconditioned calculation would imply. The conditional ratio still exceeds 1, so the observed family has a larger conditional likelihood under \(H_1\) than under \(H_0\), but the ratio is not as large as the unconditioned one.

Why the Correction Has This Form

For measurable \(A\) with \(P_j(A)>0\), the conditional density is \(f_j(y)\mathbf{1}_A(y)/P_j(A)\) almost everywhere. In this example, \(y_1=1.15>1.1095\), so \(y\) is an interior point of \(A\) and the ratio applies pointwise there.

$$ f_j(y\mid A) =\frac{\mathbf{1}_A(y)f_j(y)}{P_j(A)} \quad\text{for almost every }y. $$

The key condition is that the density point \(y\) satisfies the selection event. If \(y\notin A\), its conditional density given \(A\) is zero. The denominator is a normalizing factor: integrating \(f_j(y)/P_j(A)\) over all phenotype vectors that satisfy \(A\) gives 1, because the integral of \(f_j\) over \(A\) is \(P_j(A)\).

Result: For a family recruited through event \(A\), the conditional likelihood ratio for \(H_1\) against \(H_0\) is the unconditioned ratio multiplied by the ratio of recruitment probabilities in the opposite order: $$ \frac{L_{\mathrm{cond},1}(y)}{L_{\mathrm{cond},0}(y)} = \frac{f_1(y)}{f_0(y)} \frac{P_0(A)}{P_1(A)}. $$

In the worked example, the recruitment-probability factor is \(0.1336/0.2115\approx0.6317\). Thus the conditional ratio is the unconditioned ratio, approximately 2.980, multiplied by about 0.6317, giving approximately 1.882. Equivalently, on the log scale, the conditioning adjustment is the log of this factor.

Worked Example: The Log-Likelihood Adjustment

The unconditioned log likelihood ratio is

$$ \log\left(\frac{0.031239}{0.010484}\right) \approx 1.092. $$

The log of the ascertainment factor is

$$ \log\left(\frac{0.1336}{0.2115}\right) \approx -0.459. $$

Adding these terms gives the conditional log likelihood ratio:

$$ 1.092+(-0.459)\approx 0.633, \qquad \exp(0.633)\approx 1.883. $$

The small differences that can arise from recomputing with rounded inputs are rounding effects; the displayed conditional likelihood ratio is based on the unrounded quantities. A positive log ratio means that the conditional likelihood is larger under \(H_1\) for this family. It does not mean there is a 1.882 probability that \(H_1\) is true.

Assumptions and Interpretation

An ascertainment correction is only as appropriate as the selection mechanism used to calculate it. In the example, the assumptions are that families are recruited if and only if the proband exceeds the stated threshold, the threshold is known, and the proband phenotype has the stated marginal normal distribution under each hypothesis. The full family phenotype densities also come from the specified bivariate normal models. If recruitment also depends on another relative's phenotype, a diagnosis, or a combination of criteria, \(A\) must represent that actual rule and \(P_j(A)\) must be calculated accordingly.

The correction conditions on the fact of recruitment; it does not discard the proband's observed phenotype. The numerator still uses the joint density of the proband and relative. The denominator accounts for the fact that the family entered the study through a phenotype-based selection process. Replacing the joint likelihood with a likelihood for only the unselected relative would be a different analysis and would not implement this conditional likelihood.

A likelihood ratio compares the relative likelihoods of the observed data under specified models. It is not a posterior probability, and it does not by itself provide a calibrated significance level. In family QTL analysis, the conditional likelihood is the relevant input when the sampling design conditions on ascertainment. How a test statistic is calibrated is a separate question.

Common Pitfall

  • Wrong reasoning: “The alternative model gives the larger unconditioned density, so it must provide the same relative evidence after recruitment is accounted for.” This ignores that the models can have different probabilities of producing a recruited family.
  • Correction: Divide each model's joint density by its own recruitment probability. In this example \(P_1(A)>P_0(A)\), so conditioning multiplies the unconditioned likelihood ratio by a factor below 1 and reduces the relative evidence for \(H_1\).
  • Another error: Dividing the alternative likelihood by \(P_0(A)\), or dividing only the proband's density while leaving the rest of the family likelihood untreated. The correction is \(f_j(y)/P_j(A)\) for each model's full observed phenotype vector.
Key takeaway: For an extreme-proband family, calculate the recruitment probability under every model being compared, then use the joint phenotype density divided by that model-specific probability. The corrected likelihood ratio can differ materially from the unconditioned ratio.

Check Your Understanding

Use the definitions and calculations above to answer these questions.

  1. If the recruitment event is \(Y_1>c\), which distribution is needed to calculate \(P_j(A)\): the proband's marginal distribution or the full joint distribution of all family phenotypes?
  2. Suppose \(f_0(y)=0.020\), \(f_1(y)=0.030\), \(P_0(A)=0.10\), and \(P_1(A)=0.20\). Calculate both conditional likelihoods and their ratio.
  3. In the worked example, why is the ascertainment factor \(P_0(A)/P_1(A)\), rather than \(P_1(A)/P_0(A)\), when correcting the likelihood ratio for \(H_1\) against \(H_0\)?
  4. Does a conditional likelihood ratio of 1.882 mean that the probability \(H_1\) is true is 1.882 times the probability \(H_0\) is true? Explain.
  5. How would the event \(A\) change if recruitment required both the proband and a sibling to exceed prespecified thresholds?

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