Conditioning Likelihoods on Proband Selection

Why Recruitment Changes the Likelihood

Families in a genetic study are often recruited because a particular feature is present: for example, a designated family member has a diagnosis. The recruited pedigrees are therefore not a random sample of all pedigrees in the population. A likelihood that treats them as an unselected sample can mistake the recruitment rule for evidence about disease risk. The correction is to model the observed pedigree conditional on the event that qualified it for recruitment.

The key operation is conditioning at the family level. The numerator is the probability of the observed pedigree phenotype under the genetic model. The denominator is the probability that a pedigree would meet the study’s recruitment criterion under that same model. Because this denominator generally depends on model parameters, it can change how plausible different parameter values are.

Pedigree Probabilities and the Selection Event

Definition: Let $Y$ denote the observed phenotype pattern in a pedigree, $G$ its (possibly unobserved) genotype configuration, and $\theta$ the parameters of the segregation model. Let $S$ denote the event that the pedigree satisfies the specified recruitment rule. The ascertainment-conditioned likelihood for an observed pattern $Y=y$, when $y$ satisfies the rule, is $$ L_{\mathrm{cond}}(\theta;y)=P(Y=y\mid S;\theta) =\frac{P(Y=y,S;\theta)}{P(S;\theta)}. $$ If the recruitment event is determined by the observed phenotype pattern, then $y\in S$ implies $P(Y=y,S;\theta)=P(Y=y;\theta)$, so $$ L_{\mathrm{cond}}(\theta;y)=\frac{P(Y=y;\theta)}{P(S;\theta)}. $$

The unconditioned pedigree phenotype probability is computed using the segregation model. When genotypes are uncertain, it is obtained by summing over possible configurations:

P(Y=y;\theta)=\sum_g P(G=g;\theta)P(Y=y\mid G=g;\theta).

Here $P(G=g;\theta)$ includes the relevant genotype probabilities and transmission probabilities, while $P(Y=y\mid G=g;\theta)$ is determined by the phenotype model, such as the penetrance and phenocopy probabilities. This marginalization is important: conditioning on recruitment does not make unobserved genotypes known.

If the rule is defined by a subset $\mathcal{S}$ of possible phenotype patterns, its probability is the sum over every pattern meeting that rule:

P(S;\theta)=\sum_{y'\in\mathcal{S}}P(Y=y';\theta).

Thus the denominator is not generally the probability of the one observed pattern. It is the total probability of all outcomes that would have led to recruitment under the stated rule.

Assumptions for the Conditional Likelihood

Before using this likelihood, specify how recruitment occurred. The conditional formulation is appropriate when the study’s relevant selection event is known and the sampled pedigrees can be treated as sampled from the population of pedigrees satisfying that event. In particular, conditional on meeting the rule, selection should not further favor some eligible phenotype patterns over others in a way omitted from the model. If recruitment probabilities vary among eligible families, the sampling probabilities must also be modeled; division by $P(S;\theta)$ alone does not correct that additional selection.

The phenotype and genotype probabilities in the numerator and denominator must use the same model and parameter values. The pedigree structure, observed family members, and ascertainment rule must be represented consistently in both. For independent recruited pedigrees with selection events $S_i$ and observed patterns $y_i$, the conditional likelihood is the product of the family-specific conditional probabilities:

L_{\mathrm{cond}}(\theta)=\prod_{i=1}^{m}P(Y_i=y_i\mid S_i;\theta).

If pedigrees share population-level or study-level dependence, this product requires an appropriate joint model rather than an assumption of independence. The conditioning corrects for the stated recruitment mechanism; it does not correct an incorrect penetrance model, unmodeled dependence, or other sources of bias.

Worked Example: Conditioning on an Affected Designated Proband

Consider a hypothetical two-person pedigree with a designated proband and one relative. Suppose, for illustration, that their phenotype indicators are independent and each has probability $\theta$ of being affected. Families are recruited only when the proband is affected. A recruited family is observed with an affected proband and an unaffected relative. Find its conditional likelihood and compare it with the unconditioned probability when $\theta=0.20$.

The observed pattern has probability $P(Y=y;\theta)=\theta(1-\theta)$. The recruitment event $S$ is that the proband is affected, so $P(S;\theta)=\theta$. Since the observed pattern satisfies $S$,

L_{\mathrm{cond}}(\theta;y) =\frac{\theta(1-\theta)}{\theta} =1-\theta,\qquad \theta>0.

At $\theta=0.20$, the unconditioned pattern probability is $0.20(0.80)=0.16$, while its conditional probability is $0.16/0.20=0.80$. The latter is the probability of observing an unaffected relative among families already selected for an affected proband. The example assumes independent phenotypes and a deterministic recruitment rule; it is a simple illustration, not a realistic full genetic model.

Deriving the Correction With Uncertain Genotypes

In a pedigree, phenotype probabilities are usually mixtures over possible genotypes. The same mixture must be used to calculate the selection probability. A short derivation shows why.

Write the joint probability.
For an observed pattern $y$ that qualifies, $P(Y=y,S;\theta)=\sum_g P(G=g;\theta)P(Y=y,S\mid G=g;\theta)$. The sum accounts for every possible genotype configuration.

Sum over qualifying patterns.
The probability of recruitment is $P(S;\theta)=\sum_g P(G=g;\theta)P(S\mid G=g;\theta)$, where $P(S\mid G=g;\theta)$ sums the phenotype probabilities that meet the rule for that genotype configuration.

Apply the definition of conditional probability.
Divide the joint probability for the observed pattern by the total recruitment probability. If $y$ necessarily satisfies $S$, the numerator reduces to $P(Y=y;\theta)$, but the denominator still sums over all qualifying outcomes.

In general, one must not first average the genotypes only among observed cases and then use a denominator from a different model. The numerator and denominator describe the same population of pedigrees before selection; conditioning then restricts that population to those satisfying $S$.

Worked Example: Marginalizing a Latent Family Risk State

Suppose a hypothetical pedigree model has a latent family state $Z$, representing a simplified grouping of genotype configurations. Let $P(Z=1)=0.25$ and $P(Z=0)=0.75$. Conditional on $Z=1$, each of the proband and one relative is affected with probability $0.80$; conditional on $Z=0$, each is affected with probability $0.10$. Phenotypes are conditionally independent given $Z$. Families are recruited when the proband is affected. Calculate the conditional probability that the relative is unaffected.

First calculate the probability of recruitment by summing across the two latent states:

$$ P(S)=0.25(0.80)+0.75(0.10)=0.20+0.075=0.275. $$

The joint probability that the proband is affected and the relative is unaffected is

P(S,\text{relative unaffected}) =0.25(0.80)(0.20)+0.75(0.10)(0.90) =0.0400+0.0675=0.1075.

Therefore,

P(\text{relative unaffected}\mid S) =\frac{0.1075}{0.275} =0.3909.

The answer is not the population-average unaffected probability $0.25(0.20)+0.75(0.90)=0.725$. Conditioning on an affected proband changes the mix of latent states among recruited families, so the relative’s phenotype distribution changes too. This model is a hypothetical mixture illustrating marginalization and conditioning.

Checking the Denominator

A useful check is to enumerate the phenotype patterns that meet the rule and confirm that their probabilities sum to the denominator. In the preceding two-person example, the qualifying patterns are “proband affected, relative affected” and “proband affected, relative unaffected.” Their probabilities are, respectively,

P(S,\text{relative affected}) =0.25(0.80)(0.80)+0.75(0.10)(0.10) =0.1675,

and $0.1075$ for an unaffected relative. Their sum is $0.1675+0.1075=0.275=P(S)$. After conditioning, the two outcomes have probabilities $0.1675/0.275=0.6091$ and $0.1075/0.275=0.3909$, which sum to 1. This normalization check can reveal omitted qualifying patterns or an incorrectly calculated denominator.

Worked Example: How the Correction Can Change a Comparison

Return to the hypothetical independent-phenotype model with an affected proband required for recruitment. Suppose the observed recruited family has an affected proband and an unaffected relative. Compare the conditional likelihoods at $\theta=0.20$ and $\theta=0.40$.

The conditional likelihood is $1-\theta$, so the values are $0.80$ and $0.60$. Their likelihood ratio, comparing $\theta=0.20$ to $\theta=0.40$, is

\frac{L_{\mathrm{cond}}(0.20)}{L_{\mathrm{cond}}(0.40)} =\frac{0.80}{0.60}=1.333.

For these conditional data, the observation is 1.333 times as likely under $\theta=0.20$ as under $\theta=0.40$. If one instead compared the unconditioned pattern probabilities, they would be $0.20(0.80)=0.16$ and $0.40(0.60)=0.24$, giving $0.16/0.24=0.6667$. That comparison points in the opposite direction because it incorrectly treats the selected proband’s affected status as if it had been observed in an unselected family.

Common Pitfall and Practical Interpretation

A common mistake is to divide by the probability that the observed pattern occurred, or to divide each individual’s penetrance by a selection probability. Neither operation is the conditional likelihood. The denominator is the probability of the entire recruitment event for a pedigree, calculated under the same model and summed over all qualifying outcomes. The numerator remains the probability of the particular observed pedigree pattern.

The correction matters because families selected for a phenotype are enriched for outcomes associated with that phenotype, and potentially for latent genotype configurations that make it more likely. Consequently, the selected families do not generally follow the population distribution of family phenotypes. Conditioning accounts for this known feature of sampling; it does not establish that the genetic model is correct, nor does it make observational associations causal.

Key takeaway: For a pedigree recruited under event $S$, use the probability of its observed phenotype pattern divided by $P(S;\theta)$, with genotype uncertainty marginalized in both calculations. Define the recruitment event precisely, and verify that its probability includes every outcome that would have qualified a family.

Check Your Understanding

For each question, identify the conditioning event and explain which probabilities belong in the numerator and denominator.

A study recruits a family whenever a designated member is diagnosed. What event defines $S$, and why does the denominator include more than the one observed family pattern?
Explain why an unobserved genotype configuration must be marginalized when calculating both a pedigree phenotype probability and a recruitment probability.
In a model where the proband is affected with probability $\theta$, families are selected for an affected proband, and a relative is unaffected, derive the conditional likelihood.
What does it indicate if the probabilities of all qualifying phenotype patterns do not sum to the calculated $P(S;\theta)$?
Name one recruitment feature for which conditioning only on $S$ may be inadequate.