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Likelihoods for Genetic Data · Tutorial 51 of 989

Factorizing Likelihoods Across Individuals and Families

Learn to identify the correct independent sampling units and construct likelihoods that preserve dependence among relatives.

Beginner 9 min read

What You'll Learn

  • State the conditional-independence assumptions needed to factor a likelihood across individuals or families.
  • Distinguish independent family-level contributions from dependent observations within a family.
  • Construct a joint family likelihood by summing over shared latent genotype configurations.
  • Check why multiplying separately marginalized relative-level likelihoods can give the wrong result.
  • Calculate likelihoods for hypothetical unrelated individuals and related family observations.

Why the Unit of Factorization Matters

A likelihood for genetic data is often written as a product. The product is justified only when its factors correspond to observations that are independent under the stated model. For unrelated individuals sampled independently, individual contributions may factor. For relatives, inheritance connects their genotypes, so treating each person as an independent observation can discard information about how their data arose jointly.

The key is to decide what the model says is independent, conditional on which parameters or latent genotypes. A family can be one independent sampling unit even though its members are dependent. Its contribution to the likelihood must then describe the observations of all included relatives together.

Definition: Let \(D_i\) denote the observed genetic data for individual \(i\), and let \(\theta\) denote the model parameters. The likelihood \(L(\theta;D)\) is the probability of the observed data, viewed as a function of \(\theta\). If individuals are independent conditional on \(\theta\), then \(L(\theta;D_1,\ldots,D_n)=\prod_{i=1}^{n}P(D_i\mid\theta)\). If the observations are grouped into families, the corresponding factorization is \(L(\theta;D_1,\ldots,D_F)=\prod_{f=1}^{F}P(D_f\mid\theta)\) only when the family data \(D_f\) are independent across families conditional on \(\theta\).

Sharing the same parameter \(\theta\) does not prevent conditional independence. For example, independent families can all have genotype probabilities determined by the same allele-frequency parameter. But relatives within one family usually share inheritance paths or unobserved genotypes. Their data cannot generally be treated as independent after those shared quantities have been left unobserved.

From Individuals to Family Contributions

Write \(D_f\) for all observed data in family \(f\), including whatever genotype calls, read data, phenotypes, or other measurements the model uses. Let \(G_f\) denote the family’s relevant true genotypes or other unobserved genetic states. A general family contribution is obtained by summing over the possible latent states:

$$ L_f(\theta;D_f) = P(D_f\mid\theta) = \sum_{g_f} P(D_f\mid G_f=g_f,\theta)\,P(G_f=g_f\mid\theta). $$

For continuous latent quantities, the sum is replaced by an integral. The distribution \(P(G_f\mid\theta)\) must represent the model for the family’s genetic states, including any founder-genotype probabilities and transmission probabilities that apply. The observation term \(P(D_f\mid G_f,\theta)\) describes how the observed data arise given those states.

Sometimes the observations within a family are independent conditional on the complete latent genotype configuration. In that case, the first probability in the sum can be written as a product over people or measurements. The sum over \(G_f\), however, must remain outside that product: the same latent family configuration governs multiple relatives.

$$ L_f(\theta;D_f) = \sum_{g_f} \left\{ \prod_{i\in f}P(D_{fi}\mid G_{fi}=g_{fi},\theta) \right\} P(G_f=g_f\mid\theta). $$

This is different from multiplying separately marginalized individual likelihoods. Such a multiplication would sum over each person’s latent genotype separately, thereby removing the shared family structure. It is appropriate only if the individual data are in fact independent under the model, or if the model’s conditional structure otherwise makes that factorization valid.

Worked Example: Factorizing Across Unrelated Individuals

Worked Example: Factorizing Across Unrelated Individuals

Suppose four unrelated individuals are sampled independently from a population in Hardy–Weinberg equilibrium at a biallelic locus. Let \(p=0.30\) be the frequency of allele \(A\). Their observed genotypes, in order, are \(AA,Aa,Aa,aa\). Calculate the likelihood of these observations under the stated model.

Under Hardy–Weinberg equilibrium, the genotype probabilities are \(P(AA)=p^2\), \(P(Aa)=2p(1-p)\), and \(P(aa)=(1-p)^2\). The assumptions relevant to the factorization are that the individuals were sampled independently and that their genotype probabilities are governed by the same fixed value of \(p\). Thus the joint likelihood is the product of the four genotype probabilities:

$$ L(p;D) = p^2\{2p(1-p)\}\{2p(1-p)\}(1-p)^2. $$

Substitute \(p=0.30\): \(P(AA)=0.30^2=0.09\), \(P(Aa)=2(0.30)(0.70)=0.42\), and \(P(aa)=0.70^2=0.49\). Therefore,

$$ L(0.30;D) = (0.09)(0.42)(0.42)(0.49) = 0.00777924. $$

This is the probability of this particular ordered sequence of four genotype observations under the model. The product is justified by independent sampling of the individuals, not simply by the fact that each person has a genotype. If the sampled people were relatives, the independence assumption would need to be reconsidered.

Worked Example: Multiplying Independent Family Contributions

Worked Example: Multiplying Independent Family Contributions

Consider two hypothetical, independently sampled families from a population in Hardy–Weinberg equilibrium with \(p=0.30\). Assume the families are unrelated to each other, are sampled without phenotype-based selection, and have no unmodeled shared source of dependence. In Family 1, both observed parents are \(Aa\), and their two children are \(Aa\) and \(aa\). In Family 2, the observed parents are \(AA\) and \(Aa\), and their child is \(Aa\). Calculate the likelihood of all observed genotypes.

For Family 1, each \(Aa\) founder has probability \(2p(1-p)=0.42\). Given two \(Aa\) parents, each child has probabilities \(1/4,1/2,1/4\) for \(AA,Aa,aa\), respectively. The two child transmissions are independent conditional on the parental genotypes. Thus the family contribution, including the observed founder genotypes, is:

$$ L_1 = (0.42)(0.42)(0.50)(0.25) = 0.02205. $$

For Family 2, the founder genotype probabilities are \(P(AA)=0.09\) and \(P(Aa)=0.42\). Given an \(AA\) parent and an \(Aa\) parent, the child has probability \(1/2\) of being \(Aa\). Therefore,

$$ L_2 = (0.09)(0.42)(0.50) = 0.01890. $$

The family records are independent by the assumptions in the example, so their contributions multiply:

$$ L = L_1L_2 = (0.02205)(0.01890) = 0.000416745. $$

This joint probability includes the observed founder genotypes as well as the children’s genotypes. If an analysis instead conditions on the observed parental genotypes, the founder-probability factors are omitted; that is a different likelihood, because it conditions on part of the data. Within either formulation, related children are modeled through their parental genotypes and transmission probabilities rather than treated as unrelated population draws.

A Shared Latent Genotype Prevents Naive Factorization

Uncertain genotype calls make the distinction especially visible. Suppose a parent’s true genotype is not observed, and two children’s data both carry information about it. Even when the children’s observations are independent conditional on that parental genotype, they may be dependent after the genotype is averaged over.

Worked Example: Marginalizing a Shared Family State

Use a deliberately simple hypothetical model. A family has one unobserved parental genetic state \(G\), with \(P(G=0)=P(G=1)=0.50\). For each of two children, \(Y_i=1\) denotes a specified observed genotype-call category and \(Y_i=0\) denotes another category. Conditional on \(G\), the children’s calls are independent, with \(P(Y_i=1\mid G=0)=0.20\) and \(P(Y_i=1\mid G=1)=0.80\). Suppose both calls are in category 1. Find their joint probability.

The correct calculation first forms the joint conditional probability for both calls, then averages over the shared state:

$$ P(Y_1=1,Y_2=1) = (0.50)(0.20)(0.20) + (0.50)(0.80)(0.80) = 0.34. $$

For either child separately, the marginal probability of category 1 is \((0.50)(0.20)+(0.50)(0.80)=0.50\). Multiplying those marginal probabilities would give \(0.50\times0.50=0.25\), not \(0.34\). The product is wrong because both children’s calls depend on the same unobserved \(G\). This toy model illustrates the shared-state calculation; its probabilities are stipulated for demonstration and are not a particular Mendelian cross.

Why the Family Formula Works

The family likelihood formula follows from the law of total probability. The possible values of \(G_f\) divide the ways the family data could arise into mutually exclusive cases. For each case, multiply the probability of the observed data conditional on that state by the probability of the state, then sum over all cases:

1
List the latent family states.
Include the genotype configurations or other unobserved states needed to represent the family’s inheritance and observation process.
2
Write the joint observation probability.
For each state, calculate the probability of all the family’s observed data. Factor this conditional probability across observations only when the model makes them conditionally independent given that state.
3
Weight and sum.
Multiply each conditional observation probability by the probability of its latent state, then sum over the possible states to obtain \(L_f(\theta;D_f)\).
4
Multiply independent families.
If family records are independent conditional on \(\theta\), multiply the completed family contributions. Do not split a family into person-level factors after marginalizing shared states unless that factorization is justified.

The derivation of the overall family product is direct. Conditional independence across families means the joint probability of their observed data equals the product of their probabilities:

$$ P(D_1,\ldots,D_F\mid\theta) = \prod_{f=1}^{F}P(D_f\mid\theta) = \prod_{f=1}^{F}L_f(\theta;D_f). $$

This result requires independence across the family units as modeled. For example, if two recorded families are connected through unrecorded relatives, or if all families share an unmodeled sampling mechanism that affects their data, the product may not be valid as written. A shared fixed parameter is not itself a problem: the product remains conditional on that same parameter value. If a shared latent quantity is instead integrated out, the resulting family records can become dependent, just as the two children did in the previous example.

Common Mistake and Practical Checks

  • Wrong reasoning: “There are four people, so the likelihood is the product of four individual probabilities.” The number of people does not establish independence.
  • Correction: Identify the sampling units and the model’s conditional dependencies. If relatives share latent genotypes or transmission pathways, construct their joint family contribution first.
  • Wrong reasoning: “The children are independent given the parents, so their probabilities can be multiplied after the parents’ genotypes are made uncertain.” Conditional independence justifies a product before averaging over shared latent states, not generally after separate marginalization.
  • Correction: Keep the sum over the shared state outside the product of conditionally independent observation probabilities.
  • Wrong reasoning: “Every family contribution has to be identical.” Family sizes and observed genotypes can differ. Independence justifies multiplying their potentially different contributions; it does not make those contributions equal.
  • Correction: Define \(D_f\) for each family and calculate its contribution from the data and model for that family.
Key takeaway: Factor a likelihood across the units that are independent under the model. For related individuals, preserve their joint inheritance structure within a family contribution; sum over shared latent genotypes before multiplying contributions from independent families.

Check Your Understanding

For each question, state the relevant independence assumption or likelihood operation.

  1. Two unrelated individuals are sampled independently under a fixed genotype-frequency model. What condition permits their likelihood contributions to be multiplied?
  2. Two siblings’ observed genotype calls are independent conditional on their parents’ true genotypes. Why might their marginal call probabilities fail to multiply?
  3. A family has uncertain parental genotypes and observed data from three children. Where should the sum over parental genotype configurations appear relative to the product of conditional child-data probabilities?
  4. Several families share the same fixed allele-frequency parameter but are independently sampled. Does sharing that parameter alone prevent their family likelihoods from being multiplied? Explain.
  5. An analysis conditions on observed parental genotypes rather than modeling their population probabilities. Which factors in a family likelihood change, and why?

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