Bernoulli Likelihoods for Genotype-Specific Penetrance

From Genotype-Specific Risk to a Likelihood

When disease status is binary, each individual’s outcome can be represented as $Y_i=1$ for affected and $Y_i=0$ for unaffected. If genotype is observed, the penetrance for genotype category $g$ is the probability of disease among individuals with that genotype. A likelihood combines these probabilities across the observed individuals to describe how compatible a proposed set of penetrances is with the data.

This tutorial conditions on the observed genotypes. It does not model how common each genotype is in the population. That distinction matters: the likelihood here concerns the phenotype outcomes given the genotypes, not the joint probability of genotypes and phenotypes.

Definition: Let $G_i$ be individual $i$'s observed genotype category and $Y_i\in\{0,1\}$ the binary phenotype. The genotype-specific penetrance is $p_g=P(Y_i=1\mid G_i=g)$. Thus $P(Y_i=0\mid G_i=g)=1-p_g$. Under independent Bernoulli outcomes conditional on the observed genotypes, the phenotype likelihood is $$ L(\{p_g\};\mathbf{y}\mid\mathbf{G}) =\prod_{i=1}^{n}p_{G_i}^{Y_i}(1-p_{G_i})^{1-Y_i}. $$

The formula gives a factor $p_g$ for each affected individual with genotype $g$, and a factor $1-p_g$ for each unaffected individual with that genotype. A likelihood is a function of the unknown penetrance parameters for the observed data; it is not a probability distribution over parameter values.

Likelihood From Genotype and Outcome Counts

For each genotype category $g$, let $n_{g1}$ be the number of affected individuals and $n_{g0}$ the number of unaffected individuals. The total with that genotype is $n_g=n_{g1}+n_{g0}$. Grouping individuals by genotype rewrites the likelihood as

L(\{p_g\}) =\prod_g p_g^{n_{g1}}(1-p_g)^{n_{g0}}.

If the data are recorded only as counts, the probability of those counts under independent Bernoulli outcomes includes a binomial coefficient for each genotype: $\prod_g {n_g\choose n_{g1}}p_g^{n_{g1}}(1-p_g)^{n_{g0}}$. The coefficient does not depend on the penetrances, so it does not change which penetrance values maximize the likelihood. The likelihood kernel, without these coefficients, is often the convenient form for estimation.

Taking logarithms turns the product into a sum. The log likelihood kernel is

\ell(\{p_g\}) =\sum_g\left[n_{g1}\log(p_g)+n_{g0}\log(1-p_g)\right].

For the separate-risk model, each genotype category has its own $p_g$. The likelihood then separates into one term per genotype, so the data for one genotype determine that category’s estimate without changing the estimate for another category.

Worked Example: Building a Likelihood From Individual Outcomes

Suppose four hypothetical participants have observed genotype and disease-status pairs $(G,Y)=(0,0),(1,1),(1,0),(2,1)$. For illustration, evaluate the likelihood at proposed penetrances $p_0=0.10$, $p_1=0.30$, and $p_2=0.60$. Assume each outcome is independent of the others conditional on the observed genotypes, and that the genotype and phenotype measurements are correct.

The first participant is unaffected with genotype 0, contributing $1-p_0=0.90$. The second is affected with genotype 1, contributing $p_1=0.30$. The third is unaffected with genotype 1, contributing $1-p_1=0.70$. The fourth is affected with genotype 2, contributing $p_2=0.60$. Therefore,

$$ L=(1-p_0)p_1(1-p_1)p_2 =(0.90)(0.30)(0.70)(0.60)=0.1134. $$

This value is the likelihood of the observed ordered outcome pattern at these proposed penetrances, conditional on the four observed genotypes. It is not the probability that these penetrances are correct. Other penetrance values would produce other likelihood values for the same data.

Estimating Separate Penetrances

When every genotype category has its own penetrance, the likelihood is maximized by the observed affected fraction within each category. This estimate has a direct interpretation: among the sampled individuals with genotype $g$, it is the proportion who were affected. The derivation makes clear both the result and its boundary cases.

Result: In the separate-risk model, if $n_g>0$, the maximum-likelihood estimate of $p_g$ is $$ \hat p_g=\frac{n_{g1}}{n_{g1}+n_{g0}}=\frac{n_{g1}}{n_g}. $$ If all individuals in a category are unaffected, the estimate is $0$; if all are affected, it is $1$.

To justify the result for a category with both affected and unaffected individuals, consider its log likelihood contribution, $n_{g1}\log(p_g)+n_{g0}\log(1-p_g)$. For $0<p_g<1$, its derivative is $n_{g1}/p_g-n_{g0}/(1-p_g)$. Setting the derivative to zero gives $n_{g1}(1-p_g)=n_{g0}p_g$, and hence $p_g=n_{g1}/(n_{g1}+n_{g0})$. Its second derivative is $-n_{g1}/p_g^2-n_{g0}/(1-p_g)^2$, which is negative, so this stationary point is a maximum. If one count is zero, the likelihood is maximized at the corresponding boundary, as described above.

Worked Example: Estimating Three Genotype-Specific Risks

In a hypothetical sample, genotype categories 0, 1, and 2 have the following binary outcomes. Assume that outcomes are independent conditional on genotype, penetrance is constant within each category, and participants were sampled in a way that permits these within-genotype disease proportions to estimate the risks of interest.

Genotype category	Affected, $n_{g1}$	Unaffected, $n_{g0}$	Total, $n_g$
0	8	32	40
1	15	25	40
2	24	16	40

The likelihood kernel is

L(p_0,p_1,p_2) =p_0^8(1-p_0)^{32}p_1^{15}(1-p_1)^{25}p_2^{24}(1-p_2)^{16}.

Applying the within-category case fraction gives $\hat p_0=8/(8+32)=8/40=0.200$, $\hat p_1=15/(15+25)=15/40=0.375$, and $\hat p_2=24/(24+16)=24/40=0.600$. The estimated disease probabilities are therefore 20.0%, 37.5%, and 60.0% for categories 0, 1, and 2, respectively.

Substituting these estimates into the likelihood kernel gives

\hat L =(0.200)^8(0.800)^{32}(0.375)^{15}(0.625)^{25} (0.600)^{24}(0.400)^{16}.

For a more readable numerical summary, the log likelihood kernel at these estimates is approximately $-20.016-26.463-26.920=-73.399$. The binomial coefficients are omitted from this log likelihood kernel; including them would change its numerical value but not the maximizing penetrance estimates. These estimates describe the sampled data under the stated sampling assumptions; they do not by themselves establish that genotype causes disease.

When a Model Shares Penetrance Parameters

A scientific model may constrain two or more genotype categories to have the same disease probability. For example, a dominant-effect model might assign one penetrance to a reference genotype and a shared penetrance to two other categories. The shared value must be estimated from the combined outcomes in all categories assigned to it; separate within-category fractions will generally differ.

If categories in a set $S$ share a common penetrance $p$, their likelihood contribution is $p^{\sum_{g\in S}n_{g1}}(1-p)^{\sum_{g\in S}n_{g0}}$. Thus the shared estimate is the affected count pooled across those categories divided by their pooled total. This follows by the same derivative argument, applied to the combined counts.

Worked Example: Estimating a Shared Dominant-Effect Penetrance

Consider hypothetical counts with 4 affected and 36 unaffected individuals in genotype category 0, 12 affected and 28 unaffected in category 1, and 18 affected and 22 unaffected in category 2. Suppose the specified model assigns $p_0$ to category 0 and a common penetrance $p_D$ to categories 1 and 2. This equality is a model assumption, not a fact established by the counts. Assume independent outcomes conditional on genotype and valid sampling for estimating the risks.

The likelihood kernel under this constraint is

L(p_0,p_D) =p_0^4(1-p_0)^{36} p_D^{12+18}(1-p_D)^{28+22}.

The reference estimate is $\hat p_0=4/(4+36)=4/40=0.100$. The shared group contains $12+18=30$ affected individuals and $28+22=50$ unaffected individuals, so

\hat p_D=\frac{30}{30+50}=\frac{30}{80}=0.375.

To verify the pooled estimate, the shared group’s log likelihood is $30\log(p_D)+50\log(1-p_D)$. Its derivative is $30/p_D-50/(1-p_D)$. Setting this equal to zero gives $30(1-p_D)=50p_D$, hence $p_D=30/80=0.375$; the second derivative is negative for $0<p_D<1$, confirming a maximum. The model’s fitted penetrance for both categories 1 and 2 is 37.5%, even though their separate observed fractions are $12/40=30.0\%$ and $18/40=45.0\%$. Pooling follows from the equality constraint, not from treating those observed fractions as identical.

Assumptions and Interpretation

A Bernoulli likelihood is useful only when its assumptions match the data structure. In the model above, each outcome is binary; the disease probability is constant within each genotype category; and outcomes are independent conditional on the observed genotypes and model parameters. The analysis also treats genotypes as observed correctly. If genotype calls are uncertain, directly assigning each participant to a single category does not represent that uncertainty.

The sampling design is also important. If individuals are sampled from a population in a way that represents disease prevalence within each genotype, the case fraction can estimate that genotype’s disease probability. In a case-control study, however, the numbers of cases and controls are often selected by design. The observed case fraction within a genotype then reflects the sampling scheme and is not, in general, the population penetrance. A likelihood for that design must account for how participants entered the sample before interpreting estimates as absolute risks.

Likelihoods also make the effect of sample size visible. A genotype category with few observations may have a case fraction that changes greatly when one outcome changes. A category with no affected individuals has an estimated penetrance of zero under the unconstrained model, but that boundary estimate does not demonstrate that the true disease probability is exactly zero. It indicates that the observed outcomes place the likelihood maximum at the boundary.

Common Mistake

Common mistake: Treating the likelihood as the probability that a proposed penetrance is true, or interpreting a case-control sample fraction as a population risk without considering the sampling design. Correction: The likelihood evaluates how well parameter values account for the observed outcomes; it does not assign probabilities to parameters. A within-genotype case fraction estimates population penetrance only when the sampling and model assumptions support that interpretation.

The central construction is simple: each affected outcome contributes its genotype-specific disease probability, and each unaffected outcome contributes its complement. Counts compress these contributions into powers, and the unconstrained estimate within a genotype category is its affected fraction. Any shared-risk estimate follows from the groups the model explicitly constrains to share a probability.

Check Your Understanding

Answer each question using the likelihood definitions and assumptions in this tutorial.

For a genotype category with 9 affected and 21 unaffected individuals, write its likelihood contribution and calculate the separate penetrance estimate.
Why do binomial coefficients appear when counts are treated as the data, and why do they not change the separate penetrance estimates?
A genotype category has 7 unaffected individuals and no affected individuals. What is the unconstrained maximum-likelihood estimate, and what should you avoid concluding from it?
Two genotype categories are constrained to share a penetrance. Explain which counts determine its estimate.
Why can a case fraction within genotype fail to estimate population penetrance in a case-control study?