Likelihood Contributions from Genotype Calls

From Genotype Calls to a Likelihood

At a genetic marker with two alleles, an individual’s genotype can be one of three categories: $AA$, $Aa$, or $aa$. If the genotype is observed exactly, a model assigns a probability to each possible category. The likelihood describes how compatible the observed calls are with those model probabilities. Its essential building block is simple: for each individual, use the model probability of the genotype category that was actually observed.

The likelihood depends on the assumed probability model. It does not, by itself, describe uncertainty about the parameter after seeing the data. Keeping these roles separate is important: a model gives probabilities to possible data, while the observed data give a likelihood function over possible parameter values.

Definition: Let $G_i$ be the observed genotype category for individual $i$, and let $\theta$ denote a parameter or vector of parameters in a model. Write $\pi_g(\theta)=P(G_i=g\mid\theta)$ for the model probability of genotype category $g$. For the observed calls $g_1,\ldots,g_n$, the likelihood is the joint probability of those calls, viewed as a function of $\theta$. If individuals’ calls are independent under the model and share the same category probabilities, then $$ L(\theta;g_1,\ldots,g_n)=\prod_{i=1}^{n}\pi_{g_i}(\theta). $$

The model probabilities must be valid: each must be between 0 and 1, and the probabilities of all possible genotype categories must sum to 1. The likelihood is zero at a parameter value if the model assigns probability zero to any observed call. Otherwise, it is the product of the probabilities assigned to the observed categories.

In this tutorial, a genotype call is treated as exact: the recorded category is assumed to be the individual’s genotype. The probability model must also specify how those categories arise. A common starting assumption is that individuals are independent under the model and share the same genotype-category probabilities. Independence is a substantive assumption, not a mathematical consequence of having separate rows in a data table. Related individuals, population structure, or other dependence may require a different joint model.

Writing One Contribution at a Time

For a particular individual, the contribution is the probability of that individual’s observed category under the model. For example, if the observed call is $Aa$, the contribution is $\pi_{Aa}(\theta)$, not the sum of the probabilities of all three genotypes. The other categories were possible outcomes, but they were not observed for that individual.

With a shared probability model, each individual’s contribution uses the same set of probabilities, but the contribution selected depends on that person’s call. If a call is repeated, its probability appears repeatedly in the product. This product represents the probability of the complete ordered sequence of calls under independence. It is not a probability that the parameter equals a particular value.

Worked Example: Four Exact Calls Under a One-Parameter Model

Consider four hypothetical individuals with observed calls $AA$, $Aa$, $aa$, and $Aa$, in that order. Suppose a model with parameter $\theta$, where $0\leq\theta\leq1$, assigns probabilities $$ P(AA\mid\theta)=\theta,\qquad P(Aa\mid\theta)=\frac{1-\theta}{2},\qquad P(aa\mid\theta)=\frac{1-\theta}{2}. $$ These probabilities are nonnegative and sum to 1 for every allowed $\theta$. Assume the four calls are independent and exactly observed.

The contribution of the first call is $\theta$. Each of the next three calls has probability $(1-\theta)/2$. Therefore the likelihood is $$ L(\theta)=\theta\left(\frac{1-\theta}{2}\right)^3. $$

At $\theta=0.4$, the model assigns probabilities $0.4$, $0.3$, and $0.3$ to $AA$, $Aa$, and $aa$, respectively. Substituting these values gives $$ L(0.4)=0.4\times0.3\times0.3\times0.3=0.0108. $$ At $\theta=0.5$, the probabilities are $0.5$, $0.25$, and $0.25$, so $$ L(0.5)=0.5\times0.25\times0.25\times0.25=0.0078125. $$ The ratio $L(0.4)/L(0.5)=0.0108/0.0078125=1.3824$. For these calls, the likelihood at $0.4$ is 1.3824 times the likelihood at $0.5$. This is a relative comparison of the two model settings, not a statement that there is a 1.3824 probability that $\theta=0.4$.

Likelihood Is Not Probability of the Parameter

Before observing the calls, the model specifies probabilities for possible data at a chosen parameter value. After observing the calls, the data are fixed, and the likelihood records how their probability varies as the parameter varies. Thus, $L(\theta;g_1,\ldots,g_n)$ and $P(\theta\mid g_1,\ldots,g_n)$ are different objects. The former is a likelihood; the latter is a Bayesian posterior distribution and requires a prior distribution as well as a likelihood.

A likelihood need not sum or integrate to 1 over parameter values. Its numerical scale is also not intrinsically meaningful on its own: multiplying every likelihood value by the same positive constant leaves relative likelihood comparisons unchanged. For this reason, likelihood values are often interpreted comparatively, using ratios at specified parameter values or models.

Worked Example: Comparing Two Probability Models

Suppose the possible categories are $AA$, $Aa$, and $aa$. Four hypothetical individuals have calls $AA$, $aa$, $Aa$, and $AA$. Assume independent, exact calls and compare two fully specified models. Model A assigns probabilities $(0.5,0.3,0.2)$ to $(AA,Aa,aa)$; Model B assigns probabilities $(0.4,0.4,0.2)$. Each vector consists of nonnegative probabilities that sum to 1.

Under Model A, the four individual contributions are $0.5$, $0.2$, $0.3$, and $0.5$. Hence $$ L_A=0.5\times0.2\times0.3\times0.5=0.015. $$ Under Model B, the corresponding contributions are $0.4$, $0.2$, $0.4$, and $0.4$, giving $$ L_B=0.4\times0.2\times0.4\times0.4=0.0128. $$ The likelihood ratio is $L_A/L_B=0.015/0.0128=1.171875$. For these observed calls, Model A assigns a joint probability about 1.172 times that assigned by Model B. This comparison does not establish that Model A is true or give the probability that either model is true.

When Individuals Have Different Model Probabilities

The shared-probability formula is not the only possible setup. If the model includes known individual characteristics, such as a stratum label, the genotype probabilities may differ by individual. Write $\pi_{ig}(\theta)$ for the probability that individual $i$ has category $g$, conditional on the model and that individual’s characteristics. If calls remain conditionally independent, the likelihood is $$ L(\theta;g_1,\ldots,g_n)=\prod_{i=1}^{n}\pi_{i g_i}(\theta). $$ For each individual, the probabilities across all possible categories must still sum to 1. The contribution is the probability assigned to that individual’s observed call, using the probabilities appropriate to that individual.

Worked Example: Calls With Stratum-Specific Probabilities

Consider three hypothetical individuals. Individuals 1 and 3 are in stratum X; individual 2 is in stratum Y. The calls are $Aa$, $aa$, and $AA$, respectively. Suppose calls are independent conditional on stratum and parameter $\theta$. In stratum X, the probabilities for $(AA,Aa,aa)$ are $(0.5,\theta,0.5-\theta)$, with $0\leq\theta\leq0.5$. These sum to 1 and are nonnegative in the stated range. In stratum Y, the probabilities for the same ordered categories are $(0.2,0.5,0.3)$, which also sum to 1.

Individual 1 contributes $\theta$, individual 2 contributes $0.3$, and individual 3 contributes $0.5$. Thus $$ L(\theta)=\theta\times0.3\times0.5=0.15\theta. $$ At $\theta=0.2$, the likelihood is $$ L(0.2)=0.2\times0.3\times0.5=0.03. $$ At $\theta=0.3$, it is $$ L(0.3)=0.3\times0.3\times0.5=0.045. $$ For these data, the likelihood at $0.3$ is $0.045/0.03=1.5$ times the likelihood at $0.2$. The calculation uses each person’s stratum-specific probabilities; applying one shared probability vector to all three calls would not represent the model described.

Why the Product Formula Holds

The product follows directly from the definition of independence. For calls that are independent under the model, the probability of a particular sequence of calls equals the product of the probabilities for the calls in that sequence. Substituting the observed sequence gives the likelihood.

Specify the event.
The observed data are the joint event $G_1=g_1,\ldots,G_n=g_n$.

Apply independence.
Under independent calls with a shared category model, the joint probability factors as $P(G_1=g_1,\ldots,G_n=g_n\mid\theta)=\prod_{i=1}^{n}P(G_i=g_i\mid\theta)$.

Use the category probabilities.
By definition, $P(G_i=g_i\mid\theta)=\pi_{g_i}(\theta)$, so the joint probability becomes $\prod_{i=1}^{n}\pi_{g_i}(\theta)$.

View it as a function of the parameter.
Once the calls are observed, hold $g_1,\ldots,g_n$ fixed and let $\theta$ vary. The resulting function is $L(\theta;g_1,\ldots,g_n)$.

This justification relies on the stated independence assumption. If calls are dependent, the joint probability generally cannot be replaced by this product. One must instead specify and use the appropriate joint distribution. Likewise, if a genotype is uncertain rather than recorded as an exact category, the observed datum is not simply one of $AA$, $Aa$, or $aa$; its likelihood contribution must reflect the uncertainty in the observation.

Common Mistakes and Practical Checks

Genotype-call likelihoods matter because they connect a probability model to the actual pattern of individual observations. They also make assumptions visible: each contribution identifies what the model says about one person’s call, and the product makes clear where independence is being used.

Calling the likelihood a probability that a parameter is true. A likelihood value measures the probability assigned to the observed calls at a parameter value. It is not a posterior probability for that parameter; a posterior also requires a prior.
Using probabilities for unobserved categories in an individual’s contribution. If the call is $Aa$, use the model probability of $Aa$ for that individual. Do not add in the probabilities for $AA$ or $aa$.
Forgetting repeated contributions. If two individuals have the same call, each contributes its own probability. Under a shared model this means that the same factor appears twice in the product.
Assuming independence without stating it. Multiplication is justified by independence under the model. If individuals’ genotypes are dependent, the product may be wrong.
Confusing exact calls with uncertain observations. A product of genotype-category probabilities applies when the observed categories are treated as exact. If the data provide evidence about a genotype rather than an exact call, a different observation model is needed.

Key takeaway: For exact, independent genotype calls, multiply the model probability of each individual’s observed category. The resulting likelihood is a function of the model parameter for the fixed observed data; it is not a probability distribution for that parameter.

Check Your Understanding

Answer each question using the distinction between a model probability and a likelihood.

A model assigns probabilities $0.6$, $0.25$, and $0.15$ to $AA$, $Aa$, and $aa$. What is the contribution for an individual whose exact call is $Aa$?
Three independent calls are $AA$, $aa$, and $AA$. Write their likelihood using category probabilities $\pi_{AA}$ and $\pi_{aa}$.
In one or two sentences, explain why a likelihood value of $0.03$ is not a 3% probability that the parameter is correct.
What assumption justifies multiplying individual contributions, and what must change if the calls are dependent?
Suppose model probabilities vary by individual stratum. Which probability should be used for an individual’s contribution?