Association Likelihoods With Uncertain Structural-Variant Genotypes

Why Association Likelihoods Must Retain Genotype Uncertainty

A structural-variant call is not always a known copy-number state or carrier status. Sequencing evidence may support several states with different probabilities. Replacing that uncertainty with one selected state can change both the association estimate and its uncertainty. A likelihood can instead average the phenotype model over the possible genotypes, weighted by their posterior probabilities.

The weights must match the conditioning in the phenotype model. In particular, if the phenotype model conditions on covariates $C_i$, the relevant weight is the probability of genotype given both the calling evidence and those covariates. A posterior given calling evidence alone is not generally interchangeable with this conditional posterior.

Definition: Let $G_i$ be the structural-variant genotype or copy-number state for individual $i$, $R_i$ the calling evidence, and $C_i$ the modeled covariates. Define $q_{ig}=P(G_i=g\mid R_i,C_i)$, the posterior probability that individual $i$ has state $g$, conditional on the evidence and covariates. The weights satisfy $q_{ig}\geq 0$ and $\sum_g q_{ig}=1$. If the phenotype density under state $g$ is $f(Y_i\mid G_i=g,C_i;\theta)$, the conditional phenotype likelihood contribution is $$ L_i(\theta)=\sum_g q_{ig} f(Y_i\mid G_i=g,C_i;\theta). $$

This is a likelihood for the phenotype conditional on the observed calling evidence and covariates. Its use assumes that, given genotype and covariates, the phenotype is conditionally independent of the calling evidence: $Y_i\perp R_i\mid G_i,C_i$. The posterior weights must come from an appropriate genotype-calling model and the relevant covariate-conditional genotype distribution. In this likelihood, they are treated as fixed while estimating the phenotype-model parameters $\theta$.

If individuals are independent under the model, the likelihood for the sample is the product of the individual contributions. Its log is the sum of their log likelihoods. For a continuous phenotype, each density value is evaluated at the observed $Y_i$; for a binary phenotype, the density is replaced by the appropriate probability mass.

Mixture Likelihood, Not a Certain Genotype

Each possible genotype state implies a phenotype distribution. The observed phenotype likelihood is a weighted sum of those distributions, not a choice of the most likely state. For example, a carrier probability of 0.6 does not make the individual a certain carrier: both the carrier and noncarrier phenotype models contribute to the likelihood.

Worked Example: Evaluating One Uncertain-Call Likelihood Contribution

Suppose a hypothetical individual has posterior probabilities $0.2, 0.5,$ and $0.3$ for copy-number states $g=0,1,2$, respectively. Let the phenotype model be $Y_i\mid G_i=g\sim N(\beta g,1)$, with $\beta=1$, and suppose the observed phenotype is $y_i=1$. The variance is one, so the normal density is $\phi(y-\beta g)$, where $\phi$ is the standard normal density.

The likelihood contribution is $$ L_i(1)=0.2\phi(1)+0.5\phi(0)+0.3\phi(-1). $$

Since $\phi(1)=\phi(-1)\approx 0.2420$ and $\phi(0)\approx 0.3989$, substitution gives $$ L_i(1)=0.2(0.2420)+0.5(0.3989)+0.3(0.2420)\approx 0.3205. $$

The posterior mean dosage is $0(0.2)+1(0.5)+2(0.3)=1.1$. Substituting that dosage as if it were known would give $\phi(1-1.1)=\phi(-0.1)\approx 0.3970$, which is not the mixture likelihood $0.3205$. The difference arises because the phenotype distribution is averaged over distinct genotype states; evaluating it at an average dosage is a different calculation.

Covariates Must Be Included in the Genotype Weights

When covariates enter the phenotype model, the relevant posterior is $P(G_i=g\mid R_i,C_i)$. The genotype posterior from calling evidence alone, $P(G_i=g\mid R_i)$, can differ if the covariates carry information about genotype. Bayes' rule makes the dependence explicit: the covariate-conditional posterior combines the calling evidence with the genotype distribution conditional on $C_i$.

Worked Example: A Covariate Changes the Posterior Weight

Consider a hypothetical binary genotype $G\in\{0,1\}$. Suppose the calling evidence $R$ is uninformative, and the population has $P(G=1)=0.5$. A covariate $C$ is associated with genotype, with $G=C$ in this simplified example. For an individual with $C=1$, the posterior based on $R$ alone is $P(G=1\mid R)=0.5$, but the posterior conditional on both $R$ and $C$ is $P(G=1\mid R,C=1)=1$.

To see why this matters, use the illustrative phenotype model $Y\mid G\sim N(G,1)$, and suppose $y=1$. Using the correct conditional weight gives the likelihood contribution $\phi(1-1)=\phi(0)\approx 0.3989$. Using the posterior based on $R$ alone gives $$ 0.5\phi(1)+0.5\phi(0)\approx 0.3205. $$

These are different phenotype likelihoods. This example illustrates the weighting principle, not a setting in which a genotype effect could be separated from a perfectly aligned covariate effect. In an association analysis, the model and data must also permit the genotype coefficient to be identified.

The example also shows why conditional independence alone does not justify omitting covariates from the posterior weights. The assumption $Y_i\perp R_i\mid G_i,C_i$ permits the phenotype density to be written without $R_i$; it does not imply $P(G_i\mid R_i,C_i)=P(G_i\mid R_i)$.

Estimating an Association Effect With Uncertain Calls

A simple continuous-outcome model makes the estimation procedure concrete. Suppose $Y_i\mid G_i=g\sim N(\beta g,1)$, where the intercept is fixed at zero, the residual variance is known to be one, and $\beta$ is the phenotype change per copy of the modeled genotype dosage. For an uncertain genotype, its contribution is a mixture of normal densities. The likelihood estimate maximizes the product of these contributions over $\beta$.

For a likelihood with fixed posterior weights, differentiation gives an interpretable score. Define $w_{ig}(\theta)=q_{ig}f(Y_i\mid G_i=g,C_i;\theta)/L_i(\theta)$. These quantities sum to one and are the relative contributions of the genotype states to the phenotype likelihood at the current parameter value.

\frac{\partial \log L_i(\theta)}{\partial\theta} = \sum_g w_{ig}(\theta) \frac{\partial\log f(Y_i\mid G_i=g,C_i;\theta)}{\partial\theta}.

To justify this expression, differentiate the sum defining $L_i$. Since the weights $q_{ig}$ are fixed with respect to $\theta$, $\partial L_i/\partial\theta=\sum_g q_{ig}\,\partial f_g/\partial\theta$, where $f_g=f(Y_i\mid G_i=g,C_i;\theta)$. Then use $\partial f_g/\partial\theta=f_g\,\partial\log f_g/\partial\theta$ and divide by $L_i$. This yields the weighted score above. Thus the score averages the genotype-specific scores using weights that also depend on the observed phenotype and current model parameters.

The likelihood curvature also reflects uncertainty. The observed information for an individual is the posterior-weighted average of the genotype-specific information minus the posterior-weighted variance of their scores. Consequently, uncertainty does not imply one universal adjustment to a standard error: the fitted likelihood and its curvature determine the result.

-\frac{\partial^2\log L_i}{\partial\theta^2} = \sum_g w_{ig}\left(-\frac{\partial^2\log f_g}{\partial\theta^2}\right) - \left[ \sum_g w_{ig}s_{ig}^2- \left(\sum_g w_{ig}s_{ig}\right)^2 \right],

where $s_{ig}=\partial\log f_g/\partial\theta$. The bracketed term is the weighted variance of the genotype-specific scores. For a regular one-parameter likelihood, an approximate standard error based on observed information is the reciprocal square root of the negative second derivative of the total log likelihood at its maximum.

Worked Example: Comparing an Uncertain Call With a Hard Call

Consider two hypothetical individuals under the model $Y_i\mid G_i=g\sim N(\beta g,1)$, with known variance one and intercept fixed at zero. Their observed phenotypes are $y_1=1$ and $y_2=0$. The second individual is a known carrier, $q_{2,1}=1$. The first has $q_{1,1}=0.5$ and $q_{1,0}=0.5$. Other genotype states have probability zero.

For the first individual, the likelihood contribution is $0.5\phi(1)+0.5\phi(1-\beta)$; for the second, it is $\phi(-\beta)$. Let $w$ be the relative likelihood weight of the carrier component for the first individual: $$ w(\beta)=\frac{0.5\phi(1-\beta)} {0.5\phi(1)+0.5\phi(1-\beta)}. $$

The total log-likelihood score is $w(\beta)(1-\beta)-\beta$. Solving the score equation numerically gives $\hat{\beta}\approx 0.365$. At this estimate, $w\approx 0.574$, and $0.574(1-0.365)-0.365\approx 0$, as required.

For the uncertain individual, the observed information is $w-w(1-w)(1-\beta)^2$. At the estimate this is approximately $0.574-(0.574)(0.426)(0.635)^2\approx 0.475$. The known carrier contributes information one, so total observed information is about $1.475$. The approximate standard error is therefore $1/\sqrt{1.475}\approx 0.823$ phenotype units per copy.

Now replace the uncertain call by a hard carrier call, using the threshold convention that a posterior probability of 0.5 is called carrier. Both individuals are then coded $G=1$. The log likelihood is proportional to $-\{(1-\beta)^2+(0-\beta)^2\}/2$; its maximum is $\hat{\beta}=(1+0)/2=0.500$, and its standard error is $1/\sqrt{2}\approx 0.707$ phenotype units per copy. In this example, hard calling changes the estimate from 0.365 to 0.500 and makes the estimated standard error smaller. The direction and size of these differences depend on the data and model; the example does not establish a universal direction of bias.

Practical Interpretation and Common Pitfalls

The likelihood approach retains information in the full posterior distribution over genotype states. It is especially relevant when several structural-variant states are plausible or when a hard threshold would discard meaningful uncertainty. The fitted coefficient still describes an association under the specified phenotype model; incorporating genotype uncertainty does not, by itself, establish a causal effect.

Using the wrong conditional posterior. If the phenotype model conditions on $C_i$, use $P(G_i=g\mid R_i,C_i)$, unless a justified assumption makes it equal to $P(G_i=g\mid R_i)$.
Using the most likely state as certainty. A maximum-posterior call discards the other states and their probabilities. That is a different likelihood analysis.
Plugging in mean dosage as though known. In general, evaluating a phenotype density at posterior mean dosage does not equal averaging that density over genotype states.
Assuming uncertainty always inflates standard errors by a fixed amount. The relevant standard error comes from the curvature of the fitted likelihood; it depends on the observed phenotypes, posterior weights, and model.

Key takeaway: For phenotype association conditional on calling evidence and covariates, sum the genotype-specific phenotype likelihoods using $q_{ig}=P(G_i=g\mid R_i,C_i)$. Estimate effects and their uncertainty from this mixture likelihood rather than treating uncertain calls as known states.

Check Your Understanding

Try these questions using the definitions and likelihoods above.

Why can $P(G_i=g\mid R_i,C_i)$ differ from $P(G_i=g\mid R_i)$, even when phenotype and calling evidence are conditionally independent given genotype and covariates?
An individual has posterior probabilities 0.7 for state 0 and 0.3 for state 1. Write the phenotype likelihood contribution for a model with density $f(Y_i\mid G_i,C_i;\theta)$.
In what way does evaluating a phenotype density at posterior mean dosage differ from averaging that density over the possible genotypes?
For the two-person worked example, what estimate and standard error result from replacing the uncertain call by a carrier hard call?
What additional assumption allows the phenotype likelihood to omit calling evidence from the density after conditioning on genotype and covariates?