Posterior Genotypes For Uncertain Structural-Variant Calls

Why an Uncertain Genotype Is a Distribution

A structural-variant call need not identify one genotype with certainty. Sequencing evidence may be compatible with several copy-number states: for example, a sample may plausibly have one deleted copy or two. Reporting only the most likely state hides that ambiguity. A posterior distribution retains it and supports summaries that account for uncertainty, including the expected copy number and the variance of copy number under the stated genotype probabilities.

The calculations are simple, but their interpretation depends on the states being defined consistently. The examples below consider a diploid locus with baseline copy number two and a simple event on either homolog. A deletion can remove zero, one, or two copies; a duplication can add zero, one, or two copies. More complex structural events may require a different set of genotype states, but the same probability-weighted calculations apply once each state has a defined copy number.

Definition: Let $G$ denote the genotype state for one sample and one structural-variant event, and let $R$ denote that sample's sequencing or assay evidence. A posterior genotype probability for state $g$ is $P(G=g\mid R,M)$, conditional on the evidence and the specified genotype model $M$. For mutually exclusive and exhaustive states $g_1,\ldots,g_k$, write $q_j=P(G=g_j\mid R,M)$. Then $q_j\geq 0$ and $\sum_{j=1}^k q_j=1$.

The posterior probabilities are probabilities over possible genotype states for the sample under the stated model; they are not frequencies of those states in a population. They also depend on the evidence and model used to produce them. This tutorial takes the probabilities as given and focuses on interpreting and summarizing them. A probability such as 0.70 means that, conditional on the evidence and model, the posterior probability assigned to that state is 70%.

From Genotype States to Copy Number

Let $C=c(G)$ be the copy number associated with the genotype state. In the simple diploid deletion model, let $D$ be the number of homologs carrying the deletion. Then $D$ can be 0, 1, or 2, and the copy number is $C=2-D$. In the corresponding simple duplication model, let $A$ be the number of homologs carrying a one-copy gain. Then $A$ can be 0, 1, or 2, and $C=2+A$. These mappings are assumptions about the event representation; they are not universal mappings for every complex or multiallelic structural variant.

The expected copy number is the probability-weighted average of the copy numbers in the possible states. It summarizes the posterior distribution, but it need not itself be a possible observed copy-number state. The genotype variance measures how spread out the copy-number distribution is around that expected value.

Definition: Given posterior probabilities $q_j$ and copy numbers $c_j=c(g_j)$, the posterior expected copy number and posterior genotype variance are $$ E(C\mid R,M)=\sum_{j=1}^k q_jc_j,\qquad \operatorname{Var}(C\mid R,M)=\sum_{j=1}^k q_j\{c_j-E(C\mid R,M)\}^2. $$ Expected copy number is measured in copies; genotype variance is measured in copies squared. If the genotype is summarized instead by an event dosage that is a constant shift or sign reversal of copy number, its variance is unchanged.

For a deletion, expected deleted-homolog dosage is $E(D\mid R,M)=2-E(C\mid R,M)$. For the simple duplication defined above, expected gain dosage is $E(A\mid R,M)=E(C\mid R,M)-2$. These identities follow from the state mappings. For example, if $C=2-D$, then $E(C\mid R,M)=2-E(D\mid R,M)$. Subtracting a constant and changing a sign do not change variance, so $\operatorname{Var}(D\mid R,M)=\operatorname{Var}(C\mid R,M)$. The same reasoning gives $\operatorname{Var}(A\mid R,M)=\operatorname{Var}(C\mid R,M)$ for $C=2+A$.

Worked Example: An Uncertain Deletion Call

Suppose a hypothetical sample has posterior probabilities 0.10, 0.70, and 0.20 for deletion states with copy numbers 0, 1, and 2, respectively. The probabilities are nonnegative and sum to one: $0.10+0.70+0.20=1.00$. The state-to-copy-number mapping is the simple diploid deletion mapping described above.

Expected copy number. Multiply each possible copy number by its posterior probability and add:

E(C\mid R,M) =(0.10)(0)+(0.70)(1)+(0.20)(2) =0+0.70+0.40 =1.10\text{ copies}.

The expected deleted-homolog dosage is $2-1.10=0.90$. This is a probability-weighted dosage, not a claim that the sample physically has 1.10 copies or 0.90 deleted homologs. The modeled states remain 0, 1, and 2 copies.

Genotype variance. Use the expected copy number, 1.10, as the center:

\begin{aligned} \operatorname{Var}(C\mid R,M) &=(0.10)(0-1.10)^2+(0.70)(1-1.10)^2+(0.20)(2-1.10)^2\\ &=(0.10)(1.21)+(0.70)(0.01)+(0.20)(0.81)\\ &=0.121+0.007+0.162\\ &=0.290\text{ copies}^2. \end{aligned}

Thus, the posterior mean copy number is 1.10 copies and the posterior variance is 0.290 copies squared. Because deletion dosage is $D=2-C$, its expected value is 0.90 deleted homologs and its variance is also 0.290 copies squared.

A Useful Identity for Checking the Calculation

A second expression for the variance can make arithmetic easier, especially when there are several states. It also shows why calculating the weighted squared deviations is equivalent to subtracting the square of the mean from the weighted second moment.

Derivation: Let $\mu_C=E(C\mid R,M)=\sum_j q_jc_j$. Expand the squared deviation and use $\sum_jq_j=1$: $$ \begin{aligned} \operatorname{Var}(C\mid R,M) &=\sum_jq_j(c_j-\mu_C)^2\\ &=\sum_jq_j(c_j^2-2c_j\mu_C+\mu_C^2)\\ &=\sum_jq_jc_j^2-2\mu_C\sum_jq_jc_j+\mu_C^2\sum_jq_j\\ &=\sum_jq_jc_j^2-2\mu_C^2+\mu_C^2\\ &=\sum_jq_jc_j^2-\mu_C^2. \end{aligned} $$

The identity provides a convenient check, but it does not change the interpretation: variance is posterior spread around the posterior mean. In calculations, retain enough precision in the mean and intermediate values to avoid rounding error; round the final reported result to a sensible number of significant figures.

Worked Example: An Uncertain Duplication Call

For a hypothetical sample, suppose the posterior probabilities are 0.20, 0.50, and 0.30 for duplication states with 0, 1, and 2 gained homologs. Under the assumed simple duplication model, the corresponding copy numbers are 2, 3, and 4. The probabilities sum to $0.20+0.50+0.30=1.00$, and the mapping adds one copy for each affected homolog.

Expected copy number.

E(C\mid R,M) =(0.20)(2)+(0.50)(3)+(0.30)(4) =0.40+1.50+1.20 =3.10\text{ copies}.

The expected gain dosage is $3.10-2=1.10$ gained homologs. Again, this is an average over possible states, not a possible integer state for a single sample.

Genotype variance using deviations.

\begin{aligned} \operatorname{Var}(C\mid R,M) &=(0.20)(2-3.10)^2+(0.50)(3-3.10)^2+(0.30)(4-3.10)^2\\ &=(0.20)(1.21)+(0.50)(0.01)+(0.30)(0.81)\\ &=0.242+0.005+0.243\\ &=0.490\text{ copies}^2. \end{aligned}

Check with the second-moment identity. The weighted second moment is $(0.20)(2^2)+(0.50)(3^2)+(0.30)(4^2)=0.80+4.50+4.80=10.10$. Subtract the squared mean: $10.10-(3.10)^2=10.10-9.61=0.490$ copies squared, matching the direct calculation. Since gain dosage is copy number minus two, it has the same variance.

Equal Expected Copy Number Does Not Mean Equal Certainty

An expected copy number alone does not reveal how concentrated the posterior probabilities are. Two samples can have the same expected copy number while their probability distributions differ substantially. The variance distinguishes a posterior concentrated at one state from one spread across separated states.

Worked Example: Comparing Two Deletion Posteriors

Consider two hypothetical posterior distributions over deletion copy numbers 0, 1, and 2. In Sample A, the probabilities are 0.50, 0, and 0.50. In Sample B, they are 0, 1.00, and 0. The probabilities in each sample are nonnegative and sum to one, and both use the same diploid deletion states.

Sample A. Its expected copy number is $(0.50)(0)+(0)(1)+(0.50)(2)=1.00$ copy. Its variance is $(0.50)(0-1)^2+(0)(1-1)^2+(0.50)(2-1)^2=0.50+0+0.50=1.00$ copy squared.

Sample B. Its expected copy number is $(0)(0)+(1.00)(1)+(0)(2)=1.00$ copy. Its variance is $(0)(0-1)^2+(1.00)(1-1)^2+(0)(2-1)^2=0$ copies squared.

Both posterior means equal 1.00 copy, but they express very different uncertainty. Sample A assigns all probability to the two-copy-separated states 0 and 2, whereas Sample B assigns all probability to state 1. A summary that reports only the expected copy number would conceal this difference.

What the Variance Does—and Does Not—Measure

The posterior genotype variance describes uncertainty about one sample's copy number, conditional on that sample's evidence and the specified model. It is not the variation in copy number across a cohort, and it is not a standard error for a population mean. For cohort-level summaries, each participant has a posterior distribution; variation among participants and uncertainty within each participant are distinct concepts.

A high variance indicates that appreciable posterior probability lies on copy-number states away from the posterior mean. A variance of zero means the posterior places probability one on a single copy-number state. A low variance does not by itself establish that the call is correct: the posterior is conditional on the model and evidence, so limitations in either can affect its interpretation. Nor does a fractional expected copy number describe a fractional physical chromosome; it is a weighted summary of discrete possibilities.

The probabilities must refer to mutually exclusive, exhaustive states for the calculations above to be valid as stated. If a caller reports probabilities that do not sum to one because some states are omitted or because values are unnormalized scores, they cannot be used directly as a complete posterior distribution. First clarify what the reported values represent and whether they cover all modeled states. Also ensure that state labels and copy-number mappings refer to the same event representation; for example, do not combine probabilities for different structural events as if they were alternative genotypes of one event.

Common Mistake: Replacing the Distribution With Its Most Likely State

A common error is to choose the genotype with the largest posterior probability and then treat it as certain. In the deletion example with probabilities 0.10, 0.70, and 0.20 for copy numbers 0, 1, and 2, the most likely state is one copy. But that hard call discards the remaining 0.30 posterior probability and cannot convey the calculated variance of 0.290 copies squared. The correction is to retain the posterior distribution when uncertainty matters, and to report the expected dosage and variance as summaries alongside any hard-call label.

Do not call the posterior mean a physical state. A mean of 1.10 copies summarizes uncertainty across states; the modeled copy number for an individual is one of the discrete states.
Do not confuse posterior variance with cohort variation. The former is uncertainty for one sample conditional on its evidence and model.
Do not use a genotype code without a copy-number mapping. Define what each state means before calculating a weighted mean or variance.
Do not treat an incomplete probability list as exhaustive. Check that the state probabilities are nonnegative and sum to one.

Key takeaway: For a defined set of structural-variant genotype states, map each state to its copy number, weight those copy numbers by their posterior probabilities to obtain expected copy number, and calculate the probability-weighted squared deviations to obtain posterior genotype variance. The mean summarizes dosage; the variance preserves information about uncertainty that a hard call or mean alone can hide.

Check Your Understanding

Use the stated state mappings and show the arithmetic for each calculation.

A deletion has posterior probabilities 0.20, 0.60, and 0.20 for copy numbers 0, 1, and 2. What are its expected copy number and posterior genotype variance?
For a simple duplication, posterior probabilities for 0, 1, and 2 gained homologs are 0.40, 0.40, and 0.20. What is the expected copy number, and what is the expected gain dosage?
Explain why a posterior expected copy number of 1.35 does not mean the sample physically has 1.35 copies.
Two samples have the same expected copy number. What additional posterior summary can show that their genotype uncertainty differs?
Why is the genotype variance unchanged when a deletion is summarized by deleted-homolog dosage $D=2-C$ instead of copy number $C$?