Propagating Mapping Uncertainty into Gene Scores

Why Mapping Uncertainty Changes a Gene Score

A locus can contain several variants with evidence of causality and several plausible target genes. Even when the variant-level evidence is summarized by posterior probabilities, the link from each variant to a gene may remain uncertain. Assigning every variant only to its strongest-linked gene can hide that uncertainty and make a close, ambiguous mapping look more decisive than it is.

A probability-weighted score retains the full mapping distribution: each variant contributes to each gene in proportion to the probability of that mapping, while its contribution is also weighted by the variant’s causal probability. The result is an expected attribution score, not automatically a gene-level association test or a probability that the gene is causal.

Definition: For variant $i$, let $C_i=1$ indicate that the variant is causal, and let $M_i$ denote its mapped gene under a mapping model. Let $D$ denote the relevant association data and $A$ the relevant mapping evidence. Define $p_i=P(C_i=1\mid D,A)$ and $q_{ig}=P(M_i=g\mid C_i=1,D,A)$. The probability-weighted gene score is $$ T_g=\sum_{i=1}^{m}p_iq_{ig}. $$ When these probabilities are coherently conditioned on the same relevant evidence, $p_iq_{ig}=P(C_i=1,M_i=g\mid D,A)$. Thus $T_g$ is the expected number of causal variants assigned to gene $g$ under this model.

The conditioning matters. If a causal posterior $p_i$ is estimated using $D$ alone, while $q_{ig}$ is estimated using mapping evidence $A$, then their product is not automatically the joint probability given both sources of evidence. It is a product-based score unless the assumptions connecting the two analyses justify that interpretation. One coherent approach is to define both probabilities conditional on $D,A$, as above. Another is to state and defend an approximation under which the mapping evidence does not update the causal posterior and the mapping conditional is appropriate given the evidence used.

For each variant, the mapping probabilities should sum to one across the candidate genes if the mapping model assigns the variant to exactly one of those genes. If some variants may be unassigned, include an explicit unassigned category, such as $g=U$, rather than silently dropping that probability mass. Without an unassigned category, $\sum_g q_{ig}=1$ is an assumption about the mapping model, not a consequence of the score formula.

What the Score Means

Linearity of expectation gives the score a direct interpretation. For a given gene, the contribution from variant $i$ is the joint probability that it is causal and maps to that gene; summing those contributions gives the expected count of causal variant-to-gene assignments. This interpretation does not require causal statuses or mapping assignments across variants to be independent.

The scale depends on the causal model. Under a model that permits exactly one causal variant and assigns all mapping probability to candidate genes, $\sum_i p_i=1$ and $\sum_g T_g=1$. In that case, $T_g$ is the posterior probability that the causal variant maps to gene $g$. Under a model allowing multiple causal variants, $T_g$ is an expected count and can exceed one. It is not, in general, the probability that at least one causal variant maps to $g$; calculating that probability requires the relevant joint distribution of causal and mapping assignments.

Worked Example: One Variant With an Ambiguous Mapping

Suppose one variant has posterior causal probability $p_1=0.30$, conditional on both the association and mapping evidence. Conditional on being causal, its mapping probabilities are $0.60$ for gene A, $0.30$ for gene B, and $0.10$ for gene C. These probabilities sum to one, so the mapping model assigns the variant among the three genes.

The gene contributions are the causal probability multiplied by the corresponding mapping probability:

T_A=0.30(0.60)=0.18,\qquad T_B=0.30(0.30)=0.09,\qquad T_C=0.30(0.10)=0.03.

For example, $T_A=0.18$ is the joint posterior probability that this variant is causal and maps to gene A, under the stated conditioning. The three scores sum to $0.30$, the variant’s causal probability. The remaining $0.70$ is posterior probability that the variant is not causal; it is not missing mapping probability.

Deriving the Probability-Weighted Score

The construction can be justified by first considering a random attribution indicator. Let $Z_{ig}=1$ if variant $i$ is both causal and assigned to gene $g$, and $Z_{ig}=0$ otherwise. Then $E(Z_{ig}\mid D,A)=p_iq_{ig}$, using the conditional definitions above. The total number of causal variants assigned to gene $g$ is $N_g=\sum_i Z_{ig}$. Therefore,

E(N_g\mid D,A) =E\left(\sum_{i=1}^{m}Z_{ig}\,\middle|\,D,A\right) =\sum_{i=1}^{m}E(Z_{ig}\mid D,A) =\sum_{i=1}^{m}p_iq_{ig} =T_g.

The second equality uses linearity of expectation, which holds whether or not the indicators $Z_{ig}$ are dependent. This is why dependence across variants does not invalidate the expected score. It can, however, matter for the uncertainty around a random count or score.

A common alternative is the best-link score. Let $b_i$ be the gene with the largest $q_{ig}$ for variant $i$, using a prespecified rule to resolve ties. The best-link score for gene $g$ is

H_g=\sum_{i=1}^{m}p_i\,\mathbf{1}(b_i=g),

where $\mathbf{1}(b_i=g)$ is one if the best-linked gene is $g$ and zero otherwise. This score preserves the causal weights but replaces each mapping distribution with a single chosen gene. It is useful as a simple summary, but it discards the distinction between a nearly certain link and a weak plurality.

Worked Example: Comparing Weighted and Best-Link Scores at a Locus

Consider a hypothetical locus with three variants and three candidate genes. The causal posterior probabilities are $p_1=0.50$, $p_2=0.30$, and $p_3=0.20$, which sum to one under a single-causal-variant model. Each row below gives the conditional mapping probabilities for a causal variant.

Variant	$p_i$	$q_{iA}$	$q_{iB}$	$q_{iC}$	Best-linked gene
1	0.50	0.50	0.30	0.20	A
2	0.30	0.10	0.80	0.10	B
3	0.20	0.20	0.20	0.60	C

Each row of mapping probabilities sums to one. For gene A, the probability-weighted score is

$$ T_A=(0.50)(0.50)+(0.30)(0.10)+(0.20)(0.20) =0.25+0.03+0.04=0.32. $$

For genes B and C, the calculations are

$$ T_B=(0.50)(0.30)+(0.30)(0.80)+(0.20)(0.20) =0.15+0.24+0.04=0.43, $$

$$ T_C=(0.50)(0.20)+(0.30)(0.10)+(0.20)(0.60) =0.10+0.03+0.12=0.25. $$

The scores sum to $0.32+0.43+0.25=1.00$, consistent with the single-causal-variant model and exhaustive mapping. The best-link scores instead assign variant 1 to A, variant 2 to B, and variant 3 to C, giving $H_A=0.50$, $H_B=0.30$, and $H_C=0.20$. The ranking changes: the weighted score ranks B above A, while the best-link score ranks A above B. This happens because variant 2 has a strong mapping to B, and the other variants retain some probability of mapping to B as well. Neither score proves that a gene is causal; they summarize the specified posterior and mapping model.

Mapping-Uncertainty Variance

A score can also be used to describe variability across possible mapping assignments. To isolate mapping uncertainty, hold the weights $p_i$ fixed and let $I_{ig}$ indicate whether variant $i$'s sampled mapping assignment is gene $g$. Define the random mapping score $\widetilde T_g=\sum_i p_i I_{ig}$, where $P(I_{ig}=1\mid D,A)=q_{ig}$. This is a distribution over scores induced by mapping assignments; its expectation is $T_g$.

E(\widetilde T_g\mid D,A)=\sum_i p_iq_{ig}=T_g.

The variance is

\operatorname{Var}(\widetilde T_g\mid D,A) =\sum_i p_i^2q_{ig}(1-q_{ig}) +2\sum_{i<j}p_ip_j\operatorname{Cov}(I_{ig},I_{jg}\mid D,A).

The first sum follows because a binary indicator with probability $q_{ig}$ has variance $q_{ig}(1-q_{ig})$. The second sum is the covariance contribution from pairs of mapping indicators. If mapping assignments are independent across variants, these covariances are zero. If they are dependent, dropping the covariance terms is not justified. This statement concerns dependence among the mapping indicators in this random mapping score; dependence among causal statuses does not, by itself, establish dependence among those indicators.

Worked Example: Why Mapping Dependence Affects Variance

Suppose two variants have fixed weights $p_1=0.60$ and $p_2=0.40$. For a particular gene G, each variant has mapping probability $q_{iG}=0.50$. First assume their mapping indicators are independent. The expected mapping score is $0.60(0.50)+0.40(0.50)=0.50$, and its variance is

\operatorname{Var}(\widetilde T_G) =(0.60)^2(0.50)(0.50)+(0.40)^2(0.50)(0.50) =0.09+0.04=0.13.

Now suppose mapping evidence makes the indicators perfectly positively dependent: both variants map to G together with probability $0.50$, and neither maps to G with probability $0.50$. Each marginal mapping probability remains $0.50$, but $\operatorname{Cov}(I_{1G},I_{2G})=0.25$. The variance is then

\operatorname{Var}(\widetilde T_G) =0.13+2(0.60)(0.40)(0.25) =0.13+0.12=0.25.

The expected score remains $0.50$, but the variance increases from $0.13$ to $0.25$, with standard deviation $0.50$. The change arises from dependence between the mapping indicators, not from any claim about dependence between the variants’ causal statuses.

Practical Interpretation and Common Pitfalls

The probability-weighted score is most informative when the causal probabilities and mapping probabilities have clear, compatible conditioning and when the candidate-gene set represents the mapping possibilities adequately. Report how the probabilities were constructed, whether an unassigned category was allowed, and whether the causal model permits one or multiple causal variants. When comparing scores from alternative mapping models, keep the causal weights and candidate-gene universe fixed if the goal is to isolate the effect of mapping uncertainty.

Do not treat a product as a joint probability without compatible conditioning. If $p_i$ conditions only on association data and $q_{ig}$ on separate mapping data, their product may be an approximation. Define the joint conditioning or label the result a product-based score.
Do not interpret an expected count as a probability. With multiple causal variants, $T_g$ can exceed one. A probability that at least one causal variant maps to a gene requires a joint probability calculation.
Do not confuse best-link assignment with certainty. Choosing the largest $q_{ig}$ does not make that mapping certain; it only selects the most probable single link.
Do not omit covariance automatically from a variance calculation. Covariance terms can be dropped only when the mapping indicators in question are independent, or when a justified model otherwise makes those terms zero.

A useful reporting practice is to present the probability-weighted ranking alongside the best-link ranking and the per-variant mapping probabilities that drive their differences. A disagreement between the rankings is a diagnostic of mapping ambiguity, not necessarily an error in either calculation. It identifies genes whose relative support depends on whether uncertainty is retained or collapsed to one link.

Key takeaway: Weight each variant’s causal probability by its conditional mapping probability and sum across variants. The resulting score is an expected causal attribution under the specified model; its probability interpretation and uncertainty depend on the causal model, the conditioning assumptions, and the dependence structure of mapping assignments.

Check Your Understanding

Use the definitions and assumptions in the tutorial to answer these questions.

A variant has $p_i=0.40$, with mapping probabilities $0.50$ to gene A and $0.50$ to gene B. What are its contributions to the two probability-weighted gene scores, and what assumptions permit interpreting them as joint probabilities?
Under a single-causal-variant model with exhaustive mapping, why do the gene scores sum to one?
In a model allowing multiple causal variants, what does $T_g=1.3$ mean, and why is it not automatically a 1.3 probability?
What information does a best-link score discard compared with a probability-weighted score?
For the mapping-score variance, which indicators must be independent to set the covariance terms to zero?