Why the Number-of-Variants Prior Matters
A fine-mapping analysis compares possible configurations of causal variants in a candidate region. Evidence from the data helps distinguish those configurations, but it does not determine which configurations were plausible before the data were observed. A prior on the number of causal variants makes this part of the model explicit: it assigns prior probability to configurations with zero, one, two, or more causal variants.
This is not the same choice as specifying prior probabilities for individual variants. A count prior first assigns probability to the total number of causal variants; a second rule then distributes that probability among configurations of that size. The two choices together determine the configuration prior used to combine with Bayes factors and calculate posterior inclusion probabilities.
A common choice is a uniform conditional prior: given \(K=k\), each of the \(\binom{m}{k}\) configurations has equal probability, so \(r(C\mid K=k)=1/\binom{m}{k}\). This makes the prior exchangeable across variants: before considering variant-specific information, relabeling the variants does not change their prior probabilities. Uniformity within each count is an assumption, not a consequence of specifying \(q_k\).
Turning a Count Prior Into Configuration Probabilities
The factor \(q_k\) is the total prior probability assigned to all configurations with \(k\) causal variants. Under the uniform conditional prior, that mass is divided equally among the \(\binom{m}{k}\) configurations. The count prior therefore controls how much probability is assigned to each model size, while the conditional prior determines how that probability is allocated within a model size.
Worked Example: A Count Prior With Four Candidate Variants
Suppose a hypothetical analysis has \(m=4\) candidate variants and uses \(q_0=0.20\), \(q_1=0.50\), \(q_2=0.30\), with \(q_3=q_4=0\). Assume configurations are uniform conditional on their count. There is one configuration of size zero, four of size one, and six of size two.
The empty configuration has prior probability \(0.20\). Each singleton configuration has probability \(0.50/\binom{4}{1}=0.50/4=0.125\). Each pair has probability \(0.30/\binom{4}{2}=0.30/6=0.0500\). For example, \(P(\{1,3\})=0.0500\), while \(P(\{2\})=0.125\).
The probabilities sum to \(0.20+4(0.125)+6(0.0500)=0.20+0.50+0.30=1.00\). Thus the prior assigns 20% probability to no causal variants, 50% to exactly one, and 30% to exactly two. It does not assign probability 50% to each individual variant being causal.
For a uniform conditional prior, the marginal prior inclusion probability of any particular variant is \(E[K]/m\). To see this, conditional on \(K=k\), each of the \(m\) variants has inclusion probability \(k/m\): the \(k\) included positions are spread symmetrically across the \(m\) variants. Averaging over \(K\) gives \(P(j\in C)=\sum_k q_k(k/m)=E[K]/m\).
From Configuration Priors to Posterior Inclusion Probabilities
Let \(D\) denote the observed data, and let \(B_C\) be the Bayes factor for configuration \(C\) against a common reference model. Take the reference configuration’s Bayes factor to be one. Assume the configurations being compared form an exhaustive, mutually exclusive set under the model, and that each \(B_C\) is calculated using the same data and reference model. Bayes’ rule then gives each configuration posterior probability in proportion to its prior probability times its Bayes factor.
The posterior probability of a count is the sum of the posterior probabilities of all configurations with that count. The posterior inclusion probability for variant \(j\), or PIP, is the sum over all configurations that include \(j\). A variant can therefore have posterior support across multiple configuration sizes; its PIP is not the posterior probability of just one configuration.
A useful way to understand the count prior’s influence is to group the evidence by count. Define the average Bayes factor among size-\(k\) configurations as \(A_k=\sum_{C:|C|=k}r(C\mid K=k)B_C\). Then the unnormalized posterior mass at count \(k\) is \(q_kA_k\). The data contribute \(A_k\), while the prior contributes \(q_k\); neither should be confused with the other.
Worked Example: How Two Count Priors Change PIPs
Consider a hypothetical region with \(m=2\) variants. Use four configurations: the empty set, \(\{1\}\), \(\{2\}\), and \(\{1,2\}\). Suppose their Bayes factors are, respectively, \(1, 8, 2,\) and \(4\). Assume a uniform conditional prior within each count.
First use \(q_0=0.50\), \(q_1=0.40\), and \(q_2=0.10\). The configuration priors are \(0.50\) for the empty set, \(0.20\) for each singleton, and \(0.10\) for the pair. Multiplying each prior by its Bayes factor gives unnormalized posterior weights \(0.50\), \(1.60\), \(0.40\), and \(0.40\). Their sum is \(2.90\).
The PIP for variant 1 is \((1.60+0.40)/2.90=2.00/2.90=0.690\). The PIP for variant 2 is \((0.40+0.40)/2.90=0.800/2.90=0.276\). The posterior count probabilities are \(0.50/2.90=0.172\) for \(K=0\), \((1.60+0.40)/2.90=0.690\) for \(K=1\), and \(0.40/2.90=0.138\) for \(K=2\).
Now change only the count prior to \(q_0=0.10\), \(q_1=0.50\), and \(q_2=0.40\). The configuration priors become \(0.10\), \(0.25\) for each singleton, and \(0.40\). The unnormalized weights are \(0.10\), \(2.00\), \(0.50\), and \(1.60\), with total \(4.20\). The PIPs are \(P(1\mid D)=(2.00+1.60)/4.20=0.857\) and \(P(2\mid D)=(0.50+1.60)/4.20=0.500\). The data evidence is unchanged, but assigning more prior mass to two-variant configurations raises both PIPs, especially the PIP of variant 2, which has less support from its singleton Bayes factor.
Alternative Prior Constructions
A familiar alternative is to assign each variant an independent Bernoulli prior with inclusion probability \(\pi\). This implies a count prior \(q_k=\binom{m}{k}\pi^k(1-\pi)^{m-k}\), and each particular size-\(k\) configuration has probability \(\pi^k(1-\pi)^{m-k}\). In this model \(E[K]=m\pi\), but the count probabilities are constrained to be binomial. An arbitrary count prior need not have that form.
The distinction matters because a count prior can concentrate probability on a chosen range or count without requiring the binomial shape. For example, a prior can assign most of its mass to one or two causal variants while keeping the probability of very large counts small. Conversely, a fixed-count prior sets \(q_k=1\) for one chosen \(k\), and zero elsewhere. That can be useful when the count is genuinely known, but it rules out every other count by assumption.
Worked Example: A Nonuniform Prior Within a Count
Suppose a hypothetical three-variant region has \(q_0=0.10\), \(q_1=0.60\), and \(q_2=0.30\). Conditional on \(K=1\), assign relative weights \(3:2:1\) to variants 1, 2, and 3; conditional on \(K=2\), use a uniform prior over the three pairs.
The singleton probabilities are \(0.60(3/6)=0.30\) for \(\{1\}\), \(0.60(2/6)=0.20\) for \(\{2\}\), and \(0.60(1/6)=0.10\) for \(\{3\}\). Each pair has probability \(0.30/3=0.10\), and the empty configuration has probability \(0.10\). The total is \(0.10+0.30+0.20+0.10+3(0.10)=1.00\).
The prior inclusion probabilities are \(0.30+0.10+0.10=0.50\) for variant 1, \(0.20+0.10+0.10=0.40\) for variant 2, and \(0.10+0.10+0.10=0.30\) for variant 3. Their sum is \(1.20\), which equals \(E[K]=1(0.60)+2(0.30)=1.20\). Unlike the uniform conditional prior, this construction gives variants different prior inclusion probabilities. Such unequal probabilities should be justified by appropriate prior information, not introduced accidentally.
Choosing and Checking the Prior
A defensible count prior makes its assumptions visible. Choose the candidate region and its \(m\) variants first, then specify which counts are possible and how much prior mass each receives. Consider whether the prior predictive implication is plausible for the region and analysis: for example, whether it gives substantial probability to implausibly large counts or rules out a count that the scientific question permits. If the prior is uniform within counts, state that assumption; if it is not, state the conditional allocation rule as well.
A few checks catch common implementation errors. The count probabilities must be nonnegative and sum to one. For each count with positive prior mass, the conditional configuration probabilities must sum to one. Summing configuration probabilities across all counts must then also give one. For a uniform conditional prior, every configuration of the same size must have probability \(q_k/\binom{m}{k}\). Finally, posterior configuration probabilities must normalize to one, and each PIP must lie between zero and one.
One common mistake is to take a desired count probability \(q_k\) and assign that probability to each configuration of size \(k\). That multiplies the intended mass by \(\binom{m}{k}\): the total prior mass at that count becomes \(\binom{m}{k}q_k\), not \(q_k\). The correction is to divide \(q_k\) among the size-\(k\) configurations, using \(q_k/\binom{m}{k}\) when the conditional prior is uniform.
Another mistake is to treat the expected count as a complete description of the prior. It is only one summary. Posterior count probabilities and PIPs depend on how prior mass is distributed across counts and on how configurations within each count are weighted. Comparing the prior and posterior count distributions, and recomputing PIPs under scientifically plausible alternative count priors, reveals whether conclusions depend strongly on this assumption. Such a comparison diagnoses sensitivity; it does not identify which prior is correct.
| Prior construction | What it specifies | Main implication |
|---|---|---|
| Uniform conditional count prior | \(q_k\), then equal probability among size-\(k\) configurations | Exchangeable variants; an arbitrary count distribution is allowed |
| Independent Bernoulli prior | One inclusion probability \(\pi\) for each variant | Induces a binomial count distribution and equal probabilities within each count |
| Fixed-count prior | Probability one on a chosen count | All other counts are excluded from posterior consideration |
No count prior compensates for a candidate set that omits relevant variants or a configuration model that does not represent the data-generating possibilities. Nor does a high PIP by itself establish causation: it is a posterior probability conditional on the data, likelihood, candidate set, and stated prior model. The count prior is consequential precisely because it is one part of that full specification.
Check Your Understanding
Work each question from the configuration probabilities, then compare with the solutions.
- With \(m=5\), \(q_1=0.40\), and a uniform conditional prior, what is the prior probability of a particular singleton configuration?
- For \(m=4\), a uniform conditional prior, and \(q_0=0.20\), \(q_1=0.50\), \(q_2=0.30\), find \(E[K]\) and the prior inclusion probability of each variant.
- Why does an independent Bernoulli prior with inclusion probability \(\pi\) imply a binomial distribution for \(K\)?
- In the two-variant worked example, why can changing \(q_2\) affect the PIP of variant 2 even though its singleton Bayes factor stays fixed?
Solutions. (1) There are \(\binom{5}{1}=5\) singleton configurations, so the probability of each is \(0.40/5=0.0800\). (2) \(E[K]=0(0.20)+1(0.50)+2(0.30)=1.10\). Uniformity gives each variant prior inclusion probability \(E[K]/4=1.10/4=0.275\). (3) Each of the \(m\) variants is included independently with probability \(\pi\); the probability of any particular size-\(k\) configuration is \(\pi^k(1-\pi)^{m-k}\), and there are \(\binom{m}{k}\) such configurations, giving \(P(K=k)=\binom{m}{k}\pi^k(1-\pi)^{m-k}\). (4) A PIP sums posterior probability across every configuration containing the variant. Changing \(q_2\) changes the prior weight of the pair configuration, and its Bayes factor contributes to both variants’ PIPs, even though the singleton evidence is unchanged.