Why Priors Must Specify More Than a Single Variant
In fine-mapping, a candidate locus may contain several variants, and the statistical model may allow more than one of them to be causal. A prior must therefore answer two related questions: how many causal variants are plausible, and which particular configurations of variants are plausible? These choices matter because posterior comparisons combine the evidence in the data with the prior probability assigned to each configuration.
The focus here is the construction and consequences of these priors. We will treat the configuration-specific marginal likelihood, or its Bayes factor relative to a null configuration, as an input. The likelihood must be appropriate for the data and model; the prior formulas below do not determine it.
Separating the Count Prior From the Configuration Prior
A convenient way to construct a prior is in two stages. First assign probability \(\pi_k\) to each allowed count \(k\). Then, conditional on that count, assign probability \(q(S\mid k)\) to each configuration of size \(k\). The resulting prior probability for a specific configuration is
For this to be a probability distribution, \(\pi_k\geq 0\), the count probabilities must sum to one over the allowed counts, and for each \(k\), the conditional probabilities \(q(S\mid k)\) must sum to one over all configurations of size \(k\). If at most \(K_{\max}\) causal variants are allowed, then configurations with more than \(K_{\max}\) variants receive prior probability zero.
A common choice is to treat all configurations of a given size equally. There are \(\binom{m}{k}\) subsets of size \(k\), so this choice gives
This is a uniform-within-count prior, not a uniform prior over all configurations. The probability of each configuration depends both on the count prior and on how many configurations share its count. Assigning a large total probability to \(K=k\) spreads that probability across all \(\binom{m}{k}\) configurations of that size.
Worked Example: Assigning Prior Probability to Configurations
Suppose a hypothetical locus has \(m=4\) candidate variants and allows at most two causal variants. Let the count prior be \(\pi_0=0.50\), \(\pi_1=0.30\), and \(\pi_2=0.20\), with configurations equally likely within each count. Find the prior probability of the null configuration, a specified singleton, and a specified pair.
The null configuration is the only configuration with count zero, so \(P(\varnothing)=\pi_0=0.50\). There are \(\binom{4}{1}=4\) singleton configurations, so each receives \(P(\{1\})=0.30/4=0.0750\). There are \(\binom{4}{2}=6\) pairs, so a specified pair, such as \(\{1,2\}\), receives \(P(\{1,2\})=0.20/6=0.0333\).
These probabilities account for all allowed configurations: \(0.50+4(0.0750)+6(0.0333)\approx 1.00\), with the small discrepancy due to rounding. The prior expected count is \(E[K]=0(0.50)+1(0.30)+2(0.20)=0.700\) causal variants. By symmetry, the prior inclusion probability for any one variant is \(0.700/4=0.175\).
Alternative Ways to Construct the Prior
A fixed-count prior sets \(\pi_k=1\) for one specified value of \(k\) and zero for all others. For example, a single-causal-variant model fixes \(K=1\), while a two-causal-variant model fixes \(K=2\). Such a prior can be useful when the count is deliberately fixed for a particular analysis, but it rules out every other count by assumption. It does not express uncertainty about the count.
Another construction starts by assigning each variant an independent probability \(\rho\) of inclusion. Under independent Bernoulli inclusion, each specific configuration \(S\) has probability
The induced count distribution is binomial: \(P(K=k)=\binom{m}{k}\rho^k(1-\rho)^{m-k}\). Conditional on \(K=k\), configurations are uniform because each has the same probability \(\rho^k(1-\rho)^{m-k}\). Thus independent inclusion is one particular count prior combined with a uniform-within-count rule. It is not equivalent to choosing an arbitrary count prior. For instance, assigning equal probabilities to the counts \(0,1,\ldots,K_{\max}\) generally does not give independent variant-inclusion indicators.
When configurations are not to be treated equally, a conditional prior can instead assign different probabilities \(q(S\mid k)\), provided they are specified and normalized. The key requirement is transparency: the count prior and the within-count allocation together must give a proper probability to every allowed configuration. A prior should not be described merely as “sparse” without stating how that description translates into probabilities.
How Priors Enter Posterior Odds
Let \(D\) denote the observed genetic data. For each configuration \(S\), let \(m(D\mid S)\) be the marginal likelihood, obtained by integrating over the effect parameters under the model and their specified priors. Define its Bayes factor against the null configuration as \(BF_S=m(D\mid S)/m(D\mid\varnothing)\), so \(BF_{\varnothing}=1\). The posterior probability of a configuration is its prior probability multiplied by its marginal likelihood, then divided by the sum of those products over all allowed configurations. Equivalently, using Bayes factors,
For two configurations \(S\) and \(T\) with positive prior probabilities, their posterior odds have a particularly useful form:
The data contribute the Bayes-factor ratio; the prior contributes the prior-odds ratio. This identity follows directly by dividing the two posterior expressions: their common normalization cancels, as does the common null marginal likelihood used to define both Bayes factors. A large Bayes factor for \(S\) relative to \(T\) does not by itself guarantee that \(S\) has larger posterior probability: sufficiently small prior odds can offset that evidence.
Worked Example: Comparing a Singleton With a Pair
Consider a hypothetical locus with three variants, \(A\), \(B\), and \(C\), and allow counts zero, one, or two. Use the count prior \(\pi_0=0.50\), \(\pi_1=0.40\), and \(\pi_2=0.10\), uniform within each count. Suppose the configuration Bayes factors, relative to the null, are \(BF_A=12\), \(BF_B=4\), \(BF_C=1.5\), \(BF_{AB}=30\), \(BF_{AC}=2\), and \(BF_{BC}=1\).
There are three singletons and three pairs. Thus the prior probability of each singleton is \(0.40/3=0.1333\), and that of each pair is \(0.10/3=0.0333\). The prior odds of \(\{A,B\}\) against \(\{A\}\) are \(0.0333/0.1333=0.250\). The Bayes-factor ratio is \(30/12=2.50\), so
The pair has a larger Bayes factor, but its posterior odds against the singleton are only 0.625 to 1 because its prior probability was lower. To check by normalization, the unnormalized posterior weights \(P(S)BF_S\) are: null \(0.500\), \(A\) \(1.600\), \(B\) \(0.5333\), \(C\) \(0.2000\), \(AB\) \(1.000\), \(AC\) \(0.0667\), and \(BC\) \(0.0333\). Their sum is \(3.9333\). Therefore \(P(\{A\}\mid D)=1.600/3.9333=0.4068\) and \(P(\{A,B\}\mid D)=1.000/3.9333=0.2542\); their ratio is approximately \(0.625\), as calculated above.
Count-Prior Choices Can Change the Comparison
The effect of the count prior is especially clear when two competing configurations have different sizes. Under a uniform-within-count prior, the prior odds for a specified configuration \(S\) of size \(k\) against one \(T\) of size \(\ell\) are
The combinatorial factor is necessary: the count probability is divided among all configurations of that size. This is one reason why comparing only \(\pi_k\) and \(\pi_\ell\) does not give the prior odds of two particular configurations.
Worked Example: Changing the Count Prior Changes Posterior Odds
Suppose a hypothetical locus has five variants, and compare a specified singleton \(S\) with a specified pair \(T\). Let their Bayes factors be \(BF_S=10\) and \(BF_T=8\). The evidence ratio \(BF_S/BF_T=1.25\) is held fixed. Compare two count priors, each allowing counts zero, one, and two, with configurations uniform within count.
Under Prior I, let \((\pi_0,\pi_1,\pi_2)=(0.70,0.20,0.10)\). The specified singleton has prior probability \(0.20/\binom{5}{1}=0.0400\), and the specified pair has prior probability \(0.10/\binom{5}{2}=0.0100\). The prior odds are \(4.00\), so the posterior odds for \(S\) against \(T\) are \(1.25(4.00)=5.00\).
Under Prior II, let \((\pi_0,\pi_1,\pi_2)=(0.70,0.05,0.25)\). Now the specified singleton has prior probability \(0.05/5=0.0100\), and the specified pair has prior probability \(0.25/10=0.0250\). The prior odds are \(0.400\), giving posterior odds \(1.25(0.400)=0.500\) for \(S\) against \(T\). The same data favor the singleton by a Bayes-factor ratio of 1.25, but under Prior II the pair has twice the posterior probability of the singleton. The difference comes from the count prior, not from a change in the observed evidence.
Short Derivation and Practical Checks
The posterior-odds identity is a direct consequence of Bayes’ rule. For any allowed configuration \(S\), \(P(S\mid D)=m(D\mid S)P(S)/\sum_U m(D\mid U)P(U)\), where the sum is over all allowed configurations \(U\). For two configurations, dividing these expressions cancels the denominator:
The final equality holds because both Bayes factors use the same null marginal likelihood as their reference, so that reference cancels in their ratio. This argument requires that the compared configurations have well-defined marginal likelihoods and positive prior probabilities. A configuration assigned prior probability zero cannot acquire positive posterior probability from the data under that prior.
Before interpreting results, check that the count prior sums to one, that each conditional configuration prior sums to one within its count, and that the allowed configuration space is explicit. Then check whether the reported comparison is between counts or between particular configurations. A posterior probability for a count is obtained by summing configuration posterior probabilities of that size; it is not the posterior probability of any one configuration.
Common Pitfall: Confusing Count Probability With Configuration Probability
A common error is to say, “The prior probability of two causal variants is 0.20, so this particular pair has prior probability 0.20.” That assigns the entire probability for count two to one configuration while also implicitly assigning it to the other pairs. With \(m=4\) and a uniform-within-count rule, a count-two prior probability of \(0.20\) is divided among six pairs, giving \(0.20/6=0.0333\) to each pair.
The correction is to state both levels of the prior: first the probability of the count, then the rule for allocating that probability across configurations of that count. Likewise, a larger Bayes factor for one configuration is evidence in its favor relative to another, not a posterior conclusion by itself; posterior odds also include prior odds.
Check Your Understanding
Work through these questions using the definitions and formulas above.
- A locus has six candidate variants, and the prior probability of exactly one causal variant is 0.30. Under a uniform-within-count prior, what prior probability is assigned to a specified singleton?
- In your own words, distinguish a count prior from a configuration prior. Why does a count probability generally need to be divided among configurations?
- Suppose two configurations have prior odds of 2 to 1 and a Bayes-factor ratio of 0.40 for the first against the second. What are their posterior odds in that same order?
- How does independent Bernoulli inclusion with probability \(\rho\) induce a distribution on the number of causal variants?
- Why can a configuration with a larger Bayes factor have a smaller posterior probability than a competing configuration?