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Allele Frequencies Across Populations · Tutorial 115 of 989

Inverse-Probability Correction for Variant Ascertainment

Use known variant-selection probabilities to recover a target frequency summary, and assess the uncertainty and sensitivity introduced by weighting.

Intermediate 10 min read

What You'll Learn

  • Define a target mean allele frequency over a fixed set of variants
  • Construct an inverse-probability estimator from selected variants
  • State the conditions that make the estimator unbiased
  • Derive variance contributions from ascertainment and frequency estimation
  • Evaluate why rare selection probabilities can produce unstable estimates
  • Recognize when the correction does not apply

Why Correct for Variant Ascertainment?

A frequency summary calculated from selected variants describes those variants as observed, not necessarily the full set of variants of interest. If selection favors variants that are common in a discovery sample, for example, the selected set may have a different average frequency from the target set. When each variant’s probability of selection is known, inverse-probability weighting can correct the summary for unequal selection chances.

The correction has a clear target: a summary over a defined set of variants. It does not recover variants that had no chance of selection, and it does not remove bias caused by using the same noisy frequency estimate both to select variants and to estimate their frequencies.

Definition: Suppose the target set contains \(M\) variants. For variant \(i\), let \(p_i\) be its allele frequency in the target population, \(S_i=1\) if it is selected and \(S_i=0\) otherwise, and \(\pi_i=P(S_i=1)\) its known selection probability. The target mean frequency is \(\theta=M^{-1}\sum_{i=1}^{M}p_i\). An inverse-probability, or Horvitz–Thompson, estimator is \(\hat{\theta}_{HT}=M^{-1}\sum_{i=1}^{M}S_i\tilde{p}_i/\pi_i\), where \(\tilde{p}_i\) estimates \(p_i\) for a selected variant.

The target mean gives every variant in the defined set equal importance. The weight \(1/\pi_i\) compensates for the fact that a variant with a smaller selection probability represents more variants like itself in the target set. The denominator is the known number \(M\), not the number selected. This makes the estimator an estimate of a fixed-set mean rather than a mean normalized by the realized selected sample.

Conditions for the Correction

The estimator’s basic unbiasedness statement relies on a sampling design, not just on the presence of weights. First, the target set and its size \(M\) must be specified. Second, every variant in that set must have a positive, known selection probability: \(\pi_i>0\). Third, the selection probabilities used in the calculation must describe the actual ascertainment process.

For the simple variance formulas below, assume variants are selected independently, so the \(S_i\) are independent Bernoulli variables with probabilities \(\pi_i\). Also assume that \(\tilde{p}_i\) is unbiased for \(p_i\), with variance \(\tau_i^2\), and is obtained from data independent of the selection event. Errors in the frequency estimates for different variants are taken to be independent for these formulas. If variants are selected dependently or frequency errors are correlated, covariance terms are needed.

A usual way to support the independence condition is to use one data set to ascertain variants and an independent validation sample to estimate their frequencies. If selection depends on a noisy discovery estimate and that same estimate is then reported as \(\tilde{p}_i\), the selected estimates can be systematically high or otherwise distorted. Weighting by \(\pi_i\) alone does not generally repair this selection-induced dependence.

Worked Example: Correcting a Mean Across Five Variants

Consider a hypothetical target set of \(M=5\) variants. Their selection probabilities are \((0.50,1.00,0.25,0.50,0.25)\). In an independent validation sample, variants 1, 3, and 5 are selected and have estimated frequencies \(0.10\), \(0.20\), and \(0.08\), respectively. Calculate the inverse-probability estimate and compare it with the unweighted selected mean.

The estimator uses only selected variants, with each selected estimate divided by its selection probability:

$$ \hat{\theta}_{HT} =\frac{1}{5}\left(\frac{0.10}{0.50} +\frac{0.20}{0.25} +\frac{0.08}{0.25}\right) =\frac{0.20+0.80+0.32}{5} =0.264. $$

The unweighted mean among the three selected variants is \((0.10+0.20+0.08)/3=0.1267\). In this hypothetical realization, the correction is larger because variants with lower selection probabilities receive larger weights. The two results are not competing estimates of the same selected-set mean: the unweighted calculation summarizes the selected variants, whereas the weighted estimator targets the mean over all five variants. A single realization need not equal the target mean.

Why the Estimator Is Unbiased

For a fixed variant \(i\), selection occurs with probability \(\pi_i\). Under the stated assumptions, its frequency estimate has expectation \(p_i\) and is independent of selection. Therefore:

$$ E\left(\frac{S_i\tilde{p}_i}{\pi_i}\right) =\frac{E(S_i)E(\tilde{p}_i)}{\pi_i} =\frac{\pi_i p_i}{\pi_i} =p_i. $$

Averaging these unbiased contributions over the \(M\) variants gives \(E(\hat{\theta}_{HT})=M^{-1}\sum_i p_i=\theta\). This is a design-based statement: the frequencies \(p_i\) are fixed, while which variants are selected is random. If frequency estimates are also random but unbiased and independent of selection, the same expectation still holds.

Variance and the Cost of Small Probabilities

For independent selection and independent frequency-estimation errors, the contribution for variant \(i\) has variance

$$ \operatorname{Var}\left(\frac{S_i\tilde{p}_i}{\pi_i}\right) =\left(\frac{1-\pi_i}{\pi_i}\right)p_i^2 +\frac{\tau_i^2}{\pi_i}. $$

The first term is variation from whether the variant is selected. The second is variation from estimating its frequency, amplified because that estimate contributes only when the variant is selected. Thus the variance of the mean estimator is

$$ \operatorname{Var}(\hat{\theta}_{HT}) =\frac{1}{M^2}\sum_{i=1}^{M} \left[ \left(\frac{1-\pi_i}{\pi_i}\right)p_i^2 +\frac{\tau_i^2}{\pi_i} \right]. $$

The formula assumes independent selections and errors across variants. If those assumptions do not hold, covariances between variant contributions must also be included. In particular, linkage or shared sampling effects can make frequency-estimation errors correlated, so the sum of individual variances alone may not describe total uncertainty.

Derivation: Since \(S_i\) is Bernoulli with probability \(\pi_i\), \(S_i^2=S_i\). Independence of \(S_i\) and \(\tilde{p}_i\) gives \(E(S_i\tilde{p}_i/\pi_i)=p_i\) and \(E[(S_i\tilde{p}_i/\pi_i)^2]=E(\tilde{p}_i^2)/\pi_i=(p_i^2+\tau_i^2)/\pi_i\). Subtracting the squared mean yields \((1-\pi_i)p_i^2/\pi_i+\tau_i^2/\pi_i\). Independence across variants lets us add these variances and divide by \(M^2\).

Worked Example: Calculating the Estimator’s Variance

Suppose a hypothetical set has \(M=3\) variants with frequencies \(p_1=0.10\), \(p_2=0.20\), and \(p_3=0.05\), and selection probabilities \((0.50,0.25,1.00)\). For illustration, assume independent, unbiased frequency estimates with variances \((0.0009,0.0016,0.000475)\), and independent selection and estimation across variants. Calculate the variance and standard error of the inverse-probability mean estimator.

For variant 1, the selection component is \(((1-0.50)/0.50)(0.10)^2=0.0100\), and the estimation component is \(0.0009/0.50=0.0018\). For variant 2, the components are \(((1-0.25)/0.25)(0.20)^2=0.1200\) and \(0.0016/0.25=0.0064\). Variant 3 is always selected, so it contributes no selection variance; its estimation component is \(0.000475/1.00=0.000475\).

Therefore,

$$ \operatorname{Var}(\hat{\theta}_{HT}) =\frac{0.0100+0.0018+0.1200+0.0064+0.000475}{3^2} =\frac{0.138675}{9} =0.01541, \qquad SE(\hat{\theta}_{HT})=\sqrt{0.01541}=0.1241. $$

This is a hypothetical variance calculation with the underlying frequencies and estimation variances specified. In an actual analysis those quantities are not all known; variance estimation must use available sample information and respect the ascertainment design. The large contribution from variant 2 reflects both its lower selection probability and its larger frequency.

Rare Selection Probabilities and Sensitivity

A small \(\pi_i\) creates a large weight. When a rare-probability variant is selected, its contribution to the estimated mean is \( \tilde{p}_i/(M\pi_i)\), which can be large even if its frequency is modest. When it is not selected, it contributes zero. This contrast creates high variability across repeated ascertainment samples.

Worked Example: A Variant With a Very Small Selection Probability

Consider a hypothetical target set of \(M=10\) variants. One has frequency \(p=0.08\), is selected with probability \(\pi=0.02\), and its frequency is treated as known for this design-variance illustration. If selected, its contribution to the estimated mean is \(0.08/(10\times0.02)=0.40\). If it is not selected, its contribution is zero.

Its contribution to the variance of the estimated mean is

$$ \frac{1}{10^2}\left(\frac{1-0.02}{0.02}\right)(0.08)^2 =\frac{49\times0.0064}{100} =0.003136. $$

If the same variant instead had \(\pi=0.20\), its selected contribution would be \(0.08/(10\times0.20)=0.040\), and its variance contribution would be

$$ \frac{1}{10^2}\left(\frac{1-0.20}{0.20}\right)(0.08)^2 =\frac{4\times0.0064}{100} =0.000256. $$

Reducing the selection probability from \(0.20\) to \(0.02\) makes this variance contribution 12.25 times as large. If the frequency itself is estimated with uncertainty, the additional term \(\tau_i^2/\pi_i\) also grows as \(\pi_i\) decreases.

Interpretation and Practical Checks

The inverse-probability estimate is most useful when the target variant set is meaningful, selection probabilities are credible, and no target variant has zero probability of inclusion. It corrects unequal representation under the specified ascertainment mechanism; it does not correct genotype-call errors, population mismatch, or an incorrectly defined target set. Allele coding must also be consistent across variants and populations for the frequency summary to have the intended interpretation.

Very large weights signal that the estimate may be sensitive to a few selected variants. Report the target set, how the probabilities were obtained, the weighted estimate, and an uncertainty calculation that matches the selection design. A sensitivity analysis can show how estimates change when uncertain selection probabilities are varied within plausible ranges. Truncating or capping weights may reduce variance, but it generally changes the estimator and can introduce bias; it is not a cost-free repair.

  • Common mistake: “The selected variants have an average frequency of 0.12, so that is the average frequency in the full target set.” This ignores unequal selection probabilities and confuses the selected-set mean with the target-set mean.
  • Correction: Define the target set and weight each selected estimate by \(1/\pi_i\), using the known frame size \(M\) for the Horvitz–Thompson mean. Then assess uncertainty, especially for small \(\pi_i\).
  • Another limitation: If \(\pi_i=0\) for a target variant, no observed data can represent that variant through inverse weighting. The target set or ascertainment design must change.
Key takeaway: Inverse-probability weighting corrects a frequency summary for known, positive variant-selection probabilities by giving each selected variant a contribution proportional to \(1/\pi_i\). Unbiasedness depends on a correctly specified selection design and unbiased frequency estimates; precision can deteriorate sharply when selection probabilities are rare.

Check Your Understanding

Answer each question using the target set and sampling assumptions stated in the problem.

  1. Why does the Horvitz–Thompson mean divide the weighted total by the known number of target variants rather than the number selected?
  2. A target variant has selection probability zero. Can inverse-probability weighting recover its contribution? Explain.
  3. What independence condition is needed for the simple unbiasedness argument when frequency estimates are noisy?
  4. How do the selection-variance and frequency-estimation-variance terms change as a selection probability becomes small?
  5. Why can using the same discovery estimate for both variant selection and frequency reporting invalidate the basic correction?
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