Deriving Wakefield Approximate Bayes Factors

Why derive the approximate Bayes factor?

For a genetic variant, an effect estimate and its standard error summarize evidence about an association. A Bayes factor compares how well that evidence is predicted by two specified models: a null model in which the effect is zero and an alternative model that assigns a probability distribution to the effect. Wakefield’s approximate Bayes factor (ABF) is useful because, under a normal likelihood and a normal prior, the comparison has a simple closed form.

The result is a comparison of models for the observed estimate, not a probability that the variant is causal. Its value depends both on the estimate’s precision and on the alternative model’s prior distribution for effect sizes. The derivation makes those dependencies explicit.

Model and assumptions

Let \(\hat{\beta}\) be an estimated variant effect, expressed in a specified allele orientation and on a specified outcome scale, and let \(SE\) be its standard error. Write \(V=SE^2\). The approximate likelihood treats the estimate as normally distributed around the true effect \(\beta\), with known sampling variance \(V\). Under the null hypothesis \(H_0\), \(\beta=0\). Under the alternative hypothesis \(H_1\), assign \(\beta\) a zero-centred normal prior with variance \(W\).

Definition: Under Wakefield’s normal approximation, \(\hat{\beta}\mid\beta\sim N(\beta,V)\), with \(V=SE^2\). The null model fixes \(\beta=0\); the alternative model assigns \(\beta\sim N(0,W)\), where \(W>0\) is the prior variance of the effect. The approximate Bayes factor in favour of the alternative is \(ABF_{10}=p(\hat{\beta}\mid H_1)/p(\hat{\beta}\mid H_0)\).

The calculation assumes that the estimate has an approximately normal sampling distribution and that its standard error adequately describes its sampling uncertainty. It also assumes that \(V\) is treated as known for this calculation, that the same estimate and outcome scale are used in both models, and that the prior is centred at zero with variance \(W\). In practice, the normal likelihood and known-variance treatment are often approximations to an analysis in which a standard error has itself been estimated.

The effect estimate and its standard error must use compatible units: if \(\hat{\beta}\) is measured in outcome units per allele, then \(SE\) has those same units and both \(V\) and \(W\) have squared units. The prior variance is not the standard error and is not a variance of the data. It describes the spread of effects that the alternative model regards as plausible before observing this estimate.

Deriving Wakefield’s formula

Under \(H_0\), setting \(\beta=0\) in the likelihood gives \(\hat{\beta}\mid H_0\sim N(0,V)\). Under \(H_1\), the marginal distribution of \(\hat{\beta}\) is obtained by integrating over the prior on \(\beta\). Because the estimate is the sum of a normally distributed effect and independent normal sampling error, the marginal distribution is normal with variance \(W+V\) and mean zero.

\begin{aligned} \hat{\beta}\mid H_0 &\sim N(0,V),\\ \hat{\beta}\mid H_1 &\sim N(0,V+W). \end{aligned}

The normal density at the observed estimate under each model is therefore

p(\hat{\beta}\mid H_0)=\frac{1}{\sqrt{2\pi V}} \exp\!\left(-\frac{\hat{\beta}^{\,2}}{2V}\right), \qquad p(\hat{\beta}\mid H_1)=\frac{1}{\sqrt{2\pi(V+W)}} \exp\!\left(-\frac{\hat{\beta}^{\,2}}{2(V+W)}\right).

Dividing the second density by the first cancels the common normalizing constant. Combining the two exponential terms yields the ABF.

\begin{aligned} ABF_{10} &=\sqrt{\frac{V}{V+W}}\, \exp\!\left\{\frac{\hat{\beta}^{\,2}}{2} \left(\frac{1}{V}-\frac{1}{V+W}\right)\right\}\\ &=\sqrt{\frac{V}{V+W}}\, \exp\!\left\{\frac{\hat{\beta}^{\,2}W}{2V(V+W)}\right\}. \end{aligned}

A convenient reparameterization uses the Wald statistic \(Z=\hat{\beta}/SE\) and \(r=W/(V+W)\). Since \(V/(V+W)=1-r\), the same expression becomes

ABF_{10}=\sqrt{1-r}\exp\!\left(\frac{rZ^2}{2}\right).

This is the approximate Bayes factor convention used here: values greater than one favour \(H_1\) over \(H_0\) for the observed estimate; values less than one favour \(H_0\) over \(H_1\). A Bayes factor is not a posterior probability. Posterior odds require prior odds as well as the Bayes factor.

Derivation check: The factor \(\sqrt{V/(V+W)}\) penalizes the alternative for spreading its predictive distribution over a wider range of estimates. The exponential factor rewards an estimate that is more compatible with the alternative’s broader distribution than with the null distribution. Both factors are necessary for the density ratio.

Worked calculations

Worked Example: An Estimate with a Standard Error of 0.10

Suppose a hypothetical study reports \(\hat{\beta}=0.24\) outcome units per allele with \(SE=0.10\) outcome units per allele. Consider the prior variance \(W=0.04\) in squared outcome units per allele. Assume the normal likelihood approximation and the zero-centred normal prior above apply. Then \(V=SE^2=0.10^2=0.01\), and \(Z=0.24/0.10=2.4\).

Here \(V/(V+W)=0.01/0.05=0.2\), so \(r=W/(V+W)=0.8\). Substitution gives

ABF_{10}=\sqrt{0.2}\exp\!\left(\frac{0.8(2.4)^2}{2}\right) =\sqrt{0.2}\exp(2.304) \approx 0.4472(10.0142) \approx 4.48.

The estimate is about 4.48 times as probable under this specified alternative model as under the zero-effect model, using the normal approximation. This is not a 4.48 probability of association; it is a ratio of two model likelihoods for the observed estimate.

Worked Example: A Smaller Prior Variance

Now suppose a hypothetical estimate is \(\hat{\beta}=0.12\) outcome units per allele with \(SE=0.05\) outcome units per allele, and take \(W=0.0025\) in squared outcome units per allele. The sampling variance is \(V=0.05^2=0.0025\), and \(Z=0.12/0.05=2.4\). Thus \(V=W\), \(r=0.5\), and the ABF is

ABF_{10}=\sqrt{0.5}\exp\!\left(\frac{0.5(2.4)^2}{2}\right) =\sqrt{0.5}\exp(1.44) \approx 0.7071(4.2207) \approx 2.98.

For this estimate and prior, the data are about 2.98 times as probable under \(H_1\) as under \(H_0\). Although its \(Z\) statistic matches that in the previous example, the ABF differs because the prior variance relative to the sampling variance differs. The Bayes factor is not determined by a \(Z\) statistic alone.

Worked Example: An Estimate Near Zero

Suppose a hypothetical estimate is \(\hat{\beta}=0.03\) outcome units per allele, with \(SE=0.10\) outcome units per allele and \(W=0.04\) in squared outcome units per allele. These values give \(V=0.01\), \(Z=0.3\), and \(r=0.04/(0.01+0.04)=0.8\). Therefore,

ABF_{10}=\sqrt{0.2}\exp\!\left(\frac{0.8(0.3)^2}{2}\right) =\sqrt{0.2}\exp(0.036) \approx 0.4472(1.0367) \approx 0.4636.

Because the ABF is below one, this estimate is about \(1/0.4636\approx 2.16\) times as probable under the null as under this alternative. The result is evidence in favour of the null relative to the specified alternative, not proof that the true effect is exactly zero.

Worked Example: Comparing Prior Variances for One Estimate

For a hypothetical estimate \(\hat{\beta}=0.12\) outcome units per allele with \(SE=0.05\) outcome units per allele, \(V=0.0025\) and \(Z=2.4\). Calculate the ABF under three prior variances, all in squared outcome units per allele.

For \(W=0.000625\), \(r=0.000625/(0.0025+0.000625)=0.2\). Thus \(ABF_{10}=\sqrt{0.8}\exp(0.2(2.4)^2/2)=0.8944\exp(0.576)\approx1.59\).

For \(W=0.0025\), \(r=0.5\), giving \(ABF_{10}=\sqrt{0.5}\exp(1.44)\approx2.98\), as calculated above. For \(W=0.01\), \(r=0.8\), giving \(ABF_{10}=\sqrt{0.2}\exp(2.304)\approx4.48\).

The same estimate can therefore yield different Bayes factors under different alternative models. Each result compares the data with a different prior predictive distribution; none should be reported without identifying the prior variance used.

Interpretation, checks, and limitations

The formula separates two contributions. The square-root term is at most one and reflects the cost of the alternative’s larger predictive variance. The exponential term increases with \(Z^2\), reflecting how far the estimate lies from zero relative to its standard error. A sign change in \(\hat{\beta}\) leaves the ABF unchanged because the prior is symmetric about zero and the formula depends on \(\hat{\beta}^{\,2}\). The ABF therefore measures evidence for a nonzero effect under this two-sided alternative, not the direction of the effect.

Before interpreting a result, check that \(V=SE^2\), that \(W\) and \(V\) have the same squared units, and that \(Z=\hat{\beta}/SE\) uses the reported estimate and standard error. It is also useful to compute the logarithm of the ABF when values are extreme:

\log(ABF_{10})=\frac{1}{2}\log\!\left(\frac{V}{V+W}\right) +\frac{\hat{\beta}^{\,2}W}{2V(V+W)}.

This form makes the two contributions visible and can avoid numerical overflow when the exponential is very large. If the estimate and standard error have been reported on mismatched scales, or if the prior variance is given on a different scale, the resulting calculation is not meaningful. In genetic analyses, the effect allele and outcome scale must also be consistent with the estimate being evaluated.

The ABF is approximate when the normal likelihood with a fixed sampling variance is only an approximation to the data-generating analysis. Strong departures from normality, poorly estimated standard errors, or a likelihood that cannot be summarized well by one estimate and one standard error can undermine it. The normal prior is also a substantive modelling choice: it determines which effect sizes the alternative predicts. The ABF does not itself account for linkage disequilibrium among variants or decide how many variants are causal.

A frequentist \(Z\)-test answers a different question: it evaluates how unusual a test statistic would be under a null sampling model, and its \(p\)-value is not a likelihood ratio. A full Bayes factor instead integrates the likelihood over the chosen prior; in this normal-normal setting, Wakefield’s formula is that integration under the stated likelihood approximation. If the likelihood or prior is not well represented by these normal distributions, a direct integration using an appropriate model may be preferable. The calculation should not be used to turn a \(p\)-value into a posterior probability.

Common mistakes

Using \(SE\) where \(V\) is required. The formula uses \(V=SE^2\). Squaring the standard error is essential, and \(W\) must be a variance on the same squared scale.
Reversing the ratio. Here \(ABF_{10}\) is alternative over null. An ABF below one favours the null relative to the specified alternative; it does not mean the alternative is more supported.
Treating the ABF as a probability. It is a ratio of marginal likelihoods. It is neither the probability that a variant is causal nor the probability that \(H_1\) is true.
Assuming a \(Z\) statistic determines the ABF. The prior variance enters through \(r=W/(V+W)\). Report \(W\) and its scale alongside the estimate and standard error.
Reading direction from a two-sided ABF. Squaring \(Z\) removes its sign. Report the estimate itself when describing the direction and size of association.

Key takeaway: With a normal estimate likelihood and a zero-centred normal effect prior, \(ABF_{10}=\sqrt{V/(V+W)}\exp\{\hat{\beta}^{\,2}W/[2V(V+W)]\}\). The result compares the observed estimate under two specified models; its interpretation depends on the likelihood approximation, the prior variance, and consistent units.

Check Your Understanding

Work through the questions, then compare your reasoning with the solutions.

An estimate is \(0.20\) outcome units per allele with \(SE=0.10\) outcome units per allele. If \(W=0.01\) in squared outcome units per allele, find \(V\), \(Z\), \(r\), and \(ABF_{10}\).
If an ABF is \(0.4\), which model is better supported by the observed estimate, and by what factor?
Why must a prior variance \(W\) be expressed in squared units matching \(V=SE^2\)?
Would changing an estimate from \(0.12\) to \(-0.12\), while keeping the same standard error and prior variance, change this two-sided ABF? Explain.

Solutions. (1) \(V=0.10^2=0.01\), \(Z=0.20/0.10=2\), and \(r=0.01/(0.01+0.01)=0.5\). Hence \(ABF_{10}=\sqrt{0.5}\exp(0.5(2)^2/2)=0.7071\exp(1)\approx1.92\). (2) Since \(0.4<1\), the observed estimate is \(1/0.4=2.5\) times as probable under \(H_0\) as under \(H_1\). (3) \(V\) and \(W\) are added in the marginal variance \(V+W\); they must have matching units, which are squared effect units. (4) No. The formula depends on \(\hat{\beta}^{\,2}\), so reversing the sign leaves the ABF unchanged; the effect direction must be read from the signed estimate.