Wald Inference for Genetic Parameters

Why quantify uncertainty around a genetic estimate?

A maximum-likelihood estimate gives a value for an unknown genetic parameter, but an estimate alone does not show how precisely the data determine that value. Wald inference combines a maximum-likelihood estimate (MLE) with an estimated standard error (SE) to produce a confidence interval or test a specified value. For example, an estimated allele frequency of 0.20 has a different interpretation when its SE is 0.01 than when its SE is 0.08.

The method is based on an approximate normal distribution for the MLE in sufficiently large samples. It is convenient and applies to many likelihood models, including allele-frequency and penetrance models. Its normal approximation can be poor, however, especially when the sample is small or the parameter is near a boundary such as zero or one.

Wald intervals and tests

Definition: Let $\theta$ be a scalar parameter and let $\hat{\theta}$ be its MLE. Let $\widehat{SE}(\hat{\theta})$ be an estimated standard error, calculated from the estimated information or another appropriate likelihood-based variance estimate. A two-sided Wald confidence interval with approximate confidence level $100(1-\alpha)\%$ is $$ \hat{\theta} \ \pm\ z_{1-\alpha/2}\widehat{SE}(\hat{\theta}), $$ where $z_{1-\alpha/2}$ is the corresponding quantile of the standard normal distribution. To test $H_0:\theta=\theta_0$ against $H_a:\theta\ne\theta_0$, the Wald statistic is $$ Z_W=\frac{\hat{\theta}-\theta_0}{\widehat{SE}(\hat{\theta})}. $$ Under $H_0$, $Z_W$ is approximately standard normal when the large-sample conditions hold. The two-sided p-value is $2P(Z\geq |Z_W|)$ for $Z\sim N(0,1)$.

For a 95% interval, $z_{0.975}$ is approximately 1.96. The interval uses the estimate and its SE; the test compares the estimate with the null value using that SE. In the usual Wald test, the SE is estimated at the unrestricted MLE, not by fitting the model with the null value imposed. A small two-sided p-value indicates that the observed estimate is far from the null value relative to its estimated uncertainty. It is not the probability that the null hypothesis is true.

In likelihood models, the estimated variance is often obtained from the inverse of the estimated Fisher information. For a scalar parameter, if the estimated information is $\widehat{I}(\hat{\theta})$, then $\widehat{SE}(\hat{\theta})=\sqrt{1/\widehat{I}(\hat{\theta})}$. For multiple parameters, the estimated covariance matrix is the inverse information matrix, and its diagonal entries give estimated variances. The specific likelihood determines how the information and SE are calculated.

Why the normal approximation supports Wald inference

The large-sample justification comes from the approximate normality of a regular MLE. In a model with independent observations and a correctly specified, identifiable likelihood, the score is the derivative of the log-likelihood. Near the MLE, a Taylor expansion of the score around the true parameter gives, to first order,

0 \approx U(\theta)-I(\theta)(\hat{\theta}-\theta), \qquad \hat{\theta}-\theta \approx I(\theta)^{-1}U(\theta).

Here $U(\theta)$ is the score at the true parameter and $I(\theta)$ is the Fisher information. Under standard regularity conditions, the score is approximately normal with mean zero and variance $I(\theta)$. The displayed approximation therefore gives $\hat{\theta}$ an approximate normal distribution with mean $\theta$ and variance $I(\theta)^{-1}$. Replacing the unknown parameter and information with their estimates yields the SE used in Wald inference.

This justification requires more than a large sample count by itself. The parameter should be identifiable and in the interior of its allowable range, the information should be nonsingular, and the likelihood model and independence assumptions should be appropriate. If these conditions fail or the sample provides little information, the estimated normal approximation may be inaccurate.

Wald inference for an allele frequency

Suppose $n$ unrelated individuals are sampled from a population in Hardy–Weinberg equilibrium at a biallelic locus. Let $p$ be the frequency of one allele, and assume genotypes are correctly observed and individuals are sampled independently. Under these assumptions, the count $K$ of copies of the allele among the $2n$ sampled chromosomes has a binomial distribution with size $2n$ and probability $p$. Its MLE is $\hat p=K/(2n)$, and the estimated SE is $\sqrt{\hat p(1-\hat p)/(2n)}$.

Worked Example: A Wald Interval for an Allele Frequency

Suppose a hypothetical sample contains $n=200$ unrelated individuals in Hardy–Weinberg equilibrium. Genotypes are assumed to be called without error. The allele of interest is observed $K=80$ times among the $2n=400$ chromosomes. Estimate its frequency and construct a 95% Wald interval.

The MLE is $\hat p=80/400=0.200$. The independent sampling and Hardy–Weinberg assumptions make the binomial allele-count model appropriate here. The estimated SE is

\widehat{SE}(\hat p) =\sqrt{\frac{\hat p(1-\hat p)}{2n}} =\sqrt{\frac{0.200(0.800)}{400}} =\sqrt{0.000400} =0.0200.

Using $z_{0.975}=1.96$, the interval is

0.200 \pm 1.96(0.0200) =0.200\pm0.0392 =(0.1608,\ 0.2392).

Thus the 95% Wald interval is approximately 0.161 to 0.239. In repeated samples satisfying this model, intervals constructed this way would contain the population allele frequency about 95% of the time under the approximation. This statement does not assign a 95% probability to the fixed population frequency being inside this particular interval.

A Wald interval for a frequency is computed on the original probability scale. Because its endpoints are constructed by adding and subtracting a multiple of the SE, they can sometimes fall outside the possible range from zero to one. That is a warning that the normal approximation may be unsuitable, rather than a reason to interpret an impossible frequency as valid.

Worked Example: A Wald Test of an Allele Frequency

Use the same hypothetical allele-count data: $K=80$ copies among $400$ sampled chromosomes. Test $H_0:p=0.250$ against $H_a:p\ne0.250$ at $\alpha=0.05$. The allele-count model and assumptions are as stated in the preceding example.

The unrestricted MLE is $\hat p=0.200$, and its estimated SE is $0.0200$. The Wald statistic is

Z_W=\frac{0.200-0.250}{0.0200}=-2.50.

The two-sided p-value is $2P(Z\geq2.50)\approx0.0124$. Since $0.0124<0.05$, reject $H_0$ at the 5% significance level. Under the assumed model, these data provide evidence that the allele frequency differs from 0.250; the test does not establish why it differs or imply a causal explanation.

Wald inference for penetrance

For a binary phenotype $Y$, the penetrance for genotype $g$ is $\pi_g=P(Y=1\mid G=g)$. If $n_g$ independent individuals with genotype $g$ have observed phenotypes, and each phenotype is a Bernoulli observation with the same probability $\pi_g$, then the number with the phenotype has a binomial likelihood. If $y_g$ of the $n_g$ individuals have the phenotype, the MLE is $\hat{\pi}_g=y_g/n_g$, with estimated SE $\sqrt{\hat{\pi}_g(1-\hat{\pi}_g)/n_g}$. This model conditions on genotype; it does not by itself describe how genotype frequencies arise.

Worked Example: A Wald Interval and Test for Penetrance

In a hypothetical study, 30 of 100 independently sampled individuals with a specified genotype show a binary trait. Assume phenotype outcomes are independent Bernoulli observations with a common genotype-specific penetrance $\pi_g$, and that genotype and phenotype are measured correctly. Estimate $\pi_g$, calculate a 95% Wald interval, and test $H_0:\pi_g=0.20$ against a two-sided alternative.

The MLE is $\hat{\pi}_g=30/100=0.300$. Its estimated SE is

\widehat{SE}(\hat{\pi}_g) =\sqrt{\frac{0.300(1-0.300)}{100}} =\sqrt{0.00210} =0.04583.

The 95% Wald interval is

0.300\pm1.96(0.04583) =0.300\pm0.08983 =(0.2102,\ 0.3898).

For the test, use the SE estimated at the unrestricted MLE:

Z_W=\frac{0.300-0.200}{0.04583}=2.182.

The two-sided p-value is approximately 0.0291. Because this is below 0.05, the data provide evidence against penetrance equal to 0.20 under the stated model. The interval describes uncertainty in the penetrance for this genotype, measured as a probability; it does not describe the probability that a particular individual will show the trait after their outcome is known.

Contrasts between genetic parameters

Wald inference can also address a scalar contrast involving more than one parameter. Let $\boldsymbol{\theta}$ be a vector of parameters with MLE $\hat{\boldsymbol{\theta}}$ and estimated covariance matrix $\widehat{\Sigma}$. For a linear contrast $c^\mathsf{T}\boldsymbol{\theta}$, its estimate is $c^\mathsf{T}\hat{\boldsymbol{\theta}}$, and its estimated variance is $c^\mathsf{T}\widehat{\Sigma}c$. The SE is the square root of that variance. This calculation accounts for covariance between estimates when they are not independent.

Worked Example: Comparing Two Genotype-Specific Penetrances

In a hypothetical study, 40 of 100 individuals with genotype $g_1$ and 20 of 100 individuals with genotype $g_2$ show the trait. Assume independent individuals, independent groups, and a common Bernoulli penetrance within each genotype group. Estimate $\pi_{g_1}-\pi_{g_2}$, form a 95% Wald interval, and test whether the penetrances are equal.

The estimates are $\hat{\pi}_{g_1}=40/100=0.400$ and $\hat{\pi}_{g_2}=20/100=0.200$. The estimated difference is $0.400-0.200=0.200$. Because the groups are independent, the estimated variance of the difference is the sum of the two estimated variances:

\widehat{\operatorname{Var}}(\hat{\pi}_{g_1}-\hat{\pi}_{g_2}) =\frac{0.400(0.600)}{100}+\frac{0.200(0.800)}{100} =0.00240+0.00160 =0.00400.

Thus the SE is $\sqrt{0.00400}=0.06325$. The 95% interval for the difference in penetrance is

0.200\pm1.96(0.06325) =0.200\pm0.1240 =(0.0760,\ 0.3240).

To test equality, use $H_0:\pi_{g_1}-\pi_{g_2}=0$. The statistic is $Z_W=0.200/0.06325=3.162$, giving a two-sided p-value of approximately 0.00157. The interval and test both indicate evidence of a difference in penetrance between these genotype groups under the stated assumptions. This is an association between genotype group and the trait in this model; it does not alone establish a causal effect.

When the Wald approximation can mislead

A Wald interval or test is most dependable when the likelihood is regular, the MLE is well inside the allowed parameter space, and the data contain enough information for the sampling distribution to be close to normal. It can be unreliable when counts are sparse, penetrance estimates are near zero or one, or an allele frequency is close to a boundary. For example, if no individuals in a small genotype group show a trait, the estimated penetrance is zero and the plug-in Bernoulli SE is also zero. The resulting Wald calculation cannot provide a useful assessment of uncertainty.

Wald intervals are also symmetric around the estimate on the parameter scale, even when the parameter is restricted to $[0,1]$ or its likelihood is asymmetric. An endpoint outside that range is not a biologically possible value; it signals a poor fit between the normal approximation and the parameter space. The interval’s nominal coverage can then differ substantially from its stated level. A large-sample justification should therefore be considered alongside the observed counts and the location of the estimate, not treated as automatic.

Key takeaway: Wald inference uses an MLE and its estimated SE to measure how far an estimate is from a null value or to form an approximate interval. Its validity depends on a suitable likelihood model and a credible large-sample normal approximation, particularly for parameters constrained to a range such as allele frequencies and penetrances.

Common mistake

Wrong reasoning: “The 95% Wald interval contains 0.20, so there is a 95% probability that the allele frequency is 0.20.” Correction: The interval procedure has approximate 95% coverage over repeated samples under its assumptions. The parameter is fixed; the interval is random before data are collected. An interval containing a null value also does not assign that value a probability.
Wrong reasoning: “A Wald interval is valid because the sample size is large.” Correction: Sample size alone does not ensure an accurate approximation. Check model assumptions, information, and whether the estimate is near a boundary.
Wrong reasoning: “The Wald test SE should always be calculated under the null.” Correction: The usual Wald statistic uses the estimated SE at the unrestricted MLE. Its numerator measures the difference between that estimate and the null value.

Check Your Understanding

Use the definitions and assumptions above to answer the following questions.

In one sentence, explain what the SE in a Wald statistic measures for an allele-frequency estimate.
A penetrance estimate is 0.10 with estimated SE 0.03. Write the 95% Wald interval using $z_{0.975}=1.96$, and calculate its endpoints.
Why does the usual Wald test use the estimated SE at the unrestricted MLE rather than automatically calculating it under the null?
Give two reasons a Wald interval for a penetrance near zero might be unreliable.
For two independent genotype groups, why are the estimated variances added when calculating the variance of the difference in penetrances?