Wald test

The Wald test is a parametric statistical test named after the Hungarian statistician Abraham Wald. Whenever a relationship within or between data items can be expressed as a statistical model with parameters to be estimated from a sample, the Wald test can be used to test the true value of the parameter based on the sample estimate.

Suppose an economist, who has data on social class and shoe size, wonders whether social class is associated with shoe size. Say $\theta$ is the average increase in shoe size for upper-class people compared to middle-class people: then the Wald test can be used to test whether $\theta$ is 0 (in which case social class has no association with shoe size) or non-zero (shoe size varies between social classes). Here, $\theta$, the hypothetical difference in shoe sizes between upper and middle-class people in the whole population, is a parameter. An estimate of $\theta$ might be the difference in shoe size between upper and middle-class people in the sample. In the Wald test, the economist uses the estimate and an estimate of variability (see below) to draw conclusions about the unobserved true $\theta$. Or, for a medical example, suppose smoking multiplies the risk of lung cancer by some number R: then the Wald test can be used to test whether R = 1 (i.e. there is no effect of smoking) or is greater (or less) than 1 (i.e. smoking alters risk).

A Wald test can be used in a great variety of different models including models for dichotomous variables and models for continuous variables.[1]

Mathematical details

Under the Wald statistical test, the maximum likelihood estimate $\hat\theta$ of the parameter(s) of interest $\theta$ is compared with the proposed value $\theta_0$, with the assumption that the difference between the two will be approximately normally distributed. Typically the square of the difference is compared to a chi-squared distribution.

Test on a single parameter

In the univariate case, the Wald statistic is

$\frac{ ( \widehat{ \theta}-\theta_0 )^2 }{\operatorname{var}(\hat \theta )}$

which is compared against a chi-squared distribution.

Alternatively, the difference can be compared to a normal distribution. In this case the test statistic is

$\frac{\widehat{\theta}-\theta_0}{\operatorname{se}(\hat\theta)}$

where $\operatorname{se}(\widehat\theta)$ is the standard error of the maximum likelihood estimate (MLE). A reasonable estimate of the standard error for the MLE can be given by $\frac{1}{\sqrt{I_n(MLE)}}$, where $I_n$ is the Fisher information of the parameter.

Test(s) on multiple parameters

The Wald test can be used to test a single hypothesis on multiple parameters, as well as to test jointly multiple hypotheses on single/multiple parameters. Let $\hat{\theta}_n$ be our sample estimator of P parameters (i.e, $\hat{\theta}_n$ is a Px1 vector), which is supposed to follow asymptotically a normal distribution with covariance matrix V, $\sqrt{n}(\hat{\theta}_n-\theta)\xrightarrow{\mathcal{D}} N(0, V)$. The test of Q hypotheses on the P parameters is expressed with a Q x P matrix R:

$H_0: R\theta=r$
$H_1: R\theta\neq r$

The test statistic is:

$(R\hat{\theta}_n-r)^{'}[R(\hat{V}_n/n)R^{'}]^{-1}(R\hat{\theta}_n-r) \quad \xrightarrow{\mathcal{D}}\quad \Chi^2_Q$

where $\hat{V}_n$ is an estimator of the covariance matrix.[2]

Nonlinear hypothesis

In the standard form, the Wald test is used to test linear hypotheses, that can be represented by a single matrix R. If one wishes to test a non-linear hypothesis of the form:

$H_0: c(\theta)=0$
$H_1: c(\theta)\neq 0$

The test statistic becomes:

$c(\hat{\theta}_n)^{'}[c^{'}(\hat{\theta}_n)(\hat{V}_n/n)c^{'}(\hat{\theta}_n)^{'}]^{-1}c(\hat{\theta}_n) \quad \xrightarrow{\mathcal{D}}\quad \Chi^2_Q$

where $c^{'}(\hat{\theta}_n)$ is the derivative of c evaluated at the sample estimator. This result is obtained using the delta method, which uses a first order approximation of the variance.

Non-invariance to re-parametrisations

The fact that one uses an approximation of the variance has the drawback that the Wald statistic is not-invariant to a non-linear transformation/reparametrisation of the hypothesis: it can give different answers to the same question, depending on how the question is phrased.[3] For example, asking whether R = 1 is the same as asking whether log R = 0; but the Wald statistic for R = 1 is not the same as the Wald statistic for log R = 0 (because there is in general no neat relationship between the standard errors of R and log R, so it needs to be approximated).

Alternatives to the Wald test

There exist several alternatives to the Wald test, namely the likelihood-ratio test and the Lagrange multiplier test (also known as the score test). Robert F. Engle showed that these three tests, the Wald test, the likelihood-ratio test and the Lagrange multiplier test are asymptotically equivalent.[4] Although they are asymptotically equivalent, in finite samples, they could disagree enough to lead to different conclusions.

There are several reasons to prefer the likelihood ratio test or the lagrange multiplier to the Wald test:[5][6][7]

• Non-invariance: As argued above, the Wald test is not invariant to a reparametrization, while the Likelihood ratio tests will give exactly the same answer whether we work with R, log R or any other monotonic transformation of R.
• The other reason is that the Wald test uses two approximations (that we know the standard error, and that the distribution is chi-squared), whereas the likelihood ratio test uses one approximation (that the distribution is chi-squared).
• The Wald test requires an estimate under the null hypothesis. In some cases, the model is simpler under the alternative hypothesis, so that one might prefer to use the score test (also called Lagrange Multiplier test), which has the advantage that it can be formulated in situations where the variability is difficult to estimate; e.g. the Cochran–Mantel–Haenzel test is a score test.[8]