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Wald Score and Likelihood-Ratio Tests · Tutorial 263 of 981

Deriving the Wald Test from Asymptotic Normality

Learn how asymptotic normality turns a smooth restriction into a Wald statistic, and how its covariance and chi-square reference distribution are derived.

Intermediate 10 min read

What You'll Learn

  • Express a null hypothesis as a smooth vector-valued restriction on the parameter.
  • Identify the Jacobian and the covariance of the unrestricted estimator needed for the test.
  • Derive the Wald quadratic form using the delta method and Slutsky’s theorem.
  • Establish the chi-square reference distribution and its required rank condition.
  • Calculate Wald statistics for nonlinear and multiple restrictions.
  • Recognize why the Wald statistic depends on the parameterization used.

From an Estimated Parameter to a Test of a Restriction

An unrestricted maximum-likelihood estimate does not usually satisfy a null hypothesis exactly. The Wald test asks whether its estimated departure from the null restriction is large compared with the sampling uncertainty in that departure. For a smooth restriction, the delta method supplies that uncertainty: it transforms the covariance of the parameter estimate into the covariance of the restriction estimate.

This construction applies to generalized linear models (GLMs) when the unrestricted estimator has a suitable asymptotic normal distribution and its covariance can be estimated consistently. The test uses the unrestricted fit; it does not require fitting the model subject to the null.

Definition: Let \(\theta\) be a \(p\)-dimensional parameter in an open parameter space, and let \(h:\mathbb{R}^p\to\mathbb{R}^q\) be a continuously differentiable function near the true parameter \(\theta_0\). A smooth restriction is the null hypothesis \(H_0:h(\theta_0)=0\). Its Jacobian is the \(q\times p\) matrix \(H(\theta)=\partial h(\theta)/\partial\theta^\mathsf{T}\). The restriction is locally independent at \(\theta_0\) when \(H(\theta_0)\) has row rank \(q\).

For example, a single restriction on two parameters might be \(h(\theta)=\theta_1^2-\theta_2=0\). Its Jacobian is the row vector \((2\theta_1,-1)\), which has rank one. The restriction is not required to be linear; differentiability and the rank condition are what allow a local normal approximation to be used.

Deriving the Wald Statistic

Suppose the unrestricted maximum-likelihood estimator \(\hat\theta\) is consistent and, at the null parameter, satisfies \(\sqrt{n}(\hat\theta-\theta_0)\xrightarrow{d}N(0,V)\), where \(V\) is positive definite. In likelihood applications, \(V\) is often the limit of \(n\) times the estimator’s covariance, for example the inverse of a suitable limiting information matrix. Assume also that an estimator \(\hat C_n\) of \(\operatorname{Var}(\hat\theta)\) satisfies \(n\hat C_n\xrightarrow{p}V\). These are asymptotic assumptions; the resulting test is not generally an exact finite-sample test.

Under \(H_0\), the delta method gives the limiting distribution of \(h(\hat\theta)\). Since the null value is zero, the transformed estimate itself is a measure of departure from the null. Its estimated covariance is obtained by applying the Jacobian to the estimated covariance of \(\hat\theta\).

Theorem: Suppose \(\theta_0\) is an interior parameter value, \(h\) is continuously differentiable near \(\theta_0\), \(H_0:h(\theta_0)=0\) holds, \(H_0\) has \(q\) locally independent restrictions, and \(\sqrt{n}(\hat\theta-\theta_0)\xrightarrow{d}N(0,V)\) for positive-definite \(V\). If \(\hat\theta\xrightarrow{p}\theta_0\) and \(n\hat C_n\xrightarrow{p}V\), then the Wald statistic $$ W_n=h(\hat\theta)^\mathsf{T} \left\{H(\hat\theta)\hat C_n H(\hat\theta)^\mathsf{T}\right\}^{-1} h(\hat\theta) $$ converges in distribution under the null to \(\chi^2_q\).

Here \(H(\hat\theta)\hat C_nH(\hat\theta)^\mathsf{T}\) estimates the covariance of the \(q\)-vector \(h(\hat\theta)\). Its inverse exists with probability approaching one under the stated conditions. Equivalently, the statistic can be written with an explicit sample-size factor:

$$ W_n=n\,h(\hat\theta)^\mathsf{T} \left\{H(\hat\theta)(n\hat C_n)H(\hat\theta)^\mathsf{T}\right\}^{-1} h(\hat\theta). $$

To see the equivalence, for any invertible matrix \(A\) and positive scalar \(n\), \((A/n)^{-1}=nA^{-1}\). Thus the two expressions differ only in whether the factor \(n\) is written inside or outside the inverse.

Why the Reference Distribution Is Chi-Square

The derivation has two steps. First, approximate the transformed estimate by a multivariate normal vector. Second, standardize its covariance and use the distribution of a squared length of independent standard normal variables.

1
Linearize the restriction.
Because \(h(\theta_0)=0\), differentiability gives \(\sqrt{n}h(\hat\theta)=H(\theta_0)\sqrt{n}(\hat\theta-\theta_0)+o_p(1)\). The delta method therefore gives a limiting normal distribution with covariance \(A=H(\theta_0)VH(\theta_0)^\mathsf{T}\).
2
Estimate and standardize the covariance.
Consistency of \(\hat\theta\), continuity of \(H\), and \(n\hat C_n\xrightarrow{p}V\) imply \(nH(\hat\theta)\hat C_nH(\hat\theta)^\mathsf{T}\xrightarrow{p}A\). The rank condition and positive definiteness of \(V\) make \(A\) positive definite.
3
Form the quadratic statistic.
Slutsky’s theorem then gives the limiting quadratic form \(Z^\mathsf{T}A^{-1}Z\), where \(Z\sim N(0,A)\). Its distribution is \(\chi^2_q\).

The Taylor expansion in the first step is justified locally: differentiability at \(\theta_0\) means \(h(\hat\theta)=h(\theta_0)+H(\theta_0)(\hat\theta-\theta_0)+r_n\), where \(\|r_n\|/\|\hat\theta-\theta_0\|\xrightarrow{p}0\). Asymptotic normality implies \(\sqrt n(\hat\theta-\theta_0)=O_p(1)\), so \(\sqrt n r_n=o_p(1)\). This yields the claimed linearization without assuming that the restriction is exactly linear.

For completeness, \(A\) is positive definite because for any nonzero \(a\in\mathbb{R}^q\), full row rank of \(H(\theta_0)\) implies \(H(\theta_0)^\mathsf{T}a\ne0\). Hence \(a^\mathsf{T}Aa=(H(\theta_0)^\mathsf{T}a)^\mathsf{T}V(H(\theta_0)^\mathsf{T}a)>0\). By the spectral theorem, \(A\) has a symmetric positive-definite square root \(A^{1/2}\). The vector \(A^{-1/2}Z\) is therefore \(N(0,I_q)\), and \(Z^\mathsf{T}A^{-1}Z=(A^{-1/2}Z)^\mathsf{T}(A^{-1/2}Z)\), the sum of squares of \(q\) independent standard normal variables. That sum has a \(\chi^2_q\) distribution. Convergence of the covariance estimate and Slutsky’s theorem establish the theorem. \(\square\)

Worked Calculations

Worked Example: A Nonlinear Restriction on Two Parameters

Suppose an unrestricted estimator has value \(\hat\theta=(1.4,1.2)^\mathsf{T}\) and estimated covariance \(\hat C_n=\frac{1}{100}\begin{pmatrix}0.16&0.02\\0.02&0.09\end{pmatrix}\). Test \(H_0:\theta_1^2-\theta_2=0\) using the Wald statistic.

Here \(h(\theta)=\theta_1^2-\theta_2\), so \(h(\hat\theta)=1.4^2-1.2=1.96-1.2=0.76\). The Jacobian at the estimate is \(H(\hat\theta)=(2.8,-1)\). Multiplying by the covariance scale matrix gives

$$ H(\hat\theta) \begin{pmatrix}0.16&0.02\\0.02&0.09\end{pmatrix} H(\hat\theta)^\mathsf{T} =2.8^2(0.16)+2(2.8)(-1)(0.02)+(-1)^2(0.09) =1.2544-0.112+0.09=1.2324. $$

Thus the estimated variance of \(h(\hat\theta)\) is \(1.2324/100=0.012324\), and \(W_n=0.76^2/0.012324=0.5776/0.012324\approx46.868\). Under the null, the reference distribution is \(\chi^2_1\). The nonlinear restriction is handled by evaluating its derivative at the unrestricted estimate, not by pretending the restriction is linear.

Worked Example: Two Restrictions on Three Parameters

Let \(\hat\theta=(0.4,-0.2,1.1)^\mathsf{T}\) and \(\hat C_n=\frac{1}{100}\begin{pmatrix}4&1&0\\1&3&1\\0&1&2\end{pmatrix}\). Test the two restrictions \(\theta_1+\theta_2=0\) and \(\theta_2+\theta_3=1\).

Write \(h(\theta)=(\theta_1+\theta_2,\theta_2+\theta_3-1)^\mathsf{T}\). Then \(h(\hat\theta)=(0.2,-0.1)^\mathsf{T}\), and the Jacobian is \(H=\begin{pmatrix}1&1&0\\0&1&1\end{pmatrix}\).

$$ H \begin{pmatrix}4&1&0\\1&3&1\\0&1&2\end{pmatrix} H^\mathsf{T} = \begin{pmatrix}9&5\\5&7\end{pmatrix}. $$

For example, the first diagonal entry is \(\operatorname{Var}\)-scale \(4+3+2(1)=9\), and the off-diagonal entry is \(1+0+3+1=5\). The determinant of this matrix is \(9(7)-5^2=38\), so its inverse is \(\frac{1}{38}\begin{pmatrix}7&-5\\-5&9\end{pmatrix}\). The Wald statistic is therefore

$$ W_n=100\,(0.2,-0.1)\frac{1}{38} \begin{pmatrix}7&-5\\-5&9\end{pmatrix} \begin{pmatrix}0.2\\-0.1\end{pmatrix} =\frac{100}{38}\left(7(0.04)+10(0.02)+9(0.01)\right) =\frac{57}{38}=1.5. $$

The reference distribution is \(\chi^2_2\), because the two restriction gradients are linearly independent. The off-diagonal covariance matters: dropping it would not give the same quadratic form.

Worked Example: Testing an Odds-Ratio Restriction

In a logistic regression, let \(\beta\) be a coefficient whose exponent is an odds ratio. Suppose the unrestricted estimate is \(\hat\beta=0.75\), with estimated variance \(0.04\). Test the smooth restriction that the odds ratio equals 2, or \(h(\beta)=e^\beta-2=0\).

The estimated restriction is \(h(\hat\beta)=e^{0.75}-2\approx2.117000-2=0.117000\). Its derivative is \(h'(\beta)=e^\beta\), so \(h'(\hat\beta)\approx2.117000\). The delta-method variance estimate is \(h'(\hat\beta)^2\operatorname{Var}(\hat\beta)=e^{1.5}(0.04)\approx4.48169(0.04)=0.179268\).

The statistic is \(W_n\approx(0.117000)^2/0.179268=0.013689/0.179268\approx0.0764\). The null reference distribution is \(\chi^2_1\). This calculation tests the odds-ratio restriction on its stated scale while using the estimated covariance of the regression coefficient.

Conditions, Interpretation, and a Common Pitfall

The chi-square calibration is a large-sample result under the null. It relies on a regular, interior parameter value, asymptotic normality, a consistent covariance estimate, and a locally full-rank Jacobian. If the restriction is redundant, or the parameter lies on a boundary with a nonstandard estimator limit, these conditions may fail and the stated reference distribution need not apply. In a GLM with nuisance parameters, the covariance used must correspond to the full estimator or its appropriate marginal covariance; omitting uncertainty in nuisance parameters can give the wrong variance for \(h(\hat\theta)\).

A frequent mistake is to compare \(h(\hat\theta)^2\) directly with a chi-square distribution, as though every restriction estimate had variance one. The correction is to scale the departure by its estimated covariance, including the Jacobian transformation. For multiple restrictions, that means the full covariance matrix and its inverse, not separate unadjusted squared departures.

A second caution is that the Wald test is built around the unrestricted estimate and its local linear approximation. A nonlinear transformation can change the numerical Wald statistic, even when it describes the same null set. For that reason, the parameterization and the restriction function should be chosen deliberately, and the result should be understood as an asymptotic test rather than an invariant measure of evidence.

Key takeaway: For a smooth \(q\)-dimensional null restriction, the Wald statistic is the squared, covariance-standardized departure \(h(\hat\theta)\). Under regularity conditions and the null, its asymptotic reference distribution is \(\chi^2_q\).

Check Your Understanding

Use the derivation and examples above to answer these questions.

  1. For \(h(\theta)=\theta_1\theta_2-1\), what is the Jacobian row vector, and what condition makes this a locally independent single restriction?
  2. Why does a \(q\)-component restriction with full-row-rank Jacobian lead to a \(\chi^2_q\) reference distribution rather than a \(\chi^2_p\) distribution?
  3. In the two-restriction example, what role does the off-diagonal entry of the transformed covariance matrix play in the Wald statistic?
  4. State two conditions needed for the chi-square approximation that may fail for a boundary parameter.
  5. Why is the raw squared departure \(h(\hat\theta)^\mathsf{T}h(\hat\theta)\) generally not the Wald statistic?
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