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Wald, Score, and Likelihood-Ratio Tests · Tutorial 884 of 970

Wald Tests with Nuisance Parameters

Build and interpret Wald tests for target parameters while correctly accounting for nuisance parameters estimated in the same model.

Advanced 10 min read

What You'll Learn

  • Partition a parameter vector into target and nuisance components for a Wald test.
  • Extract the target covariance block from the full estimated covariance matrix.
  • Explain why nuisance estimation changes the target block of the covariance matrix.
  • Derive the chi-square calibration of the Wald statistic under the null.
  • Relate the target covariance block to the Schur complement of the information matrix.
  • Identify the error in treating nuisance parameters as known when calculating a Wald statistic.

Why Nuisance Parameters Belong in a Wald Test

A hypothesis may concern only one parameter, or a small group of parameters, even though a model estimates many others. The parameters named in the hypothesis are the target parameters; the remaining parameters are nuisance parameters. “Nuisance” does not mean irrelevant: estimating them can affect the uncertainty of the target estimates.

For a Wald test, the relevant uncertainty is the covariance of the target estimator when the model’s parameters are estimated jointly. It is obtained by taking the target block of the full estimated covariance matrix. In general, it is not the variance that would apply if nuisance parameters were known and held fixed.

Definition: Partition a parameter vector as \(\theta=(\psi^\mathsf{T},\lambda^\mathsf{T})^\mathsf{T}\), where \(\psi\) is a \(q\)-dimensional target parameter and \(\lambda\) is a nuisance parameter. For the null hypothesis \(H_0:\psi=\psi_0\), the Wald statistic is
$$ W=(\hat\psi-\psi_0)^\mathsf{T} \left\{\widehat{\operatorname{Cov}}(\hat\psi)\right\}^{-1} (\hat\psi-\psi_0). $$
Here \(\widehat{\operatorname{Cov}}(\hat\psi)\) is the target block of the estimated covariance matrix for the full estimator \(\hat\theta\), not a covariance calculated as though \(\lambda\) were known.

If the estimated covariance of \(\hat\theta\) is reported directly, take its rows and columns corresponding to \(\psi\). If instead a procedure reports an estimate \(\hat V\) of the asymptotic covariance of \(\sqrt n(\hat\theta-\theta)\), then the estimated covariance of \(\hat\theta\) is \(\hat V/n\), and its target block is \(\hat V_{\psi\psi}/n\). Keeping track of this scaling prevents a common factor-of-\(n\) error.

The Wald Statistic and Its Null Calibration

For a \(q\)-dimensional target, a large value of \(W\) indicates that the estimate is far from the null value relative to its estimated uncertainty. Under regularity conditions, the null reference distribution is chi-square with \(q\) degrees of freedom. The result below states the conditions needed for that calibration.

Theorem: Suppose that under \(H_0\), \(\sqrt n(\hat\theta-\theta_0)\xrightarrow{d}N(0,V)\), where \(\theta_0=(\psi_0^\mathsf{T},\lambda_0^\mathsf{T})^\mathsf{T}\), and \(V\) is positive definite. Suppose also that \(\hat V\xrightarrow{p}V\), and that the estimated covariance of \(\hat\theta\) is \(\hat V/n\). Then the Wald statistic for \(H_0:\psi=\psi_0\), using the target block \(\hat V_{\psi\psi}/n\), satisfies \(W\xrightarrow{d}\chi^2_q\).

Proof. Let \(Z_n=\sqrt n(\hat\psi-\psi_0)\), the target coordinates of \(\sqrt n(\hat\theta-\theta_0)\), and let \(V_{\psi\psi}\) be the corresponding block of \(V\). Marginal convergence gives \(Z_n\xrightarrow{d}N(0,V_{\psi\psi})\). Consistency of \(\hat V\) implies \(\hat V_{\psi\psi}\xrightarrow{p}V_{\psi\psi}\). Rewrite the statistic as

$$ W=Z_n^\mathsf{T}\hat V_{\psi\psi}^{-1}Z_n. $$

Because \(V_{\psi\psi}\) is positive definite, matrix inversion is continuous at \(V_{\psi\psi}\), so \(\hat V_{\psi\psi}^{-1}\xrightarrow{p}V_{\psi\psi}^{-1}\). Slutsky’s theorem and continuity of the quadratic-form map give the limiting distribution of \(Z_n^\mathsf{T}V_{\psi\psi}^{-1}Z_n\). If \(Z\sim N(0,V_{\psi\psi})\), then \(V_{\psi\psi}^{-1/2}Z\sim N(0,I_q)\); its squared Euclidean length has a \(\chi^2_q\) distribution. Therefore \(W\xrightarrow{d}\chi^2_q\). \(\square\)

At significance level \(\alpha\), reject when \(W\) exceeds the \(1-\alpha\) quantile of \(\chi^2_q\). Equivalently, the large-sample p-value is the upper-tail probability \(P(\chi^2_q\ge W)\). This calibration is asymptotic; the theorem does not assert that the statistic has an exact chi-square distribution in a finite sample.

Why the Full Covariance Matrix Matters

In a regular likelihood model, the covariance block used in the Wald statistic can be understood through the information matrix. Let the per-observation Fisher information for \(\theta=(\psi^\mathsf{T},\lambda^\mathsf{T})^\mathsf{T}\) be partitioned as

$$ I= \begin{pmatrix} I_{\psi\psi}&I_{\psi\lambda}\\ I_{\lambda\psi}&I_{\lambda\lambda} \end{pmatrix}. $$

When nuisance parameters are estimated, the asymptotic covariance of the full maximum likelihood estimator is \(I^{-1}/n\). Thus the target covariance is the \(\psi\psi\) block of \(I^{-1}/n\). It is generally not \(I_{\psi\psi}^{-1}/n\), which corresponds to using the target information as if the nuisance parameters were fixed.

Result: If \(I\) is positive definite and \(I_{\lambda\lambda}\) is invertible, then
$$ (I^{-1})_{\psi\psi} = \left(I_{\psi\psi}-I_{\psi\lambda}I_{\lambda\lambda}^{-1}I_{\lambda\psi}\right)^{-1}. $$
The matrix inside the inverse is the information for \(\psi\) after accounting for estimation of \(\lambda\).

Proof. Write the first block column of \(I^{-1}\) as \(\begin{pmatrix}A\\B\end{pmatrix}\), so that

$$ I\begin{pmatrix}A\\B\end{pmatrix} = \begin{pmatrix}I_q\\0\end{pmatrix}. $$

The lower block equation gives \(I_{\lambda\psi}A+I_{\lambda\lambda}B=0\), hence \(B=-I_{\lambda\lambda}^{-1}I_{\lambda\psi}A\). Substituting this into the upper block equation yields

$$ \left(I_{\psi\psi}-I_{\psi\lambda}I_{\lambda\lambda}^{-1}I_{\lambda\psi}\right)A=I_q. $$

The matrix in parentheses is invertible because \(I\) is positive definite. Therefore \(A\) equals its inverse. Since \(A\) is the upper-left block of \(I^{-1}\), the identity follows. \(\square\)

The subtracted term reflects the information about the target that is entangled with the nuisance parameters. When the target and nuisance scores are not orthogonal, estimating the nuisance parameters changes the target’s uncertainty. The block-inverse identity makes precise why one must use the full information matrix before extracting the target covariance.

Worked Calculations

Worked Example: A Target Mean with a Correlated Nuisance Mean

Suppose \(X_1,\ldots,X_n\) are independent observations from a bivariate normal distribution with mean \((\psi,\lambda)\) and known covariance

$$ \Sigma= \begin{pmatrix} 4&3\\ 3&9 \end{pmatrix}. $$

Test \(H_0:\psi=0\), treating \(\lambda\) as a nuisance parameter. With \(n=100\), suppose the sample mean is \((\hat\psi,\hat\lambda)=(0.35,1.2)\). The joint maximum likelihood estimator is the sample mean, and its covariance matrix is \(\Sigma/n\). The target block is the first diagonal entry:

$$ \widehat{\operatorname{Var}}(\hat\psi)=\frac{4}{100}=0.04. $$

Consequently,

$$ W=\frac{(0.35-0)^2}{0.04} =\frac{0.1225}{0.04} =3.0625. $$

This is compared with \(\chi^2_1\); its upper-tail probability is approximately \(0.080\). The off-diagonal covariance does not enter this scalar target block directly, but the block must still come from the joint covariance matrix. To see the danger of treating \(\lambda\) as known, the per-observation information is

$$ \Sigma^{-1} =\frac{1}{27} \begin{pmatrix} 9&-3\\ -3&4 \end{pmatrix}. $$

Using only the target information \(I_{\psi\psi}=1/3\) would give variance \(3/100=0.03\), rather than the correct \(4/100=0.04\). That incorrect fixed-nuisance calculation would produce \(W=0.1225/0.03\approx4.083\), a materially different test.

Worked Example: Two Target Parameters and One Nuisance Parameter

Suppose an estimator of \(\theta=(\psi_1,\psi_2,\lambda)\) has estimated covariance matrix

$$ \widehat{\operatorname{Cov}}(\hat\theta) = \frac{1}{100} \begin{pmatrix} 4&1&1\\ 1&9&2\\ 1&2&4 \end{pmatrix}. $$

Test \(H_0:(\psi_1,\psi_2)=(0,0)\). The target block, including the covariance between the two target estimates, is

$$ \widehat{\operatorname{Cov}}(\hat\psi) = \frac{1}{100} \begin{pmatrix} 4&1\\ 1&9 \end{pmatrix}. $$

Suppose \(\hat\psi=(0.5,-1)^\mathsf{T}\). The determinant of the unscaled target block is \(4(9)-1(1)=35\), so its inverse is

$$ \left[ \begin{pmatrix} 4&1\\ 1&9 \end{pmatrix} \right]^{-1} = \frac{1}{35} \begin{pmatrix} 9&-1\\ -1&4 \end{pmatrix}. $$

The factor \(1/100\) in the covariance matrix makes the inverse covariance \(100/35\begin{pmatrix}9&-1\\-1&4\end{pmatrix}\). Therefore

$$ W = \begin{pmatrix}0.5&-1\end{pmatrix} \frac{100}{35} \begin{pmatrix}9&-1\\-1&4\end{pmatrix} \begin{pmatrix}0.5\\-1\end{pmatrix} = \frac{100}{35} \left(0.5(5.5)+(-1)(-4.5)\right) = \frac{100(7.25)}{35} \approx20.714. $$

The reference distribution is \(\chi^2_2\). Using only the two diagonal variances would discard the covariance \(1/100\) between the target estimates and would not give this Wald statistic.

Worked Example: Testing a Regression Slope with an Unknown Intercept

Consider the normal linear model \(Y_i=\alpha+\beta x_i+\varepsilon_i\), with independent errors \(\varepsilon_i\sim N(0,4)\). The target is the slope \(\beta\); the intercept \(\alpha\) is a nuisance parameter. Use the four fixed design values \(x_i=1,2,3,4\). For the parameter order \((\beta,\alpha)\), the information matrix is

$$ I= \frac{1}{4} \begin{pmatrix} \sum_i x_i^2&\sum_i x_i\\ \sum_i x_i&4 \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 30&10\\ 10&4 \end{pmatrix}. $$

Its inverse is

$$ I^{-1} = \begin{pmatrix} 0.8&-2\\ -2&6 \end{pmatrix}, $$

as can be checked by multiplying it by \(I=\begin{pmatrix}7.5&2.5\\2.5&1\end{pmatrix}\). The slope variance is therefore \(0.8\). If the fitted slope is \(1.4\) and the null is \(H_0:\beta=1\), the Wald statistic is

$$ W=\frac{(1.4-1)^2}{0.8}=\frac{0.16}{0.8}=0.2. $$

For contrast, using only \(I_{\beta\beta}=7.5\) as though the intercept were known gives the incorrect variance \(1/7.5=2/15\), and hence the incorrect statistic \(0.16/(2/15)=1.2\). This small design illustrates the covariance calculation; chi-square calibration is a large-sample result, not an exact finite-sample claim here.

Common Pitfall and Practical Checks

Wrong reasoning: “The null concerns only \(\psi\), so I should use the variance for \(\hat\psi\) with \(\lambda\) held fixed.” This changes the estimation problem. In the actual fit, \(\lambda\) is estimated, and its uncertainty and association with \(\hat\psi\) are reflected in the target block of the full covariance matrix.

Correction: Fit the model with nuisance parameters estimated as specified, obtain the full covariance matrix for that fit, and extract the target rows and columns. Then invert that target block to form the Wald statistic. For multiple target parameters, keep their off-diagonal covariances; for any target dimension, check whether the reported matrix is the covariance of \(\hat\theta\) or the asymptotic covariance of \(\sqrt n(\hat\theta-\theta)\).

The main practical checks are: the null value and target coordinates have the same dimension; the extracted covariance block is symmetric and positive definite; and the inverse is applied to the entire target block, not separately to its entries. The number of chi-square degrees of freedom here is the number \(q\) of target coordinates in \(H_0:\psi=\psi_0\).

Check Your Understanding

Use the definitions and calculations above to answer these questions.

  1. In a model with target \(\psi\) and nuisance parameter \(\lambda\), which part of the full covariance matrix is used for a scalar test of \(H_0:\psi=\psi_0\)?
  2. For a two-dimensional target, why can the covariance between its two estimates affect the Wald statistic?
  3. What covariance matrix is used if a procedure reports \(\hat V\) as an estimate of the asymptotic covariance of \(\sqrt n(\hat\theta-\theta)\)?
  4. In the information-matrix identity, what does the term \(I_{\psi\lambda}I_{\lambda\lambda}^{-1}I_{\lambda\psi}\) account for?
  5. Under the stated theorem, what is the null limiting distribution of a Wald statistic for \(q\) target coordinates?
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