Joint Contrast Tests in Quasi-GLMs

Why Test Several Coefficients Together?

A coefficient-by-coefficient review can miss the question a scientific analysis actually asks. For example, a categorical predictor with four levels has three treatment coefficients under reference-level coding. The question “Does this factor contribute anything to the model?” concerns all three coefficients together, not three separate decisions. A joint test accounts for their estimated covariances and assesses the full set of restrictions in one calculation.

In a quasi-binomial or quasi-Poisson model, the estimated dispersion affects the coefficient covariance matrix. A joint contrast test therefore uses that dispersion-adjusted covariance. Its familiar \(F\) reference distribution is a finite-sample approximation: quasi-likelihood specifies mean and variance relationships, but does not in general supply a full probability model that makes the \(F\) distribution exact.

Linear Restrictions and the Joint Wald Statistic

Definition: Let \(\hat{\beta}\) be a \(p\)-dimensional coefficient estimator, and let \(\widehat V\) estimate its covariance matrix. A joint linear hypothesis is written \(H_0:C\beta=d\), where \(C\) is a known \(q\times p\) matrix of rank \(q\), and \(d\) is a known \(q\)-vector. The estimated contrast vector is \(C\hat{\beta}-d\), with estimated covariance \(\widehat S=C\widehat V C^\mathsf{T}\). If \(\widehat S\) is nonsingular, the joint Wald statistic is \(W=(C\hat{\beta}-d)^\mathsf{T}\widehat S^{-1}(C\hat{\beta}-d)\).

The rank \(q\) is the number of independent restrictions, not necessarily the number of rows originally written down. If some rows of \(C\) are redundant, remove them or use an equivalent full-rank representation. The estimate \(\widehat V\) should include the estimated quasi-dispersion. For instance, under the usual model-based quasi-GLM covariance calculation it has the form \(\hat{\phi}(X^\mathsf{T}WX)^{-1}\), with the appropriate design and working-weight matrices for the fitted model.

A common reporting convention converts the joint Wald statistic to \(F_{\mathrm{obs}}=W/q\), and compares it with an \(F_{q,\nu}\) distribution. Here \(\nu\) is a specified denominator degrees of freedom, commonly the residual degrees of freedom \(n-p\) for a model with \(n\) independent observational units and \(p\) fitted coefficients. In grouped-binomial data, \(n\) in this convention counts grouped observations, not the total number of trials. The degrees-of-freedom convention should be stated, since the \(F\) reference is approximate and practices can differ.

Worked Example: Two Restrictions on Three Coefficients

Suppose a quasi-binomial fit estimates \(\beta=(\beta_1,\beta_2,\beta_3)^\mathsf{T}\) as \(\hat{\beta}=(0.4,-0.1,0.7)^\mathsf{T}\). The scientific question is whether \(\beta_1+\beta_2=0\) and \(\beta_3=1\). Thus \(C=\begin{pmatrix}1&1&0\\0&0&1\end{pmatrix}\) and \(d=(0,1)^\mathsf{T}\). Assume the fitted, dispersion-adjusted covariance of these two contrasts is \(\widehat S=\begin{pmatrix}0.09&0\\0&0.16\end{pmatrix}\).

The estimated contrast vector is \(C\hat{\beta}-d=(0.4-0.1,\;0.7-1)^\mathsf{T}=(0.3,-0.3)^\mathsf{T}\). The inverse covariance is \(\widehat S^{-1}=\operatorname{diag}(1/0.09,1/0.16)\), so \(W=0.3^2/0.09+(-0.3)^2/0.16=1+0.5625=1.5625\). There are \(q=2\) restrictions, giving \(F_{\mathrm{obs}}=1.5625/2=0.78125\). With denominator degrees of freedom \(\nu=30\), compare this with \(F_{2,30}\). The statistic is well below the usual 5% upper critical value, so this test does not reject the joint null at that level.

Why the Wald Statistic Has a Chi-Square Limit

The chi-square result describes large-sample calibration of the Wald statistic. It requires more than merely inserting an estimated covariance matrix: the estimator must have an appropriate joint normal limit, and the estimated contrast covariance must consistently represent its true covariance on the relevant scale.

Theorem: Suppose \(q\) is fixed, \(H_0:C\beta_0=d\) holds, and \(S_n\) is a positive-definite covariance matrix for the contrast estimator at sample size \(n\). Assume \(Z_n=S_n^{-1/2}(C\hat{\beta}-d)\xrightarrow{d}N(0,I_q)\). Let \(\widehat S_n\) be a positive-definite estimator of \(S_n\), and assume every eigenvalue of \(\widehat S_n^{-1/2}S_n\widehat S_n^{-1/2}\) converges in probability to \(1\). Then \(W_n=(C\hat{\beta}-d)^\mathsf{T}\widehat S_n^{-1}(C\hat{\beta}-d)\xrightarrow{d}\chi_q^2\).

Proof. Write \(X_n=C\hat{\beta}-d\), so \(Z_n=S_n^{-1/2}X_n\), and define \(K_n=S_n^{1/2}\widehat S_n^{-1}S_n^{1/2}\). Direct substitution gives \(W_n=Z_n^\mathsf{T}K_nZ_n\). The matrices \(K_n=AA^\mathsf{T}\) and \(\widehat S_n^{-1/2}S_n\widehat S_n^{-1/2}=A^\mathsf{T}A\), where \(A=S_n^{1/2}\widehat S_n^{-1/2}\), have the same eigenvalues. They are positive definite, so the assumed eigenvalue convergence implies \(\lVert K_n-I_q\rVert\to_p0\), where \(\lVert\cdot\rVert\) is the spectral norm.

Consequently, \(\left|W_n-Z_n^\mathsf{T}Z_n\right|\leq\lVert K_n-I_q\rVert\lVert Z_n\rVert^2\). Since \(Z_n\xrightarrow{d}N(0,I_q)\), the sequence \(\lVert Z_n\rVert^2\) is bounded in probability. The product on the right therefore converges in probability to zero. Also \(Z_n^\mathsf{T}Z_n\xrightarrow{d}\chi_q^2\), by the continuous mapping theorem applied to the squared Euclidean norm. Slutsky’s theorem now gives \(W_n\xrightarrow{d}\chi_q^2\). \(\square\)

This result supports the large-sample chi-square calibration. It does not, by itself, establish an exact \(F\) distribution for a quasi-GLM in a finite sample. Using \(W/q\) with a residual-degrees-of-freedom \(F\) reference is a practical adjustment when dispersion is estimated, not a finite-sample theorem for quasi-likelihood.

Testing a Multi-Level Factor

For a factor with \(K\) levels and a reference level, treatment coding represents its main effect with \(K-1\) coefficients. The joint null that the factor has no effect is that all \(K-1\) coefficients equal zero, holding the other model terms as specified. The test is a single \((K-1)\)-restriction contrast test. It is not equivalent to requiring every separate coefficient test to be nonsignificant: separate tests do not use the full covariance structure or address the joint hypothesis in one calibrated statistic.

Worked Example: A Four-Level Treatment Factor

A quasi-binomial model uses treatment A as the reference and includes indicators for treatments B, C, and D. The fitted logit coefficients are \((\hat{\beta}_B,\hat{\beta}_C,\hat{\beta}_D)=(0.20,-0.10,0.30)^\mathsf{T}\). The estimated dispersion is \(\hat{\phi}=1.5\), and the fitted covariance matrix of the three factor coefficients, already adjusted for dispersion, is \(\widehat S=\begin{pmatrix}0.04&0.008&0\\0.008&0.04&0\\0&0&0.06\end{pmatrix}\). Test whether treatment has any effect by testing \(H_0:\beta_B=\beta_C=\beta_D=0\).

Here \(C=I_3\), \(d=0\), and \(q=3\). The determinant of the upper-left \(2\times2\) covariance block is \(0.04(0.04)-0.008^2=0.001536\), so its inverse is \(\begin{pmatrix}26.0417&-5.2083\\-5.2083&26.0417\end{pmatrix}\) (rounded). The inverse variance for the third coefficient is \(1/0.06=16.6667\). Thus \(W\approx26.0417(0.20^2+(-0.10)^2)+2(-5.2083)(0.20)(-0.10)+16.6667(0.30^2)\) \(=1.3021+0.2083+1.5000=3.0104\).

The dispersion adjustment is already present in \(\widehat S\); multiplying the statistic by \(\hat{\phi}\) again would be an error. The observed value is \(F_{\mathrm{obs}}=3.0104/3\approx1.0035\). With, for example, \(\nu=40\) residual degrees of freedom, compare with \(F_{3,40}\). This value is below the usual 5% upper critical value, so the joint test does not reject at that level. The conclusion concerns the three treatment contrasts together; it is not proof that the treatments have identical response probabilities.

Equivalent Ways to Write the Same Restrictions

A hypothesis can be expressed using different, but equivalent, full-rank contrast matrices. For example, multiplying all rows of \(C\) and the vector \(d\) by the same nonsingular \(q\times q\) matrix changes the coordinates used to describe the restrictions, not the null set. The joint Wald statistic is unchanged.

To verify this, let \(A\) be nonsingular and replace \(C\) by \(AC\), and \(d\) by \(Ad\). Put \(x=C\hat{\beta}-d\). The new contrast covariance is \(A\widehat S A^\mathsf{T}\), and its inverse is \(A^{-\mathsf{T}}\widehat S^{-1}A^{-1}\). Therefore the new statistic is \((Ax)^\mathsf{T}A^{-\mathsf{T}}\widehat S^{-1}A^{-1}(Ax)=x^\mathsf{T}\widehat S^{-1}x=W\). So the result does not depend on the particular basis chosen for the same restrictions.

Worked Example: Joint Test of Two Quasi-Poisson Slope Terms

Suppose a quasi-Poisson log-mean model contains two slope coefficients \(\beta_1,\beta_2\) for terms whose joint contribution is of interest. Their estimates are \((0.25,-0.15)^\mathsf{T}\), and their dispersion-adjusted covariance matrix is \(\widehat S=\begin{pmatrix}0.04&0.01\\0.01&0.04\end{pmatrix}\). The test is \(H_0:\beta_1=\beta_2=0\), so \(q=2\).

The determinant of \(\widehat S\) is \(0.04^2-0.01^2=0.0015\), giving \(\widehat S^{-1}=\begin{pmatrix}26.6667&-6.6667\\-6.6667&26.6667\end{pmatrix}\) (rounded). Hence \(W\approx26.6667(0.25^2+(-0.15)^2)+2(-6.6667)(0.25)(-0.15)\) \(=26.6667(0.085)+0.5000=2.7667\). Thus \(F_{\mathrm{obs}}=2.7667/2=1.3834\). Using a prespecified denominator \(\nu=24\), compare this with \(F_{2,24}\). The test addresses whether the two slopes are jointly zero, accounting for their positive covariance; two separate one-coefficient tests would not be the same calculation.

Common Pitfalls and Practical Interpretation

Testing each coefficient and calling that a joint test. Separate tests do not form the joint Wald statistic. Use the full covariance matrix \(C\widehat V C^\mathsf{T}\).
Counting restrictions incorrectly. Use the rank \(q\) of the restriction matrix. Redundant equations do not increase the numerator degrees of freedom.
Applying dispersion twice. If the coefficient covariance matrix already includes \(\hat{\phi}\), do not multiply the Wald statistic by \(\hat{\phi}\) again.
Describing the \(F\) reference as exact. In quasi-likelihood, the \(F_{q,\nu}\) calibration is generally approximate. State the denominator degrees of freedom and interpret the resulting evidence with that limitation in mind.
Interpreting non-rejection as equality. Failure to reject the joint null means the data do not provide sufficient evidence against those restrictions at the chosen level; it does not establish that all effects are exactly zero.

Joint contrast tests matter because many scientific questions concern a set of model terms rather than one coefficient in isolation. In quasi-GLMs, the appropriate covariance scaling carries the estimated extra variability into the joint calculation. The result is a coherent test of the stated restrictions, while the chosen reference distribution and its approximation remain explicit.

Check Your Understanding

Answer these questions by identifying the restrictions, the covariance needed, and the calibration being used.

A five-level factor is coded with one reference level. How many independent restrictions express the null that its main effect is zero?
Why must the joint Wald statistic use the covariance between estimated contrasts, rather than only their individual variances?
If a restriction matrix has four rows but rank three, what is the numerator degrees of freedom for the joint test?
A quasi-Poisson covariance matrix already includes the estimated dispersion. Should the Wald statistic be multiplied by the dispersion estimate again? Explain.
What does the large-sample chi-square result establish, and why does it not make the residual-df \(F\) reference exact in every quasi-GLM?