Comparing Tests for a Joint Poisson Regression Effect

Why Compare the Tests for a Joint Poisson Effect?

In a Poisson regression, a coefficient describes how a covariate changes the log of the expected count or rate. Often the scientific question concerns several covariates together: for example, whether two treatment indicators contribute anything beyond an intercept. The joint null hypothesis sets both coefficients to zero, and the Wald, score, and likelihood-ratio tests each compare that null with the same alternative.

The tests use different information. The Wald test evaluates the unrestricted estimate and its covariance; the score test evaluates the slope and information at the restricted fit; the likelihood-ratio test compares the two maximized log-likelihoods. Under standard regularity conditions they have the same limiting chi-square distribution under the null, but their finite-sample statistics need not agree.

When the model is identifiable, the true parameter is an interior point, the relevant information is nonsingular, and the sample grows under the null with fixed parameter dimension, each statistic has an asymptotic \(\chi^2_q\) distribution. The same degrees of freedom apply to the joint test because there are \(q\) independent restrictions. In a finite sample, the chi-square p-values are approximations; they do not imply that the three observed statistics must match.

Set Up the Joint Test

Suppose independent observations satisfy \(Y_i\sim\operatorname{Poisson}(\mu_i)\), with \(\log(\mu_i)=\alpha+\beta_1x_{i1}+\beta_2x_{i2}\). The joint test of no covariate effect is \(H_0:\beta_1=\beta_2=0\). For this two-restriction test, the reference distribution is \(\chi^2_2\). A useful comparison keeps the null hypothesis, data, and reference distribution fixed, and changes only which fitted model supplies the statistic.

A Four-Cell Poisson Example

Consider four independent counts, one for each combination of two covariates coded \(-1\) and \(+1\). Let \(x_1\) index rows and \(x_2\) index columns, and use the additive log-mean model \(\log(\mu)=\alpha+\beta_1x_1+\beta_2x_2\). There are no offsets. The observed counts are:

We test \(H_0:\beta_1=\beta_2=0\), so \(q=2\). The unrestricted model fits the four observed means exactly: \(60\times30=40\times45\), which makes the four log-counts additive in \(x_1\) and \(x_2\). The fitted coefficients are \(\hat{\beta}_1=\tfrac12\log(4/3)\) and \(\hat{\beta}_2=\tfrac12\log(3/2)\). The unrestricted intercept is \(\hat{\alpha}=\tfrac14\log(60\cdot40\cdot45\cdot30)\), and the restricted fit has common fitted mean \(\tilde{\mu}=(60+40+45+30)/4=43.75\).

Calculate the Score, Wald, and Likelihood-Ratio Statistics

For the score test, use the restricted fit. The design vector is \((1,x_1,x_2)^\mathsf{T}\). Since the four covariate combinations are balanced, both covariates sum to zero. The intercept score is zero at the restricted estimate, while the two slope-score components are the signed count contrasts:

At the restricted fit, the Poisson information matrix is \(\tilde{\mu}X^\mathsf{T}X\). The intercept is orthogonal to both slope columns, and the two slope columns are orthogonal to each other. Each slope-information entry is \(4(43.75)=175\). Thus the efficient information for \(\boldsymbol{\beta}\), after accounting for the intercept, is \(175I_2\).

For two degrees of freedom, the chi-square upper-tail probability is \(\exp(-S/2)\). Here the score-test p-value is \(\exp(-10.5714/2)\approx0.00506\).

For the Wald test, use the unrestricted coefficient estimates and their covariance matrix. The expected information at the unrestricted fit, with rows ordered as \((1,x_1,x_2)\), is

Removing the nuisance intercept by taking the Schur complement gives the information for the two coefficients:

The corresponding covariance estimate is the inverse of this matrix. Therefore

Its \(\chi^2_2\) p-value is \(\exp(-10.4518/2)\approx0.00538\). This calculation uses the unrestricted estimate and covariance; substituting the restricted information from the score calculation would not give the Wald statistic.

Finally, the likelihood-ratio statistic follows from the likelihood derivation above:

The likelihood-ratio p-value is \(\exp(-10.6311/2)\approx0.00491\). Thus, at the 0.01 level all three tests reject. At the 0.005 level, only the likelihood-ratio test rejects for these data: the score p-value is just above 0.005, and the Wald p-value is also above it. This is a genuine finite-sample difference, not a change in the hypothesis or degrees of freedom.

Why the Statistics Differ

The score statistic asks how strongly the likelihood initially points away from the null, measured using information at the restricted fit. The Wald statistic measures the unrestricted estimate’s distance from the null using covariance at the unrestricted fit. The likelihood-ratio statistic measures the total gain in maximized log-likelihood. These are distinct calculations, so finite-sample differences are expected; their asymptotic equivalence does not make their observed values identical.

Common Mistakes

• Using the wrong fit for a statistic. The score test uses the restricted fit and its efficient information; the Wald test uses the unrestricted estimate and covariance.
• Using one degree of freedom because there is one scientific question. The reference degrees of freedom are determined by the number of independent restrictions. Here there are two.
• Assuming asymptotic equivalence means identical results. The three procedures can produce different finite-sample statistics and p-values. Near a chosen significance threshold, report the test and its numerical result rather than claiming the methods must agree.
• Reading a joint rejection as evidence that every coefficient matters. A joint test rejects the hypothesis that all tested coefficients are zero; it does not identify which coefficients drive the result.