Poisson Loglinear Deviance Test Worked Example

Introduction

A contingency table records counts for combinations of categories. A Poisson loglinear model treats those cell counts as responses and represents patterns in their expected counts through main effects and interactions. Comparing a model with main effects only to one that includes interactions gives a direct test of whether the two categorical variables are associated.

The key is to keep the statistical question and the calculation connected: the added interaction terms specify the alternative hypothesis, and the difference in deviances measures how much the fitted likelihood improves when those terms are included. The test is approximate in general, so its chi-square interpretation depends on the usual regularity and large-sample conditions for a likelihood-ratio test.

Independence and Association in Loglinear Models

Let \(Y_{ij}\) be the count in row category \(i\) and column category \(j\) of an \(r\)-by-\(c\) table, where \(i=1,\ldots,r\) and \(j=1,\ldots,c\). In the Poisson sampling formulation, the cell counts are independent Poisson random variables with means \(\mu_{ij}>0\). Write \(y_{ij}\) for the observed count.

The parameters in a loglinear model need constraints for unique parameter labels. For example, one may impose sum-to-zero constraints on the row and column effects, and corresponding margin constraints on interaction effects. These constraints identify the parameters; they do not change the fitted means or the hypotheses being tested.

For a two-by-two table, the independence hypothesis is \(\mu_{11}\mu_{22}=\mu_{12}\mu_{21}\), or, equivalently, \(\mu_{11}\mu_{22}/(\mu_{12}\mu_{21})=1\). In larger tables, independence requires this cross-product equality for every choice of two rows and two columns. Thus, adding the interaction block tests whether the table needs association terms beyond its row and column main effects.

Fitting the Independence Model

Let \(y_{i+}=\sum_j y_{ij}\) be the observed row totals, \(y_{+j}=\sum_i y_{ij}\) the observed column totals, and \(n=\sum_i\sum_j y_{ij}\) the grand total. The independence-model fitted means are the products of the corresponding margins divided by the grand total.

To verify the result, write the Poisson log-likelihood, omitting no terms that depend on the means, as \(\ell=\sum_{i,j}\{y_{ij}\log(\mu_{ij})-\mu_{ij}-\log(y_{ij}!)\}\). Under independence, \(\mu_{ij}=\exp(\lambda+\alpha_i+\beta_j)\). The score for a row effect is the observed row total minus its fitted row total, and the score for a column effect is the observed column total minus its fitted column total. The score for the intercept is the observed grand total minus the fitted grand total.

Now set \(\hat{\mu}_{ij}=y_{i+}y_{+j}/n\). Summing over columns gives \(\sum_j\hat{\mu}_{ij}=y_{i+}(\sum_j y_{+j})/n=y_{i+}\), because \(\sum_j y_{+j}=n\). Similarly, \(\sum_i\hat{\mu}_{ij}=y_{+j}\), and summing all fitted cells gives \(n\). Every score is therefore zero. The log-likelihood is concave in the canonical parameters: its Hessian is \(-X^\mathsf{T}\operatorname{diag}(\mu_{ij})X\), where \(X\) is the model matrix. After imposing identifying constraints, \(X\) has full column rank, so this Hessian is negative definite for positive fitted means. The stationary fit is consequently the maximum-likelihood fit. \(\square\)

If a row or column total is zero, the displayed fitted means include zeros and the maximum can lie on the boundary rather than at finite loglinear parameter values. The worked tables below have positive margins, so the interior argument applies directly.

From Deviance to the Test Statistic

For any fitted model \(M\), its deviance is twice the log-likelihood difference between the saturated model and \(M\). In a table with positive observed counts, the saturated model fits \(\hat{\mu}_{ij}=y_{ij}\). Subtracting the two Poisson log-likelihoods gives \(D(M)=2\sum_{i,j}\{y_{ij}\log(y_{ij}/\hat{\mu}_{ij})-(y_{ij}-\hat{\mu}_{ij})\}\).

Both the independence fit and the saturated fit have fitted grand total \(n\). Therefore the linear difference terms sum to zero for their deviance difference, giving the familiar likelihood-ratio statistic for independence:

For the saturated model, \(D(\text{saturated})=0\). A zero observed count contributes zero to the sum, using the convention \(0\log(0/x)=0\) for \(x>0\). The statistic is also \(2\{\ell(\text{saturated})-\ell(\text{independence})\}\), because the Poisson factorial terms cancel in the likelihood ratio.

The independence model has \(r+c-1\) identifiable parameters: one overall level and \(r-1\) and \(c-1\) independent main-effect contrasts. The saturated model has \(rc\) cell means. Their dimension difference is \(rc-(r+c-1)=(r-1)(c-1)\). Under the null hypothesis, with independent Poisson sampling, positive means, an identifiable regular model, and a large-sample regime in which expected counts grow, the likelihood-ratio statistic is approximately chi-square with \((r-1)(c-1)\) degrees of freedom. This is the standard nested-model likelihood-ratio calibration.

Common Pitfall and Interpretation

A frequent error is to use the deviance of the smaller model alone as though it were the comparison statistic. The test statistic is the difference in deviances, with the reduced-model deviance first: \(D(\text{independence})-D(\text{saturated})\). Since the saturated model has zero deviance here, the difference equals the independence-model deviance, but that shortcut depends on this specific full model. In a comparison of two nonsaturated nested models, both deviances must be fitted and subtracted.

Another error is to treat an interaction test as a test of causation. The null hypothesis is a statement about association in the modeled cell means. Study design and other assumptions determine whether a causal interpretation is warranted. Likewise, a small \(p\)-value indicates incompatibility with the null model under the reference distribution; it does not quantify the practical importance of the association.