Log-Linear Model

Study guide: Exam MAS-I

A Log-Linear Model for a contingency table treats each cell count as Poisson and models the log of its expected count as a sum of row, column and interaction effects. It is Poisson Regression applied to cross-classified counts, and its terms correspond to independence and association between the classifying variables.

$\ln \mu_{ij} = \eta + \alpha_i + \beta_j$

$\ln \mu_{ij} = \eta + \alpha_i + \beta_j + (\alpha\beta)_{ij}$

$\hat\mu_{ij} = \frac{y_{i\cdot}\,y_{\cdot j}}{n}$

$\mu_{ij}$ is the expected count in row $i$ , column $j$ of an $I \times J$ table, and $y_{i\cdot}$ , $y_{\cdot j}$ and $n$ are the row totals, column totals and grand total. The first block is the additive model, which corresponds to independence, $\pi_{ij} = \pi_{i\cdot}\pi_{\cdot j}$ . Its fitted counts are the third block. The second block is the saturated model, which reproduces every count.
Sampling scheme doesn't matter. The counts may be independent Poisson, multinomial with $n$ fixed, or product multinomial (binomial) with one margin fixed. Conditioning Poisson counts on their total gives the multinomial, and a model that includes those margins reproduces them exactly. So the Poisson log-linear fit gives the same fitted values and tests in each case.
Testing independence. The Deviance of the additive model, $D = 2\sum y_{ij}\ln(y_{ij}/\hat\mu_{ij})$ , is approximately $\chi^2$ with $(I-1)(J-1)$ degrees of freedom. Pearson's $X^2 = \sum (y_{ij} - \hat\mu_{ij})^2/\hat\mu_{ij}$ is the traditional alternative.
Three-way tables. Terms are added hierarchically. $A + B + C$ is complete independence, and $AB + C$ means $A$ and $B$ are associated but jointly independent of $C$ . $AB + AC$ and $AB + AC + BC$ allow more association, and $ABC$ is saturated. Nested models are compared by the drop in deviance (Likelihood Ratio Test).
When one classifying variable is really a response, the log-linear model can stand in for a logistic or multinomial logit model for it. It needs a parameter for each predictor group (so the group totals are reproduced), a main effect for the response, and response × predictor interactions. The logit coefficients are then differences of those interaction terms: main effects in the logit model become interactions in the log-linear one.
On MAS-I this is Dobson & Barnett Chapter 9 (§§9.3–9.7: contingency tables, their probability models and log-linear models).

Is Litigation Independent of Claim Size? $600$ closed bodily-injury claims are cross-classified:

Large Not large Total

Litigated 24 76 100

Not litigated 66 434 500

Total 90 510 600

Fit the independence log-linear model and test it at the $5\%$ level.
Answer Expected counts are row total × column total ÷ $600$ : $15$ , $85$ , $75$ and $425$ .
 $\begin{align*} D &= 2\Big[24\ln\tfrac{24}{15} + 76\ln\tfrac{76}{85} \\ &\qquad + 66\ln\tfrac{66}{75} + 434\ln\tfrac{434}{425}\Big] \\ &= 2\,[11.280 - 8.506 - 8.437 + 9.095] \\ &= 6.86 \end{align*}$ 
 $\begin{align*} X^2 &= \frac{9^2}{15} + \frac{9^2}{85} + \frac{9^2}{75} + \frac{9^2}{425} \\ &= 5.400 + 0.953 + 1.080 + 0.191 \\ &= 7.62 \end{align*}$ 
With $(2-1)(2-1) = 1$ degree of freedom the $5\%$ critical value is $3.841$ . Both statistics exceed it, so independence is rejected: litigation and claim size are associated.

	Large	Not large	Total
Litigated	24	76	100
Not litigated	66	434	500
Total	90	510	600

Reading the Coefficients For the same table, use reference-cell coding (not litigated, not large). Find $\eta$ , $\alpha_{\text{lit}}$ and $\beta_{\text{large}}$ in the independence model, and the interaction term $(\alpha\beta)$ in the saturated model.
Answer Under independence, every fitted count is a product of margins, so each coefficient is a log ratio of margins:
 $\begin{align*} \eta &= \ln\frac{500 \times 510}{600} \\ &= \ln 425 \\ &= 6.052 \\[4pt] \alpha_{\text{lit}} &= \ln\frac{100}{500} \\ &= -1.609 \\[4pt] \beta_{\text{large}} &= \ln\frac{90}{510} \\ &= -1.735 \end{align*}$ 
Check: $e^{6.052 - 1.609 - 1.735} = e^{2.708} = 15$ , the expected count for litigated and large.

The saturated model fits each cell exactly. With reference-cell coding, its interaction is the log odds ratio of the table:
 $\begin{align*} (\alpha\beta) &= \ln\frac{24 \times 434}{76 \times 66} \\ &= \ln 2.077 \\ &= 0.731 \end{align*}$ 
The odds of a large claim are about twice as high when the claim is litigated. Dropping this one term is exactly the $1$ -df deviance test of the previous example ( $D = 6.86$ ).