3
$\begingroup$

I'm confused about the interpretation of interaction terms in poisson regressions. Here's a hypothetical dataset

Id  Group  Condition   Error rate
1   A      X           1
1   A      Y           2
2   B      X           3
2   B      Y           6

Imagine that individuals 1 and 2 are duplicated a number times (with random noise) to make two groups. If we analyse this data using a poisson regression (as it's count data) there will be no Group by Condition interaction (in both groups, error rates increase by 100% in condition Y, i.e. an Incidence Rate Ratio of 2). If we analyse this data using an ANOVA (and don't transform data) it is likely to yield a significant Group by Condition interaction (I've simulated this). While the poisson regression looks at proportional differences across conditions, the ANOVA is more sensitive to absolute differences.

It seems to me that because the poisson regression uses a multiplicative scale (via log transformation), an additive effect under the poisson regression will become multiplicative when the same data is analysed using OLS (or ANOVA).

So which interaction term is more "truthful"? Is it just a matter of interpretation? Or are there any a priori reasons for choosing one over the other. For example, could you argue that because count data exhibits ratio properties we should be analysing ratios and not intervals (according to Stevens)?

Thanks and hope this makes sense.

$\endgroup$
  • 1
    $\begingroup$ (1) The ANOVA analysis is suspect due to the strong heteroscedasticity in the response (the variance of a count is proportional to its square root), so obtaining a significant result with this method means little. (2) Which better describes such data--the Poisson model without interaction or the ANOVA one with an interaction? Many people would say the former, because it's more parsimonious. (3) Stevens' taxonomy is not helpful here because "ratio" is a misleading characterization of what counts are and how they can appropriately be analyzed. $\endgroup$ – whuber Apr 9 '13 at 18:15
  • $\begingroup$ Thanks for the comments. I hadn't thought about the issue of parsimony. I guess the model without the interaction term is more parsimonious. $\endgroup$ – Nick Riches Apr 11 '13 at 9:32
3
$\begingroup$

Interaction terms in Poisson regression models are interpreted as a ratio of ratios of rates. With an interaction term, your model's interpretation of that parameter would be, "a rate ratio comparing condition Y to X among individuals of type 2 relative to rate ratio comparing condition Y to X among individuals of type 1".

OLS and Poisson regression in this case will give the same fitted means in models adjusting for interaction since you have a fully specified log-linear model for the table of fitted means. Having a different working model for the distribution of the data will lead to different inference of course (so OLS and Poisson GLMs will have different P-values across the board). However, the fitted means for the models without interaction will be different between OLS and Poisson. This is because the model is not fully specified and the difference between mean rates is taken to be constant in the model.

\begin{equation} \begin{array}{c} \log(\lambda_{ij}) = \beta_0 + \beta_1 i + \beta_2 j \qquad (\mbox{Poisson})\\ \mu_{ij} = \gamma_0 + \gamma_1 i + \gamma_2 j \qquad (\mbox{OLS}) \end{array} \end{equation}

Taking $i = 0, 1$ to denote individual membership and $j=0 ,1$ to denote condition,

Poisson \begin{equation} \begin{array}{ccc} & j=0 & j=1 \\ i=0 & \lambda_{11} = \exp(\beta_0) & \lambda_{12} = \exp(\beta_0 + \beta_2) \\ i=1 & \lambda_{21} = \exp(\beta_0 + \beta_1) & \lambda_{22} =\exp(\beta_0 + \beta_1 + \beta_2) \\ \end{array} \end{equation}

OLS \begin{equation} \begin{array}{ccc} & j=0 & j=1 \\ i=0 & \mu_{11} = \gamma_0 & \mu_{12} = \gamma_0 + \gamma_2 \\ i=1 & \mu_{21} = \gamma_0 + \gamma_1 & \mu_{22} =\gamma_0 + \gamma_1 + \gamma_2 \\ \end{array} \end{equation}

$\endgroup$
  • 2
    $\begingroup$ Thanks. I guess it all boils down to an issue of interpretation, and whether it makes sense to conceptualise the interactions as ratios of ratios of rates. $\endgroup$ – Nick Riches Apr 11 '13 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.