I have some data I am trying to analyze of how often someone has chosen a given option (which I'll call choice A) vs not choosing it (which I'll call choice B) during a 6 month period. Each subject has had a different level of exposure, and therefore a different number of opportunities to choose between A and B (lets say subject 1 had 4 decisions to make, choosing A once and B 3 times, where subject 2 had 20 decisions to make, choosing A 7 times and B 13). I'm trying to estimate the effects of various covariates on how often they chose A vs. B.

My colleague suggested I use logistic regression, by turning each decision into a binary variable, with 1 representing choice A and 0 choice B. I was concerned about independence between the trials and that individuals with lots of exposures would bias the estimates. My colleague thinks I could solve this by either adding a bucketed (1-5 exposures, 6-10 exposures, etc.) variable on how many decisions they've made as a covariate, or by using a mixed model with a random effect for the individual.

I on the other hand had originally thought to use use poisson regression, with an offset term to account for exposure.

Looking at the link functions and the generalized models for these two methods, they clearly are modeling different things, albeit very similar:


$$E(Y_i) = \mu_i= \eta_ie^{\bf{x}_i^T\beta}$$

where $\eta_i$ is the offset term, which in this case would be the number of decisions. So the equation simplifies to: $$ \text{log}\Big(\frac{\mu_i}{\eta_i}\Big) = \bf{x}_i^T\beta$$


$$E(Y_i) = \pi_i= (1-\pi_i)e^{\bf{x}_i^T\beta}$$

where $\pi_i$ is the probability of choosing A. The equation then simplifies to

$$ \log\Big(\frac{\pi_i}{1-\pi}\Big) = \bf{x}_i^T\beta$$

So in a poisson regression I would be modeling the log-rate where with Logistic I would be modeling the log-odds.

What are the implications of this difference? What would make poisson regression (with an offset) a more or less valid approach than logistic regression (accounting for exposure)? Or are they simply different paths to arrive at the same destination?

I've looked around here and through Google and while I've seen the topic touched on a little bit, I haven't really found any satisfactory answers.

  • $\begingroup$ Possibly related answer here. $\endgroup$ – AdamO Jun 1 '18 at 17:44

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