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The way I understand a Poisson regression is that we model $y|x \sim \text{Poisson}(\exp(x'\theta))$ so that $E[y|x]=\exp(x'\theta)$ (e.g. http://en.wikipedia.org/wiki/Poisson_regression).

My question is whether we can reformulate the model so that we relate $y$ to $x$ and unobservables (or latent variables) $\varepsilon$ and then make assumptions about $\varepsilon$.

As motivation, consider a linear regression; we can either model the problem as $E[y|x]=x'\beta$ or as $y=x'\beta+\varepsilon$ and $E[\varepsilon|x]=0$.

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While you could write it as a model with an additive error, the problem is that with Poisson regression the error terms have shifted Poisson distributions, each different -- they have different variances and different skewness.

So unlike linear regression, where the errors have a common distribution, in Poisson regression they don't. This makes an additive-error formulation unproductive for most purposes. (There are some other issues, as well.)

The reason why many people write their Poisson regression in forms like $\underline{y}|X\sim\text{Pois}(e^{X\beta})$ is because that's actually a better way to deal with it.

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  • $\begingroup$ Thanks very much. Do you think it is possible to setup a nonlinear (perhaps complicated) model in combination with some other distribution on $\varepsilon$ that after manipulation leads to this Poisson regression or do you think it's unlikely? It's just very unnatural for me to have the modeling assumptions on $y|x$ as the starting point without a model in the background with unobservables. (for example in some multinomial choice models the errors are assumed to be extreme value type I and then after manipulation of the model this leads to a logistic regression). $\endgroup$ – user103828 Feb 12 '15 at 9:30
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    $\begingroup$ (for example in some multinomial choice models the errors are assumed to be extreme value type I and then after manipulation of the model this leads to a logistic regression). $\:$ Ah! You mean as a latent variable. Hmm. That might be possible. I'll see if I can find one. $\endgroup$ – Glen_b Feb 12 '15 at 9:46

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