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Given a Poisson regression model as $y = E(y\mid x) + ε$ where $λ = E(y\mid x) = \exp(x'β)$ with $y$ from the Poisson distribution ($\operatorname{Poisson}(λ)$) I am trying to understand the distribution of $ε$. I tried writing the error term as $ε = y - E(y\mid x) = y - λ$ and obtained the PMF of $ε$ as $P(ε = k) = P(y = k+λ) = {λ^{k+λ}e^{-λ}}/{(k+λ)!}$. So can it be concluded that the error term of the Poisson model follow a distribution with the obtained PMF or there is a known distribution that the error term follows?

Michael Hardy
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Dick
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    This is not a Poisson regression model. There is no "error term." The correct model is $E[y\mid x] = \exp(x^\prime \beta).$ – whuber Jun 17 '22 at 19:50
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    @whuber I would say the model is $$ y\mid x \sim \operatorname{Poisson}(\exp(x'\beta)). $$ – Michael Hardy Jun 17 '22 at 20:42
  • @Michael Thank you, that is more explicit. I chose the form I did in order to emulate the OP's attempt. – whuber Jun 17 '22 at 21:12
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    A similar question with a similar answer: https://stats.stackexchange.com/questions/124818/logistic-regression-error-term-and-its-distribution – kjetil b halvorsen Jun 17 '22 at 23:19

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@whuber's comment is the answer. Typically we don't assume an independent error term with non-normal generalized linear models. The response distribution already contains the theorized stochastic elements of the situation. In the context of a Poisson GLM, we assume that the expected value of the inverse transformed linear predictor (the most common link would be the log) is equal to the parameter ($\lambda$) that controls the behavior of the conditional response distribution. The observed datum is seen as a realized value drawn from that distribution.

gung - Reinstate Monica
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  • Yes. Thanks for your response. But I am wondering about the error terms if we write the model in a traditional way as indicated in my question. – Dick Jun 17 '22 at 19:58
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    That's not a useful way to analyze the model or even to think about it, because you get a different conditional distribution for every explanatory value! – whuber Jun 17 '22 at 20:00
  • @Dick, it is incorrect to write this model like that. That is a "tradidional way" to write a linear model, but it is not a traditional (or correct) way to write a Poisson GLM. – gung - Reinstate Monica Jun 17 '22 at 20:01
  • I totally understand you points. I am just following the lecture notes of my course and it is defined in the way presented in my question which seems confusing to me too. – Dick Jun 17 '22 at 20:04
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    Hmmm, that's a tough one, @Dick. You'll need to ask the lecturer what they have in mind. The traditional formulation of the Poisson GLM doesn't work like that, so I can't say what they're thinking. – gung - Reinstate Monica Jun 17 '22 at 20:08
  • You're welcome, @Dick. – gung - Reinstate Monica Jun 17 '22 at 20:14
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    I have used this linear form with error term $\epsilon$ as well in an answer here. For me it is sort of like a multistable perception where there is a transition between two images that is difficult to see, the conditional probability expression $\pi(z;\theta,\phi) = \exp \left[ \alpha(\phi) \lbrace z\theta - g(\theta) +h(z)\rbrace +\beta(\phi,z) \right]$ and the itterated reweighted regression. Thinking of the Poisson or other regression, in terms of a linear added term helps me to intuitively transition between those two images. – Sextus Empiricus Jun 18 '22 at 06:31