I was just wondering if Poisson regression has an error term? Can a Poisson regression have random effects and an error term? I am confused about this point. In logistic regression, there is no error term because your outcome variable is binary. Is that the only glm model that doesn’t have a residual term?

**Answer**

I think the problem that’s confusing you is that you’re used to having an additive error. Most models won’t.

Think of linear regression not as a linear mean with an additive error, but as the response being conditionally normal:

$$(Y|X) \sim \operatorname{N}(X\beta,\sigma^2I)$$

Then the similarities to GLMs, in particular, to Poisson regression and logisitic regression are more clear.

Because of the nice properties of the normal, the normal case can be written in terms of the mean and an additive error. This is not always so nice with other models and it makes sense to stick to the distributional form for the model, or at least to write about the mean and variance of $(Y|X)$ rather than writing a model for $\operatorname{E}(Y|X)$ and trying to describe the characteristics of $Y-\operatorname{E}(Y|X)$.

[You can take any particular combination of predictors and write the response variable in terms of its expectation and a deviation from that – an ‘error’ if you will – but it’s not particularly enlightening when its a different object from every other combination of predictors. It’s usually more informative and more intuitive to just write the response as a distribution that is a function of the predictors than in deviation-from-expectation form.]

So while you *can* write it ‘with an error term’ it’s just less convenient and conceptually harder to do so than to do other things.

**Attribution***Source : Link , Question Author : phil12 , Answer Author : Glen_b*