Why do we assume a distribution for the dependent variable and not the error when using GLMs? OLS: $y = X\beta+ \epsilon$ with $\epsilon \sim N(0,\sigma^2)$
GLM: $g(\mu) = X\beta$ with $y \sim $<Distribution from exponential family>
I have knowledge of OLS and am trying to understand the more general case of GLM. I noticed that for GLMs the assumption of the distribution is made on $y$ and $\epsilon$ is never explicitly shown in any of the equations. I am aware that in OLS the assumption $\epsilon \sim N(0,\sigma^2)$ implies $y \sim N(X\beta,\sigma^2I)$, but I want to know:

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*Why do we not explicitly show $\epsilon$ in the GLM model (or probit & logit)?

*Is $\epsilon$ even present in these models? The model is not perfect and therefore there has to be an error, but where exactly is it? Can you write the GLM model with $\epsilon$ so I can see how it is incorporated in the model?




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*Why do we not explicitly show $\epsilon$ in the GLM model (or probit & logit)?


It can surely be shown, but does not lead to any new insight. For example, for the Bernoulli GLM model, we have an assumption on the $y$, that it is Bernoulli-distributed. The residuals, however, are not (since they are continuous) and should have a more complicated distribution.



*Is $\epsilon$ even present in these models? The model is not perfect and therefore there has to be an error, but where exactly is it? Can you write the GLM model with $\epsilon$ so I can see how it is incorporated in the model?


If $E(y|X) = \mu = g^{-1}(\eta)$, where $\eta = \beta X$, then $y_i = \hat{\mu_i} + \epsilon_i$.
