# 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:

1. Why do we not explicitly show $$\epsilon$$ in the GLM model (or probit & logit)?
2. 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?

1. 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.

1. 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 $$y_i = \hat{\mu_i} + \epsilon_i$$.

• But in $y = \hat{\mu} + \epsilon$ you claim that the observation $y_i$ equals the mean (or technically an estimator of the mean) plus an error. Shouldn't it be the case that $E(y) = \hat{\mu} + E(\epsilon)$ but then again $E(\epsilon)=0$ and we no longer show $\epsilon)$ explicitly in the equations? Jan 19, 2023 at 9:46
• @Xtiaan Why would you do the expected value of $y$? Jan 19, 2023 at 13:46
• Because according to the current equation we say that y equals the the predictor of the mean of y (+ an error). Should it not be the case that we say that y equals a predictor of y (+ an error) or that the expectation of y equals a the predictor of the mean of y (+ an error)? Jan 19, 2023 at 14:08
• @Xtiaan the predictor and the conditional expectation are the same thing. Jan 19, 2023 at 14:23
• So $y_i = \hat{\mu_i} + \epsilon_i$ says that its observation equals $E(y_i)$ plus an error. However, what is $E(y_i)$? Isn't $y_i$ just an observation. Isn't $y$ stochastic and not $y_i$? Jan 19, 2023 at 14:28