Residuals for Logistic Regression I have read that there are multiple ways to look at residuals for logistic regression. I am not necessarily that interested in analyzing the residuals for a logit model, but my main question is rather:
Where are residuals in the specification of the model?
For linear models, it's common to write
$$y=X\beta+\varepsilon \ \ \text{ or } \ \   y = \beta_0+\beta_1x_1 + \dots + \beta_kx_k+\varepsilon$$
However, for the logit model I keep finding
$$l=\log\frac{p}{1-p}=\beta_0+\beta_1x_1 + \dots + \beta_kx_k  \ \ \text{ or } \ \    p = \frac{1}{1+e^{-(\beta_0+\beta_1x_1 + \dots + \beta_kx_k)}}$$
How come the disturbances are not part of the model in this case?
Edit:
I have thought of something but am not sure if this is the complete reasoning behind it. In my understanding, a linear model assumes that data have been generated according to $y=X\beta$ with random disturbances, as nothing is perfectly linear.
And for a logit model this disturbance is not needed, as the data are assumed to be generated(?) according to $p$: at every point $x_i$ there is a certain probability $p$ that the outcome will be $1$, and the points $y=0,1$ are generated according to this, rather than a set linear relation with disturbances that make it random.

You can have residuals, or not, in either case. But to start with, you don't need residuals at all to specify the model: In either case, you can write $$Y|X=x \ \ \sim \ \  p(y|x),$$ where $p(y|x)$ is some conditional distribution. In the case of logistic regression, $p(y|x)$ is a Bernoulli distribution. In the case of classical regression, $p(y|x)$ is a normal distribution, but this can be relaxed to allow other forms, like Poisson, negative binomial, multinomial, non- or semi-parametric variants, etc.
If you want residuals, you can also have them, in either case: For an observation where $X=x$, the residual is $\epsilon = Y - E(Y|X=x)$.  In the case of standard regression, under the usual linearity assumption, this gives you $$\epsilon = Y - (\beta_0 + \beta_1 x).$$  In the case of logistic regression, under the usual linearity (of logit) assumption, this gives you $$\epsilon = Y - \frac{\exp{(\beta_0 + \beta_1 x)}}{1+\exp{(\beta_0 + \beta_1 x)}},$$ where $Y$ is either 0 or 1.
It is just not common to use this form in the case of logistic regression because the salient feature of the model is the form of the conditional Bernoulli distribution of $Y|X=x$.
