In a logistic regression, given that $\pi_i=Pr(Y_i=1|X_i=x_i)=\dfrac{\text{exp}(\beta_0+\beta_1 x_i)}{1+\text{exp}(\beta_0+\beta_1 x_i)}$, is it correct to say that logistic regression assumes that the conditional distribution $f(Y|X) \sim Logistic$?
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1$\begingroup$ The assumed distribution in logistic regression is binomial. $f(Y_i|X_i = x_i) \sim B(1, \pi_i(x_i))$ $\endgroup$– AlexRCommented Jun 26, 2018 at 11:06
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$\begingroup$ Thanks for the response. Does the logistic distribution only apply to the errors then? It just seems intuitive that because we estimate probabilities using a logistic function, this follows from the conditional being logistically distributed- do you see where I am coming from? $\endgroup$– cdDCCommented Jun 26, 2018 at 11:11
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1$\begingroup$ The logistic distribution doesn't have much to do with logistic regression - except that the link function is the same as the CDF for $\mu=0, s=1$ (wikipedia notation) $\endgroup$– AlexRCommented Jun 26, 2018 at 11:49
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2$\begingroup$ @AlexR Actually, there is a connection to the logistic distribution. $\endgroup$– Glen_bCommented Jun 26, 2018 at 12:54
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4$\begingroup$ @Glen_b You're right, cf. Latent variable interpretation $\endgroup$– AlexRCommented Jun 26, 2018 at 13:00
1 Answer
I am going to flesh out some of the comments to this post into an answer (with thanks to Glen_b and AlexR for their excellent comments). The first thing to note is that the response variable in a logistic regression is binary, with allowable values $y_i = 0,1$, so any distribution that does not accord with that support is incorrect. The distribution for the response variable in a logistic regression is Bernoulli, with a logistic mean.
Response distribution: The response distribution for a logistic regression is:
$$Y_i | \mathbf{x} \sim \text{Bern}(\pi(x_i)) \quad \quad \quad \pi(x_i) = \frac{\exp(\beta_0+\beta_1 x_i)}{1 + \exp(\beta_0+\beta_1 x_i)}.$$
The function $\pi$ is a logistic function of an affine transformation of the argument value $x_i$.
The logistic regression model can be represented using various (equivalent) model forms that yield the above response distribution. One of these model formulations uses a pseudo-error term with a standard logistic distribution. Taking $\varepsilon_i \sim \text{Logistic}(0, 1)$ we have:
$$\begin{equation} \begin{aligned} \mathbb{P}(\beta_0+\beta_1 x_i + \varepsilon_i > 0) &= \mathbb{P}(\varepsilon_i > -(\beta_0+\beta_1 x_i)) \\[16pt] &= 1- F_{\varepsilon_i}(-(\beta_0+\beta_1 x_i)) \\[12pt] &= 1 - \frac{1}{1 + \exp(\beta_0+\beta_1 x_i)} \\[6pt] &= \frac{\exp(\beta_0+\beta_1 x_i)}{1 + \exp(\beta_0+\beta_1 x_i)} \\[8pt] &= \pi(x_i). \\[6pt] \end{aligned} \end{equation}$$
Since this is the probability of a positive response outcome in the above distribution, we can formulate the logistic regression model in a way that is similar to a standard linear regression model, but with a pseudo-error term that is only used to classify the response into two categories:
$$Y_i = \mathbb{I}(\beta_0+\beta_1 x_i + \varepsilon_i > 0) \quad \quad \quad \varepsilon_1, ..., \varepsilon_n | \mathbf{x} \sim \text{IID Logistic}(0, 1).$$
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$\begingroup$ +1. But maybe it's worth pointing out that logistic regression does not need to assume the explanatory variable is random, so there is no needed for conditioning in the standard probabilistic sense: the explanatory variable's role is that of a parameter of the response distribution, nothing more. $\endgroup$– whuber ♦Commented Sep 11 at 14:42