Is there a connection between the normal and the logistic distribution?

Question

Regarding Bayesian statistics I found in a script that there is such link, and the logistic arises in context of a normal distribution and a "binary state". However, I have no idea what is the meaning behind this. And I found no further hints. The Student-t is highly connected to the confidence interval for mean of a normal distribution, so I wonder if there is something similar regarding the logistic distribution.

Answer 1

The only link I can think of is that...

...the probit is to the Normal what the logit is to the logistic...

Namely that the probit regression model where a conditional distribution of a binary variable is defined by $$\mathbb{P}(Y=1|X) = \Phi(\beta^\text{T}X)$$ and can be interpreted as $$Y=\mathbb{I}_{Y^*\ge 0}\qquad Y^*|X\sim\mathcal{N}(\beta^\text{T}X,\sigma^2)$$ (interpreted meaning that the latent variable $Y^*$ does not need to exist "for real"). Similarly for the logit regression model where a conditional distribution of a binary variable is defined by $$\mathbb{P}(Y=1|X) = \dfrac{1}{1+\exp(-\beta^\text{T}X)}$$ which can be interpreted as$$Y=\mathbb{I}_{Y^*\ge 0}\qquad Y^*|X\sim\mathcal{L}(\beta^\text{T}X,\sigma^2)$$ where $\mathcal{L}(\beta^\text{T}X,\sigma^2)$ denotes the logistic distribution.

Answer 2

Without the precise quote and a source we can only guess what the author meant. Are we sure that we are talking about the logistic distribution, and not logistic function? Because, from your paraphrase

the logistic arises in context of a normal distribution and a "binary state"

I'd guess the latter. Logistic function arises when you have two normally distributed classes, $A$ and $B$, with an equal variance $\sigma$ and means $\mu_A$ and $\mu_B$, respectively. Then, the posterior probability of an observation with a value $x$ to belong to a class, say, $B$ is given by the logistic function

$$ P(B | x) = \frac{1}{1 + \exp \left(-\beta_0 - \beta_1 x \right) }. $$

where

$$ \beta_0 = \frac{\mu_A^2 - \mu_B^2} {2 \sigma^2} - \ln \frac{P(A)}{P(B)} ~ ~ ~ ~ ~ ~ ~ ~ \text{and} ~ ~ ~ ~ ~ ~ ~ ~  \beta_1 = \frac{\mu_B - \mu_A} {\sigma^2} $$

See this answer for details.