log-odds and it's standard error as priors in logistic regression

Intro

Yes this is correct. Since you asked for a complete answer, I'll start by setting up notation and establishing preliminaries.

Notation

It sounds like your goal is to understand the relationship between driving at night and crashing a car. Let's denote the binary dependent variable of whether a car crash occurred as $y = {0,1}$, and the binary independent variable of driving at night as $x= {0,1}$. Furthermore, we'll denote the probability $P[y|x] = p(x)$. We'll estimate $p(x)$ using a logistic regression

As your sources note the motivation of logistic regression is a linear model for the log-odds: $$ logleft[frac{p(x)}{1-p(x)}right] = alpha + beta x $$
Since $x$ has only two levels, we can make this notation a little simpler by defining $beta_0 = alpha$ (the log-odds of a crash during the day) and $beta_1 = alpha + beta$ (the log-odds of a crash at night). It will also help if we define the logit function: $$ text{logit}(z) = logleft[frac{z}{1-z}right] $$ Which allows us to easily write: $$ p(x) = begin{cases} text{logit}^{-1}(beta_0) & x=0\ text{logit}^{-1}(beta_1) & x=1\ end{cases} $$

Posteriors and Priors

In the Bayesian methodology this model would be fit to the datapoints $(x_1,y_1),...,(x_n,y_n)$ by looking at the posterior distribution: $$ P[beta_0,beta_1|x_1,...,x_n,y_1,...,y_n] = prodlimits_{i=1}^n p(x_i)^{y_i} (1-p(x_i))^{1-y_i} P[beta_0,beta_1] $$ Where $P[beta_0,beta_]$ is the prior distribution over the parameters, typically assumed to have prior independence in the parameters: $$ P[beta_0,beta_1] = P[beta_0] P[beta_1] $$

Answer

The paper has provided you the 95% quantiles, mean, and standard deviation of the posterior distribution of the value $text{logit}(p(1)) = beta_1$. Say the mean here is $m_1$ and the standard deviation is $s_1$. A standard result in Bayesian analysis is that, with sufficiently many datapoints, the posterior distribution is approximately normal (the Laplace approximation). Thus $m_1$ and $s_1$ are sufficient to characterize the posterior distribution (approximately), and it is the normal distribution $N(m_1,s_1)$. In general, variance is standard deviation squared, so an alternate parameterization of their posterior/your prior would be the normal distribution $N(m_1,s_1^2)$, which is what you have here:

$beta_k = (1.12,.04)$

PS

• Note that the variance of the prior equaling $.04 = .02^2$ is not unique to log-odds. For any distribution, variance equals standard deviation squared (this is just the definition of standard deviation). Your source 3 is actually providing a proof of the Laplace Approximation, ie. the fact that the previous posterior is approximately normal.

• In general you will want to also perform a sensitivity analysis on your choice of prior. $N(1.12,.04)$ is very tight around a rather large value of $m_1$ (it implies that the probability of crashing at night is like ~75%). It would be smart to re-run your analysis with multiple priors with increasing variances, to see what happens to your results when you loosen up your prior confidence.