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

Question

I'm attempting to complete a Bayesian logistic regression with the outcome of whether or not a crash occurred. I have various covariates in my model that are widely used to predict crash occurrence. As such, I'm using informed priors from prior publications that report the odds ratio and it's 95% C.I for each covariate.

Here's an example of a prior provided by the model I'm pulling from

crash at night (OR 13.1; 95% CI 5.0 to 31.5) : log-odds (1.12,.20) from $$ \frac{log(31.5-5)}{3.92}$$

I wanted to apply the log-odds of these results and their standard error in my updated model as priors. My first thought was to apply the log-odds and it's a standard error on a normal prior. I'm using logic from the sources 1 & 2 listed at the end of the post.

My question, if my assumptions about applying these log-odds and SE's on a normal prior are correct, can I simply transform the SE of the log odds to variance and implement?

a normal prior:

β_k = (μ_βk,σ²_βk)

requires a variance rather than an SE. According to citation 3 log-odds SE and be transformed into log-odds VAR:

$$SE[log(OR)] = \sqrt{VAR[log(OR)]} => SE^2 = VAR[log(OR)]$$

therefore, if I square the standard error x then I should be able to apply this as my final prior:

β_k = (1.12,.04)

Is this assumption correct or am I way off? Is there a better way of implementing log-odd priors and their SE's into a logistic regression model?

Thanks!

1. AdamO (https://stats.stackexchange.com/users/8013/adamo), Prior for Bayesian multiple logistic regression, URL (version: 2016-03-16): https://stats.stackexchange.com/q/202046

"Basically, you have the flexibility to parametrize estimation however you see fit, but using a model which is linear on the log odds scale makes sense for many reasons. Furthermore, using a normal prior for log odds ratios should give you very approximately normal posteriors."

2. Sander Greenland, Bayesian perspectives for epidemiological research: I. Foundations and basic methods, International Journal of Epidemiology, Volume 35, Issue 3, June 2006, Pages 765–775, https://doi.org/10.1093/ije/dyi312

"To start, suppose we model these a priori ideas by placing 2:1 odds on a relative risk (RR) between ½ and 2, and 95% probability on RR between ¼ and 4. These bets would follow from a normal prior for the log relative risk ln (RR) that satisfies..."

3. StatsStudent (https://stats.stackexchange.com/users/7962/statsstudent), How do I calculate the standard deviation of the log-odds?, URL (version: 2020-04-19): https://stats.stackexchange.com/q/266116

Answer 1

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: $$ \log\left[\frac{p(x)}{1-p(x)}\right] = \alpha + \beta x $$
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: $$ \text{logit}(z) = \log\left[\frac{z}{1-z}\right] $$ 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: $$ P[\beta_0,\beta_1|x_1,...,x_n,y_1,...,y_n] = \prod\limits_{i=1}^n p(x_i)^{y_i} (1-p(x_i))^{1-y_i} P[\beta_0,\beta_1] $$ 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.