Hi all - I’m a newcomer to stan and Bayesian statistics so bear with me…

I’m looking to find a way to code a prior for linear regression such that

Suppose that

p \sim N(0, 1)\\
q = \beta_0 + \beta_1 p + \beta_2 p^2\\
\beta = [\beta_0, \beta_1, \beta_2], P = [1, p, p^2]

I want to find a \beta such that \frac{\partial q}{\partial p} = 2\beta_2 p_i + \beta_1 < 0 for as many p_i, in addition to the standard bayesian linear regression constraints.

I know for the latter, my prior for \beta is q \sim N(\beta'P, \sigma). However, how would I encode the former (\frac{\partial q}{\partial p} = 2\beta_2 p_i + \beta_1 < 0) as a prior?

Usually, a regression involves predictors x_n \in \mathbb{R}^P and outcomes y_n \in \mathbb{R} and a parameter vector \beta \in \mathbb{R}^P and scale parameter \sigma > 0, with likelihood y_n \sim \textrm{normal}(x_n \beta, \sigma). I don’t see where the observations (equivalent of y) come in for your model.

A prior for \beta needs to be a distribution over \beta, but this is a distribution over q,.

To answer your question, if you have a bunch of observed p_1, \ldots, p_N, and a coefficient vector \beta and you want to ensure 2 \beta_2 p_n < \beta_1 for all n, then you can do this:

```
data {
int N;
vector[N] p;
...
parameters {
real beta_2
real<lower = max(2 * beta_2 * p)> beta_1;
...
```

Ah, I used the incorrect terminology. My observations q have likelihood q \sim N(\beta' P, \sigma) - and this informs how my observations come in for my model. Thanks for the correction!

As stated earlier, in In my example, I believe dq/dp = b1 + 2b2*p > 0. However, is there another way to do this, that doesn’t necessarily force the inequality for all n, but ensures that it will be true for a high percentage of n? I’ve read of https://en.m.wikipedia.org/wiki/Cromwell%27s_rule which suggests that perhaps imposing this absolute constraint would not be the best solution to this problem.

Thank you for the response! This is really helpful.