# How to prove equivalence between models, and show that one (Bayesian hierarchical) model is an extension of another?

Is there such a thing as equivalence between statistical models (in my particular case, Bayesian hierarchical models) ? If so, how to prove it ?

Let me explain myself with an example. Consider a Gaussian linear regression model $$y\sim \mathcal{N}(X\beta^T, \sigma^2)$$. I parametrized the normal distribution with the variance $$\sigma^2$$ but I could use the standard deviation, the precision, even the log variance... but the models would all be equivalent Gaussian models. From this example, I guess that one has to show that the density from the first model is equal to the density of the second model, using a change of variables ( https://en.wikipedia.org/wiki/Probability_density_function#Vector_to_vector ). Is this a correct approach ? Some references for a principled definition of model equivalence would be welcome.

However, my original problem is not exactly this one. I am working on Bayesian hierarchical models, and I have two models : a ''basic'' one, and a ''richer'' one with more parameters. I would like to show that the second model is an extension of the first model. My approach is :

1. Make some assumptions to harmonize the hyperpriors of the complex and simple model
2. Condition the complex model by an event $$E$$ that "freezes" the additional variables of the complex model with respect to the simple model
3. Show that, conditionally on $$E$$, the densities are the same

What happens in step 3 is that I do not find strict equality of the distributions but equality within a very simple bijective transformation of the parameters. This is my first question (I could tinker either model to make the problem vanish, but it would be a crooked solution imo). Additionaly, any comments on my approach ?