# What are the optimized “data” given parameter? Use of “symmetry” of Bayesian theorem

I am interested in "reverse" way.
Suppose you already managed to get the model parameter$\theta$, given data $D$ through optimization. What you did is exactly below.

$p(\theta|D)=\frac{p(D|\theta)p(\theta)}{p(D)}$ What are the likely "model parameter $\theta$ " given data $D$?

Now, by using "Symmetry of Bayesian theorem", one can also say,

$p(D|\theta)=\frac{p(\theta|D)p(D)}{p(\theta)}$ What are the likely "data $D$ " given model parameter $\theta$ ?

So, after calculated parameter $\theta$, now you are in search for Data $D$,... better data,.. data that explains a given parameter/Model best.
I am wondering if there is any research or paper talking about this by using the symmetry of Bayesian theorem.

Thank you

## 1 Answer

There is no particular reason to create a reversal of Bayes' theorem in this case, since specification of a statistical model is usually a direct specification of the sampling distribution (i.e., the distribution of the data conditional on the parameters). Bayes' rule exists because we do not directly specify the posterior, but this is instead derived from the prior and sampling density.

Specification of a model for the sample data $x \in \mathscr{X}$ involves direct specification of a family of density functions $\{ p_\theta(x) | \theta \in \Theta \}$ conditional on a parameter $\theta$. The data that "fits" this model best is the mode of the sampling distribution, which is given by:

$$\text{mode}(\theta) = \underset{x \in \mathscr{X}}{\arg \max} \text{ } p_\theta(x).$$

Searching for the mode of the sampling density is something that can be done directly from the specified statistical model. It is not necessary to try to derive this from reversal of Bayes' theorem using a prior distribution and the corresponding posterior.