# Reasonable assumption in Bayesian linear regression

I'm learning bayesian linear regression from the book Bayesian Data Analysis. Here is my questions.

Notation (follow the book): $X$ stands for explanatory variables, with parameter $\psi$; $\bf{y}$ for responsible variables, conditional on $X$ and regression parameter $\theta$.

1. Why do we need linear regression? It seems we can make predictions on $\tilde{y}$ directly, given the past values of $\bf{y}$. i.e. build a posterior predictive distribution $p(\tilde{y} | \bf{y})$.
2. In the book, one of the assumptions is (Pg.354 Ch.14):

The distribution of $X$ is assumed to provide no information about the conditional distribution of $\bf{y}$ given $X$; that is, the parameters $\theta$ determining $p(y|X,\theta)$ and the parameters $\psi$ determining $p(X|\psi)$ are assumed independent in their prior distributions.

Is this a reasonable assumption? I think we can't say $\psi$ and $\theta$ are independent in the prior, but conditional independent after we observed the variables $X$ and ${y}$. i.e. they are independent in the posterior.

3. The last question is about the factorization of joint posterior. $$p(\psi, \theta |X, y) = p(\psi|X)p(\theta|X,y)$$

According to the author, we can analyze the second factor by itself, with no loss of information: (Pg.355 Ch.14) $$p(\theta|X,y) \propto p(\theta)p(y|X,\theta)$$

How to get the above formula? My way: \begin{align} p(\theta|X,y) &= \frac{p(X,y|\theta)p(\theta)}{p(X,y)} \\ &\propto p(X,y|\theta)p(\theta) \end{align}

In addition, what does the "information loss" refer to?

1. Yes, you could "forget", that you have explanatory variables, i.e. you could model just the response data $Y$. But why to neglect the information about what influence your response? Suppose you observe a failure times of some device. You can model just those failure times and obtain (maybe) reasonable results. But if you have information about at what age each device failed you could use this information to obtain more accurate predictions, e.g. you have 3 other devices with different ages, then you could predict how much are left for them to operate depending on their specific ages.
3. You can get it by writing out all the details: the likelihood under stated assumptions is $$p(X,y|\psi, \theta)=p(X|\psi)p(y|\theta,X)$$ Hence: $$p(\psi, \theta|X, y)=\frac{p(X,y|\psi, \theta)p(\psi, \theta)}{p(X,y)}=\frac{p(X|\psi)p(y|\theta,X)p(\theta)p(\psi)}{p(X)p(y|X)}=p(\psi|X)p(\theta|y,X)$$
Since $p(X)$ and $p(y|X)$ are just constant, we can write (up to a constant) $$p(\psi, \theta|X, y)\propto p(X|\psi)p(y|\theta, X)p(\theta)p(\psi)$$ Or $$p(\psi|X)\propto p(X|\psi)p(\psi)$$ $$p(\theta|y,X)\propto p(y|\theta,X)p(\theta)$$