The following is from a textbook one bayesian stats. that I can't understand some deduction. It is relevant about multiple parameters to be estimated.
The jth observation in the ith group is denoted by $y_{ij}$,
where
$$(y_{ij}|\mu_i,\sigma)\sim N(\mu_i,\sigma^2) \quad j=1,2, \dots, n_i \quad i= 1,2, \dots, m$$
Also the $y_{ij}$ are independent from each other.
Suppose $\mu_i \sim N(\mu,\tau^2)$ and denote
$$\theta= (\mu, \log(\sigma),\log(\tau))$$
$$Y=\{y_{ij}: j=1,\dots, n_i, i=1,\dots, n\}$$
$$Z=(\mu_1,\dots, \mu_m)$$
$$n=n_1+n_2+\cdots +n_m$$
So $\theta$ is the unknown parameters interested. Take its prior distribution as $p(\theta) \propto \tau$. Then by Bayes rule, it is not difficult to get the posterior distribution:
$$p(Z,\theta|Y) \propto p(\theta) \prod\limits_{i = 1}^m {p(\mu_i|\mu,\tau)} \prod\limits_{i = 1}^m \prod\limits_{j = 1}^{n_i} {p(y_{ij}|\mu_i,\sigma)}$$
This is the place I can't understand. How to get this formula if No.3 formula is not correct in this thread: I am confused about Bayes' rule in MCMC
Could someone explain it in detail? If there are any excellent books that could help me, please list them.
AFAICT, the "trick" is that, by definition, $y_{ij}$ depends on $\mu$ and $\tau$ only via $\mu_i$. Thus, $$p(y_{ij}\mid\mu_i,\sigma) = p(y_{ij}\mid\mu_i,\sigma,\mu,\tau) = p(y_{ij}\mid\mu_i,\theta).$$
Similarly, $\mu_i$ does not depend on $\sigma$, so $$p(\mu_i\mid\mu,\tau) = p(\mu_i\mid\mu,\tau,\sigma) = p(\mu_i\mid\theta).$$ In particular, this means that we can rewrite your equation as
$$ \begin{aligned} p(Z,\theta\mid Y) \propto& p(\theta) \prod_{i = 1}^m p(\mu_i\mid\mu,\tau) \prod_{i = 1}^m \prod_{j = 1}^{n_i} p(y_{ij}\mid\mu_i,\sigma) \\ =& p(\theta) \prod_{i = 1}^m p(\mu_i\mid\sigma,\mu,\tau) \prod_{i = 1}^m \prod_{j = 1}^{n_i} p(y_{ij}\mid\mu_i,\sigma,\mu,\tau) \\ =& p(\theta) \prod_{i = 1}^m p(\mu_i\mid\theta) \prod_{i = 1}^m \prod_{j = 1}^{n_i} p(y_{ij}\mid\mu_i,\theta) \\ =& p(\theta)\, p(Z\mid\theta)\, p(Y\mid Z,\theta) \\ =& p(Y,Z,\theta) \\ =& p(Z,\theta\mid Y)\, p(Y). \end{aligned} $$