Suppose you are measuring $n$ quantities with error. Let $\beta_1,\ldots, \beta_n$ represent the true values and $X_1, \ldots, X_n$ represent the measured values of those quantities. Assume that the errors are centered normal. Let $\sigma_i^2\,, i=1, \ldots, n$ represent the known standard deviation of each measurement. So that the measurements are $$ X_i | \beta_i \sim N(\beta_i, \sigma_i^2)\,.$$
I can recover the above model by writing $$ X_i = \beta_i + \varepsilon_i\,,$$ where $\varepsilon_i \sim N(0, \sigma_{i}^2)$.
Now I make the following extension to the model after which I get confused.
Suppose that $\beta_i \sim N(\mu, \sigma_b^2)$, with known parameters $\mu, \sigma_b^2$. I want to write down form of the posterior distribution $p(\beta_i | X)$.
On the one hand, if the relationship $X_i = \beta_i + \varepsilon_i $ is still in force, then $$ \beta_i = X_i - \varepsilon_i $$ so the posterior is $p(\beta_i | \{X_i\}) \sim N(X_i, \sigma_i^2)$ In particular it does not depend on $\sigma_b, \mu$.
On the other hand, if I just proceed by Bayesian theorem then $$ p(\beta_i | \{X_i\}) = \frac{p(\beta_i, \{X_i\})}{P(\{X_i\})} = \frac{p(\{X_i\}|\beta_i) p(\beta_i)}{P(\{X_i\})} \propto p(\{X_i\}|\beta_i) p(\beta_i) = f_1(\{X_i\}| \beta_i) f_2(\beta)\,, $$
with $f_1(\{X_i\}| \beta_i) = \prod_{i=1}^n f(X_i| \beta_i)$ where $ f(X_i; \beta_i)$ is density of $N(\beta_i, \sigma_i^2)$ and $f_2(\beta_i)$ is the density of $N(\mu, \sigma_b^2)$.
The results of those two approaches differ, what am I confusing here?
Added late: As one of the comments suggested my question is related to this question, but the refereed question asks about the specific form of posterior distribution (why the posterior is normally distributed), this is different from what I was trying to figure out.
