Interpretation of Bayesian vs Frequentist statement Although I am completely new to Bayesian Analysis I struggle sometimes when trying to investigate some intersections between Bayesian and Frequentist analysis.
I would like to discuss the different implications based on the following simple example:
Let $y=(y_1,\ldots,y_N)$ were $y_i|\mu \sim N(\mu,1)$, $\mu\in\mathbb{R}$. Hereby $\mu$ is not known to us. A Frequentist would compute $\hat{\mu}=\frac{1}{N}\sum_{i=1}^{N}y_i$. For the estimator $\hat{\mu}$ it holds that $\hat{\mu} \sim N(\mu,\frac{1}{N})$.
In a Bayesian approach we could assign a normal prior for $\mu$ with infinite variance $\mu\sim N(\mu_0,\lambda^{-1}), \lambda\rightarrow \infty$ and obtain a normal distributed posterior density
$\mu|y\sim N(\hat{\mu},\frac{1}{N})$.
I am wondering how $\hat{\mu} \sim N(\mu,\frac{1}{N})$ and $\mu|y\sim N(\hat{\mu},\frac{1}{N})$ can be interpreted and what different implications you can draw out of those formulas. Is it a correct Bayesian interpretation to state that after observing our data $y$ the variance term $\frac{1}{N}$ reflects the 'uncertainty' that is incorporated in choosing $\hat{\mu}$ as our decision rule? On the other hand side what does the Frequentist approach tells us regarding the 'true' parameter? Each and every comment is welcome!
 A: So, I welcome any comments or corrections, it's been a while since I've sat in front of a textbook.
Insofar as I've always thought of it, the frequentist isn't as interested in the distribution of $\hat{\mu}$ as the Bayesian is of $\mu \mid y$ (I'm totally going to get flamed for that statement). Why do I say this?

*

*In the frequentist view of things, $\mu$ is a degenerate random variable, and $\hat{\mu}$ is our best guess at $\mu$. Hence, we may use the (implicit) distribution of the estimator as a means of handling the error of our guess, but we can't really interpret $$P( \hat{\mu} \in [a,b]) = p(a,b)$$ as something like 

The probability that $\mu$ lies in $[a,b]$ is $p(a,b)$.

Instead, $p(a,b)$ encodes our uncertainty in $\hat{\mu}$ and -- this is my opinion -- this doesn't really help us get a handle on $\mu$...

*In the bayesian view, however (as you've identified), $\mu$ is a (nondegenerate) random variable, and our collection of data, $y$, affords us the ability to say things like $$P(\mu \in [a,b] \mid y) = p(a,b)$$


The probability that $\mu$ lies in $[a,b]$ is $p(a,b)$.
  
  ... I'm not sure if this helps, but leave a comment and I'll cleanup/elaborate as necessary!

