Are normal prior and posterior random variables dependent? Consider the usual normal Bayesian model. The prior $X=\mu_0+\sigma_0\epsilon_0$, where $\epsilon_0$ follows a standard normal distribution. The data $Y=X+\sigma_1\epsilon_1$, where $\epsilon_1$ follows a standard normal distribution. We know that the posterior $(X|Y=y)$ is in the format of $a\mu_0+(1-a)y+b\epsilon_2$ for some $a$ and $b$, where $\epsilon_2$ follows a standard normal distribution. My question is: what is the relationship between $\epsilon_2$ and $\epsilon_0$? Are they independent, or are they the same with probability one? 
All current books state that the distribution of $(X|Y=y)$ is $N(a\mu_0+(1-a)y,b)$. However, none of them discusses the relationship between $\epsilon_2$ and $\epsilon_0$.
 A: This is an interesting question! However, you cannot use these representations for fear of falling into fiducial statistics and their difficulties. When you write [with an abuse of notations]
\begin{align*}
X&=\mu_0+\sigma_0\epsilon_0\\
Y|X&=X+\sigma_1\epsilon_1\\
Y&=\mu_0+\{\sigma_0^2+\sigma_1^2\}^{1/2}\epsilon_3\\
X|Y&=\{\sigma_0^{-2}\mu_0+\sigma_1^{-2}Y\}\{\sigma_0^{-2}\mu_0+\sigma_1^{-2}\}^{-1}+\sigma_2\epsilon_2\\
\end{align*}
the random variables $\epsilon_i$ do not live in the same probabilistic universe (aka $\sigma$-algebra), even though they all are $\mathcal{N}(0,1)$ variates. Some are conditionally normal while others are marginally normals. For one thing, if you start substituting terms in these equations, you end up with the nonsensical
$$X|Y = \{\sigma_0^{-2}\mu_0+\sigma_1^{-2} [X+\sigma_1\epsilon_1]\}\{\sigma_0^{-2}\mu_0+\sigma_1^{-2}\}^{-1}+\sigma_2\epsilon_2$$
which sees an X on both sides of the equal sign! For another thing, if you agree that $X$ has to be produced or generated before $Y$ is produced or generated conditional on $X$, $\epsilon_0$ has to be generated first, followed by $\epsilon_1$, while the order of generation is reverted for $\epsilon_3$ and $\epsilon_2$. For yet another thing, they cannot all be produced together. Either the pair $(\epsilon_0,\epsilon_1)$ or the pair $(\epsilon_3,\epsilon_2)$ is produced. But not both.
As an extra remark, note also that, since the pair $(X,Y)$ is a bivariate Normal vector, $$(X,Y) \sim \mathcal{N}_2\left((\mu_0,\mu_0),
\left[\begin{matrix} \sigma_0^2 &-\sigma_1^2\\-\sigma_1^2 &\sigma_0^2+\sigma_1^2\end{matrix}\right]\right),$$it can also be written as 
$$(X,Y)=(\mu_0,\mu_0)+\left[\begin{matrix} \sigma_0^2 &-\sigma_1^2\\-\sigma_1^2 &\sigma_0^2+\sigma_1^2\end{matrix}\right]^{1/2}(\epsilon_4,\epsilon_5)$$where these new $\epsilon$'s live in yet another probabilistic space, while being independent Normal variates.
