Finding $\mathsf P(\textbf{Y}\mid\textbf{Z})$ where $Z_1,Z_2,\ldots,Z_n$ are conditionally independent As an extension of my previous question which is described below...

Assume $Y_1$, $Y_2$, $\ldots$ ,$Y_n$ are random variables over a
  regular lattice indexed by $i= 1,2,\ldots,n$ where
  $Y_i\in\{1,2,...,K\}$.  Let the probability of a particular
  configuration $\textbf{y}= (y_1,y_2,...,y_n)$ be given by
$$\mathsf P(\textbf{Y}=\textbf{y})
 =C\cdot\text{exp}\left(\sum_{i=1}^n\alpha_{i,y_i}+\frac{1}{2}\beta\sum_{i=1}^n\sum_{j\in
 N(i)}1(y_i=y_j)\right)$$
where $C$ is the normalizing constant, $N(i)$ is the set of neighbor
  points of $i$ and $1(.)$ is the indicator function.  This model is
  known as Potts model and is popular in image analysis.

Let us instead of observing $\textbf{Y}$,

we observe
  $Z_1,Z_2,\ldots,Z_n$ which are conditionally (on $\textbf{Y}$)
   independent with  $$Z_i\mid\textbf{Y}\stackrel{d}{=}Z_i\mid Y_i\sim
 f_{Y_i}(.)$$ where $f_k(.)$ are known distributions for $k=
 1,2,\ldots,K$. Show that
$$\mathsf P(\textbf{Y}\mid\textbf{Z})
 =\tilde{C}\cdot\text{exp}\left(\sum_{i=1}^n\tilde{\alpha}_{i,y_i}+\frac{1}{2}\beta\sum_{i=1}^n\sum_{j\in
 N(i)}1(y_i=y_j)\right)$$
for appropriate choices of $\tilde{\alpha}_{i,k}$ and a new normalizing
  constant $\tilde{C}$.

My thoughts:
It's not clear to me what it means, in words, for $Z_1,Z_2,\ldots,Z_n$ to be conditionally independent (on $\textbf{Y}$). Does this mean $Y_i$ will give us just as much information about $Z_i$ as $\textbf{Y}$?
I thought it may be useful to use Bayes' Theorem. We have 
$$\begin{align*}
\mathsf P(\textbf{Y}\mid\textbf{Z})
&=\frac{\mathsf P(\textbf{Z}\mid\textbf{Y})\cdot\mathsf P(\textbf{Y})}{P(\textbf{Z})}\\\\
&\propto\mathsf P(\textbf{Z}\mid\textbf{Y})\cdot\mathsf P(\textbf{Y})\\\\
&=\left(\prod_{i=1}^n \mathsf P(Z_i\mid Y_i)\right)\mathsf P(\textbf{Y})
\end{align*}$$
However, I'm not sure if this is correct or if this is even the right approach. Any suggestions on how I can proceed or alternative approaches would be greatly appreciated.
 A: 
It's not clear to me what it means, in words, for Z1,Z2,…,Zn to be conditionally independent (on Y).

It means that given $Y=y$ all your $Z_i's $ are statistically independent of each other. Otherwise they could be correlated. An example could be :
Suppose there is an algorithm which classifies if an image has a car in it or not. Every time it runs on a bunch of images you generate a score for it represented by the variable $Z_i$. Let $Y$ be 1 if the algorithm learns over time after knowing the correct results (a reinforcement machine learning kind of an algorithm) and let $Y$ be 0 otherwise. 
Now given $Y=0$ you really won't expect your algorithm to improve in performance over time, or in other words $Z_1$ wouldn't have much to say about $Z_2$. However, for a reinforcement type of an algorithm (when $Y=1 $), you would expect the scores to improve over time in general $Z_1 < Z_{1000}$ is more likely compared to $Z_{1000} < Z_1$.
Hence conditional on $Y=0$, $ \quad Z_i's$ are independent. 
A: We have that
$$\begin{align*}
\mathsf{P}(\textbf{Y}\mid\textbf{Z})
&=\frac{\mathsf P(\textbf{Z}\mid\textbf{Y})\cdot\mathsf P(\textbf{Y})}{\mathsf P(\textbf{Z})}\\\\
&=C_0\cdot \mathsf P(\textbf{Z}\mid\textbf{Y})\cdot\mathsf P(\textbf{Y})\\\\
&=C_0\prod_{i=1}^n f_{Y_i}(\cdot)\cdot C\cdot\text{exp}\left(\sum_{i=1}^n \alpha_{i,y_i}+\frac{1}{2}\beta\sum_{i=1}^n \sum_{j\in N(i)}1(y_i=y_j)\right)\\\\
&=C_0\cdot\text{exp}\left(\sum_{i=1}^n \text{log}\left(f_{Y_i}(\cdot)\right)\right)\cdot C\cdot\text{exp}\left(\sum_{i=1}^n \alpha_{i,y_i}+\frac{1}{2}\beta\sum_{i=1}^n \sum_{j\in N(i)}1(y_i=y_j)\right)\\\\
&=\tilde{C}\cdot\text{exp}\left(\sum_{i=1}^n\left(\alpha_{i,y_i}+\text{log}\left(f_{Y_i}(\cdot)\right)\right)+\frac{1}{2}\beta\sum_{i=1}^n\sum_{j\in N(i)}1(y_i=y_j)\right)\\\\
&=\tilde{C}\cdot\text{exp}\left(\sum_{i=1}^n\tilde\alpha_{i,y_i}+\frac{1}{2}\beta\sum_{i=1}^n\sum_{j\in N(i)}1(y_i=y_j)\right)
\end{align*}$$
for appropriate choices of $\tilde\alpha_{i,k}$ and a new normalizing constant $\tilde{C}$.
