How to correctly interpret the likelihood function of a Markov Chain, $P_{\theta}(X_1, ..., X_T)$ vs $P(X_1, ..., X_T|\theta)$? Suppose $ \{ X_i \}_{i=1}^T$ are states of a markov chain and let $P_{\theta}(X_1, \ldots, X_T)$ be the probability of observing the path when $\theta$ is the true parameter value (a.k.a. the likelihood function for $\theta$). Using the definition of conditional probability, we know 
$$ P_\theta(X_1, \ldots, X_T)  = P_\theta(X_T \mid X_{T-1}, \ldots, X_1) \cdot P_\theta(X_1, \ldots, X_{T-1})$$ 
Since this is a markov chain, we know that $P_\theta(X_T \mid X_{T-1}, \ldots, X_1) = P_\theta(X_T \mid X_{T-1} )$, so this simplifies this to 
$$ P_\theta(X_1, \ldots, X_T)  = P_\theta(X_T \mid X_{T-1}) \cdot P_\theta (X_1, \ldots, X_{T-1})$$
Then, the likelihood function is then:
$$ P_\theta(X_1, \ldots, X_T)  = \prod_{i=1}^T P_\theta(X_i \mid X_{i-1} ) $$ 
where $X_0$ is to be interpreted as the initial state of the process. 
My question is how the above analysis works if we use the different notation of $$P(X_1, \ldots, X_T\mid\theta)$$
being the likelihood function for $\theta$ instead. Then, won't be end up with:
$$ P(X_1, \ldots, X_T\mid\theta)  = P(X_T \mid X_{T-1}, \ldots, X_1, \theta) \cdot P(X_1, \ldots, X_{T-1}\mid\theta)\text{ ?}$$
Then, we will have:
$$
P(X_1, \ldots, X_T\mid\theta) = P(X_T \mid X_{T-1}, \ldots, X_1, \theta) \cdots P(Y_1\mid Y_0,\theta)
$$
But, I am not sure how to find $P(Y_1\mid Y_0,\theta)$ since it now has $\theta$ in the conditional part?
In other words, if we were to treat the likelihood as a conditional probability distribution, do we now have a different problem of trying to find the joint distribution of the states and $\theta$? What is the correct notation for using the conditional distribution notation for the likelihood?
 A: The two descriptions you provide are indeed equivalent. Let me try to shine some light on what's really going on here:
Consider your second setup, $P(X^n \mid \theta)$ (we use $X^k$ to denote $(X_1,X_2, \ldots, X_k)$, with theta as a realization of some random variable linked to $X^n$.The sequence $X_1 \to X_2 \to \cdots \to X_n$ supposedly still forms a Markov chain, yet all the variables depend on $\theta$, so how come no dependence is introduced, i.e., how come we are still allowed to express
\begin{equation}
  P(X^n \mid \theta) = \prod_{i=1}^n P(X_i \mid X_{i-1}, \theta)?
\end{equation}
Well, that's precisely the notion of conditional independence. If $\theta$ was unknown, Markovity would indeed be lost, in fact
\begin{equation}
  P(X_k \mid X^{k-1}) \neq P(X_k \mid X_{k-1}),
\end{equation}
as more past samples ($X^{k-1}$) yield more information about $\theta$, which in turn yields more information about $X_k$.
On the other hand, once you fix $\theta$, i.e., once you condition on a certain value for it, you introduce your desired Markov chain as you cap the indirect link between the $\{X_k\}$ via $\theta$.
In other words, the first scenario you describe is merely a shorthand notation for the second one; one might use this to elevate $\theta$ to some kind of "meta-parameter".
A practical situation for the second scenario is if $\theta$ is known to originate according to some prior distribution $P_\theta$. Can you formulate the MAP estimate for $\theta$ if


*

*$\theta \sim P_\theta$

*Conditioned on $\theta$ $X^n$ forms a Markov chain?

