For a Markov Chain is $(_|_{+1},…,_)$ equals to $(_| _{+1})$? I am new to Markov Chains and using this concept in statistics.
For a Markov Chain, may I say that $(_|_{+1},…,_)$ equals to $(_| _{+1})$?
If yes, how can I prove that?
 A: The Markov property applies both forwards and backwards
The Markov property holds that $p(q_{t+k}|q_t,...,q_{t+k-1}) = p(q_{t+k}|q_{t+k-1})$ for all $t \in \mathbb{Z}$ and $k \in \mathbb{N}^+$.  This property is usually stated as a forward equation (i.e., the probability density of an outcome conditional on past values), but it also implies the result in your question, which is the same essential property as a backward equation (i.e., the probability density of an outcome conditional on future values).  This is shown as follows:
$$\begin{equation} \begin{aligned}
p(q_t| q_{t+1},...,q_T) 
&\overset{q_t}{\propto} p(q_t, q_{t+1},...,q_T) \\[12pt]
&= p(q_t) \prod_{i=1}^{T-t} p(q_{t+i}| q_{t},...,q_{t+i-1}) \\[6pt]
&= p(q_t) \prod_{i=1}^{T-t} p(q_{t+i}| q_{t+i-1}) \\[6pt]
&\overset{q_t}{\propto} p(q_t) p(q_{t+1}|q_t) \\[12pt]
&= p(q_t,q_{t+1}) \\[12pt]
&\overset{q_t}{\propto} p(q_t|q_{t+1}). \\[6pt]
\end{aligned} \end{equation}$$
(The third step in this working uses the forward equation for the Markov property.)  So, the take-away message is that the Markov property applies both forward and backward --- one implies the other.
A: Ben correctly points out that a Markov chain is Markovian both forwards and backwards (+1). This is always true, but it is not the same thing as reversibility. In particular, reversibility requires the existence of a stationary distribution, call it $\pi$. 
For a chain with a stationary distribution (aka a marginal distribution that doesn't change depending on what time point you're at), 
$$
p(q_{t-1} \mid q_t) = \frac{p(q_t \mid q_{t-1})\pi(q_{t-1})}{\pi(q_t)} \tag{1}.
$$
This is the definition of a reversible Markov chain. If you multiply both sides of the above by the denominator on the right hand side, you will get the more familiar definition of reversibility:
$$
p(q_{t-1} \mid q_t)\pi(q_t) = p(q_t \mid q_{t-1})\pi(q_{t-1})
$$
which says being in state $q_t$ and then $q_{t-1}$ a moment later has the same chances as being in state $q_{t-1}$ and then state $q_t$ a moment later.
You'll notice that sub-scripting elements of the state space with a time index isn't great for expressing this idea very well. 
