Is AR(1) a Markov process? Is AR(1) process such as
$y_t=\rho y_{t-1}+\varepsilon_t$ a Markov process? 
If it is, then VAR(1) is the vector version of Markov process?
 A: A process $X_{t}$ is an AR(1) process if
$$X_{t} = c + \varphi X_{t-1} + \varepsilon_{t} $$ 
where the errors, $\varepsilon_{t}$ are iid. A process has the Markov property if 
$$P(X_{t} = x_t | {\rm entire \ history \ of \ the \ process }) = P(X_{t}=x_t| X_{t-1}=x_{t-1})$$
From the first equation, the probability distribution of $X_{t}$ clearly only depends on $X_{t-1}$, so, yes, an AR(1) process is a Markov process. 
A: The following result holds: If $\epsilon_1, \epsilon_2, \ldots$ are independent taking values in $E$ and $f_1, f_2, \ldots $ are functions $f_n: F \times E \to F$ then with $X_n$ defined recursively as
$$X_n = f_n(X_{n-1}, \epsilon_n), \quad X_0 = x_0 \in F$$
the process $(X_n)_{n \geq 0}$ in $F$ is a Markov process starting at $x_0$. The process is time-homogeneous if the $\epsilon$'s are identically distributed and all the $f$-functions are identical. 
The AR(1) and VAR(1) are both processes given in this form with  
$$f_n(x, \epsilon) = \rho x + \epsilon.$$
Thus they are homogeneous Markov processes if the $\epsilon$'s are i.i.d.
Technically, the spaces $E$ and $F$ need a measurable structure and the $f$-functions must be measurable. It is quite interesting that a converse result holds if the space $F$ is a Borel space. For any Markov process $(X_n)_{n \geq 0}$ on a Borel space $F$ there are i.i.d. uniform random variables $\epsilon_1, \epsilon_2, \ldots$ in $[0,1]$ and functions $f_n : F \times [0, 1] \to F$ such that with probability one 
$$X_n = f_n(X_{n-1}, \epsilon_n).$$
See Proposition 8.6 in Kallenberg, Foundations of Modern Probability.
A: What is a Markov process? (loosely speeking) A stochastic process is a first order Markov process if the condition
$$P\left [ X\left ( t \right )= x\left ( t \right ) | X\left ( 0 \right )= x\left ( 0 \right ),...,X\left ( t-1 \right )= x\left ( t-1 \right )\right ]=P\left [ X\left ( t \right )= x\left ( t \right )  | X\left ( t-1 \right )= x\left ( t-1 \right )\right ]$$
holds.
Since next value (i.e. distribution of next value) of $AR(1)$ process only depends on current process value and does not depend on the rest history, it is a Markov process. When we observe the state of autoregressive process, the past history (or observations) do not supply any additional information. So, this implies that probability distribution of next value is not affected (is independent on) by our information about the past.
The same holds for VAR(1) being first order multivariate Markov process.
