Is AR(1) process such as

yt=ρyt−1+εt a Markov process?If it is, then VAR(1) is the vector version of Markov process?

**Answer**

The following result holds: If ϵ1,ϵ2,… are *independent* taking values in E and f1,f2,… are functions fn:F×E→F then with Xn defined recursively as

Xn=fn(Xn−1,ϵn),X0=x0∈F

the process (Xn)n≥0 in F is a Markov process starting at x0. The process is time-homogeneous if the ϵ‘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

fn(x,ϵ)=ρx+ϵ.

Thus they are homogeneous Markov processes if the ϵ‘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 (Xn)n≥0 on a Borel space F there are i.i.d. uniform random variables ϵ1,ϵ2,… in [0,1] and functions fn:F×[0,1]→F such that with probability one

Xn=fn(Xn−1,ϵn).

See Proposition 8.6 in Kallenberg, *Foundations of Modern Probability*.

**Attribution***Source : Link , Question Author : Flying pig , Answer Author : NRH*