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?


3 Answers 3


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.

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    $\begingroup$ The proposition mentioned in the answer is Proposition 11.6 in the third edition of Foundations of Modern Probability by Kallenberg. $\endgroup$
    – Cm7F7Bb
    Apr 27, 2022 at 14:19

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.

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    $\begingroup$ -1, the same reason as for another poster. The answer implies that it is easy to check the cited Markov property. It is not, unless demonstrated otherwise. Note also that AR(1) processes can be defined with $\varepsilon_t$ being non-iid, so this should be adressed also. $\endgroup$
    – mpiktas
    Mar 2, 2012 at 4:25
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    $\begingroup$ The main problem is that we can easily write $X_t=c+\phi c+\phi^2X_{t-2}+\phi\varepsilon_{t-1}+\varepsilon_{t}$ and then the last sentence would imply that $P(X_t=x_t|\text{entire history})=P(X_{t}=x_t|X_{t-2}=x_{t-2})$. $\endgroup$
    – mpiktas
    Mar 2, 2012 at 4:29
  • $\begingroup$ Well, markov processes do depend on $X_{t-2}$ when you haven't also conditioned on $X_{t-1}$. I suppose a more formal argument would assume you're conditioning sequentially (i.e. you don't include $X_{t-2}$ unless you've already conditioned on $X_{t-1}$). $\endgroup$
    – Macro
    Mar 2, 2012 at 12:58
  • $\begingroup$ and what you've written there actually depends on both $X_{t-2}$ and $X_{t-1}$ (through the error term $\varepsilon_{t-1}$). The bottom line is that joint likelihood can be written easily as a product of conditional likelihoods which only require conditioning on the previous time point. Through parameter redundancies you can make it look like the distribution of $X_t$ depends on $X_{t-2}$ but, once you've conditioned on $X_{t-1}$, it clearly doesn't. (p.s. I was using a standard definition of an AR(1) process per Shumway and Stoeffer's time series book) $\endgroup$
    – Macro
    Mar 2, 2012 at 13:18
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    $\begingroup$ Note I do not say that the answer is incorrect. I am just nitpicking at the details, i.e. that the second equality is intuitively evident, but if you want to prove it formally it is not so easy, IMHO. $\endgroup$
    – mpiktas
    Mar 2, 2012 at 16:32

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.

  • $\begingroup$ Hm, if $\varepsilon_t$ are not iid, I do not think it holds. Also you did not give a proof, only cited the Markov property. $\endgroup$
    – mpiktas
    Feb 28, 2012 at 11:18
  • $\begingroup$ I thought that Markov Process refers to the continuous case. Usual AR time series are discrete, so it should correspond to a Markov Chain instead of a Markov Process. $\endgroup$
    – joint_p
    Feb 28, 2012 at 13:31
  • $\begingroup$ So we observe state of autoregressive process, $X_t$. The past history is $X_{t-1},X_{t-2},...$. This does not supply any additional information? $\endgroup$
    – mpiktas
    Feb 28, 2012 at 13:44
  • $\begingroup$ @joint_p, the terminology is not completely consistent in the literature. Historically, as I see it, the usage of "chain" instead of "process" was typically a reference to the state space of the process being discrete but occasionally also time being discrete. Today many use "chain" to refer to discrete time but allowing for a general state space, as in Markov Chain Monte Carlo. However, using "process" is also correct. $\endgroup$
    – NRH
    Feb 28, 2012 at 14:06
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    $\begingroup$ -1, since the proof of Markovian property is not given. Also the hand waving argument is not consistent with formula given. current state = $t$, past means $t-1, t-2,...$, next means $t+1$, but the formula does not involve $t+1$. $\endgroup$
    – mpiktas
    Mar 2, 2012 at 4:21

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