A stochastic process constitutes a discrete Markov Chain of order 1 if it has the memoryless property, in the sense that the probability that the chain will be in a particular state i, of a finite set of possible states, at time t+1 depends only on the state of the chain at time t.

Is an AR(1) memoryless? When written in the form below, it may seem reasonable to guess that an AR(1) model is also memoryless to the same extent as a Markov chain of order 1, because it depends only on its one preceding lag.

$$x_t=a+bx_{t-1}+\epsilon_t \quad \text{ and assume }|b|<1.$$

However, when $|b|<1$ holds and we write the AR(1) model in MA($\infty$) form, we end up with $x_t$ as a function of innovations going back infinitely far in the past. Now it looks like the AR(1) model has "infinite" memory of all the innovations going back infinitely far in the past, albeit as a function of innovations and not of own past values.

What exactly is the memoryless property? Is AR(1) memoryless or of "infinite" memory?

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    $\begingroup$ The crucial condition here is the state given the condition at time $t$. The MA($\infty$) representation does not preclude memorylessness. $\endgroup$ Commented Nov 9, 2018 at 18:39

1 Answer 1


In your question, you switched from conditioning on prior values of the process $x_t$ to conditioning on prior values of the shocks $\epsilon_t$. Those are different! The AR(1) process $y_t = b y_{t-1} + \epsilon_t$ satisfies the Markov property while the MA(1) process $y_t = -c \epsilon_{t-1} + \epsilon_{t} $ does not.

The Markov property basically says that the future is independent of the past given the present value of the process $y_{t}$.

In the AR(1) case, the current state $y_{t}$ can be written as an infinite sum of prior shocks: $y_{t} = \sum_{j=0}^\infty b^j \epsilon_{t-j}$. Knowing the history $\epsilon_t, \epsilon_{t-1}, \epsilon_{t-2}, \ldots$ basically tells you what $y_{t}$ is. An AR(1) is Markov so all that matters for the future distribution is the current state. That you can write down how we got to the current state in terms of prior shocks doesn't imply the process isn't Markov.

Regarding your question about an MA(1)

Let's assume we have the MA(1) process $y_t = -c \epsilon_{t-1} + \epsilon_t$. Let $L$ denote the lag operator. $$ y_t = (1 - cL) \epsilon_t $$

Assuming $|c| < 1$ and the process is invertible we can write: $$ (1 - cL)^{-1} y_t = \epsilon_t$$ As discussed in this answer, this turns out to be: $$ (1 + cL + (cL)^2 + (cL)^3 + \ldots ) y_t = \epsilon_t $$ Hence:

$$ y_t = -\sum_{j=1}^\infty c^j y_{t-j} + \epsilon_t$$

The MA(1) can be written as an AR($\infty$). You can see this process isn't Markov (it depends on an infinite history of past states).

  • $\begingroup$ Thank you, this is helpful. In your answer, you state that the MA(1) process does not satisfy the Markov property. Couldn't MA(1) be thought of as a zero-order Markov Chain? $\endgroup$ Commented Nov 9, 2018 at 20:01
  • $\begingroup$ @ColorStatistics No. It would be an infinite order auto-regressive process. $\endgroup$ Commented Nov 9, 2018 at 20:45
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    $\begingroup$ @ColorStatistics Something that I originally found a little bit tricky in the MA(1) context is that at time $t$, you don't exactly know $\epsilon_t$. Writing it in terms of observables: $\epsilon_t = \sum_{j=0}^\infty c^j y_{t-j}$. As a practical matter, $c^j$ gets small quite fast, but in a technical sense, $\epsilon_t$ depends on an infinite history. $\endgroup$ Commented Nov 9, 2018 at 21:43

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