# Writing AR(1) as a MA($\infty$) process

The AR(1) process is

$$X_t = \phi X_{t-1} + \varepsilon_t$$

if we use this formula recursively, we get $$X_t = \phi(\phi X_{t-2} + \varepsilon_{t-1}) + \varepsilon_t = \phi^2X_{t-2} + \phi\varepsilon_{t-1} + \varepsilon_t = \cdots = \phi^k X_{t-k} + \sum_{j=0}^k \phi^j\varepsilon_{t-j}$$

If we let $k\to\infty$, we get $$X_t = \lim_{k\to\infty}(\phi^k X_{t-k} + \sum_{j=0}^k \phi^j\varepsilon_{t-j}) = \lim_{k\to\infty}(\phi^k X_{t-k}) + \sum_{j=0}^\infty \phi^j\varepsilon_{t-j}$$ The duality between AR(1) and MA($\infty$) states that there is an equivalence between the two, and that we can write $X_t$ as

$$X_t = \sum_{j=0}^\infty \phi^j\varepsilon_{t-j}$$

The difference between the two results is the term $\lim_{k\to\infty}(\phi^k X_{t-k})$, which should be zero, but how do I show this?

Assuming $|\phi| < 1$, we have that $\lim_{k\to\infty}\phi^k = 0$ of course, but I don't see why $\lim_{k\to\infty} X_{t-k} < \infty$? Does convergence asuume the law of large numbers, or is there another way to show equivalence?

I know there is a proof which inverts the lag operator $1-B$, but I didn't find any justification for why the operator can even be inverted, so I wanted an alternative proof, as the one above.

• There are some issues with your indices - summations should start at $j=0$ and it should then be $\epsilon_{t-j}$, not $\epsilon_{t-k}$. Jul 20, 2018 at 14:56

The usual sense in which convergence is understood in this case is in mean square:

$$E[Y_t-(\epsilon_t+\phi\epsilon_{t-1}+\phi^2\epsilon_{t-2} +\ldots+\phi^j\epsilon_{t-j})]^2=\phi^{2(j+1)} E[Y_{t-j-1}]^2$$ If $Y_t$ is stationary $$E[Y_{t-j-1}]^2=\gamma_0+\mu^2$$ Hence $$\lim_{j\to\infty}E[Y_t-(\epsilon_t+\phi\epsilon_{t-1}+\phi^2\epsilon_{t-2} +\ldots+\phi^j\epsilon_{t-j})]^2=0$$

• So we would use the weak law of large numbers, i.e. $\lim_{j\to\infty} E(\phi^j X_{t-j})^2 = 0$? Jul 20, 2018 at 21:44
• The WLLN usually refers to sample size going to infinity, whereas in this situation, the number of lags goes to infinity. I rather read it as saying that the mean of the squared difference vanishes. Jul 23, 2018 at 5:42

You are right to be suspicious of this step, and in fact, without further assumptions to limit the size of $X_{-\infty}$ you cannot get the required form. Remember that the recursive equation for the AR model is insufficient to yield the joint distribution of the process. (You need to impose a distribution for the error process, and even then, you either need to impose stationarity or specify an initial distribution that leads to some non-stationary model.) If you only have this recursive equation, there is no reason that the time-series values could not explode out to large values as $t \rightarrow -\infty$.

For example, the deterministic non-stationary AR process $X_t = \phi^t$ satisfies the recursive equation you have specified (with zero errors), and in this case you have $\lim_{k \rightarrow \infty} X_{t-k} = \infty$. In this model, for any $\phi \neq 0$ you also have:

$$\phi^k X_{t-k} = \phi^k \phi^{t-k} = \phi^t \neq 0.$$

Since the errors are zero in this deterministic model, this gives you the limiting result:

$$X_t = \underbrace{\sum_{k=0}^\infty \phi^k \varepsilon_{t-k}}_{0} + \underbrace{\lim_{k \rightarrow \infty} \phi^k X_{t-k}}_{\phi^t}. \\[6pt]$$

Clearly, in this case, the limiting term is non-zero, and cannot be removed from the result. If you would like to be able to remove this last term, you need to add further assumptions to your model (e.g, stationarity).

• I have the same reasoning, However, when I transform $(1 - \phi L)X_{t} = \varepsilon_t$ into $X_{t} = (1 - \phi L)^{-1} \varepsilon_{t}$ when $|\phi| < 1$, it implies that $X_{t}$ is stationary, and the limiting term you showed above converges to 0. I am sure there is something wrong with my reasoning. Can you show it to me as it is important for me? Thank you so much!
– user340483
Aug 11, 2022 at 8:23