I once heard some statements regarding the relationship between ARMA and AR process, such as

An average of severl lags of an autoregression forms an ARMA process


A weighted mixture of lags of an AR(P) model is ARMA

I am not very clear about these two statements, whether they are right? If they are right, how to get them? Thanks.


1 Answer 1


As I'm sure you are aware, ARMA is an acronym for AutoRegressive Moving Average (Stochastic Process). More fully, we use ARMA(p, q) where p is the order of the autoregressive portion and q the order of the moving average portion.

Both of the statements are saying the same thing: if you form a linear combination of AR processes an ARMA process of some order results. This is a consequence of an inversion theorem in Time Series analysis which states that all stationary ARMA processes have a representation as a (possibly infinite) stationary AR process.

The ARMA model expresses $X_t$ as the sum of polynomials in the lagged values of X_t and the lagged values of $\epsilon_t$ (the i.i.d. noise terms). So an AR(1) model looks like

$X_t = \theta X_{t-1} + \epsilon_t$.

The single lagged value of the series is

$X_{t-1} = \theta X_{t-2} + \epsilon_{t-1}$.

Now, let $0 < \phi < 1$ and consider $Y_t = \phi X_t + (1-\phi) X_{t-1}$. $Y_t$ is a weighted average of lags of an AR(1) model. Let's define $\xi = \phi \theta$

$Y_t = \phi\theta X_{t-1} + (1-\phi)\theta X_{t-2} + \phi \epsilon_t + (1-\phi) \epsilon_{t-2}$.

Some mild algebra (basically arranging things to make a $Y_{t-1}$ appear on the right hand side) puts this in the form

$Y_t - \epsilon_t = (\xi_1 Y_{t-1} + \xi_2 Y_{t-2}) + (\phi \epsilon_{t-1} + \phi^2 \epsilon_{t-2})$

Note that this is the sum of a lagged polynomial in $Y_t$ and a lagged polynomial in the noise vector, $\epsilon$. This is an ARMA process.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.