ARMA(1,1) Moments With and Without Stationarity

Summary

The document asks how to obtain the mean and variance of an ARMA(1,1) process when the autoregressive coefficient is not restricted to the stationary range. It specifies Gaussian, independent, zero-mean innovations and presents the usual lag-operator representation for the case where the coefficient’s absolute value is below one. In that stationary case, the process can be expressed as a convergent infinite sum of current and past innovations, from which moments can be calculated.

The supplied answer does not work out those moments and does not derive a general formula outside stationarity. It notes that the infinite-series representation used for the stationary case fails to converge when the condition is violated, and suggests the process is then nonstationary. Thus, the material chiefly clarifies the scope of the standard representation; it leaves the broader question of time-dependent moments and initial conditions unanswered.

Key ideas

• The process is an ARMA(1,1) driven by independent Gaussian innovations.
• When the autoregressive coefficient has absolute value below one, the lag representation yields a convergent infinite sum.
• That representation provides a route to calculating stationary moments.
• The answer does not derive the moments explicitly.
• Outside the stationary condition, the response leaves general moments unresolved.

Tags

# ARMA moments proof


# ARMA moments proof












Consider a standard ARMA(1,1) process such as

$$x_t - \beta x_{t-1} = \theta u_{t-1} + u_t$$

where $u_t$ is i.i.d. $u_t \sim N(0,\sigma^2)$. I know how to derive mean and variance with stationary condition ($|\beta| < 1$), but how can I derive mean and variance in general form for all values of $\beta$? This means without stationary or weak dependence of ARMA(1,1) process.

Thanks

## Answer by mark leeds (score 1)

https://quant.stackexchange.com/a/49456

For the first, where $|\beta| < 1.0$, you can write it using the lag operator.

$x_t (1 - \beta L) = (1 + \theta L) u_t $

$X_t = \frac{(1 + \theta L) u_t}{(1- \beta L)} $

Since $|\beta| < 1.0 $, this is an infinite sum that converges:

$X_t = \sum_{i=0}^\infty \beta^{i}( 1 + \theta L) u_{t-i}$

The $u_{t}$ are independent and normal with mean zero and variance $\sigma^2$ so you have a converging infinite sum of iid random variables so you should be able to calculate the mean and the variance. I leave that as an exercise for the reader.

I'm not sure if the second part is possible because the series doesn't converge in that case because $X_t$ is not stationary. Hopefully someone else can say something about that part.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.