I am trying to determine the values for when this ARMA model is covariance stationary.

I have the model: $z_t = a + Bz_{t-1} + u_t + u_{t-1}$

I have written it in terms of the lag operator:

(1 - BL) $z_t= a + (1 + L)u_t$

However I am unsure what to do about the constant?

  • $\begingroup$ Hi: The constant won't play a role ion the determination of covariancestationarity since its not random. To check if the process is covariance stationary, check if $cov(z_{t}, z_{t+1}) = cov(z_{t+k}, z_{t+k+1})$. It's easier to do this if you don't move the lagged $z_t$ over to the right hand side. $\endgroup$ – mark leeds May 15 at 13:19
  • $\begingroup$ Actually, my mistake: Looking at it again, it's best to do exactly what you did and then divide both sides by $(1-BL)$ in order to get $z_t$ by itself. Then, you can calc those things I mentioned in previous comment. $\endgroup$ – mark leeds May 15 at 13:21

Hi: Here's an attempt at a heuristic solution.

Dividing both sides by $(1 - \rho L)z_{t}$ ( using $\rho$ instead of B and $\epsilon_t$ instead of $u_{t}$ because that notation is easier for me ), gives

$z_{t} = \frac{a}{1-\rho L} + \frac{\epsilon_t}{1 - \rho L} + \frac{\epsilon_{t-1}}{ 1 - \rho L} $

= $\frac{a}{1-\rho L} + \sum_{i=0}^{\infty} \rho^{i} \epsilon_{t-i} + \sum_{i=0}^{\infty} \rho^{i} \epsilon_{t-i-1}$

This means that we have a similiar expression for $z_{t+1}$.

$ z_{t+1} = \frac{a}{1-\rho L} + \sum_{i=0}^{\infty} \rho^{i} \epsilon_{t-i+1} + \sum_{i=0}^{\infty} \rho^{i} \epsilon_{t-i}$

To calculate $cov(z_{t}, z_{t+1})$, the first term on the RHS doesn't play a role since it's not random. Now, since the $\epsilon_{t}$ are independent, one only needs to find the terms in the two expressions where the same $\epsilon_{t}$ occurs because that will result in a non-zero covariance term. The second infinite sum in $z_{t+1}$ coincides exactly with the first infinite sum in $z_{t}$ so all of those terms coincide without any shifting. Next, the terms in the second infinite sum of $z_{t}$, shifted over two time periods, coincide exactly with the terms in the first infinite sum of $z_{t+1}$ so all of those terms coincide.

So, now it should be clear that $cov(y_{t}, y_{t+1})$ will be equal to $cov(y_{t+k}, y_{t+k+1})$ because, with the terms in each expression be shifted over by k time units, the term terms that coincide won't change. Of course, this is a heuristic argument but hopefully you can work out the details.

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