# How to determine covariant stationary values?

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?

• 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. – mark leeds May 15 at 13:19
• 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. – 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.