<p>Hi: Here's an attempt at a heuristic solution.</p>
<p>Dividing both sides by <span class="math-container">$(1 - \rho L)z_{t}$</span> ( using <span class="math-container">$\rho$</span> instead of B and <span class="math-container">$\epsilon_t$</span> instead of <span class="math-container">$u_{t}$</span> because that notation is easier for me ),
gives</p>
<p><span class="math-container">$z_{t} = \frac{a}{1-\rho L} +  \frac{\epsilon_t}{1 - \rho L} + \frac{\epsilon_{t-1}}{ 1 - \rho L} $</span></p>
<p>= <span class="math-container">$\frac{a}{1-\rho L}  + \sum_{i=0}^{\infty} \rho^{i} \epsilon_{t-i} + \sum_{i=0}^{\infty} \rho^{i} \epsilon_{t-i-1}$</span></p>
<p>This means that we have a similiar expression for <span class="math-container">$z_{t+1}$</span>.</p>
<p><span class="math-container">$ 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}$</span></p>
<p>To calculate <span class="math-container">$cov(z_{t}, z_{t+1})$</span>, the first term on the RHS doesn't play a role since it's not random.  Now,  since the <span class="math-container">$\epsilon_{t}$</span> are independent, one only needs to find the terms in the two expressions where the same <span class="math-container">$\epsilon_{t}$</span> occurs because that will result in a non-zero covariance term. The second infinite sum in <span class="math-container">$z_{t+1}$</span> coincides exactly with the first infinite sum in <span class="math-container">$z_{t}$</span> so all of those terms coincide without any shifting.  Next,  the terms in the second infinite sum of <span class="math-container">$z_{t}$</span>, shifted over two time periods, coincide exactly with the terms in the first infinite sum of <span class="math-container">$z_{t+1}$</span> so all of those terms coincide.</p>
<p>So, now it should be clear that <span class="math-container">$cov(y_{t}, y_{t+1})$</span> will be equal to <span class="math-container">$cov(y_{t+k}, y_{t+k+1})$</span> 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.</p>

CC BY-SA 4.0
2020-05-17T01:40:20.390Z
2020-06-21T13:42:26.237Z
2020-06-21T13:42:26.237Z
mark leeds
<p>I am trying to determine the values for when this ARMA model is covariance stationary.</p>

<p>I have the model: <span class="math-container">$z_t = a + Bz_{t-1} + u_t + u_{t-1}$</span></p>

<p>I have written it in terms of the lag operator:</p>

<p>(1 - BL) <span class="math-container">$z_t= a + (1 + L)u_t$</span></p>

<p>However I am unsure what to do about the constant?</p>

CC BY-SA 4.0
2020-05-15T13:19:19.203Z
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
CC BY-SA 4.0
2020-05-15T13:21:22.730Z
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
<econometrics><time-series>
How to determine covariant stationary values?
Katie
# How to determine covariant stationary values?

(Asked by: username_0 on 2020-05-15)

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?

**Tags:** econometrics, time-series

> **Comments:**
> **username_1**: 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.  
> **username_1**: 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.  

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## 1 answers

### Answer by username_1 on 2020-05-17. Score: 2

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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