<p>The process in question is a Moving-Average process of order one,</p>
<p><span class="math-container">$$\varepsilon_t = \frac {1} {1+\theta L} X_t = \frac {1} {1-(-\theta) L} X_t, \qquad  \theta =4.$$</span></p>
<p>Here <span class="math-container">$L$</span> is the lag operater, <span class="math-container">$L^k(z_t) = z_{t-k}$</span>. The process is covariance-stationary irrespective of the value of <span class="math-container">$\theta$</span>, but it is not invertible since <span class="math-container">$\theta \geq 1$</span>, so it does not have an autoregressive <span class="math-container">$AR(\infty)$</span> representation.</p>
<p>Since <span class="math-container">$\theta \geq 1$</span> we can solve it &quot;forward&quot;, as follows:
First, define the forward operator <span class="math-container">$F^k \equiv L^{-k}, F^k(z_t) = z_{t+k}$</span>.</p>
<p>Second, note that</p>
<p><span class="math-container">$$\frac {1} {1-(-\theta) L} = \frac{1}{\theta L[-(-\theta^{-1})L^{-1} +1]} = \frac{\theta^{-1}L^{-1}}{1-(-\theta)^{-1}L^{-1}]} = \frac{\theta^{-1}F}{1-(-\theta^{-1})F}$$</span></p>
<p>Then, since <span class="math-container">$\theta^{-1} &lt;1$</span>,</p>
<p><span class="math-container">$$\frac {1} {1-(-\theta) L} = \theta^{-1}F\cdot [1-\theta^{-1}F + \theta^{-2}F^{2}-\theta^{-3}F^{3}+...]$$</span></p>
<p>So
<span class="math-container">$$\varepsilon_t = X_t \cdot \theta^{-1}F\cdot [1-\theta^{-1}F + \theta^{-2}F^{2}-\theta^{-3}F^{3}+...]$$</span></p>
<p><span class="math-container">$$\implies \varepsilon_t = \sum_{j=1}^{\infty}\frac {(-1)^{j+1}}{\theta^j}X_{t+j}$$</span></p>
<p>This is the &quot;forward&quot; solution, when the MA process is not invertible.</p>
<p>I guess the OP can now determine whether this solution is covariance-stationary.</p>

CC BY-SA 4.0
2024-01-26T04:08:02.093Z
Thanks Alecos. That's the solution that I was thinking of in my mind but the gory details are appreciated.
mark leeds
CC BY-SA 4.0
2024-01-26T08:39:18.120Z
@markleeds You're welcome Mark.
Alecos Papadopoulos
CC BY-SA 4.0
2024-01-26T02:04:00.077Z
2024-01-26T02:04:00.077Z
Alecos Papadopoulos
<p>Is there a covariance stationary solution to the equation?</p>
<p><span class="math-container">$X_t = 4 \varepsilon_{t-1} + \varepsilon_t, \quad \text{where } \{ \varepsilon_t \} \in WN(0, \sigma^2), t \in \mathbb{Z}$</span>.</p>
<p>Unfortunately I completely do not know how to handle this exercise. My only thought was to check if <span class="math-container">$X_t$</span> is covariance stationary. So is it true that</p>
<ol>
<li><span class="math-container">$E(X_t) = const$</span>,</li>
<li><span class="math-container">$\gamma(t,t+l) = Cov(X_t, X_{t+1}) = \gamma(s, s+l)$</span>.</li>
</ol>
<p>But is a good idea?</p>

CC BY-SA 4.0
2024-01-24T05:17:10.940Z
Hi: that's an MA(1) with $\theta = 4$. So, I would look up info on the MA(1) because I could try to explain it ( obtaining a solution ) here but it won't be as good as what you should find. The short of it is that, $\theta=4$ means that you can't write $\epsilon_t$ as a function of the past ( which is what I think they mean by covariance stationary solution ) but you can write it as a function of the future ( but this solution I don't think is considered covariance stationary ) If you can't find anything that explains this, let me know and I'll try to find something.
mark leeds
CC BY-SA 4.0
2024-01-24T08:21:13.447Z
Not sure if this is relevant, but all MA processes are covariance stationary. I wonder what the definition of "covariance stationary solution" is. Could you recommend a textbook?
Richard Hardy
<econometrics><stationary>
Covariance stationary solution to the equation
nodis6
# Covariance stationary solution to the equation

(Asked by: username_0 on 2024-01-23)

Is there a covariance stationary solution to the equation?

$X\_t = 4 \varepsilon\_{t-1} + \varepsilon\_t, \quad \text{where } \{ \varepsilon\_t \} \in WN(0, \sigma^2), t \in \mathbb{Z}$.

Unfortunately I completely do not know how to handle this exercise. My only thought was to check if $X\_t$ is covariance stationary. So is it true that

1. $E(X\_t) = const$,
2. $\gamma(t,t+l) = Cov(X\_t, X\_{t+1}) = \gamma(s, s+l)$.

But is a good idea?

**Tags:** econometrics, stationary

> **Comments:**
> **username_1**: Hi: that's an MA(1) with $\theta = 4$. So, I would look up info on the MA(1) because I could try to explain it ( obtaining a solution ) here but it won't be as good as what you should find. The short of it is that, $\theta=4$ means that you can't write $\epsilon_t$ as a function of the past ( which is what I think they mean by covariance stationary solution ) but you can write it as a function of the future ( but this solution I don't think is considered covariance stationary ) If you can't find anything that explains this, let me know and I'll try to find something.  
> **username_2**: Not sure if this is relevant, but all MA processes are covariance stationary. I wonder what the definition of "covariance stationary solution" is. Could you recommend a textbook?  

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## 1 answers

### Answer by username_3 on 2024-01-26. Score: 3

The process in question is a Moving-Average process of order one,

$$\varepsilon\_t = \frac {1} {1+\theta L} X\_t = \frac {1} {1-(-\theta) L} X\_t, \qquad \theta =4.$$

Here $L$ is the lag operater, $L^k(z\_t) = z\_{t-k}$. The process is covariance-stationary irrespective of the value of $\theta$, but it is not invertible since $\theta \geq 1$, so it does not have an autoregressive $AR(\infty)$ representation.

Since $\theta \geq 1$ we can solve it "forward", as follows:
First, define the forward operator $F^k \equiv L^{-k}, F^k(z\_t) = z\_{t+k}$.

Second, note that

$$\frac {1} {1-(-\theta) L} = \frac{1}{\theta L[-(-\theta^{-1})L^{-1} +1]} = \frac{\theta^{-1}L^{-1}}{1-(-\theta)^{-1}L^{-1}]} = \frac{\theta^{-1}F}{1-(-\theta^{-1})F}$$

Then, since $\theta^{-1} <1$,

$$\frac {1} {1-(-\theta) L} = \theta^{-1}F\cdot [1-\theta^{-1}F + \theta^{-2}F^{2}-\theta^{-3}F^{3}+...]$$

So
$$\varepsilon\_t = X\_t \cdot \theta^{-1}F\cdot [1-\theta^{-1}F + \theta^{-2}F^{2}-\theta^{-3}F^{3}+...]$$

$$\implies \varepsilon\_t = \sum\_{j=1}^{\infty}\frac {(-1)^{j+1}}{\theta^j}X\_{t+j}$$

This is the "forward" solution, when the MA process is not invertible.

I guess the OP can now determine whether this solution is covariance-stationary.

> **Comments:**
> **username_1**: Thanks Alecos. That's the solution that I was thinking of in my mind but the gory details are appreciated.  
> **username_3**: @markleeds You're welcome Mark.  

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