The autocovariance function of ARMA(1,1)

Question

So I am reading Brockwell and Davis introduction to Time Series analysis on page 89 where he derives the ACVF of an $ARMA(1,1)$ given by:

$X_t - \phi X_{t-1}=Z_t+\theta Z_{t-1}$ with ${Z_t}$ is $WN(0,\sigma^2)$ and $\mid \phi \mid < 1$

What is first told is that by causality assumption, the autocovariance at lag $h$ is:

$\gamma(h)=\sigma^2\sum_{j=0}^\infty\psi_j \psi_{j+\mid h \mid}$

So this at lag $h = 0$ is becomes:

$\gamma (0) = \sigma^2 \sum_{j=0}^\infty \psi_j^2$

How can this then be shown that $\sigma^2 \sum_{j=0}^\infty \psi_j^2 = \sigma^2 \Big[ 1 + \frac{(\theta+\phi)^2}{1-\phi^2} \Big]$ ?

And in the same way for $\gamma(1) = \sigma^2 \Big[ \theta + \phi + \frac{(\theta+\phi)^2\phi}{1-\phi^2} \Big]$?

I know that there is a definition of the function $\psi (z) = \sum_{j=0}^\infty\psi_j z^j = \frac{\theta(z)}{\phi(z)}$, $\mid z \mid\leq 1$. In what can this be applied here?

Answer 1

Eventhough this question is old, here is the answer:

The Autocovariance function for a causal time series is: $\\$

$\gamma(h) = \sigma^2 \sum_{j=0}^{\infty}\psi_j \psi_{j+|h|}$ $\\$

The MA($\infty)$ representation of $X_t$ is:

$X_t = Z_t + \sum_{j=1}^{\infty} (\phi+\theta)\phi^{j-1} Z_{t-j}$ $\\$ ,where $\psi_j = (\phi+\theta)\phi^{j-1}$

Note that I will write the white noise $Z_{t-j}$ and not directly $\sigma^2$ to hopefully generate a better understanding of what is going on

$\\$

Case h = 0:

$Cov(X_t,X_t) = E(X_t X_t) = \gamma(0) = \sigma^2 \sum_{j=0}^{\infty}\psi_j^2$ $\\$

$E(X_t X_t)$ $= E(( Z_t + (\phi+\theta)\sum_{j=1}^{\infty} \phi^{j-1} Z_{t-j}) ^2) = E(Z_t^2) + E( ((\phi+\theta)\sum_{j=1}^{\infty} \phi^{j-1} Z_{t-j})^2 ) +$ $2(\phi+\theta) \sum_{j=1}^{\infty} \phi^{j-1}E( Z_{t-j}Z_t)$

since $Z_t$ is a White noise; $E( Z_{t-1}Z_t) = 0$, hence:

$E(X_t X_t)$ $= E(Z_t^2) + (\phi+\theta)^2 \sum_{j=1}^{\infty} \phi^{2j-2} E(Z_{t-j}^2)= \sigma^2 + (\phi+\theta)^2 \sum_{j=1}^{\infty} \phi^{2j-2} \sigma^2= \sigma^2 ( 1 + (\phi+\theta)^2 \sum_{j=1}^{\infty} \phi^{2j-2})= \sigma^2 ( 1 + \frac{(\phi+\theta)^2}{1-\phi^2})$

where $\sum_{j=1}^{\infty} \phi^{2j-2} = \sum_{j=0}^{\infty} \phi^{2j} = $$ \sum_{j=0}^{\infty} (\phi^{j})^2$ is a geometric series converging to $\frac{1}{1-\phi^2}$

$\\$

Case h = 1:

$Cov(X_t,X_{t-1}) = E(X_t X_{t-1}) = \gamma(1) = \sigma^2 \sum_{j=0}^{\infty}\psi_j \psi_{j-1}$ $\\$

$E(X_t X_{t-1})$ $ = E( (Z_t + (\phi+\theta)\sum_{j=1}^{\infty} \phi^{j-1} Z_{t-j}) \cdot ( Z_{t-1} + (\phi+\theta)\sum_{j=2}^{\infty} \phi^{j-2} Z_{t-j}) ) = E( (Z_t + Z_{t-1}(\phi+\theta) + (\phi+\theta)\sum_{j=2}^{\infty} \phi^{j-1} Z_{t-j}) \cdot ( Z_{t-1} + (\phi+\theta)\sum_{j=2}^{\infty} \phi^{j-2} Z_{t-j}) ) = E( (Z_t + Z_{t-1}(\phi+\theta) + (\phi+\theta)\phi\sum_{j=2}^{\infty} \phi^{j-2} Z_{t-j}) \cdot ( Z_{t-1} + (\phi+\theta)\sum_{j=2}^{\infty} \phi^{j-2} Z_{t-j}) ) = (\phi+\theta)E(Z_{t-1}^2) + (\phi+\theta)^2\phi \sum_{j=2}^{\infty} \phi^{2j-4} E(Z_{t-j}^2)= (\phi+\theta)E(Z_{t-1}^2) + (\phi+\theta)^2\phi \sum_{j=2}^{\infty} \phi^{2j-4} E(Z_{t-j}^2)= (\phi+\theta)\sigma^2 + (\phi+\theta)^2\phi \sum_{j=2}^{\infty} \phi^{2j-4} \sigma^2)= \sigma^2(\phi+\theta + \frac{(\phi+\theta)^2\phi}{1-\phi^2} )$

$\\$

Case h = 2:

$Cov(X_t,X_{t-2}) = E(X_t X_{t-2}) = \gamma(2) = \sigma^2 \sum_{j=0}^{\infty}\psi_j \psi_{j-2}$ $\\$

$E(X_t X_{t-2})$ $ = E( (Z_t + (\phi+\theta)\sum_{j=1}^{\infty} \phi^{j-1} Z_{t-j}) \cdot ( Z_{t-2} + (\phi+\theta)\sum_{j=3}^{\infty} \phi^{j-3} Z_{t-j}) ) = E( (Z_t + Z_{t-1}(\phi+\theta) + Z_{t-2}(\phi+\theta)\phi + (\phi+\theta)\phi^2\sum_{j=3}^{\infty} \phi^{j-3} Z_{t-j}) \cdot ( Z_{t-2} + (\phi+\theta)\sum_{j=3}^{\infty} \phi^{j-3} Z_{t-j}) ) = (\phi+\theta)\phi E(Z_{t-2}^2) + (\phi+\theta)^2\phi^2\sum_{j=3}^{\infty} \phi^{2j-6} E(Z_{t-j}^2)) = \sigma^2( (\phi+\theta)\phi + \frac{(\phi+\theta)^2\phi^2}{1-\phi^2})$

We see the pattern and follow that:

$\gamma(h) = \sigma^2 ( 1 + \frac{(\phi+\theta)^2}{1-\phi^2}) ~$ if h=0

$\gamma(h) = \sigma^2( (\phi+\theta)\phi^{h-1} + \frac{(\phi+\theta)^2\phi^h}{1-\phi^2}) ~$ if h>0

Answer 2

To be honest, you can get that variance just from the way they defined the model at the start. If you rearrange to make Xt the subject, and then use the fact we have no mean, one can get to the result in a few lines if we use the assumptions that future errors do not depend on past outcomes. I'll walk through it in more detail later as I'm on my phone now.