MHB Solving Autocovariance Function $\gamma(t+h,t)$

[nacho-man]

This one is bugging me!

Let ${Z_t}$ ~ $(0, \sigma^2)$

And $X_t = Z_t + \theta Z_t$

im trying to find the autocovariance function $\gamma(t+h,t)$ And nearly have it, but am struggling with some conceptual issues :S

$\gamma(t+h,t) = \text{COV}[Z_{t+h} + \theta Z_{t-1+h}, Z_t + \theta Z_t-1]$

= $\text{COV}(Z_{t+h}, Z_t) + \theta \text{COV}(Z_{t+h}, Z_{t-1}) + \theta \text{COV}(Z_{t-1+h}, Z_t) + \theta^2 \text{COV}(Z_{t-1+h}, Z_{t-1})$

$\text{COV}(Z_{t+h}, Z_t) = \sigma^2$ (at $h=0$)
$\theta \text{COV}(Z_{t+h}, Z_{t-1})$ = $\theta \sigma^2$ (at $h=-1$)
$ \theta \text{COV}(Z_{t-1+h}, Z_t)$ = $\theta \sigma^2$ (at $h=1$)
$ \theta^2 \text{COV}(Z_{t-1+h}, Z_{t-1})$ = $\theta^2 \sigma^2$ (at $h=0$)

So, to summarise,

for h = 0, autocovariance = $\sigma^2 + \theta^2 \sigma^2$
for h = |1|, autocovariance = $\theta \sigma^2 + \theta \sigma^2$ = $2 \theta \sigma^2$ <<< textbook disagrees here!
f0r h>|1|, autocovariance = 0

The answers are attached to this post, I have a discrepancy for h=|1| and cannot see why. Is there a typo in the book?I Have an additional follow up question, depending on the response i receive for this initial post!

Any help very much appreciated as always,
thank you in advance.

 

[zzephod]

This one is bugging me!

Let ${Z_t}$ ~ $(0, \sigma^2)$

And $X_t = Z_t + \theta Z_t$

im trying to find the autocovariance function $\gamma(t+h,t)$ And nearly have it, but am struggling with some conceptual issues :S

$\gamma(t+h,t) = \text{COV}[Z_{t+h} + \theta Z_{t-1+h}, Z_t + \theta Z_t-1]$

= $\text{COV}(Z_{t+h}, Z_t) + \theta \text{COV}(Z_{t+h}, Z_{t-1}) + \theta \text{COV}(Z_{t-1+h}, Z_t) + \theta^2 \text{COV}(Z_{t-1+h}, Z_{t-1})$

$\text{COV}(Z_{t+h}, Z_t) = \sigma^2$ (at $h=0$)
$\theta \text{COV}(Z_{t+h}, Z_{t-1})$ = $\theta \sigma^2$ (at $h=-1$)
$ \theta \text{COV}(Z_{t-1+h}, Z_t)$ = $\theta \sigma^2$ (at $h=1$)
$ \theta^2 \text{COV}(Z_{t-1+h}, Z_{t-1})$ = $\theta^2 \sigma^2$ (at $h=0$)

So, to summarise,

for h = 0, autocovariance = $\sigma^2 + \theta^2 \sigma^2$
for h = |1|, autocovariance = $\theta \sigma^2 + \theta \sigma^2$ = $2 \theta \sigma^2$ <<< textbook disagrees here!
f0r h>|1|, autocovariance = 0

The answers are attached to this post, I have a discrepancy for h=|1| and cannot see why. Is there a typo in the book?I Have an additional follow up question, depending on the response i receive for this initial post!

Any help very much appreciated as always,
thank you in advance.

Because we have zero means:

$$\begin{aligned}\gamma_Z(t+1,t)&=E(Z_{t+1}Z_t)\\
&=E( (X_{t+1}+\theta X_t)(X_{t}+\theta X_{t-1}))\\
&=E(X_{t+1}X_t)+E(X_{t+1}\theta X_{t-1})+E(\theta X_t X_t)+E(\theta X_t \theta X_{t-1})
\end{aligned}$$

Now as the $X_i$s are uncorrelated and independent all the expectations but the third are zero, so:

$$\gamma_Z(t+1,t)=\theta \sigma^2$$

and similarly:

$$\begin{aligned}\gamma_Z(t-1,t)&=E(Z_{t-1}Z_t)\\
&=E( (X_{t-1}+\theta X_{t-2})(X_{t}+\theta X_{t}))\\
&=E(X_{t-1}X_t)+E(X_{t-1}\theta X_{t-1})+E(\theta X_{t-2} X_t)+E(\theta X_{t-2} \theta X_{t-1})
\end{aligned}$$

Now for the same reasons as before all the expectations other than the third are zero and we have as before:

$$\gamma_Z(t-1,t)=\theta \sigma^2$$

.