How to find the Autocorrelation function and Covaraince I had to find the autocorrelation function in the following time series model
$X_t = a+ bt+ Z_t + 0.6Z_{t-1} $ where $ a $ and $ b $ are constants.
I used that $ \gamma (k) $ = $ Cov(X_t, X_{t+k})$.
Then $Cov(a+ b t+ Z_t + 0.6 Z_{t-1}, a+ b(t+k)+ Z_{t+k} + 0.6 Z_{t+k-1}) $ but I am not sure how to evaluate that please help me on that. Any help would be appreciated.
 A: Let $X, Y$ and $Z$ be any random variables; then
$$Cov(X+Y,Z)=Cov(X,Z)+Cov(Y,Z)$$
from the very definition of covariance. Let us consider the given time series model 
$$X_t=a+bt+Z_t+0.6Z_{t-1},$$
with $Z_t$ i.i.d. random variables with variance $\sigma^2$.
As preliminary result we have
$$Cov(a,a)=Cov(bt,bt)=Cov(a,X)=Cov(bt,X)=0,$$
for any random variable $X$ because $E[a]=a$ and $E[bt]=bt$. Here $a$ denotes the random variable $a\cdot \mathbf{1}$, with $\mathbf{1}$ the identity r.v.. Similar considerations hold for $bt$, i.e. $bt:=bt\cdot \mathbf{1}$. 
Using the above results we can write for $k\neq 0$
$$Cov(X_t,X_{t+k})=Cov(a,X_{t+k})+Cov(bt,X_{t+k})+Cov(Z_t,X_{t+k})+0.6Cov(Z_{t-1},X_{t+k})=\\ Cov(Z_t,X_{t+k})+0.6Cov(Z_{t-1},X_{t+k})=\\
Cov(Z_t,a)+Cov(Z_t,b(t+k))+Cov(Z_t,Z_{t+k})+0.6Cov(Z_t,Z_{t+k-1})+\\
0.6Cov(Z_{t-1},a)+0.6Cov(Z_{t-1},b(t+k))+0.6Cov(Z_{t-1},Z_{t+k})+0.36Cov(Z_{t-1},Z_{t+k-1}),$$
i.e.
$$Cov(X_t,X_{t+k})= Cov(Z_t,Z_{t+k})+0.6Cov(Z_t,Z_{t+k-1})+0.6Cov(Z_{t-1},Z_{t+k})+0.36Cov(Z_{t-1},Z_{t+k-1}).$$
If $k\pm 1$, then 
$$Cov(X_t,X_{t\pm 1})= 0.6Cov(Z_t,Z_{t})=0.6\sigma^2.$$
If $k\neq \pm 1$, then
$$Cov(X_t,X_{t+k})= 0.$$
A: $E[X_t]=a+bt$,
therefore we may find the variance as: 
$$Var[X_t]=E[(Z_t+0.6Z_{t-1})(Z_t+0.6Z_{t-1})]=\sigma^2(1+0.36),$$
and covariance as:
$$Cov[X_t,X_{t+k}]=E[(X_t-(a+bt))(X_{t+k}-(a+b(t+k)))]$$
$$=E[(Z_t+0.6Z_{t-1})(Z_{t+k}+0.6Z_{t-1+k})].$$
Let $k=0$, then$$Cov[X_t,X_t]=Var[X_t]=\sigma^2(1+0.36).$$
Let $k= + 1$, then$$Cov[X_t,X_{t+1}]=E[(Z_t+0.6Z_{t-1})(Z_{t+1}+0.6Z_{t})]$$$$=E[(Z_t\cdot0.6Z_{t})]=0.6\sigma^2.$$
You may see that the same holds for $k=-1$.
Finally for $k>1$ and $k<-1$
$$Cov[X_t,X_{t+k}]=E[(Z_t+0.6Z_{t-1})(Z_{t+k}+0.6Z_{t-1+k})]=0,$$
since, in the expansion there will be no random variable products with same time index and since different time indexes are independent and mean is zero, the result will be zero.
