Find the Autocovariance Function of process $Y_t$

Consider the processes $$X_t = \phi X_{t-1} + v_t$$ and $$Y_t = \phi Y_{t-1} + X_t + e_t$$, in which $$|\phi| < 1$$ and $$v_1$$ and $$e_t$$ are non-correlated random errors with zero mean and variances equal to $$\sigma^2$$. Based on these informations, find the autocovariance function of the process $$Y_t$$.

First, I tried to find $$E[Y_t]$$:

$$E[Y_t] = E[\phi Y_{t-1} + X_t + e_t] = \phi E[Y_{t-1}] + E[X_t] = \phi \mu_Y + \mu_X$$

And then, I tried to find $$Var(Y_t)$$:

$$Var(Y_t) = Var(\phi Y_{t-1} + X_t + e_t)$$

But I am stuck at this point.

• Are there initial conditions for $X_t$, $Y_t$, or are we to assume that $t\in\mathbb Z$? Oct 12, 2019 at 16:55
• there are not any initial conditions Oct 12, 2019 at 17:10
• I guess we can assume $t$∈ℤ Oct 12, 2019 at 17:12

The autocovariance between $$Y_{t+h}$$ and $$Y_t$$ is defined as: $$\gamma(h)=cov(Y_{t+h},Y_t)=E(Y_{t+h}-E(Y_{t+h}))(Y_t-E(Y_t)).$$ If $$Y_t$$ is a stationary process, then $$E(Y_{t+h})=E(Y_t)=\mu_Y$$, therefore $$\gamma(h)=E(Y_{t+h}Y_t)-\mu_Y^2.$$
• How do we know that $Y_t$ is stationary? Oct 12, 2019 at 20:59
• @Math1000 For an autoregressive model, if $|\phi|<1$ then the process is stationary (this condition is given in the question). Oct 12, 2019 at 21:10
• In order to express autocovariance in terms of $X$, you need to plug in into $\gamma(h)$ the functional form of $Y_{t}$, i.e. $\gamma(h)=E((\phi Y_{t+h-1}+X_{t+h}+e_{t+h})(\phi Y_{t-1}+X_t+e_t))-\mu_Y^2$. Oct 13, 2019 at 9:06
• but how to go further after plugging the functional form? How one could say about $E((\phi Y_{t+h-1})(\phi Y_{t-1}))+...$ when we multiply inside the expectation? Oct 14, 2019 at 18:09