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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.

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  • $\begingroup$ Are there initial conditions for $X_t$, $Y_t$, or are we to assume that $t\in\mathbb Z$? $\endgroup$
    – Math1000
    Oct 12, 2019 at 16:55
  • $\begingroup$ there are not any initial conditions $\endgroup$
    – motipai
    Oct 12, 2019 at 17:10
  • $\begingroup$ I guess we can assume $t$∈ℤ $\endgroup$
    – motipai
    Oct 12, 2019 at 17:12

1 Answer 1

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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 $$\gamma(h)=E(Y_{t+h}Y_t)-\mu_Y^2.$$

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  • $\begingroup$ How do we know that $Y_t$ is stationary? $\endgroup$
    – Math1000
    Oct 12, 2019 at 20:59
  • $\begingroup$ @Math1000 For an autoregressive model, if $|\phi|<1$ then the process is stationary (this condition is given in the question). $\endgroup$
    – ssane
    Oct 12, 2019 at 21:10
  • $\begingroup$ how would it be in terms of X? $\endgroup$
    – motipai
    Oct 12, 2019 at 23:47
  • $\begingroup$ 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. $\gamma(h)=E((\phi Y_{t+h-1}+X_{t+h}+e_{t+h})(\phi Y_{t-1}+X_t+e_t))-\mu_Y^2$. $\endgroup$
    – ssane
    Oct 13, 2019 at 9:06
  • $\begingroup$ 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? $\endgroup$
    – motipai
    Oct 14, 2019 at 18:09

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