I have some problems computing the autocovariance in the above exercise. Especially when given different lags, I do not understand why the number of lags is in the exponent of $\phi$.
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1$\begingroup$ Did you try writing $C(X_t, X_{t-1})$ and substitute in the equation you've written above for $X_t$ and then apply rules of covariance? For rules of covariance see, en.wikipedia.org/wiki/Covariance#Properties here you've got a lag 1 but a lag 2, lag 3 approach follows similar logic except the number of substitutions you'll need is the lag. Logic is the same here with $t$ and $t+1$. $\endgroup$– Lucas RobertsCommented Feb 22, 2020 at 18:44
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You'll first find $\operatorname{cov}(X_t,X_t)=\rho_0$: $$\rho_0=\operatorname{cov}(c+\phi X_{t-1}+Z_t,c+\phi X_{t-1}+Z_t)$$ Covariance with a constant is $0$, and $\operatorname{cov}(X_{t-1},Z_t)=0$ (no corr. between previous output and current noise), and $\operatorname{cov}(X_{t-1},X_{t-1})=\rho_0$ given stationarity (i.e. $|\phi|<1$). Solve the equation for $\rho_0$, and substitute below:
$$\rho_1=\operatorname{cov}(X_t,X_{t-1})=\operatorname{cov}(c+\phi X_{t-1}+Z_t,X_{t-1})=\phi\rho_0$$