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I have a time seres {$X_t$} that is defined by the following recursive relation: $X_t=\phi_1X_{t-1}+\phi_2X_{t-2}+Z_t$ with {$X_t$} being stationairy with $\mathbb{E}(X_t)=0$

Now apparently you are able to rewrite the autocorrelation function of this series to the following: $\rho(h)=\phi_1\rho(h-1)+\phi_2\rho(h-2)$

When trying to reproduce this, I get the following:

$\rho_x(h)=\frac{\gamma_x(h)}{\gamma_x(0)}= \frac{\operatorname{Cov}(X_{t+h},X_t)}{\operatorname{Cov}(X_{t},X_t)}=\frac{\operatorname{Cov}(\phi_1X_{t-1+h}+\phi_2X_{t-2+h}+Z_{t+h}, \phi_1X_{t-1}+\phi_2X_{t-2}+Z_t)}{\sigma^2}$

I have no idea how I can rewrite this to $\rho(h)=\phi_1\rho(h-1)+\phi_2\rho(h-2)$, do you guys know what I am doing wrong?

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Look up Yule-Walker equations in Google, there's a ton of references with derivations.

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  • $\begingroup$ Thanks for the help! So your linked helped me with the following steps: I started by multiplying boys sides of the timeseries with $X_{t-h}$, then I took the expectation. This gave the following result: $X_tX_{t-h}=\phi_1X_{t-1}X_{t-h}+\phi_2X_{t-2}X_{t-h}+Z_tX_{t-h}$ $\mathbb{E}[X_tX_{t-h}]=\mathbb{E}[\phi_1X_{t-1}X_{t-h}]+\mathbb{E}[\phi_2X_{t-2}X_{t-h}]+\mathbb{E}[Z_tX_{t-h}]$ $=\phi_1\mathbb{E}[X_{t-1}X_{t-h}]+\phi_2\mathbb{E}[X_{t-2}X_{t-h}]+0$ However I'm stil not sure how to go on from here, could you give me another hint? $\endgroup$ – Alfons Ingomar Oct 6 '15 at 20:41

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