What is the PACF(1) of the following AR(2) process?

$ y_t = \phi y_{t-2}+\epsilon_t $ with $\epsilon_t \sim WN(0, \sigma^2)$

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    $\begingroup$ If this is homework or self-study, please add the self-study tag and read its wiki. Thank you! $\endgroup$ – S. Kolassa - Reinstate Monica May 25 '16 at 8:56
  • $\begingroup$ (Plus: what does the subscript $y$ refer to in the $\epsilon_y$ error term?) $\endgroup$ – S. Kolassa - Reinstate Monica May 25 '16 at 8:58
  • $\begingroup$ I don't see any options to choose from, so why do you ask which? $\endgroup$ – Richard Hardy May 25 '16 at 13:55

Since $PAC(K) = Corr(Y_t, Y_{t-K}|Y_{t-K-1}, ..., Y_{t-1})$, $PAC(1)$ is equal to $\rho(1)$, i.e. the autocorrelation between $Y_t$ and $Y_{t-1}$ (there are no observations between $Y_t$ and $Y_{t-1}$, since they are two consecutive observations).

It is easy to see that $PAC(1)=0$. That's because if you compute the autocovariance function $Cov(Y_t, Y_{t-1})$, the two observations are not correlated if you have defined $y_t$ as $y_t = \phi y_{t-2} + \epsilon_t$; there is no correlation between $y_{t-2}$, $\epsilon_t$ and $y_{t-1}$. Thus, also the autocorrelation function $\rho(1) = \frac{\gamma(1)}{\gamma(0)}$ is equal to zero and the $PAC$ is zero too for the reason above.


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