Let a autoregressive process $AR(1)$ given by $$X_t = \phi_1X_{t-1}+\epsilon_t$$ where $|\phi_1|<1$. I am trying to show that $$\sum_{k=-\infty}^\infty \gamma(k) = \frac{\sigma_\epsilon^2}{(1-\phi_1)^2}.$$ I understand the result of $$\gamma(k) = \frac{\sigma_\epsilon^2\phi^{|k|}}{1-\phi^2},$$ but I cannot transform that into what I want.
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You need to clean up symbolism. $\phi_1=\phi$? $\epsilon_t$ relationship to $\sigma_\epsilon$ (without sub t). Is $\gamma (k)=$ (last equation) the definition? – herb steinberg Apr 20 '19 at 21:54
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For AR(1) process we have $\gamma(k) = \frac{\phi_1^{|k|}\sigma_\epsilon^2}{1-\phi_1^2}$. Hence, $$\sum_{k=-\infty}^\infty \gamma(k)\\ =\sum_{k=0}^\infty \gamma(k) - \gamma(0)\\ =\frac{\sigma_\epsilon^2}{1-\phi_1^2}\Big(\frac{2}{1-\phi_1}-1\Big)\\ =\frac{\sigma_\epsilon^2}{(1-\phi_1)(1+\phi_1)}\Big(\frac{1+\phi_1}{1-\phi_1}\Big)\\ =\frac{\sigma_\epsilon^2}{(1-\phi_1)^2}.$$
VincentN
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