I'm studying the derivation of the spectral density function of an AR(1) process. Starting from its autocovariance function, we have that:
$$\gamma_0 = \frac{\sigma^2}{1-\alpha_1 ^2}$$ and $\gamma_k = \rho^{|k|}\gamma_0$ for $k\neq 0$
we have then
$$f(\omega) = \frac{1}{2 \pi}\gamma_0 \sum_{k=-\infty}^{+\infty} \alpha^{|k|} e^{i\omega k}$$
$$=\frac{\gamma_0}{2\pi} \left[1 +\sum_{k=1}^{+\infty} \alpha^k e^{i\omega k}+ \sum_{k=1}^{+\infty} \alpha^k e^{-i\omega k}\right]$$
Now, I understand that the summation can be split due to the properties of symmetry of the autocovariance function, but I don't get why I have a minus in the argument of the exponential in the second summation. Can anybody give me an answer? Thanks a lot!
Best Answer
HINT:
$$\sum_{k=-\infty}^{+\infty} \alpha^{|k|} e^{i\omega k} = \sum_{k=-\infty}^{-1} \alpha^{-k} e^{i\omega k} + 1 + \sum_{k=1}^{+\infty} \alpha^k e^{i\omega k}$$
Since:
$$|k| = \begin{cases} k,& \text{if } k\geq 0\\ -k, & \text{if } k\lt 0 \end{cases}$$
And from the first summation you have this equivalence:
$$\sum_{k=-\infty}^{-1} \alpha^{-k} e^{i\omega k} = \sum_{k=1}^{\infty} \alpha^{k} e^{-i\omega k}$$