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 $kneq 0$
we have then
$$f(omega) = frac{1}{2 pi}gamma_0 sum_{k=-infty}^{+infty} alpha^{|k|} e^{iomega k}$$
$$=frac{gamma_0}{2pi} left[1 +sum_{k=1}^{+infty} alpha^k e^{iomega k}+ sum_{k=1}^{+infty} alpha^k e^{-iomega k}right]$$
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:
$$sum_{k=-infty}^{+infty} alpha^{|k|} e^{iomega k} = sum_{k=-infty}^{-1} alpha^{-k} e^{iomega k} + 1 + sum_{k=1}^{+infty} alpha^k e^{iomega k}$$
Since:
$$|k| = begin{cases} k,& text{if } kgeq 0\ -k, & text{if } klt 0 end{cases}$$
And from the first summation you have this equivalence:
$$sum_{k=-infty}^{-1} alpha^{-k} e^{iomega k} = sum_{k=1}^{infty} alpha^{k} e^{-iomega k}$$