# Spectral density function of AR(1) process

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!

• The summation can be split without any assumption of symmetry, into negative values of the index, and into positive values of the index. Then, in the first, "transfer" the minus sign of the index to where $k$ appears in the summand. Oct 7, 2015 at 17:52
• I don't think so, you can't simplify further if you have $\alpha^{-k}$ Oct 7, 2015 at 18:18
• sorry, I just deleted the comment. Can you double-check your last equation ? Oct 7, 2015 at 18:22
• @ale42 The parameter $k$ is in absolute value when exponentiating $\alpha$, so what you write does not arise. Oct 7, 2015 at 18:22
• But if you assume that the summand terms have modulus smaller than one, you can use geometric series and get finally the spectrum density... Oct 7, 2015 at 18:25

$$\sum_{k=-\infty}^{+\infty} \alpha^{|k|} e^{i\omega k} = \sum_{k=-\infty}^{-1} \alpha^{-k} e^{i\omega k} + 1 + \sum_{k=1}^{+\infty} \alpha^k e^{i\omega k}$$
$$|k| = \begin{cases} k,& \text{if } k\geq 0\\ -k, & \text{if } k\lt 0 \end{cases}$$
$$\sum_{k=-\infty}^{-1} \alpha^{-k} e^{i\omega k} = \sum_{k=1}^{\infty} \alpha^{k} e^{-i\omega k}$$