# Solved – 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 \$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!

Contents

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}\$\$

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