Why is the Spectral Density of a stationary sequence real? I am trying to understand (and prove) why the spectral density $\Phi_s \in L^1([- \pi, \pi])$ of a stationary sequence $s = \{s_n\}_{n \in \mathbb{Z}}$ is real.

*

*I wanted to argue via the auto-correlation function but got stuck somewhere in between.

*I know that the auto-correlation in this case is defined as $R_s(n) = \langle s_n, s_0 \rangle = \frac{1}{2 \pi} \int_{- \pi}^{\pi} \Phi_s(\theta) \cdot e^{- i n \theta} d \theta$ and must be non-zero and real-valued since the inner product is given this way.

*Further, I know that the spectral density (or any function $\Phi \in L^1$) is uniquely determined by its Fourier Coefficients.

But how do I use this information to show that $\Phi_s$ is real-valued? Maybe the solution is very obvious and I just cannot seem to see it!
Thanks in advance! :-)
 A: Suppose $(\gamma_\nu)_{\nu \in \mathbb{Z}}$ is a summable sequence (i.e. integrable wrt the counting measure, i.e. $\sum_{\nu \in \mathbb{Z}}|\gamma_\nu|<\infty$). Now suppose that $(\gamma_\nu)_{\nu \in \mathbb{Z}}$ is symmetric around $0$, that is $\gamma_{\nu}=\gamma_{-\nu},\,\forall \nu \in \mathbb{Z}$. Then $S(\xi):=\sum_{\nu \in \mathbb{Z}}\gamma_\nu e^{-i\xi\nu}$ is real valued $\forall \xi \in \mathbb{\mathbb{R}}$. To see this, consider that by definition of complex integral we have
$$S(\xi)=\sum_{\nu \in \mathbb{Z}}\cos(\xi \nu)\gamma_\nu+i\sum_{\nu \in \mathbb{Z}}(-\sin(\xi \nu))\gamma_\nu$$
and by dominated convergence $\sum_{\nu \in \mathbb{Z}}(-\sin(\xi \nu))\gamma_\nu=\lim_{N \to \infty}\sum_{\nu=-N}^{N}(-\sin(\xi \nu))\gamma_\nu=0$ as the terms cancel out (recall $\sin(-x)=-\sin(x)$). So $S(\xi)$ is entirely real. The autocorrelation of a sequence of weakly stationary real random variables with zero mean is symmetric around zero: $E[X_{k}X_{k-\nu}]=E[X_{k}X_{k+\nu}]$ as it depends only on $|\nu|$, the distance between the times. If $\nu \mapsto E[X_{k}X_{k+\nu}],\,\nu \in \mathbb{Z}$ is summable, the claim follows.
