In [1, p. 62 exercise 3.13] it is requested to prove that the PSD of a real WSS random process is a real even function of frequency. Solution: The PSD of a WSS process is given by the Fourier transform of its autocorrelation function:
 $P_{xx}(f)=\sum\limits_{k=-\infty}^{\infty}r_{xx}[k]e^{-j2\pi f k}.$ (1)

We note that for a real process $x[n]$, $r_{xx}[k]$ will be also real thus we obtain the relations:
 $r^{\ast}_{xx}[k]=r_{xx}[-k]=r_{xx}[k].$ (2)

First let us prove that for a real WSS this function is even. Having the previous relations in mind:
 $P_{xx}(-f)$ $=$ $\sum\limits_{k=-\infty}^{\infty}r_{xx}[k]e^{j2\pi f k}$ $=$ $\sum\limits_{k=-\infty}^{\infty}r_{xx}[-k]e^{j2\pi f k}$

Let $u=-k$ then
 $P_{xx}(-f)$ $=$ $\sum\limits_{u=-\infty}^{\infty}r_{xx}[u]e^{-j2\pi f u}$ $=$ $P_{xx}(f)$ (3)

Thus we have shown that the PSD is symmetric $P_{xx}(f)=P_{xx}(-f)$. The next step is to show that the PSD is also real. We know that the real part of a complex function is given as one half of the sum of the complex function with its complex conjugate form, that is $Re\left\{P_{xx}(f)\right\}=\frac{1}{2}\left(P_{xx}(f)+P^{\ast}_{xx}(f)\right)$. Let’s determine the complex conjugate of the PSD:
 $P_{xx}^{\ast}(f)$ $=$ $\sum\limits_{k=-\infty}^{\infty}r^{\ast}_{xx}[k]e^{j2\pi f k}$ $=$ $\sum\limits_{k=-\infty}^{\infty}r_{xx}[-k]e^{j2\pi f k}=P_{xx}(-f)=P_{xx}(f)$ (4)

Again relation (2) was used in order to obtain the last result. The real part of the PSD is thus given by
 $Re\left\{P_{xx}(f)\right\}$ $=$ $\frac{1}{2}\left(P_{xx}(f)+P^{\ast}_{xx}(f)\right)$ $=$ $Re\left\{P_{xx}(f)\right\}=\frac{1}{2}\left(P_{xx}(f)+P_{xx}(f)\right)$ $=$ $P_{xx}(f),$

because of (4). So the real part of the PSD $P_{xx}(f)$ is the PSD itself, or stating it in another way: the PSD of a WSS process is real. Thus together with (3) we have shown that the PSD of a real WSS random process is a real even function of frequency. QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.