In [1, p. 62 exercise 3.15] it is requested to verify the ACF and PSD relationships given in [1, p. 53 eq. (3.50)] and [1, p. 54 eq. (3.51)].
Solution: Starting from the first relationship of [1, p. 53 eq. (3.50)] and the definition of the cross correlation function we can derive
 $r_{xy}[n,k]$ $=$ $E\{x^{\ast}[n]y[n+k]\}$ $=$ $E\{x^{\ast}[n]\sum\limits_{l}x[n+k-l]h[l]\}$ $=$ $\sum\limits_{l}E\{x^{\ast}[n]x[n+k-l]\}h[l]$

Using the starting assumption that $x[n]$ is WSS we can derive the final result, that the cross correlation is also independent of the observation instance $n$ and depends only on the lag $k$:
 $r_{xy}[n,k]=r_{xy}[k]$ $=$ $\sum\limits_{l}r_{xx}[k-l]h[l]$

Similar the second relation can be obtained by:
 $r_{yx}[n,k]$ $=$ $E\{y^{\ast}[n]x[n+k]\}$ $=$ $E\{\sum\limits_{l}x^{\ast}[n-l]h^{\ast}[l]x[n+k]\}$ $=$ $\sum\limits_{l}E\{x^{\ast}[n-l]x[n+k]\}h^{\ast}[l]$ $=$ $\sum\limits_{l}r_{xx}[k+l]h^{\ast}[l]$

and setting $l^{\prime} = -l$:
 $r_{yx}[n,k]$ $=$ $\sum\limits_{l^{\prime}}r_{xx}[k-l^{\prime}]h^{\ast}[-l^{\prime}]$ $=$ $r_{xx}[k]\star h^{\ast}[-k] =r_{yx}[k]$

The third equation can be derived by:
 $r_{yy}[n,k]$ $=$ $E\{y^{\ast}[n]y[n+k]\}$ $=$ $E\{\sum\limits_{l}h^{\ast}[l]x^{\ast}[n-l]\sum\limits_{\lambda}h[\lambda]x[n+k-\lambda] \}$ $=$ $\sum\limits_{l}\sum\limits_{\lambda}h^{\ast}[l]h[\lambda]E\{x^{\ast}[n-l]x[n+k-\lambda] \}$ $=$ $\sum\limits_{l}h^{\ast}[l]\sum\limits_{\lambda} h[\lambda]r_{xx}[k+l-\lambda]$ (1)

By setting $\rho[k+l]=\sum_{\lambda} h[\lambda]r_{xx}[k+l-\lambda] =h[k+l]\ast r_{xx}[k+l]$ we can rewrite (1) by
 $r_{yy}[n,k]$ $=$ $\sum\limits_{l}h^{\ast}[l] \rho[k+l]$ $=$ $h^{\ast}(-k)\star \rho(k)$ $=$ $h^{\ast}(-k)\star h[k] \ast r_{xx}[k]$ (2)

which proves the last equation of [1, p. 53 eq. (3.50)] . We can now proceed to prove the relations for [1, p. 54 eq. (3.51)]
 $P_{xy}(z)$ $=$ $\mathcal{Z}\{r_{xy}[k]\}$ $=$ $\mathcal\{r_{xx}[k]\star h[k]\}$ $=$ $H(z)R_{xx}(z)$ (3)

The second relation $P_{yx}(z)$ can similar be proven by:
 $P_{yx}(z)$ $=$ $\mathcal{Z}\{r_{yx}[k]\}$ $=$ $\mathcal{Z}\{h^{\ast}[-k]\star r_{xx}[k]\}$ $=$ $\mathcal{Z}\{h^{\ast}[-k]\}R_{xx}(z)$ $=$ $R_{xx}(z)\sum\limits_{k=-\infty}^{\infty}h^{\ast}[-k]z^{-k}$

By change of variables $z=(u^{-1})^{\ast}$ and $k=-l$ for the sum of the previous equation can be written as:
 $P_{yx}(z)$ $=$ $R_{xx}(z)\sum\limits_{l=-\infty}^{\infty}h^{\ast}[l](u^{-l})^{\ast}$ $=$ $R_{xx}(z)(\sum\limits_{l=-\infty}^{\infty}h[l]u^{-l})^{\ast}$ $=$ $R_{xx}(z)(H(u))^{\ast}$ $=$ $R_{xx}(z) \left[H(u=\frac{1}{z^{\ast}})\right]^{\ast}$ $=$ $R_{xx}(z) \left[H(\frac{1}{z^{\ast}})\right]^{\ast}$

The last equation proves the second part of [1, p. 54 eq. (3.51)]. By similar reasoning the third relation can also be shown to be
 $P_{yy}(z)$ $=$ $\mathcal{Z}\{r_{yy}[k]\}$ $=$ $\mathcal{Z}\{h^{\ast}[-k]\star h[k]\star r_{xx}[k]\}$ $=$ $\mathcal{Z}\{h^{\ast}[-k]\}\mathcal{Z}\{ h[k]\}\mathcal{Z}\{r_{xx}[k]\}$ $=$ $\left[H(\frac{1}{z^{\ast}})\right]^{\ast}H(z)R_{xx}(z),$

which concludes the proof. QED.

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