In [1, p. 62 exercise 3.19] we are asked to find for the multiple sinusoidal process
 $x[n]=\sum\limits_{i=1}^{P}A_{i}\cos(2\pi f_{i}n+\phi_{i})$

the ensemble ACF and the temporal ACF as $M\rightarrow \infty$, where the $\phi_{i}$‘s are all uniformly distributed random variables on $[0, 2 \pi)$ and independent of each other. We are also asked to determine if this random process is autocorrelation ergodic.
Solution: We note that the joint p.d.f of the uniformly random variables $\phi_{i}$ is given by $f(\boldsymbol{\phi})=\prod_{i=1}^{P}f(\phi_{i})=\frac{1}{(2\pi)^{P}}$, with domain $\mathbb{D}= \mathbb{A}^{P}, \; \mathbb{A} =[0, 2 \pi)$ we can proceed to calculate the ensemble autocorrelation function which is defined as:
 $r_{xx}[k]$ $=$ $E \left\{ x^{\star}[n] x[n+k] \right\}$ $=$ $\int\limits_{\mathbb{D}}\sum\limits_{i=1}^{P}A_{i}\cos(2\pi f_{i}n+\phi_{i})\sum\limits_{\ell=1}^{P}A_{\ell}\cos(2\pi f_{\ell}(n+k)+\phi_{\ell}) f(\boldsymbol{\phi})d\boldsymbol{\phi}$ $=$ $\sum\limits_{i=1}^{P}\sum\limits_{\ell=1, \ell\neq i}^{P} \int_{0}^{2\pi} \int_{0}^{2\pi}\frac{A_{i}A_{\ell}}{(2\pi)^{2}}\cos(2\pi f_{i}n+\phi_{i})\cos(2\pi f_{\ell}(n+k)+\phi_{\ell}) d\phi_{i}d\phi_{\ell}$  $+ \sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(2\pi)}\cos(2\pi f_{i}n+\phi_{i})\cos(2\pi f_{i}(n+k)+\phi_{i})d\phi_{i}$ $=$ $\sum\limits_{i=1}^{P} \sum\limits_{\ell=1, \ell\neq i}^{P} \int_{0}^{2\pi} \frac{A_{i}A_{\ell}}{(2\pi)^{2}} \underbrace{\left[ \sin(2\pi f_{i}n+\phi_{i})\right]_{0}^{2\pi}}_{=0}\cos(2\pi f_{\ell}(n+k)+\phi_{\ell}) d\phi_{\ell}$  $+ \sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(2\pi)}\cos(2\pi f_{i}n+\phi_{i})\cos(2\pi f_{i}(n+k)+\phi_{i})d\phi_{i}$ $=$ $\sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(2\pi)}\cos(2\pi f_{i}n+\phi_{i})\cos(2\pi f_{i}(n+k)+\phi_{i})d\phi_{i}$

As in exercise [2] we can use the trigonometric relation [3, p. 810] :
 $2\cos\frak{A}\cos\frak{B} = \cos(\frak{A-B}) + \cos(\frak{A+B})$ (1)

($\frak{A}=2\pi f_{i}(n+k)+\phi_{i}$, $\frak{B}=2\pi f_{i}n+\phi_{i}$) to further simplify the expression for the ensemble autocorrelation function:
 $r_{xx}[k]$ $=$ $\sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(4\pi)}\cos(2\pi f_{i}k)d\phi_{i}$  $+\sum\limits_{i=1}^{P}\int_{0}^{2\pi}\frac{A_{i}^{2}}{(4\pi)}\cos(2\pi f_{i}(2n+k)+2 \phi_{i})d\phi_{i}$ $=$ $\sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(4\pi)}\cos(2\pi f_{i}k)d\phi_{i}$  $+\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{(8\pi)}\underbrace{\left[\sin(2\pi f_{i}(2n+k)+2 \phi_{i})\right]_{\phi_{i}=0}^{2\pi}}_{=0}$ $=$ $\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2}\cos(2\pi f_{i}k)$ (2)

Having obtained the ensemble autocorrelation function we can proceed to obtain the temporal autocorrelation function, which we hope will be a good approximation $\hat{r}_{xx}[k]$ to the ensemble autocorrelation function $r_{xx}[k]$. By definition the temporal autocorrelation function is given by
 $\hat{r}_{xx}[k]$ $=$ $\frac{1}{2M+1} \sum\limits_{n=-M}^{M} x[n+k]x^{\star}[n]$ $=$ $\frac{1}{2M+1} \sum\limits_{n=-M}^{M}\Bigg[ \left(\sum\limits_{i=1}^{P}A_{i}\cos(2\pi f_{i}(n+k)+\phi_{i})\right) \cdot$  $\cdot \left(\sum\limits_{\ell=1}^{P}A_{\ell}\cos(2\pi f_{\ell}n+\phi_{\ell}) \right)\Bigg]$ $=$ $\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\sum\limits_{n=-M}^{M} \frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}(n+k)+\phi_{i}) \cos(2\pi f_{\ell}n+\phi_{\ell})$  $+ \sum\limits_{i=1}^{P}\sum\limits_{n=-M}^{M}\frac{A_{i}^{2}}{2M+1}\cos(2\pi f_{i}(n+k)+\phi_{i}) \cos(2\pi f_{i}n+\phi_{i})$ (3)

The second sum of the previous relation can be simplified by one of the derived formulas of [2], for $e^{j2\pi f_{0}}\neq 1$ :
 $\frac{1}{2M+1}\sum\limits_{n=-M}^{M}A^{2}\cos(2\pi f_{0}n+\phi)\cos(2\pi f_{0}(n+k)+\phi) =$ $\frac{A^{2}}{2} \cos(2\pi f_{0}k) + \frac{A^{2}}{2(2M+1)}\cos(2\pi f_{0}k+2\phi)\frac{\sin(2 \pi f_{0}(2M+1))}{\sin(2\pi f_{0})}$

Thus the temporal autocorrelation function may be expressed, as long as $e^{j2\pi f_{i}}\neq 1, \; i=1,...,P$ as:
 $\hat{r}_{xx}[k]$ $=$ $\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\sum\limits_{n=-M}^{M} \frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}(n+k)+\phi_{i}) \cos(2\pi f_{\ell}n+\phi_{\ell})$  $+ \sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2} \cos(2\pi f_{i}k)$  $+\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2(2M+1)}\cos(2\pi f_{i}k+2\phi_{i})\frac{\sin(2 \pi f_{i}(2M+1))}{\sin(2\pi f_{i})}$ (4)

In order to simplify the relation further, especially the first sum, we will again use (1), with $\frak{A} =2\pi f_{i}(n+k)+\phi_{i}$ and $\frak{B} = 2\pi f_{\ell}n+\phi_{\ell}$, and thus :
  $\sum\limits_{n=-M}^{M}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}(n+k)+\phi_{i}) \cos(2\pi f_{\ell}n+\phi_{\ell}) =$  $\sum\limits_{n=-M}^{M}\frac{A_{i}A_{\ell}}{2M+1} \cos(2 \pi (f_{i}-f_{\ell})n+2\pi f_{i}k+\phi_{i}-\phi_{\ell}) +$  $\sum\limits_{n=-M}^{M}\frac{A_{i}A_{\ell}}{2M+1} \cos(2 \pi (f_{i} + f_{\ell})n+2\pi f_{i}k+\phi_{i}+\phi_{\ell})$ (5)

We see that both distinct parts are of the form $\sum_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n+ 2\pi \frak{f}_{2} k + \Phi)$, and thus it remains to simplify this relation. Again using a trigonometric formula [3, p. 810]:
 $\cos(\frak{A+B})=\cos\frak{A}\cos\frak{B}-\sin\frak{A}\sin\frak{B}$ (6)

with $\frak{A}= 2\pi \frak{f}_{1} n$ and $\frak{B}=2\pi \frak{f}_{2} k + \Phi$ we can simplify the relation as:
 $\sum\limits_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n+ 2\pi \frak{f}_{2} k + \Phi)$ $=$ $\cos(2\pi \frak{f}_{2} k + \Phi)\sum\limits_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n)$  $-\sin(2\pi \frak{f}_{2} k + \Phi)\underbrace{\sum\limits_{n=-M}^{M}\sin(2\pi \frak{f}_{1} n)}_{=0, \textrm{ \scriptsize{odd symmetry of sine}}}$ $=$ $\cos(2\pi \frak{f}_{2} k + \Phi)\sum\limits_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n)$

Furthermore it was shown in [2] that $\sum_{n=-M}^{M}\cos(2\pi (2 f_{0}))=\frac{\sin(\pi (2 f_{0})(2M+1))}{\sin(\pi(2f_{0})}$ , for $e^{j2\pi 2f_{0}}\neq 1$, so we can further simplify the previous relation by:
 $\sum\limits_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n+ 2\pi \frak{f}_{2} k + \Phi)=\cos(2\pi \frak{f}_{2} k + \Phi)\sum\limits_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n)$ $= \cos(2\pi \frak{f}_{2} k + \Phi)\frac{\sin(\pi \frak{f}_{1}(2M+1))}{\sin(\pi\frak{f}_{1})}, \; e^{j2\pi\frak{f}_{1}}\neq 1$ (7)

Thus finally we can simplify (5) by:
  $\sum\limits_{n=-M}^{M}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}(n+k)+\phi_{i}) \cos(2\pi f_{\ell}n+\phi_{\ell}) =$  $\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}-\phi_{\ell}) \frac{\sin(\pi (f_{i}-f_{\ell})(2M+1))}{\sin(\pi (f_{i}-f_{\ell}))} +$  $\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}+\phi_{\ell}) \frac{\sin(\pi (f_{i}+f_{\ell})(2M+1))}{\sin(\pi (f_{i}+f_{\ell}))}$ (8)

So finally the temporal autocorrelation (4) can be reduced using (8) to the following formula:
 $\hat{r}_{xx}[k]$ $=$ $\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}-\phi_{\ell}) \frac{\sin(\pi (f_{i}-f_{\ell})(2M+1))}{\sin(\pi (f_{i}-f_{\ell}))}$  $+\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}+\phi_{\ell}) \frac{\sin(\pi (f_{i}+f_{\ell})(2M+1))}{\sin(\pi (f_{i}+f_{\ell}))}$  $+ \sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2} \cos(2\pi f_{i}k)$  $+\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2(2M+1)}\cos(2\pi f_{i}k+2\phi_{i})\frac{\sin(2 \pi f_{i}(2M+1))}{\sin(2\pi f_{i})}$ (9)

Rearranging the terms on the right we see that the temporal autocorrelation function equals the ensemble autocorrelation function with an additional error $\epsilon$, which we have derived for the the case when $e^{(j2\pi f_{i})}\neq 1 \; , i=1,...,P$:
 $\hat{r}_{xx}[k]$ $=$ $\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2} \cos(2\pi f_{i}k)$  $\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}-\phi_{\ell}) \frac{\sin(\pi (f_{i}-f_{\ell})(2M+1))}{\sin(\pi (f_{i}-f_{\ell}))}$  $+\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}+\phi_{\ell}) \frac{\sin(\pi (f_{i}+f_{\ell})(2M+1))}{\sin(\pi (f_{i}+f_{\ell}))}$  $+\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2(2M+1)}\cos(2\pi f_{i}k+2\phi_{i})\frac{\sin(2 \pi f_{i}(2M+1))}{\sin(2\pi f_{i})}$ $=$ $r_{xx}[k] +\epsilon$

At once we see again that when $M\rightarrow \infty$ that the error goes to zero $\epsilon \rightarrow 0$. Thus the random process of the sum of sinusoids is also autocorrelation ergodic as long as $e^{(j2\pi f_{i})}\neq1 \; , i=1,...,P$. It is also easy to recognize, by using the argumentation of the previous exercise [2] , that this is not true if $e^{(j2\pi f_{i})}\neq1 \; , i=1,...,P$ does not hold. In this case parts of the the error of equation (3) are proportional to $M$ as was already observed in [2] . QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.
[2] Chatzichrisafis: “Solution of exercise 3.18 from Kay’s Modern Spectral Estimation - Theory and Applications”.
[3] Granino A. Korn and Theresa M. Korn: “Mathematical Handbook for Scientists and Engineers”, Dover, ISBN: 978-0-486-41147-7.