# Suppose Z(t)=sum_{k=1}^{n}Xe^{j(omega_{0}t+Phi_{k})}, t in R where omega_{0} is a constant...

Suppose $Z\left(t\right)={\Sigma }_{k=1}^{n}X{e}^{j\left({𝜔}_{0}t+{𝚽}_{k}\right)}$, $t\in R$ where ${𝜔}_{0}$ is a constant, n is a fixed positive integer, ${X}_{1},...,{X}_{n}, {𝚽}_{1},...,{𝚽}_{n}$ are mutually independent random variables, and $E{X}_{k}=0,D{X}_{k}={\sigma }_{k}^{2},𝚽$ , $U\left[0,2\pi \right],k=1,2,...,n$ . Find the mean function and correlation function of .
I have tried to solve it.
For mean function,
${m}_{Z}\left(s\right)=E\left\{{Z}_{s}\right\}=E\left\{{X}_{s}\right\}+iE\left\{{Y}_{t}\right\}$
$=E\left\{{\Sigma }_{k=1}^{s}X{e}^{j\left({𝜔}_{0}t+{𝚽}_{k}\right)}\right\}$
For correlation function,
${R}_{Z}\left(s,u\right)=E\left\{{Z}_{s},{Z}_{u}\right\}$
$=E\left\{Y\left(s\right)Y\left(u\right)\right\}$
$=E\left\{{\Sigma }_{k=1}^{s}X{e}^{j\left({𝜔}_{0}t+{𝚽}_{k}\right)}{\Sigma }_{k=1}^{u}X{e}^{j\left({𝜔}_{0}t+{𝚽}_{k}\right)}\right\}$
$=E\left\{{\Sigma }_{k=1}^{s}{\Sigma }_{k=1}^{u}X{e}^{j\left({𝜔}_{0}t+{𝚽}_{k}\right)}X{e}^{j\left({𝜔}_{0}t+{𝚽}_{k}\right)}\right\}$
I am stuck here. How to move from here ahead?
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suponeriq
Step 1
Before finding the mean and variance function, one has to fix the notation of $Z\left(t\right)$, which should be
$Z\left(t\right)=\sum _{k=1}^{n}{X}_{k}{e}^{j\left({𝜔}_{0}t+{\mathrm{\Phi }}_{k}\right)}.$ .
In order to compute the mean of $Z\left(t\right)$, it suffices to compute the expectation of ${X}_{k}{e}^{j\left({𝜔}_{0}t+{\mathrm{\Phi }}_{k}\right)}$ . To do so, use the independence between ${X}_{k}$ and ${\mathrm{\Phi }}_{k}$ and the fact that ${X}_{k}$ is centered.
For the correlation function, the most technical part is to compute $\mathrm{Cov}\left(Z\left(s\right),Z\left(t\right)\right)$ which actually reduces to
$\sum _{k=1}^{n}\sum _{\ell =1}^{n}\mathbb{E}\left[{X}_{k}{X}_{\ell }{e}^{j\left({\omega }_{0}s+{\mathrm{\Phi }}_{k}\right)}{e}^{j\left({\omega }_{0}t+{\mathrm{\Phi }}_{\ell }\right)}\right]$
(when you have to sums, it is always preferable to sum over different indices). If $k\ne \ell$ , the corresponding term vanishes.