Suppose $Z(t)=Σ_{k=1}^{n}Xe^{j(_0t+_k)}$, $t∈R$ where $_0$ is a constant, $n$ is a fixed positive integer, $X_1,...,X_n, _1,...,_n$ are mutually independent random variables, and $EX_k=0, DX_k=σ_{k}^{2}, $~$U[0,2π], k=1,2,...,n$. Find the mean function and correlation function of {Z(t), t ∈ R}.
I have tried to solve it.
For mean function,
$m_Z(s)=E\{Z_s\}=E\{X_s\}+iE\{Y_t\}$
$=E\{Σ_{k=1}^{s}Xe^{j(_0t+_k)}\}$
For correlation function,
$R_Z(s,u)=E\{Z_s,Z_u\}$
$=E\{Y(s)Y(u)\}$
$=E\{Σ_{k=1}^{s}Xe^{j(_0t+_k)}Σ_{k=1}^{u}Xe^{j(_0t+_k)}\}$
$=E\{Σ_{k=1}^{s}Σ_{k=1}^{u} Xe^{j(_0t+_k)}Xe^{j(_0t+_k)}\}$
I am stuck here. How to move from here ahead?
Before finding the mean and variance function, one has to fix the notation of $Z(t)$, which should be $$ Z(t)=\sum_{k=1}^{n}X_ke^{j(_0t+\Phi_k)}. $$ In order to compute the mean of $Z(t)$, it suffices to compute the expectation of $X_ke^{j(_0t+\Phi_k)}$. To do so, use the independence between $X_k$ and $\Phi_k$ and the fact that $X_k$ is centered.
For the correlation function, the most technical part is to compute $\operatorname{Cov}\left(Z(s),Z(t)\right)$ which actually reduces to compute $$ \sum_{k=1}^{n}\sum_{\ell=1}^n\mathbb E\left[X_kX_\ell e^{j(\omega_0s+\Phi_k)}e^{j(\omega_0t+\Phi_\ell)}\right] $$ (when you have to sums, it is always preferable to sum over different indices). If $k\neq\ell$, the corresponding term vanishes.