Mbalisikerc

2022-07-16

Suppose $Z(t)={\Sigma}_{k=1}^{n}X{e}^{j({\mathit{\omega}}_{0}t+{\mathbf{\Phi}}_{k})}$, $t\in R$ where ${\mathit{\omega}}_{0}$ is a constant, n is a fixed positive integer, ${X}_{1},...,{X}_{n},\u3000{\mathbf{\Phi}}_{1},...,{\mathbf{\Phi}}_{n}$ are mutually independent random variables, and $E{X}_{k}=0,D{X}_{k}={\sigma}_{k}^{2},\mathbf{\Phi}$ , $U[0,2\pi ],k=1,2,...,n$ . Find the mean function and correlation function of $\{Z(t),\text{}t\in R\}$ .

I have tried to solve it.

For mean function,

${m}_{Z}(s)=E\{{Z}_{s}\}=E\{{X}_{s}\}+iE\{{Y}_{t}\}$

$=E\{{\Sigma}_{k=1}^{s}X{e}^{j({\mathit{\omega}}_{0}t+{\mathbf{\Phi}}_{k})}\}$

For correlation function,

${R}_{Z}(s,u)=E\{{Z}_{s},{Z}_{u}\}$

$=E\{Y(s)Y(u)\}$

$=E\{{\Sigma}_{k=1}^{s}X{e}^{j({\mathit{\omega}}_{0}t+{\mathbf{\Phi}}_{k})}{\Sigma}_{k=1}^{u}X{e}^{j({\mathit{\omega}}_{0}t+{\mathbf{\Phi}}_{k})}\}$

$=E\{{\Sigma}_{k=1}^{s}{\Sigma}_{k=1}^{u}X{e}^{j({\mathit{\omega}}_{0}t+{\mathbf{\Phi}}_{k})}X{e}^{j({\mathit{\omega}}_{0}t+{\mathbf{\Phi}}_{k})}\}$

I am stuck here. How to move from here ahead?

I have tried to solve it.

For mean function,

${m}_{Z}(s)=E\{{Z}_{s}\}=E\{{X}_{s}\}+iE\{{Y}_{t}\}$

$=E\{{\Sigma}_{k=1}^{s}X{e}^{j({\mathit{\omega}}_{0}t+{\mathbf{\Phi}}_{k})}\}$

For correlation function,

${R}_{Z}(s,u)=E\{{Z}_{s},{Z}_{u}\}$

$=E\{Y(s)Y(u)\}$

$=E\{{\Sigma}_{k=1}^{s}X{e}^{j({\mathit{\omega}}_{0}t+{\mathbf{\Phi}}_{k})}{\Sigma}_{k=1}^{u}X{e}^{j({\mathit{\omega}}_{0}t+{\mathbf{\Phi}}_{k})}\}$

$=E\{{\Sigma}_{k=1}^{s}{\Sigma}_{k=1}^{u}X{e}^{j({\mathit{\omega}}_{0}t+{\mathbf{\Phi}}_{k})}X{e}^{j({\mathit{\omega}}_{0}t+{\mathbf{\Phi}}_{k})}\}$

I am stuck here. How to move from here ahead?

suponeriq

Beginner2022-07-17Added 10 answers

Step 1

Before finding the mean and variance function, one has to fix the notation of $Z(t)$, which should be

$Z(t)=\sum _{k=1}^{n}{X}_{k}{e}^{j({\mathit{\omega}}_{0}t+{\mathrm{\Phi}}_{k})}.$ .

In order to compute the mean of $Z(t)$, it suffices to compute the expectation of ${X}_{k}{e}^{j({\mathit{\omega}}_{0}t+{\mathrm{\Phi}}_{k})}$ . 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 $\mathrm{Cov}(Z(s),Z(t))$ which actually reduces to

$\sum _{k=1}^{n}\sum _{\ell =1}^{n}\mathbb{E}\left[{X}_{k}{X}_{\ell}{e}^{j({\omega}_{0}s+{\mathrm{\Phi}}_{k})}{e}^{j({\omega}_{0}t+{\mathrm{\Phi}}_{\ell})}\right]$

(when you have to sums, it is always preferable to sum over different indices). If $k\ne \ell $ , the corresponding term vanishes.

Before finding the mean and variance function, one has to fix the notation of $Z(t)$, which should be

$Z(t)=\sum _{k=1}^{n}{X}_{k}{e}^{j({\mathit{\omega}}_{0}t+{\mathrm{\Phi}}_{k})}.$ .

In order to compute the mean of $Z(t)$, it suffices to compute the expectation of ${X}_{k}{e}^{j({\mathit{\omega}}_{0}t+{\mathrm{\Phi}}_{k})}$ . 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 $\mathrm{Cov}(Z(s),Z(t))$ which actually reduces to

$\sum _{k=1}^{n}\sum _{\ell =1}^{n}\mathbb{E}\left[{X}_{k}{X}_{\ell}{e}^{j({\omega}_{0}s+{\mathrm{\Phi}}_{k})}{e}^{j({\omega}_{0}t+{\mathrm{\Phi}}_{\ell})}\right]$

(when you have to sums, it is always preferable to sum over different indices). If $k\ne \ell $ , the corresponding term vanishes.

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