Answer to Question #131027 in Statistics and Probability for Meenu Rani

Question

A random process has sample functions of the form X(t) = Acos(wt+Θ) where w is constant, A is a random variable that has a magnitude of +1 and −1 with equal probability, and Θ is a random variable that is uniformly distributed between 0 and 2π. Assume that the random variables A and Θ are independent. 1. Is X(t) a wide-sense stationary process?

Answer 1

1.

$E[X(t)]=E[Acos(\omega t+\Theta)].$

Since $A$ and $\Theta$ are independent, then

$E[X(t)]=E[Acos(\omega t+\Theta)]=\\ E[A]\cdot E[cos(\omega t+\Theta)]=\\ (1\cdot1/2+(-1)\cdot1/2)E[cos(\omega t+\Theta)]=\\ 0\cdot E[cos(\omega t+\Theta)]=0.$

Expectation of $A^2$:

$E[A^2]=(1^2\cdot1/2+(-1)^2\cdot1/2)=1.$

2.

Correlation function:

$R(t_1,t_2)=E[X(t_1)X(t_2)]=\\ E[Acos(\omega t_1+\Theta)\cdot Acos(\omega t_2+\Theta)]=\\ E[A^2]E[cos(\omega t_1+\Theta)\cdot cos(\omega t_2+\Theta)]=\\$

$E[cos(\omega t_1+\Theta)\cdot cos(\omega t_2+\Theta)]=\\ E[\frac{1}{2}cos(\omega t_1+\Theta+\omega t_2+\Theta)+\\ +\frac{1}{2}cos(\omega t_1+\Theta-\omega t_2-\Theta)]=\\ E[\frac{1}{2}cos(\omega (t_1+ t_2)+2\Theta)+\frac{1}{2}cos(\omega( t_1- t_2))]=\\ E[\frac{1}{2}cos(\omega (t_1+ t_2)+2\Theta)]+\\ +E[\frac{1}{2}cos(\omega( t_1- t_2))].$

Since $sin(x)$ is periodic, then

$E[\frac{1}{2}cos(\omega (t_1+ t_2)+2\Theta)]=\\ \intop_{0}^{2\pi}\frac{1}{2}cos(\omega (t_1+ t_2)+2\Theta)\frac{1}{2\pi}d\theta=\\ \frac{1}{4\pi}\frac{sin(\omega (t_1+ t_2)+2\Theta)}{2}|_0^{2\pi}=\\ \frac{1}{4\pi}\frac{sin(\omega (t_1+ t_2)+4\pi)}{2}-\frac{1}{4\pi}\frac{sin(\omega (t_1+ t_2))}{2}=0,$

hence

$R(t_1,t_2)=E[X(t_1)X(t_2)]=\\ E[\frac{1}{2}cos(\omega (t_1+ t_2)+2\Theta)]+\\ +E[\frac{1}{2}cos(\omega( t_1- t_2))]=\\ 0+\frac{1}{2}cos(\omega( t_1- t_2))=\\ \frac{1}{2}cos(\omega( t_1- t_2))$

3.

Since $E[X(t)]=0$ and $R(t_1,t_2)=\frac{1}{2}cos(\omega( t_1- t_2))$ is not random and depends only on $(t_1-t_2)$, then $X(t)$ is a wide-sense stationary process.

Answer: Yes, $X(t)$ is a wide-sense stationary process.