Question

For a stationary ergodic process $X\left(t\right),$ the autocorrelation function $R_X\left(\tau \right)$ is given by, $R_X\left(\tau \right)=\frac{{\tau }^2}{2+9{\tau }^2}$
The magnitude of the mean value of $X\left(t\right)$ i.e., $|\overline{X}|$ is ______

• A. 0.33

Answer 1

The correct option is A 0.33
$(\overline{X}{)}^2=\underset{|\tau |\to \infty }{lim}{R}_X(\tau )$
$=\underset{|\tau |\to \infty }{lim}\frac{{\tau }^2}{2+9{\tau }^2}=\frac{1}{\frac{2}{{\tau }^2}+9}=\frac{1}{9}$
So, $|\overline{X}|=\frac{1}{3}=0.33$