Friday 27 November 2015

Random process $X(t)$ with autocorrelation function given find the mean and the variance


Autocorrelation function is $$R_{xx}(\tau)=\frac{20}{1+2\tau^2}$$ So at $\tau=0$$$R_{xx}(0)=20=E[X(t)X(t)]=E[X^2(t)]$$ The variance is $$\mathrm{Var}[X(t)]=E[X^2(t)]-E^2[X(t)]=20-E^2[X(t)]$$ As $X(t)$ is WSS (wide sense stationary) the mean is a constant. Is there any way to find its numerical value?



Answer



The limit $\lim_{\tau\to\infty} R_x(\tau)$, if it exists, equals $E^2[X(t)]$ and so $E[X(t)]=0$ in this case.


More generally, the mean of a WSS process is nonzero only if the power spectral density has an impulse at the origin. This can be applied to periodic autocorrelation functions such as $\cos(\omega_0t)$ pointed out in @MattL's comment. If the Fourier series for a periodic autocorrelation function has a nonzero DC term, the mean is nonzero.


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