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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?

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  • $\begingroup$ In your second formula $E[X^2(t)]=20$, whereas in your third formula $E[X^2(t)]=6$. $\endgroup$
    – Matt L.
    Sep 28, 2016 at 7:23
  • $\begingroup$ Edited that error. $\endgroup$
    – Don
    Sep 28, 2016 at 15:52
  • $\begingroup$ Got it. As If $\lim_{\tau \to \infty} R_{xx}(\tau)=C$ Then $C=E^2[X(t)]$ where $C$ is a constant $\endgroup$
    – Don
    Sep 28, 2016 at 16:07

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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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    $\begingroup$ This is of course true for this example, but it might be useful to add that the mean cannot always be determined in this way from the autocorrelation function. E.g., the autocorrelcation function of the random process $X(t)=\cos(\omega_0t+\phi)$ (with $\phi$ a random phase uniformly distributed in $[0,2\pi)$) equals $R_X(\tau)=\frac12\cos(\omega_0\tau)$. Consequently, the limit doesn't exist. The mean, however, is $E[X(t)]=0$. $\endgroup$
    – Matt L.
    Sep 28, 2016 at 18:40

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