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Suppose a continuous time process $X(t)=A\sin(\omega t + \theta)$ with $A$, $\omega$ fixed and $\theta\sim uniform[0,2\pi]$. It is easy to see that this is a strict sense stationary process.

Now, let us have a bounded, single valued monotonic function $f(z)$ for any real number $z$. (for example, $f(z)=\tanh(z)$)

My question is that can we say that $X(t)$ and $f(X(t))$ are jointly wide sense stationary (WSS) ? PS: We know that $X(t)$ is ergodic. So, $f(X(t))$ and $X(t)$ are jointly ergodic maybe? I am stuck in this chain of thought.

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2 Answers 2

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Yes, we can.

In order to show that two random processes are jointly WSS, we need to show that

(i) the two processes, $X(t)$ and $f(t)$, are themselves WSS, and

(ii) their cross-correlation $R_{Xf}(t_1,t_2)$ depends only on $t_1-t_2$.

Now, $X(t)$ is SSS, hence it's WSS as well.

For $f(t)$ we have $E[f(t)] = \int_0^{2\pi} f(A \sin(\omega t + \theta)) P_{\Theta}(\theta) d\theta$, where $P_{\Theta}(\theta)=1/2\pi$. This can also be written in terms of the random variable $x$ as

$E[f(t)] = \int_{-A}^{A} f(x) P_{X}(x) dx$

where $P_X(x) = \frac{1}{\pi \sqrt{(A^2-x^2)}}$. Clearly, $E[f(t)]$ above does not depend on $t$.

Similarly, \begin{eqnarray*} E[f(t_1)f(t_2)] &=& \int_0^{2\pi} f(A \sin(\omega t_1 + \theta)) f(A \sin(\omega t_2 + \theta)) P_{\Theta}(\theta) d\theta\\ &=& \int_{-A}^{A} \int_{-A}^{A} f(x_1) f(x_2) \: P_{X(t_1),X(t_2)}(x_1,x_2) \: dx_1 dx_2\\ &=& \int_{-A}^{A} \int_{-A}^{A} f(x_1) f(x_2) \: P_{X(t_1)}(x_1) \: P_{X(t_2)|X(t_1)}(x_2|x_1) \: dx_1 dx_2\\ &=& \int_{-A}^{A} \int_{-A}^{A} f(x_1) f(x_2) \: P_{X(t_1)}(x_1) \: \left(\frac{1}{2} \delta(x_2-x_{21})+\frac{1}{2} \delta(x_2-x_{22}) \right) \: dx_1 dx_2\\ \end{eqnarray*} where $x_{21}$ and $x_{22}$ are the two possible values of $X(t_2)$ once $X(t_1)$ is fixed at $x_1$. enter image description here Here, \begin{eqnarray*} x_{21} &=& A \sin \left( \omega(t_2-t_1) + \sin^{-1} (x_1/A) \right)\\ \text{and} \quad x_{22} &=& -A \sin \left( \omega(t_2-t_1) - \sin^{-1} (x_1/A) \right). \end{eqnarray*}

As can be seen, $x_{21}$ and $x_{22}$ depend on only on $x_1$ and $t_1-t_2$. Therefore, $E[f(t_1)f(t_2)]$, after integrating out on the $x_1$, will depends only on $t_1-t_2$. Hence, $f(t)$ is a WSS process.

Finally, to prove that the cross-correlation $R_{Xf}(t_1,t_2)$ depend only on $t_1-t_2$ we proceed in a similar manner: \begin{eqnarray*} R_{Xf}(t_1,t_2) &=& E[X(t_1)f(t_2)]\\ &=& \int_0^{2\pi} A \sin(\omega t_1 + \theta)\: f(A \sin(\omega t_2 + \theta)) P_{\Theta}(\theta) d\theta\\ &=& \int_{-A}^{A} \int_{-A}^{A} x_1 f(x_2) \: P_{X(t_1),X(t_2)}(x_1,x_2) \: dx_1 dx_2\\ &=& \int_{-A}^{A} \int_{-A}^{A} x_1 f(x_2) \: P_{X(t_1)}(x_1) \: P_{X(t_2)|X(t_1)}(x_2|x_1) \: dx_1 dx_2\\ &=& \int_{-A}^{A} \int_{-A}^{A} x_1 f(x_2) \: P_{X(t_1)}(x_1) \: \left(\frac{1}{2} \delta(x_2-x_{21})+\frac{1}{2} \delta(x_2-x_{22}) \right) \: dx_1 dx_2\\ \end{eqnarray*} which depend only on $t_1-t_2$.

Therefore, $X(t)$ and $f(X(t))$ are jointly WSS.

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  • $\begingroup$ Thanks. There are a few typos but excellent answer nonetheless. $\endgroup$ Commented May 30, 2020 at 0:12
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For those who stumbled upon this problem, The answer given above works perfectly.I am providing some fixes to the above answer. You can also prove this with ergodicity but that is complicated I think. Anyway, the minor fixes:

\begin{eqnarray*} E[f(t_1)f(t_2)] &=& \int_0^{2\pi} \int_0^{2\pi} f(A \sin(\omega t_1 + \theta_1)) f(A \sin(\omega t_2+ \theta_2)) P_{\Theta_1\Theta_2}(\theta_1,\theta_2) d\theta_1 d\theta_2\\ &=& \int_{-A}^{A} \int_{-A}^{A} f(x_1) f(x_2) \: P_{X(t_1),X(t_2)}(x_1,x_2) \: dx_1 dx_2\\ &=& \int_{-A}^{A} \int_{-A}^{A} f(x_1) f(x_2) \: P_{X(t_1)}(x_1) \: P_{X(t_2)|X(t_1)}(x_2|x_1) \: dx_1 dx_2\\ &=& \int_{-A}^{A} \int_{-A}^{A} f(x_1) f(x_2) \: P_{X(t_1)}(x_1) \: \left(\frac{1}{2} \delta(x_2-x_{21})+\frac{1}{2} \delta(x_2-x_{22}) \right) \: dx_1 dx_2\\ \end{eqnarray*} and

\begin{eqnarray*} R_{Xf}(t_1,t_2) &=& E[X(t_1)f(t_2)]\\ &=& \int_0^{2\pi} \int_0^{2\pi} A \sin(\omega t_1 + \theta_1) f(A \sin(\omega t_2+ \theta_2)) P_{\Theta_1\Theta_2}(\theta_1,\theta_2) d\theta_1 d\theta_2\\ &=& \int_{-A}^{A} \int_{-A}^{A} f(x_1) f(x_2) \: P_{X(t_1),X(t_2)}(x_1,x_2) \: dx_1 dx_2\\ &=& \int_{-A}^{A} \int_{-A}^{A} x_1 f(x_2) \: P_{X(t_1),X(t_2)}(x_1,x_2) \: dx_1 dx_2\\ &=& \int_{-A}^{A} \int_{-A}^{A} x_1 f(x_2) \: P_{X(t_1)}(x_1) \: P_{X(t_2)|X(t_1)}(x_2|x_1) \: dx_1 dx_2\\ &=& \int_{-A}^{A} \int_{-A}^{A} x_1 f(x_2) \: P_{X(t_1)}(x_1) \: \left(\frac{1}{2} \delta(x_2-x_{21})+\frac{1}{2} \delta(x_2-x_{22}) \right) \: dx_1 dx_2\\ \end{eqnarray*} where $P_{\Theta_1\Theta_2}(\theta_1,\theta_2)$ is the joint density function.

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