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I am studying stationary from wikipedia. (https://en.wikipedia.org/wiki/Stationary_process?wprov=sfii1)

In the examples section, the following is written.

let $Y$ have a uniform distribution on $(0,2π]$ and define the time series $\{ X_t \}$ by

$X_t=\cos (t+Y) \quad \text{ for } t \in \mathbb{R}$. Then $\{ X_t \}$ is strictly stationary.

I don't understand why it is stationary.

Can someone prove this?

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    $\begingroup$ Yes. // Can you write down the property you are supposed to establish? $\endgroup$
    – Did
    Mar 19 '16 at 9:09
  • $\begingroup$ @Did I just want to understand the exact definition and differences of the 1st-order, 2nd-order, wide sense, and strict sense stationary. Some sites define them using Expectation, some using Set, some sites using Sequence, some sites using mean and variance, even some sites using signal power. I decided to believe wikipedia's definition but I do not understand why $X_t=\cos{(t+Y)}$ is strictly stationary. $\endgroup$
    – Danny_Kim
    Mar 19 '16 at 10:58
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    $\begingroup$ IOW, you have no idea what you are supposed to show? Then the mention "I don't understand why it is stationa(r)y" is misleading and should read "I don't understand what is stationarity". And to that, the best answer might be to look for a definition, no? So, back to square one: what is the definition of stationarity in this context? $\endgroup$
    – Did
    Mar 19 '16 at 11:01
  • $\begingroup$ @Did First, sorry for vague question. According to wikipedia, definition of stationary is $F_X(x_{t_1+\tau}, \cdots, x_{t_k+\tau} = F_X(x_{t_1}, \cdots, x_{t_k}$. So I tried to prove that $X_t=\cos{(t+Y)}$ is strictly stationary using this definition. But I failed. I might have been able to solve this if not sinusoidal function but common polynomial function. $\endgroup$
    – Danny_Kim
    Mar 19 '16 at 11:06
  • $\begingroup$ ?? Sorry but what is your definition of $F_X(x_{t_1},\ldots,x_{t_k})$? $\endgroup$
    – Did
    Mar 19 '16 at 15:16
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Since the cosine is periodic with period 2$\pi$, $Y$ already covers all possible inputs within a whole period of the cosine, and so does $Y+t$.

A formal proof:

We need to show that $P(\cos(Y)\le x) = P(\cos(Y+t)\le x)$. We have: $P(\cos(Y)\le x)2\pi = I_{]1, \infty]}(x)2\pi + I_{[-1,1]}(x)\mu(\{y\in]0,2\pi]|\cos(y)\le x\})$ and $P(\cos(Y+t)\le x)2\pi = I_{]1, \infty]}(x)2\pi + I_{[-1,1]}(x)\mu(\{y\in]0,2\pi]|\cos(y+t)\le x\})$.

If $x \in [-1,1]$, then $\arccos(x)$ exists and we have:

$\mu(\{y\in]0,2\pi]|\cos(y)\le x\}) = I_{[-1,1]}(x)\mu(\{]0,2\pi]\cap\bigcup_{k\in\mathbf{Z}}[\arccos(x)+2k\pi,2\pi-\arccos(x)+2k\pi]\}) = \mu(\{]0,t]\cup]t,2\pi]\cap\bigcup_{k\in\mathbf{Z}}[\arccos(x)+2k\pi,2\pi-\arccos(x)+2k\pi]\}) = \mu(\{(]t,2\pi]\cup]2\pi,t+2\pi])\cap\bigcup_{k\in\mathbf{Z}}[\arccos(x)+2k\pi,2\pi-\arccos(x)+2k\pi]\}) = \mu(\{]t,t+2\pi]\cap\bigcup_{k\in\mathbf{Z}}[\arccos(x)+2k\pi,2\pi-\arccos(x)+2k\pi]\}) = \mu(\{]0,2\pi]\cap\bigcup_{k\in\mathbf{Z}}[\arccos(x)+2k\pi-t,2\pi-\arccos(x)+2k\pi-t]\}) = \mu(\{y\in]0,2\pi]|\cos(y+t)\le x\})$

We further need to show that $P(X_{t_1+\tau} \le x_1, \ldots, X_{t_n+\tau} \le x_n) = P(X_{t_1} \le x_1, \ldots, X_{t_n} \le x_n)$ for $n \ge 2$.

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  • $\begingroup$ Is it enough to say that it is strictly stationary? I tried to change domain from $X$ to $Y$ using Jacobian matrix. but it is so hard to continue to prove because arc-sinusoidal function appeared in my process. Anyhow, thank you for giving me some intuition. $\endgroup$
    – Danny_Kim
    Mar 19 '16 at 13:14
  • $\begingroup$ @Danny_Kim, I added a formal proof. $\endgroup$
    – hkBst
    Mar 19 '16 at 15:51
  • $\begingroup$ This is weak stationarity. Strict stationarity is the fact that the distribution of $(X_{t_1+s},\ldots,X_{t_k+s})$ does not depend on $s$, for every $k$ and every $(t_1,\ldots,t_k)$. $\endgroup$
    – Did
    Mar 19 '16 at 16:21
  • $\begingroup$ @Did, you are right that it is not yet strict stationarity... $\endgroup$
    – hkBst
    Mar 19 '16 at 16:34
  • $\begingroup$ @hkBst When I saw this answer first, I felt fears. But now I am analyzing your proof. This is good for understanding about WSS. Thank you. Can i ask about ]a, b]? $\endgroup$
    – Danny_Kim
    Mar 20 '16 at 4:49

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