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Is this the correct "venn diagram" that related WSS, SSS, and Ergodic process types?

$$\text{all process types}\begin{cases}\text{WSS} \begin{cases}SSS \begin{cases}\text{ergodic} \\ \text{non-ergodic}\end{cases} \\ \text{non-SSS}\end{cases}\\ \text{non-stationary}\end{cases}$$

SSS = Strict Sense Stationary

WSS = Wide Sense Stationary

Just trying to wrap my brain around this one...without getting too confused.

I started out with this diagram:

$$\text{all process types}\begin{cases}\text{stationary}\begin{cases}\text{ergodic} \\ \text{non-ergodic}\end{cases}\\\text{non-stationary}\end{cases}$$

but, i thought I could do better by include WSS in the diagram. Of course, I don't really know 100% if I'm correct... its more of a guess.

pico
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2 Answers2

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There are several questions on this forum dealing with various aspects of strict-sense and wide-sense stationarity and ergodicity, and some of the answers give counterexamples to the Venn diagram that you have constructed, such as

Dilip Sarwate
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  • So according to what you are saying, all four combinations of SSS and WSS are possible? $$\text{all process types}\begin{cases}\text{WSS} \begin{cases}SSS \ \text{non-SSS} \end{cases}\ \text{not-WSS}\begin{cases}SSS\ \text{non-SSS} \end{cases}\\end{cases}$$ and non-WSS and non-SSS means non-stationary in any form. Let's forget about ergodic, I was only trying to understand the many flavors of stationality.. – pico Jul 09 '20 at 13:52
  • $$\begin{aligned} &\textbf{WSS + not SSS:} \text{ example?} \ \&\textbf{WSS + SSS:} \text{ mean is constant, autocovariance is-time invariant} \\ &\textbf{not-WSS + SSS:} \text{ 1st order stationary because mean is constant,} \ &~~~~~~~~~~~~\text{ but autocovariance is time-dependent} \\ &\textbf{not-WSS + not-SSS:} \text{ mean is time-dependent}\end{aligned}$$ – pico Jul 09 '20 at 14:37
  • : ???example??? Read the links provided in my answer. They provide the example you are demanding. + : mean is constant, autocovariance is-time invariant: your comment defines WSS but SSS is a far stronger and different requirement. WSS Gaussian processes are SSS as noted here. - + : Read the links provided in my answer; a SSS process consisting of independent Cauchy random variables is not WSS because the mean does not exist.
  • – Dilip Sarwate Jul 09 '20 at 14:52
  • I think i understand all of them except for "WSS + not SSS", In my opinion, its impossible to have a WSS that is "not SSS", since WSS is by definition a 2nd order SSS process... I read that example, but thought it had something wrong with it... – pico Jul 09 '20 at 14:55
  • "WSS is by definition a second-order SSS process" Your "definition" is wrong. Your wording "second-order SSS process" is nonsensical: SSS generally means **Strictly Stationary Stochastic" and has no room for second-order stationarity whatever you think it means. Also, WSS processes don't have to be second-order stationary, not even first-order stationary. If you think the example I have provided is wrong, write your own answer pointing out where it is wrong. – Dilip Sarwate Jul 09 '20 at 15:06
  • If process is WSS, then maybe its also 2nd order SSS? which i'm uncertain how to prove it though because the requirements are similar but not the same.

    $$\begin{aligned}&\textbf{SSS requires: } E[X^2(t_1)] = E[X^2(t_1+\tau)] =R_X(t_1, t_1) = R_X(t_1 +\tau, t_1+\tau)=R_X(0)\ &\textbf{WSS requires: } R_X(t_1, t_1) = R_X(t_1 - t_1) = R_X(0)\end{aligned}$$ WSS more generally requires that $R_X(t_1, t_2) = R_X(t_1 - t_2)$, but i just changed $t_2$ to $t_1$ so it would equal the 2nd order moment...to show that SSS is a special case of WSS...

    – pico Jul 09 '20 at 15:11
  • definition of auto correlation: $R_x(t_1, t_2) = E\Big[~X(t_1)~X(t_2)~\Big]$... if $t_2$ = $t_1$, then it degenerates to: $R_x(t_1, t_1) = E\Big[~X(t_1)~X(t_1)~\Big]$... $R_x(t_1, t_1) = E\Big[~X^2(t_1)~\Big]=\text{2nd order moment}$ – pico Jul 09 '20 at 15:22