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I understand $IID\subseteq SSS\subseteq WSS$. What could be an example of a stochastic process which is not iid but is strict sense stationary? I will appreciate examples for $SSS\setminus IID$ and $WSS\setminus SSS$...

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Your understanding is incorrect. Wide-sense-stationary processes are not a subclass of strictly stationary processes, while strictly stationary processes are a subclass of wide-sense-stationary processes. Thus $SSS \setminus WSS = \emptyset$. –  Dilip Sarwate Mar 17 '13 at 20:14
    
@DilipSarwate: O yes, sorry... will edit the question. –  Francis Mar 17 '13 at 20:15
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@DilipSarwate Strictly stationary processes are not necessarily second-order process whereas wide-sense-stationary processes are second-order processes. So strictly stationary processes are not a subclass of wide-sense-stationarity processes. –  saz Mar 17 '13 at 20:27
    
@saz You are correct, but the class of second-order strictly stationary processes is a subclass of WSS processes, not the other way around as the OP originally had it. –  Dilip Sarwate Mar 17 '13 at 20:37
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2 Answers

up vote 1 down vote accepted

Try $x_n=x_0$ for every $n$. This is a stationary process but, unless $x_0$ is almost surely constant, not independent.

To be weakly stationary but not strongly stationary, change the distribution without changing the mean nor the variance: for example, assume that each $x_{2n}$ is symmetric Bernoulli, each $x_{2n+1}$ is standard normal, and that $(x_{n})_{n\geqslant0}$ is independent.

In continuous time, consider $x_t=\cos(t)\xi+\sin(t)\eta$ where $\xi$ and $\eta$ are independent and centered with variance $1$, for example $\xi$ standard Bernoulli and $\eta$ standard normal.

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That's a nice example, thanks! What would the example in $WSS\setminus SSS$ be? –  Francis Mar 17 '13 at 22:08
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Well, any stationary process which has some correlation (an autocorrelation function different from a Dirac delta) would fit the bill. IID is a very special case of a stationary process (white noise, basically; or a subset of white noise, if we are dealing with strict-sense stationary).

A simple example, say $x_n$ is an IID process. Then $y_n=x_n+x_{n+1}$ (or any non trivial LTI filter) is strictly stationary, but not IID.

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