# Example of a stochastic process that is 1st and 2nd order stationary, but not strictly stationary (Round 2) [closed]

This question follows from my previous question , where Robin answered the question in the case of weak stationary processes. Here, I am asking a similar question for (strong?) stationary processes. I'll define what this means (the definition can also be found here).

Let $X(t)$ be a stochastic process. We say that $X(t)$ is Nth-order stationary if, for every $t_1, t_2, \dots, t_N$ we have that the joint cumulative density functions $$F_{X(t_1),X(t_2),\dots,X(t_N)} = F_{X(t_1 + \tau),X(t_2 + \tau),\dots,X(t_N + \tau)}$$ for all $\tau$. This is quite a strong condition, it says that the joint statistics don't change at all as time shifts.

For example, a 1st order stationary process is such that $F_{X(t_1)} = F_{X(t_2)}$ for all $t_1$ and $t_2$. That is, the $X(t)$ are all identically distributed. It is quite easy to see that a 1st order stationary process need not be 2nd order stationary. Simply assign a correlation structure to say $X(t)$, $X(t+1)$, $X(t+2)$ that does not correspond to a (symmetric) Toeplitz matrix. That is, in vector form, the covariance matrix of $[ X(t), X(t+1), X(t+3)]$ could be given as $$\left[\begin{array}{cc} \sigma^2 & a & b \newline a & \sigma^2 & c \newline b & c& \sigma^2 \end{array}\right]$$ for $a,b,c$ distinct. This is now not 2nd order stationary because $E[X(t)X(t+1)] = a$ and, time shifting by 1 we have $E[X(t+1)X(t+2)] = c \neq a$.

In a similar way (presumably) a process that is 1st and 2nd order stationary need not be 3rd order stationary and this leads to my question:

Does somebody have a nice example of a stochastic process that is both 1st and 2nd order stationary, but not 3rd order stationary?

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## closed as off topic by mbq♦Oct 14 '10 at 9:00

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I see round 2 is more difficult :) Mathoverflow may give you a faster answer than stat.overflow... –  robin girard Aug 9 '10 at 14:41
Since we cannot migrate, closed. –  mbq Oct 14 '10 at 9:00