11

Does anybody have a nice example of a stochastic process that is 2nd-order stationary, but is not strictly stationary?

Silverfish
  • 20,678
  • 23
  • 92
  • 180
Robby McKilliam
  • 2,010
  • 14
  • 14
  • See also [If a time series is second order stationary, does this imply it is strictly stationary?](http://stats.stackexchange.com/q/119845/22228) – Silverfish Sep 29 '15 at 23:20

1 Answers1

7

Take any process $(X_t)_t$ with independent components that has a constant first and second moment and put a varying third moment.

It is second order stationnary because $E[ X_t X_{t+h} ]=0$ and it is not strictly stationnary because $P( X_t \geq x_t, X_{t+1} \geq x_{t+1})$ depends upon $t$

robin girard
  • 6,335
  • 6
  • 46
  • 60
  • 2
    Perhaps I am a little confused, or we are using different definitions? Am I correct in thinking that a process is 2nd order stationary if the joint marginal cdfs $F_{X(t),X(t+τ)}$ are equal for all $\tau$? Similarly for a process to be 1st-order stationary the marginal cdfs $F_{X(t)}$ need to be the same for every $t$. So all the moments of the $X(t)$ must be equal. 2nd-order stationary implies 1st-order stationary, correct? – Robby McKilliam Aug 09 '10 at 07:44
  • 1
    Extending this, a process is $N$th order stationary if for every $t_1, t_2, \dots, t_N$ the marginal cdfs $F_{X(t_1 + \tau), X(t_2 + \tau)\dots,X(t_N+\tau)}$ are the same for all $\tau$. Strictly stationary is Nth order stationary for all N. – Robby McKilliam Aug 09 '10 at 07:45
  • see http://www.statistik.tuwien.ac.at/public/dutt/vorles/geost_03/node49.html for order 2 stationnarity. Anyway, I have tryed to clarify my answer, hope it is better now... – robin girard Aug 09 '10 at 08:05
  • 1
    I believe that article describes stationary in the weak sense and this is not what I meant. I should clarify this in the question. See http://en.wikipedia.org/wiki/Stationary_process for a description of the various types of stationarity. – Robby McKilliam Aug 09 '10 at 08:24
  • @robby OK... I didn't know this "second order stationnarity" I think you should not say second order stationnarity in the question and give the definintion instead. For more clarity you should ask another question. do you have a paper that refers to this second order stationnarity ? – robin girard Aug 09 '10 at 08:32
  • I'll put a second question together that better describes what I mean. In the mean time, your answer is correct for the question I stated! – Robby McKilliam Aug 09 '10 at 09:11