An excercise question for time series analysis asks:

Consider the process $$ y_t = 0.8y_{t-1} + 0.1y_{t-2} + u_t $$

  1. Is this process weakly stationary (I would answer this with the stability triangle)

  2. Under which assumptions does this property imply strong stationarity

I thought that strong stationarity always implies weak stationarity, but not that it can be the other way around.

  • $\begingroup$ Either a "$+$" or a "$-$" is missing in your process. Can you please edit it? Thanks! $\endgroup$ – Stephan Kolassa Nov 21 '17 at 11:44
  • $\begingroup$ Stong stationarity does not imply weak stationarity. Hint: does weak stationarity require existence of the first two moments? And does strong stationarity have anything to tell about that? $\endgroup$ – Richard Hardy Nov 21 '17 at 15:11
  • $\begingroup$ True, strong stationarity only implies weak stationarity, if the first two moments also exist $\endgroup$ – Luca Thiede Nov 21 '17 at 16:10

Hint: consider what happens when you make more assumptions about the specific distribution of the errors. Then you can write down exact conditional densities. After multiplying a few together, you will have the joint density of all the time observations, and strong stationarity deals with this joint distribution.

For your model: $$ p(y_1, y_2, \ldots , y_n) = \prod_{t=3}^n p(y_t \mid y_{t-1}, y_{t-2} ) p(y_1, y_2)\tag{1}. $$ If you assumed that the errors were Normally distributed then $$ p(y_t \mid y_{t-1}, y_{t-2} ) = N(.8 y_{t-1} +.1 y_{t-2}, \sigma^2). $$

Another hint: If this Normal distribution does lead to strong stationarity, then the joint distribution of all the observations $\{y_t\}$ should have the right means, and the right variances and (auto-)covariances. Arrange all of those autocovariances and variances into a matrix $\Gamma$. Then your joint density should be $$ (2\pi)^{-n/2}(\det\Gamma)^{-1/2}\exp\left[-\frac{1}{2}\mathbf{y}_t'\Gamma^{-1}\mathbf{y}_t \right]. $$

  • $\begingroup$ Sorry, I dont get what exactly you mean $\endgroup$ – Luca Thiede Nov 21 '17 at 16:36
  • $\begingroup$ Does the edit help? $\endgroup$ – Taylor Nov 21 '17 at 17:14
  • $\begingroup$ Mh, I guess you say, that the process ist stationary, if the errors are normally distributed. But I can't see, how that comes out of the definition of the strong stationarity (sorry for asking probably trivial questions, we never did a lot with this). $\endgroup$ – Luca Thiede Nov 21 '17 at 17:45
  • $\begingroup$ @LucaThiede it's not trivial. I just don't want to spoil the fun for you. $\endgroup$ – Taylor Nov 21 '17 at 18:31
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    $\begingroup$ @LucaThiede strong stationarity is a statement about distributions. Weak stationarity only deals with moments. When moments exist they are only summaries of a distribution. you need to assume something about the distributions or else you just can't get where you want to go. $\endgroup$ – Taylor Nov 22 '17 at 22:54

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