6
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • $\begingroup$ Either a "$+$" or a "$-$" is missing in your process. Can you please edit it? Thanks! $\endgroup$
    – Stephan Kolassa
    Commented Nov 21, 2017 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
    Commented Nov 21, 2017 at 15:11
  • $\begingroup$ True, strong stationarity only implies weak stationarity, if the first two moments also exist $\endgroup$
    – Luca Thiede
    Commented Nov 21, 2017 at 16:10

1 Answer 1

Reset to default
2
$\begingroup$

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 $$ (2\pi)^{-n/2}(\det\Gamma)^{-1/2}\exp\left[-\frac{1}{2}\mathbf{y}_t'\Gamma^{-1}\mathbf{y}_t \right]. $$

Improve this answer
$\endgroup$
12
  • $\begingroup$ Sorry, I dont get what exactly you mean $\endgroup$
    – Luca Thiede
    Commented Nov 21, 2017 at 16:36
  • $\begingroup$ Does the edit help? $\endgroup$
    – Taylor
    Commented Nov 21, 2017 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
    Commented Nov 21, 2017 at 17:45
  • $\begingroup$ @LucaThiede it's not trivial. I just don't want to spoil the fun for you. $\endgroup$
    – Taylor
    Commented Nov 21, 2017 at 18:31
  • 1
    $\begingroup$ @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
    Commented Nov 22, 2017 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.