# Under which assumptions does weak stationarity imply strong stationarity

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.

• Either a "$+$" or a "$-$" is missing in your process. Can you please edit it? Thanks! – S. Kolassa - Reinstate Monica Nov 21 '17 at 11:44
• 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? – Richard Hardy Nov 21 '17 at 15:11
• True, strong stationarity only implies weak stationarity, if the first two moments also exist – Luca Thiede Nov 21 '17 at 16:10

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