26
May 17

Stationary processes 1

Along with examples of nonstationary processes, it is necessary to know a couple of examples of stationary processes.

Example 1. In the model with a time trend, suppose that there is no time trend, that is, b=0. The result is white noise shifted by a constant a, and it is seen to be stationary.

Example 2. Let us change the random walk slightly, by introducing a coefficient \beta for the first lag:

(1) y_t=\beta y_{t-1}+u_t

where u_t is, as before, white noise:

(2) Eu_t=0Eu_t^2=\sigma^2 for all t and Eu_tu_s=0 for all t\ne s.

This is an autoregressive process of order 1, denoted AR(1).

Stability condition|\beta|<1.

By now you should be familiar with recurrent substitution. (1) for the previous period looks like this:

(3) y_{t-1}=\beta y_{t-2}+u_{t-1}.

Plugging (3) in (1) we get y_t=\beta^2y_{t-2}+\beta u_{t-1}+u_t. After doing this k times we obtain

(4) y_t=\beta^ky_{t-k}+\beta^{k-1}u_{t-k+1}+...+\beta u_{t-1}+u_t.

To avoid errors in calculations like this, note that in the product \beta^{k-1}u_{t-k+1} the sum of the power of \beta and the subscript of u is always t.

Here the range of time moments didn't matter because the model wasn't dynamic. In the other example we had to assume that in (1) t takes all positive integer values. In the current situation we have to assume that t takes all integer values, or, put it differently, the process y_t extends infinitely to plus and minus infinity. Then we can take advantage of the stability condition. Letting k\rightarrow\infty (and therefore t-k\rightarrow-\infty) we see that the first term on the right-hand side of (4) tends to zero and the sum becomes infinite:

(5) y_t=...+\beta^{k-1}u_{t-k+1}+...+\beta u_{t-1}+u_t=\sum_{j=0}^\infty\beta^ju_{t-j}.

We have shown that this representation follows from (1). Conversely, one can show that (5) implies (1). (5) is an infinite moving average, denoted MA(\infty).

It can be used to check that (1) is stationary. Obviously, the first condition of a stationary process is satisfied: Ey_t=0. For the second one we have (use (2)):

(6) Var(y_t)=Ey_t^2=E(...+\beta^{k-1}u_{t-k+1}+...+\beta u_{t-1}+u_t)(...+\beta^{k-1}u_{t-k+1}+...+\beta u_{t-1}+u_t)

=...+\beta^{2k-2}Eu^2_{t-k+1}+...+\beta^2Eu^2_{t-1}+Eu^2_t=(...+\beta^{2k-2}+...+\beta^2+1)\sigma^2=\frac{\sigma^2}{1-\beta^2},

which doesn't depend on t.

Exercise. To make sure that you understand (6), similarly prove that

(7) Cov(y_t,y_s)=\beta^{|t-s|}\frac{\sigma^2}{1-\beta^2}.

Without loss of generality, you can assume that t>s. (7) is a function of the distance in time between t,s, as required.

Leave a Reply

You must be logged in to post a comment.