May 17

Stationary processes 2

Stationary processes: Here we consider an example that appeared in the Econometrics exams of the University of London.

Example 3. Consider the process defined by

(1) y_t=u_t+\theta u_{t-1}

where u_t is white noise:

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

In Economics, the error u_t is treated as a shock in the current period. So in (1) y_t sustains a shock in period t. There is no reason to put a coefficient in front of u_t (or, you could say, the coefficient is 1). Besides, y_t remembers the shock from the previous period, and the aftereffect of this shock is measured by the coefficient \theta.

Observation. Before we proceed with the analysis of this model, it is a good idea to elucidate the techniques of working with white noise. For this, we can rewrite the conditions on the second moments of errors from (2) as

Eu_tu_s=\sigma^2  if  t=s and Eu_tu_s=0  if  t\ne s.

In words, the expected value of a product of any error with itself is \sigma^2 and expected values of products of different errors disappear. This fact is used many times in Econometrics, in particular, here. I am mentioning it here because some students have problems with it even in the second semester.

First stationarity condition. Obviously,

(3) Ey_t=0 for all t.

Second stationarity condition. Variance does not depend on time:

(4) Var(y_t)=Ey_t^2=E(u_t+\theta u_{t-1})(u_t+\theta u_{t-1})=(1+\theta^2)\sigma^2

because only products u_t^2, u_{t-1}^2 have nonzero expectations.

Third stationarity condition. Using the observation above, let us look at the expectation

(5) Ey_ty_{t+1}=E(u_t+\theta u_{t-1})(u_{t+1}+\theta u_t).

The error u_{t-1} from the first parenthesis does not have a match in the second one, while the error u_{t+1} from the second parenthesis does not have a match in the first one. Thus, (5) equals

(6) Ey_ty_{t+1}=\theta\sigma^2.


(7) Ey_ty_{t-1}=E(u_t+\theta u_{t-1})(u_{t-1}+\theta u_{t-2})=\theta\sigma^2,

Further, if the distance between points t,s is larger than one, then in

(8) Ey_ty_s=E(u_t+\theta u_{t-1})(u_s+\theta u_{s-1})=0

expected values of all products will be zero. Equations (4), (6)-(8) are summarized as follows:

Ey_ty_s=(1+\theta^2)\sigma^2 if |t-s|=0Ey_ty_s=\theta\sigma^2 if |t-s|=1, and Ey_ty_s=0 if |t-s|>1..

Recalling the first stationarity condition, we can rewrite this as

Cov(y_t,y_s)=(1+\theta^2)\sigma^2 if |t-s|=0Cov(y_t,y_s)=\theta\sigma^2 if |t-s|=1, and Cov(y_t,y_s)=0 if |t-s|>1.

Thus the covariance Cov(y_t,y_s) is indeed a function of only |t-s|.

The process we studied is called a moving average of order 1 and denoted MA(1).

One Response for "Stationary processes 2"

  1. […] A simple realization of this idea is given here. How do you generalize […]

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