Raising the bar

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

where $latex u_t$ is white noise:

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

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

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

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

In words, the expected value of a product of any error with itself is $latex \sigma^2$ 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) $latex Ey_t=0$ for all $latex t$.

Second stationarity condition. Variance does not depend on time:

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

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

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

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

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

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

Similarly,

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

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

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

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

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

Recalling the first stationarity condition, we can rewrite this as

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

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

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