1
$\begingroup$

I was trying to understand difference between drift and trend wherein I came across concepts of unit roots and trend stationary. (I haven't read any books on time series, just going through web).

This question came to my mind when I read this Wikipedia article on unit roots

It says, for a unit root process with drift "Any non-zero value of the noise term, occurring for only one period, will permanently affect the value of $y_t$". This statement looks clear from the equation of $y_t$

$y_t = y_{t-1} + c + e_t$

However, for a trend stationary process, it says "any transient noise will not alter the long-run tendency for $y_t$ to be on the trend line, as also shown in the graph." I can't figure out the use of long-run here. For example, taking the equation given in Wikipedia itself

$y_t = kt + u_t$

and considering $u_t$ to be normally distributed with 0 mean, any deviation will come back to trendline at the next time point itself. Will it not be true for all trend stationary seires.

$\endgroup$
2
$\begingroup$

The statements you refer to appear stronger than they actually are. We can rewrite the unit root process as:

$$y_t=c+y_{t-1}+e_t=cs + y_{t-s} + \sum_{j=t-s}^te_j,$$

by recursively applying the definition. If we take $s=t$ and $y_{0}=0$ we get

$$y_t = ct+\sum_{j=0}^te_j.$$

Looking at this expression it is clear that all the noise terms $e_1,...,e_t$ have an equal input into $y_t$. Furthermore if we take covariance $cov(y_t,e_j)$, for $j<t$ we get $cov(y_t,e_j)=var(e_j)$, so in a sense we can justify the statement that $e_j$ has some permanent effect on $y_t$.

However if we look at the deterministic trend process $y_t=kt+u_t$, it is clear that $cov(y_t,u_j)=0$ for $j\neq t$, so not only the past values, but the future values do not have influence on $y_t$.

Having said that we should treat the statement that the noise term has a permanent effect on $y_t$ with a grain of salt. The effect is definitely permanent, but the magnitude of the effect is not certain. For any two periods $s<t$ we have that $y_t-y_s$ is independent of $y_s$, so although effect of $e_s$ is permanent in $y_t$ it is not that revealing.

$\endgroup$
  • $\begingroup$ Thanks mpiktas, your answer is helpful, specially to verify the effect using covariances. However, I am not very clear on the last paragraph - "permanent effect on $y_t$ with a grain of salt". First, If I take $y_t = f(t) + u_t$ with/without oversimplifying $f(t)$ and $u_t$ then is it possible to see mathematically that effect of shock will stay for few observations? Second, is it possible to see the very small effect that you mentioned in long-term. $\endgroup$ – Gaurav Singhal May 27 '16 at 4:22
  • 1
    $\begingroup$ If you take $y_t=f(f)+u_t$ and $f(t)$ is deterministic, then it is clear that the effect of the shock at time $t$ does not influence $y_s$ for $s\neq t$. The behaviour of $y_t$ is determined by deterministic $f(t)$. $\endgroup$ – mpiktas May 27 '16 at 5:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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