How does trend stationary recovers from shocks in long run?

Question

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.

Answer 1

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.