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I came across Pearson’s companion site of Murray, M. P. (2005). Econometrics: A modern introduction. Pearson Higher Education. While skimming through the related lecture slides here http://wps.aw.com/wps/media/objects/2387/2445250/PPTs/ch18lectr26.ppt, there is something I don’t get.

Slide #26-7 of the aformentioned powerpoint presentation correctly states that $E(v_t) = 0$. Given this assumption I do not understand why $E(Y_t) - E({Y_{t - 1}}) = \alpha + v_t$ as stated on slide #26-5: "The trending variable changes by a random amount each period". I think $v_t$ should drop out of the equation if the general assumption $E(v_t) = 0$ holds. Notice that slide #26-7 also states that $v_t$ is a random variable with $E(v_t) = 0$. So, if $E(v_t) = 0$ holds in general, why is $v_t$ still included in $E(Y_t) - E({Y_{t - 1}})$ above?

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  • $\begingroup$ You are right. The correct formulation is on page 26-7. $\endgroup$ – Christoph Hanck Aug 16 '16 at 7:08
  • $\begingroup$ @ChristophHanck: Thanks for your comment. 26-7 refers to the difference between successive periods, not beween the their expected values. Thus, am I right that the equation E(Y_t)−E(Y_t−1)=α+v_t is simply false? $\endgroup$ – denominator Aug 16 '16 at 10:57
  • $\begingroup$ Yes, if you take expectation on the left-hand side of the equation, you do so on the right-hand, too, and then $E(v_t)=0$ applies, as you correctly pointed out. $\endgroup$ – Christoph Hanck Aug 16 '16 at 11:06
  • $\begingroup$ @ChristophHanck Thanks a million, that helps. To me it then appears that the actual difference betwen the expected values is E(Yt)−E(Yt−1) = α, which is confusing because it is pretty much in line with a deterministic trend. In other words, the book fails to show the difference between a deterministic and stochastic trend. $\endgroup$ – denominator Aug 16 '16 at 11:32
  • $\begingroup$ I now give a proper answer to your last question in the comments, see below. $\endgroup$ – Christoph Hanck Aug 16 '16 at 11:53
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Let my give an answer to your last question in the comments:

We have the stochastic trend model $Y_t=\alpha+Y_{t-1}+v_t$ for which, as discussed in the comments, $$ E(\Delta Y_t)=\alpha $$ On the other hand, the deterministic trend model has $Y_t=\beta_0+\beta_1t+v_t$ and thus $$ \Delta Y_t=\beta_0+\beta_1t+v_t-[\beta_0+\beta_1(t-1)+v_{t-1}]=\beta_1+\Delta v_t $$ so that, indeed, $$ E(\Delta Y_t)=\beta_1 $$ Hence, the two processes are indeed equivalent in terms of their expected drift.

They are not in terms of their variances, however. By recursive substitution (and assuming $Y_0=0$), you can write the stochastic trend model as $$ Y_t=t\alpha+\sum_{s=1}^tv_s, $$ which has variance $$ Var(Y_t)=t\sigma^2_v $$ whereas the deterministic trend process has variance $$ Var(Y_t)=\sigma^2_v $$ Playing around with the code

alpha <- beta1 <- .1
beta0 <- 0
t <- 100
v <- rnorm(t)
Y.ST <- 1:t*alpha+cumsum(v)
Y.DT <- beta0 + beta1*1:t + v
plot(Y.ST, type="l", col="gold", lwd=2)
lines(Y.DT, col="purple", lwd=2)

will show differences between the two processes. If you pick a large value for the drift or a small standard deviation for the errors, the differences will tend to get blurred for moderate $t$, though.

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