Random Walk with Drift: Why is the change in a trending variable also a function of a random variable when $E(\epsilon_t) = 0$? 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?
 A: 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.
