0
$\begingroup$

In relation to already posted question:

The difference between the three Augmented Dickey–Fuller test (none,drift, trend)

and specifically to the details given by Graeme Walsh's (https://stats.stackexchange.com/users/24617/graeme-walsh) answer, I still miss something about the use of the embed() function and its way to arrange data in a matrix fashion. As we can see from the following dummy example:

> s <- 1:6
> embed(s,3)
     [,1] [,2] [,3]
[1,]    3    2    1
[2,]    4    3    2
[3,]    5    4    3
[4,]    6    5    4

Considering the same piece of code shown in related question at link above:

data(sunspots)
x           <- sunspots
alternative <- "stationary"
k           <- trunc((length(x) - 1)^(1/3))

k <- k + 1          # Number of lagged differenced terms
y <- diff(x)        # First differences
n <- length(y)      # Length of first differenced series
z <- embed(y, k)    # Used for creating lagged series

yt  <- z[, 1]       # First differences
xt1 <- x[k:n]       # Series in levels - the first k-1 observations are dropped
tt  <- k:n          # Time-trend
yt1 <- z[, 2:k]     # Lagged differenced series - there are k-1 of them

Hence, yt <- z[, 1] stores observations "older" than the ones stored inside yt1 <- z[, 2:k] while comparing same row indexes between yt and yt1. At the same time, with constant and time-trend regression, the formula is:

                  yt ~ xt1 + 1 + tt + yt1

I mean yt (also) dependent upon yt1.

Would it be possible to clarify this aspect ? Thanks.

$\endgroup$
0
$\begingroup$

I think I overlooked the timeline. Indexes in time series start from old observations (first index) up to the newest one (last index). Hence yt[idx,1] stores a more recent observation than yt[idx, 2:k], for each index idx.

$\endgroup$

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.