4
$\begingroup$

When talking about variables that are I(1) (the first difference is stationary), Lutkepohl book says: "...in general, a VAR process with cointegrated variables does not admit a pure VAR representation in first differences."

And that would justify the use of VECM models, instead of simply taking the first difference and running a VAR when your time serie is I(1).

But I do not get how this is possible. Suppose a vector $x_t$ is $I(1)$ and there is some cointegration between the variables in this vector. Then, since $(1-L)x_t$ is stationary, by the Wold theorem we get that it can be represented as

$$(1-L)x_t=A(L) \epsilon_t$$

where $\epsilon_t$ white noise and $A(L)$ is invertible. Therefore, we can write

$$A(L)^{-1}(1-L)x_t = \epsilon_t$$

$$A(L)^{-1}\Delta x_t=\epsilon_t$$

Which is a pure VAR representation in first differences. What am I doing wrong?

$\endgroup$
1
$\begingroup$

I have the 2nd edition of Lutkepohl's book (1993), and the above issue is treated in ch. 11.1.2, page 354.

I believe this is partly a semantic (but not unimportant) subtle issue. What you show in the question is that "the first-differenced process has a pure VAR representation" (since it is stationary), which is what the Wold Decomposition Theorem is all about. I don't see any mistake here.

What Lutkepohl appears to mean is that "simply first-differencing the non-stationary system as originally specified does not permit us to obtain a pure VAR representation".

The difference? As can be seen in the last equation of page 354, for Lutkepohl here "first differencing the non-stationary system as originally specified" means first-differencing also the original white noise term, which produces "the non-invertible MA part, $\Delta u_t =u_t-u_{t-1}$" as Lutkepohl writes immediately before what you quoted.

Why may the distinction matter? It appears because by using the (valid) VAR representation of the first-differenced process stemming from the Wold Theorem, we can no longer estimate the co-integrating relation -it has "disappeared from view". So perhaps the more accurate statement would be "...In general, the process does not admit a cointegration preserving pure VAR representation in first differences".

The example Lutkepohl works out in this page of his book is clear. To manipulate the system he does not apply the differencing operator, but he adds and subtracts variables that leave the whole unaffected, in order to obtain a linear combination of the process in the left-hand side, and a stationary expression on the right-hand side (where now the fact the first-differences are stationary is used).

Faced with an $I(1)$ co-integrated system, one can make it stationary by first-differencing and use Wold decomposition while losing the co-integrating relation, or one can manipulate the system differently in order to transform it into a stationary system by virtue of first-differences being stationary, and bring into the surface the co-integrating relationship.

$\endgroup$
0
$\begingroup$

I think we have different editions. I get that when we have an arbitrary VAR representation of some I(1) process, it is not true that we can get the same residual in a VAR representation with just first differences, as he shows in his book.

But many authors (Cochrane notes, for example) say that the VAR in first is misspecified, as if you cannot write a I(1) cointegrated proccess in first differences. To me, doing that is the same as writing a non-linear process as a linear process: you are missing the original error terms, but it is still a valid representation.

Unless the Wold theorem does not guarantee that A(L) is invertible when variables are cointegrated. Is that correct? Or is A(L) always invertible?

$\endgroup$
0
$\begingroup$

I think I understand where the confusion comes from.

If the $I(1)$ vector $x_t$ has cointegrated variables, the polinomial $A(L)$ the Wold representation of the first difference

$$(1-L)x_t = A(L)\epsilon_t$$

will have an unit root on $A(L)$. Then, it is not invertible. So it is not always true that the Wold representation is invertible. If there is not an unit root, the polinomial $A(L)$ is invertible, then we are fine just running AR's on first differences. But it just happens if the variables are not cointegrated.

If there is cointegration, the AR representation of the first differences no longer holds. But the Engel-Granger representation theorem guarantees that in this case there is the VECM representation, so we should use it.

So, is it true that any stationary has an AR representation? No! It just happens if the polinomial of the Wold representation is invertible.

$\endgroup$

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.