A - I

[-6 - a8)

(14.11)

How do we find the cointegrating relationship from the VAR model? The procedure is as follows: Find the characteristic roots. Then corresponding to each root, find the characteristic vector. This is obtained by solving the equations (A - AI)C = 0. For instance, corresponding to the root = 1, we have

, + ap6C2 = 0 -6C, + ahCi = 0

This gives C, = -a, = 1. Similarly, for the other root (1 - - ) we get the characteristic vector as C, = -p, C2 = 1. Consider the matrix with these vectors as columns. This is

Now invert this matrix. We have

R- =

- a 1

-P 1

P - a

Then the rows in this matrix give the required linear combinations:

Xf + fix2 is nonstationary (corresponding to the unit root), - :, - (XX2 is stationary.

In this example, we started out with a VAR model with one unit root. In practice, we have to test for the unit roots. We do this as follows. Let the root closest to 1 be denoted by 7. Then we consider niy - 1) and refer to the tables for nip - 1) or nip - 1) in Fuller (1976, p. 371), depending on whether p, is known or estimated, n is the sample size. As an example, consider the VAR model with y, = income and C, = consumption, based on 53 observations (1898-1950), which produced the following resuUs:"

0.055 - 0.110 0.291 -0.371

The matrix A - I has characteristic roots -0.0424 and -0.2740. The roots of A are obtained by adding 1 to each. They are 0.9576 and 0.8260. To test whether the root 0.9576 is significantly different from 1, consider 53(0.9576 - 1) = -2.20. This is not less than the tabulated 5% value in FuUer (1976, p. 371), which is -13.29. Thus this root is not significantly different from 1. We next compute the matrix R of characteristic vectors and R (These computations are left as an exercise.) The two rows of R" give the result that (C, - 2.99y,) is (approximately) a unh root process and (C, - 0.88y,) is (approximately)

"This example is from D. A. Dickey, "Testing for Unit Roots in Vector Processes and Its Relation to Cointegration," in Fomby and Rhodes (eds.). Advances in Econometrics, pp. 87-105.

Stationary. The latter result says that 0.88 is the long-run marginal propensity to consume.

In the two-variable case, the cointegration coefficient, if it exists, is uniquely determined. Also, in this case, the matrix (A - I) for the VAR model has rank 1. As we saw in equation (14.11), it can be expressed as CB, where and are row vectors and B gives the cointegrating vector. In the case of more than two variables, there can be more than one cointegrating regressions and these need not be uniquely determined. For instance, suppose that there are n variables. Suppose there are {n - r) unit roots and r cointegrating vectors. In this case the matrix A - I will be of rank r < . As before, we can then write

A - I = CB

where and are x /• matrices. The rows of B are the r distinct cointegrating vectors. However, they may not all have meaningful economic interpretation and we have to choose the linear combinations that make economic sense. In the case where there are {n - 1) unit roots, so that r = 1, the cointegrating vector, if it exists, will be unique. The determination of the cointegrating vectors (and their number) for a general VAR model with n variables and lags is described by Johansen. This is complicated for our purpose. Instead, we consider a procedure that can be appfied for a VAR model with one lag if there is only one unit root, that is, all the other variables are driven by one stochastic trend. Thus if we have three variables, ,i = 3, (ai - r) = 1 or /• = 2. Thus there are two cointegrating vectors. The procedure of deriving them starting with a VAR model is as follows:

1. Write down the matrix of coefficients in the AR model.

2. Find the characteristic roots.

3. Find the characteristic vectors (see the Appendix to Chapter 7). These are the vectors x that solve the equation (A - XI)x = 0.

4. Let R be the matrix whose columns are these vectors. Find . Then the rows of R corresponding to the nonunit roots give the coefficients of the cointegrating regressions. The row corresponding to the unit root gives the coefficients of the unit root process.

An example will illustrate this procedure.Consider the VAR model

"S. Johansen, "Statistical Analysis of Cointegration Vectors," Journal of Economic Dynamics and Control, Vol. 12, 1988, pp. 231-254, and S. G. Hall, "Maximum Likelihood Estimation of Cointegration Vectors: An Example of the Johansen Procedure," Oxford Bulletin of Economics and Statistics, Vol. 51, No. 2, 1989, pp. 213-218, and D. A. Dickey, D. W. Jansen, and D. L. Thornton, "A Primer on Cointegration with an Application to Money and Income," Federal Reserve Bank of St. Louis, March-April 1991, pp. 58-78. "This example is adapted from Dickey, "Unit Roots."

The characteristic roots of the matrix A of the coefficients in the VAR model are 1.0, 0.66, and 0.42. The matrix R whose columns are the characteristic vectors corresponding to these roots is

-0.8165 0.1879 0.4483 0.4082 0.0167 0.4082 0.4082 0.9820 0.7952

-0.8363 0.6275 0.1473 -0.3408 -1.7957 1.1141 0.8503 1.8956 - 0.1950

Note that the AR is a diagonal matrix with diagonal elements 1, 0.66, and 0.42. Hence

/, = -0.8363 ; -f- 0.6275> + 0.1473z /2 = -0.3408JC - 1.7957 -f- 1.114k /3 = 0.8503X + 1.8956 - 0.1950

are the three linear functions we need. /, is a unit root process;/2 and/3 are stationary processes. They are cointegrating regressions.

However, note that Unear combinations of/2 and/3 are also stationary. For instance, take /2 and / and eliminate y. We get (x + 2z) as a cointegration equation. Similarly, eUminating x, we get - z as a cointegration equation, and eliminating z, we get : -f 2y as a cointegration equation. Note that in this model the matrix (A - I) has one zero root and hence rank 2. It can be written as

-0.2 -0.38 -0.2 -0.44 -0.28 -0.28

1 -1

The rows of B give the cointegrating vectors, which are x -\- 2z and - z that we derived earlier. Which of these different cointegrating equations we choose depends on which of them has a meaningful economic interpretation. There is one other thing we can do in this example. We have identified/ as the common stochastic trend. Hence if we regress x, y, and z in turn on/„ the residuals will be stationary.

To simplify the exposition, we have discussed the VAR model with only one lag. Suppose instead that we have a general VAR model with lags:

X, - AX, i -I- A2X,„2 + This can be written as

(14.12)

(14.13)

, = , ,„, - 2 ,„2 -1- • • • -1- Bt , , +, -1- + ,

where , = -I -I- , -f Aj -f • • • -f „ / = 1, 2, . . . / . If , is 1(1), then , is 1(0). If some linear combinations of x, are stationary, that is, there are some cointegrating relationships among the variables in x„ then the matrix should