Your Federal Quarterly Tax Payments are due April 15th Get Help Now >>

Stability Analysis for VAR systems by elfphabet5

VIEWS: 12 PAGES: 11

									                              Stability Analysis for VAR systems

For a set of n time series variables yt  ( y1t , y2t , ..., ynt )' , a VAR model of order p (VAR(p))
can be written as:
(1)     yt  A1 yt 1  A2 yt  2  ...  Ap yt  p  ut
where the Ai ’s are (nxn) coefficient matrices and ut  (u1t , u2t ,...,unt )' is an unobservable
i.i.d. zero mean error term.


I. Stability of the Stationary VAR system:
(Glaister, Mathematical Methods for Economists

The stability of a VAR can be examined by calculating the roots of:
( I n  A1L  A2 L2  ....) yt  A( L) yt

The characteristic polynomial is defined as:
  ( z )  ( I n  A1 z  A2 z 2  ....)

The roots of (z ) = 0 will give the necessary information about the stationarity or
nonstationarity of the process.

The necessary and sufficient condition for stability is that all characteristic roots lie
outside the unit circle. Then  is of full rank and all variables are stationary.

In this section, we assume this is the case. Later we allow for less than full rank matrices
(Johansen methodology).

Calculation of the eigenvalues and eigenvectors
Given an (nxn) square matrix A, we are looking for a scalar  and a vector c  0 such
that Ac  c then  is an eigenvalues (or characteristic value or latent root) of A. Then
there will be up to n eigenvalues, which will give up to n linearly independent associated
eigenvectors such that
or Ac  Ic  0  [ A  I ]c  0 .
For there to be a nontrivial solution, the matrix [ A  I ] must be singular. Then  must be
such that A  I  0

          3 8 1 
          0 4 3 
Ex: A=          
          0 3 4 
                
  3 8 1  0 0  3                8      1 
 0 4 3    0  0    0
                                4      3 .
                                                 
  0 3 4   0 0    0
                                  3         
                                            4  


Expanding the determinant of this matrix gives the characteristic equation:
(3   )(2  8  7)  0  1  3, 2  1, 3  7 .

                        Reminder: For a characteristic equation of the type a2  b  c  0
                                      b i b 2  4ac
                         1 , 2                      i
                                     2a       2a
                                                                Real roots: 1 , 2    
                                                                Imaginary roots: 1 , 2    i
                                                                                where i     1
                                                                Modulus =     
                                                                               2     2




Note: an eigenvector is only determined up to a scalar multiple: If c is an eigenvector,
then c is also an eigenvector where  is a scalar:  ( Ac)   (c)  A(c)   (μc) .

The associated eigenvectors are those that satisfy the equations for the three distinct
values of the eigenvalues.
The eigenvector associated with 1  3 , which satisfies the equation for this matrix is
found as
   0 8 1  c11  0
 0 1 3  c12   0 . Notice that only columns 2 and 3 are linearly independent
               
   0 3 1 c 22  0
               
(rank=2) so we can choose the first element of the c matrix arbitrarily. Set c11  1 and the
                             c11  1
other two elements are = 0  c12   0 .
                                
                             c13  0
                                

Similarly, the eigenvector associated with 2  7 , which satisfies the equation for this
matrix is found as

    4 8       1  c 21  0
  0  3 3  c   0  .
                     22   
   0
          3  3 c 23  0
                      
Notice that rk(A)=2 again because this time the last two rows are linearly dependent.
Thus only the 2x2 matrix on the LHS is nonsingular. We can delete the last row and
move c 23 multiplied by the last column to the RHS. Now the first two elements will be
expressed in terms of the last element. We can fix arbitrarily c 23 and solve for the two
others: assume c 23 =4. Then c 2  (9,4,4)' is an eigenvector corresponding to the
eigenvalue 2  7

We can find similarly the last eigenvector to be c3  (7,2,2)'


Jordan Canonical Form:
Form a new matrix C whose columns are the three eigenvectors.
    1 9 7 
C  0 4  2 . You can calculate to find that the matrix product Q 1 AQ  
             
    0 4 2 
             

       3 0 0 1 0 0 
C AC  0 7 0 =  0 2 0 
  1
                       
       0 0 1  0 0 3 
                       


Thus, for any square matrix A, there is a nonsingular matrix C such that
(i) C 1 AC is diagonal with the eigenvalues on the diagonal.
(ii) The eigenvectors corresponding to distinct eigenvalues of a symmetric matrix are
orthogonal (linearly independent).


II. Stability Conditions for Stationary and Nonstationary VAR Systems
(Johnson and DiNardo, Ch 9+Appdx)
To discuss these conditions we start with simple models and generalize. We will see:
VAR(1) with 2 variables:
VAR(2) with k variables (ex: VAR(2) with 2 variables)
VAR(p) with k variables.

       1. VAR(1) with two variables (p=1, k=2).
       (1) y1t  b10  a11 y1,t 1  a12 y2,t 1  1t
       (2) y2t  b20  a21y1t 1  a22 y2t 1   2t or:

       (3) yt  b  Ayt 1   t , which can be written with the lag operator
       (4) ( I  AL) yt  b   t

       Each variable is expressed as a linear combination of itself and all other variables
       (plus intercepts, dummies, time trends). The dynamics of the system will depend on
       the properties of the A matrix.
                                                                                    s  t
The error term is a vector white noise process with E( t )  0 and E ( t  's )  
                                                                                   0 s  t
where the covariance matrix  is assumed to be positive definite  the errors are
serially uncorrelated but can be contemporaneously correlated.

Solution to 4:
(i) Homogenous equation:
Omit the error term yt  b  Ayt 1  simplest solution: yt  yt 1  ..  y . Then,
(5) y   1b if  is nonsingular (   I  A )
As a solution try dt . Substituting it in the homogenous (trivial solution) equation
(5):
(I  A)d  0 ---eigenvalues
The nontrivial solution requires the determinant to be zero:
I  A  0
 Get the eigenvalues ( ' s ).

(ii) Substitute the eigenvalues into the homogenous system, to get the corresponding
eigenvectors ( C' s ).

(iii) After calculating the nonhomogenous solution and adding to the homogenous
equation, we obtain the complete solution (in matrix form):
(6)   yt  c11  c2 t2  y
              t



yt  y (LR value) as t rises if the two eigenvalues have the modulus<1.


We can rewrite ( I  AL) yt  b in (4) as a polynomial to see the stability conditions in
terms of the eigenvalues:
 B( L) yt  b where B(L)=I-AL,  B( L)  (1  1 L)(1  2 L) .

The stability condition:
(i) Modulus s <1   nonsingular, the determinant not 0, the system is stationary.
In (6) y converges to y .
(ii) Modulus i >1   nonsingular, but the system is explosive, no convergence.
This is because one or more of the ' s grows without bound as t increases, so does y
from (6). Not a typical process observed in the macro/finance series, therefore we do
not consider this case.
(iii) Modulus i =1  unit root,  is singular, the determinant is 0 –y is
nonstationary, we need to look into the VECM specification. A lot of the
macro/finance models fall into this category.
(iv) Modulus 1  2  1  I(2) variables, VAR is I(1). In general A is not
symmetric. Look for cointegrating vectors.
Relation between VAR variables and eigenvalues
Define the eigenvalues and the corresponding eigenvectors of the matrix A as:
   0                            c      c 
 1             and C   11 12 
   0 2                    c21 c22 
If the eigenvalues are distinct then the eigenvectors are linearly independent, and C is
nonsingular  C 1 AC   and A  CC 1 .
                                                                                              1
            Theorem: (i) for any square matrix A, there is a nonsingular matrix C such that C AC is diagonal with the
            eigenvalues on the diagonal. (ii) The eigenvectors corresponding to distinct eigenvalues of a symmetric matrix
            are orthogonal (linearly independent).



Define a new vector of variables w such that yt  Cwt or wt  C 1 yt
  each y is a linear combination of w’s (or each w is a linear combination of y’s).
Multiply (3) yt  b  Ayt 1   t by C 1 :
C 1 yt  C 1b  AC 1 yt 1  C 1 t

 wt  b * wt 1  et , b*  C 1b ,   C 1 AC ,               et  C 1 t ,
or:
(7) w1t  b1 * 1 w1,t 1  e1t
    w2t  b2 * 2 w2t 1  e2t

    (i)      i  1 for i=1,2
    Both eigenvalues have modulus < 1.
    Each w is therefore I(0), and since y’s are linear combinations of w’s, each y is
    I(0). You can therefore apply the standard inference procedures and estimate
    each equation separately. As we saw above,   I  A is nonsingular, it is full
    rank (=2 here), and a unique static equilibrium exists: yt  ( I  A) 1 b   1b . The
    values of  are such that any shock die out quickly and deviations from
    equilibrium are transitory.

    (ii)    i  1 for i=1 or 2
    One of the eigenvalues has modulus > 1. Since each y is a linear combination of
    both w’s, y is unbounded and the process is explosive.

    (iii)    1  1 and 2  1
    Now w1 is a random walk with drift, or I(1), w2 is I(0). Each y is I(1) since each y
    is a linear combination of both w’s, therefore VAR is nonstationary.


Is there a linear combination of y1t and y2t that removes the stochastic trend and makes
it I(0), i.e. both variables are cointegrated?
                                 c11 * y1,t 1  c12 * y 2,t 1
   Consider again wt  C 1 yt = 
                                                                where c* represent the coefficients in
                                              c21 * y1,t 1  c22 * y 2,t 1
                                              
   the C 1 matrix. We know that w2 is I(0), thus [c21 * c22*] is a cointegrating vector.


    Look for a Relation between the CI vector [c21 * c22*] and the  matrix such that
            
     [.] c21 c22 .
            *   *
                          
   Reparameterize equation (3) to give:
   (8) yt  b  yt 1   t where   I  A .
   The eigenvalues of  are the complements of the eigenvalues  of A: i  1  i .
   Since 1  1the eigenvalues of  are 0 and 1  2 . Thus, it is a singular matrix with
   rank 1. Let us decompose  . Since   I  A and A  CC 1 , we can write

      I  C 1 AC  (CI  C)C 1  C ( I  )C 1 .

   Thus:
              0    0  1             c11 c12  0    0  c11 c12  c12 (1  2 ) 
                                                                *   *
   (9)     C         C                               *    * 
                                                                                     c21 * c22 *   '
              0 1  2               c21 c22  0 1  2  c21 c22  c22 (1  2 )

   So  , which has a rank 1, is factorized into the product of a row vector  and a
   column vector  , called an outer product:

   The row vector =  = the cointegrating vector.
   The column vector =  = the loading matrix = the weights with which the CI vector
   enters into each equation of the VAR.

-----------
Note: compare (9) to the case where  is full rank with 1  0 :
        1  1    0  1 c11 (1  1 ) c12 (1  2 )   c11 c12 
                                                            *   *
  C                   C                             *    * 
                                                                     . You can see why   ' is
         0     1  2 
                           c21 (1  1 ) c22 (1  2 ) c21 c22 
said to be of reduced rank.
------------

   Combining (8) and (9) we get the vector error correction model of the VAR:

         y1t  b1  c12 (1  2 )(c21 * y1,t 1  c22 * y 2,t 1 )   1t
                                                                                y1t  b1  c12 (1  2 ) w2,t 1  1t
   (10)                                                                        
         y 2t  b2  c22 (1  2 )(c21 * y1,t 1  c22 * y 2,t 1 )   2t
                                                                                y2t  b2  c22 (1  2 ) w2,t 1   2t

   All variables here are I(0): y’s in first differences and w’s.
The w (EC term) measures the extent to which y’s deviate from their equilibrium LR
values.
Although all the variables are I(0), the standard inference procedures are not valid.
(similar to the univariate case where in order to test whether a series is I(1), we have
to use an ADF test and not the t statistics on the AR coefficient).
--See example below—

  (iv)    Repeated unitary eigenvalues: 1  2  1
We can no longer have a diagonal eigenvalue matrix as before. But it is possible to
                                                                                         1 
find a nonsingular matrix P such that P 1 AP  J and A  PJP1 where J                       (the
                                                                                        0  
Jordan matrix). The problem with this case is that although  is still rank 1, the
transformation of y’s into w’s leads to I(2) variables, the cointegration vector gives a
linear combination of I(2) variables and is thus I(1) and not I(0). Thus y is CI(2,1),
the variables in the VAR are all I(1) but the inference procedures are nonstandard.


Example of a case with 1  1 and 2  1
Find the matrices  and  from a VAR(1) with k=2:
(11) y1t  1.2 y1,t 1  0.2 y2,t 1  e1,t
(11) y2t  0.6 y1,t 1  0.4 y2,t 1  e2,t

Reparametrizing the VAR into a VECM gives us:
y1t  0.2 y1, t 1  0.2 y2, t 1  e1,t
y2t  0.6 y1, t 1  0.6 y2,t 1  e2,t
in matrix form:
      y  0.2  0.2  y1t 1   e1t 
(12)  1t                     
     y2t  0.6  0.6  y2t 1  e2t 

or:
Yt  Yt 1  ut

But we cannot infer the loading matrix and the cointegrating matrix separately from
this. To find  and  separately, we need to calculate the eigenvector matrix:

Get the eigenvalues from the solution to A  I  0 .
             a11          a12            1.2       0.2
 A  I                                                       0  1  1, 2  0.6
                a21      a22                0.6    0.4  

Eigenvectors corresponding to 1  1:
  1.2  1  0.2  c11  0  0.2  0.2 c11  0
   0.6 0.4  1 c   0  0.6  0.6 c   0 there is linear dependency
                   12               12   
So set c12  1  c1  1 1'

Eigenvalues corresponding to 2  0.6
0.6  0.2 c21  0
0.6  0.2 c   0 there is linear dependency So set c21  1  c2  1 3'
           22   

The eigenvector matrix and its inverse are:
    1 1            1.5  0.5
C        C 1               
    1 3            0.5 0.5 

Now we can write the VAR in VECM by decomposing  :

                       0    0  1
Yt  Yt 1  ut  C          C  ut
                       0 1  2 

        0.4          y1,t 1 
Yt     0.5 0.5           ut
        1.2          y2,t 1 

This is the same expression as in (12) but now we have both the loading and the
cointegrating matrices:

      0.4
                  and        '   0.5 0.5
      1.2 


2. VAR(2) with k variables:
(13)     yt  b  A1 yt 1  A2 yt  2   t
               Note: you can also add any deterministic terms such as trend, breaks by specifying the
               model as:
               yt  A1 yt 1  A2 yt 2  Dt   t

Set the error term=0 and examine the properties of the system.
We still have the LR solution (or the particular solution) as in (5)
y   1b but now   I  A1  A2 .


y exists if  1 is nonzero. To see this, look at the eigenvalues.
We again try the same solution for the homogenous equation yt  ct and substitute it
in to get the characteristic equation
2 I  A1  A2  0


The number of roots = pk where p=order of the VAR and k=#variables.
Here we will have 2k roots.

If all eigenvalues have modulus<1 then  is non singular and the solution
 yt  i21 ci t  y
          k


will converge to y as t grows. The analysis w.r.t the modulus of the roots (<1, =1,
>1) is the same as in the VAR(1) case.

If the process is stationary then we can invert the VAR model and express y as a
function of present and past shocks, and the exogenous (deterministic)
components=Impulse Responses:
Ex: Calculate the roots of a 2-dimensional VAR(2): n=p=2 and find the effect of a
shock on a dependent variable: (Juselius Ch. 3)
     The characteristic function of ( z )  I  A1 z  A2 z 2 where z  1 is
                   1,11 1,12   2,11  2,12  2
      ( z )  I                z               z
                   1, 21 1, 22   2, 21  2, 22 

           1,11z  1,12 z   2,11z 2                2,12 z 2 
      I                                                      
           1, 21z  1, 22 z   2, 21z               2, 22 z 2 
                                          2
                                                                  

              (1   1,11z   2,11z 2 ) ( 1,12 z   2,12 z 2 ) 
                                                                2 
               ( 1, 21z   2, 21z ) (1   1, 22 z   2, 22 z )
                                     2
                                                                   

Therefore
 ( z )  (1   1,11z   2,11z 2 )(1   1, 22 z   2, 22 z 2 )  ( 1,12 z   2,12 z 2 )( 1, 21z   2, 21z 2 )


Regrouping similar terms:
 ( z )  1  a1 z  a2 z 2  a3 z 3  a4 z 4           (1  1 z )(1  2 z )(1  3 z )(1  4 z )


The determinant is a 4th order polynomial in z giving 4 characteristic roots:
z1  1 / 1; z2  1 / 2 ; z3  1 / 3 ; z4  1 /  4

Effects of a shock (or structural change dummy) on a dependent variable:
If  is invertible (all roots in the unit circle), we can write
 yt   1 t .

We can then calculate the effect of a shock on yit :
              a ( L) jt                    a ( L)                  jt
     yit                                                                   for t=1,….T.
               ( z)            (1  2 z )(1  3 z )(1  4 z ) (1  1z )

    We are assuming that all roots have modulus less than 1. The characteristic roots
    give information about the dynamic behavior of the process. To see how the shock is
    propagated, expand the last component:
    (1  1 )1 jt  (1  1L  1 L2  ...) jt .
                                  2



    You will have to do the same thing with each root. Thus, each shock will affect
    current and future values of yi .
    The persistence of the shock depends on the magnitude of the roots. The larger they
    are the more persistent will be the shocks.
        -If the roots i are real and <1, the shock will exponentially die out.
         -If one or more root i is imaginary then a shock will be cyclical but
         exponentially declining.
         -If one or more roots i lies on the unit circle, the shock will be permanent and
         and yi will show nonstationary behavior. VAR is not invertible, then we need to
         look into VECM

         We can also calculate the roots by reformulating the VAR(p) into the companion
         matrix VAR(1) form and solve for the two eigenvalues:

Alternative approach: companion matrix.
A VAR(p) can be transformed into a VAR(1). Consider the equation (6) again. We can
rewrite it as:

     yt  A1 yt 1  A2 yt  2  ut
     yt 1  yt 1

    In matrix form:
      yt   A1 A2   yt 1  ut 
      y   I               
      t 1   p 0   yt  2  0 

Calculate the eigenvalues i from the coefficient matrix:

          I          0   A1       A2            A1  A2
I  A   2                                =0=                2  A1  A2  0
          0          I 2   I 2
                                    0            I2   
 (  1 )(  2 )  0 .

Now we get the roots directly instead of the z’s, which were the inverse of the roots,
obtained by solving the characteristic polynomial. Johansen and Juselius refer to the
s as eigenvalues roots and to z’s characteristic roots.
In the case of the companion matrix, there are two roots. If the roots to the characteristic
polynomial are outside the unit circle, then the eigenvalues of the companion matrix are
inside the unit circle and the system is stable.

To recap:
       -The solution to I  zA  0 gives the stationary roots (characteristic roots) outside
       the unit circle.
       -The solution to I  A  0 gives the stationary roots (eigenvalues) inside the unit
       circle.

       -If the roots of (z ) are all outside the unit circle or the eigenvalues of the
       companion matrix are inside the unit circle, the process is stationary.
       -If one or more of the roots of (z ) or those of the companion matrix are on the
       unit circle then the process is nonstationary.
       -If one or more roots of (z ) is inside the unit circle or the eigenvalues of the
       companion matrix are outside the unit circle, the process is explosive.

								
To top