# [R-SIG-Finance] [R-sig-finance] VAR process

Eric Zivot ezivot at u.washington.edu
Tue Jan 27 18:29:32 CET 2009

```The transformation of the VAR(p) to a VAR(1) is a mechanical device used to
help understand the dynamics of the model. The VAR(1) form is called the
companion form of the VAR(p) and is an equivalent representation. In the
VAR(1) form the model is first order markov and the eigenvalues of the
transition matrix determines the stability properties of the system. This is
a common trick used for systems of stochastic differential equations.
Lutkepohl's book on multivariate time series gives a very clear explanation
of this.

-----Original Message-----
From: r-sig-finance-bounces at stat.math.ethz.ch
[mailto:r-sig-finance-bounces at stat.math.ethz.ch] On Behalf Of
markleeds at verizon.net
Sent: Monday, January 26, 2009 9:38 PM
To: RON70
Cc: r-sig-finance at stat.math.ethz.ch
Subject: Re: [R-SIG-Finance] [R-sig-finance] VAR process

I didn't respond earlier because I'm not clear on what the problem is with
rewriting it as VAR(1) ? Lutkepohl text shows how this is done on pages 15
an 16 of his text. Except for the first row, the rest of the A matrix is
composed of identity matrices. They y_t* below the first element play no
role essentially because they are already known because they are in
{t-1,t-2,t-3.... }. The only noise term is the first element, u_t associated
with the first element y_t.

I agree that ithe Cov is not of full rank when you write it that way but I
don't know of any negative repurcussions of that. I think it's more of a
tool that he uses  to show what the stability condition reduces to for a
VAR(p) and nothing more than that. This same type of technique is used when
writing AR models in state space form.

Hopefully Eric or Bernhard or someone else can say more but I think it's
just used for deriving the stability condition in a easier way.

On Mon, Jan 26, 2009 at  9:42 PM, RON70 wrote:

> Hi,
>
> More than one week, still no suggestion. Is my question not
> understandable
> Regards,
>
>
> RON70 wrote:
>>
>> Hi,
>>
>> In every book on VAR [Vector auto regression] I see that, any VAR [p]
>> process can be expressed as a VAR [1] process. Here my question is
>> how it
>> can be possible? When you change it to a VAR [1] process, the VCV
>> matrix
>> of Innovations contains zero and hence it is not of full rank.
>> Therefore
>> it is not a PD matrix, you cannot decompose that according cholesky
>> decomposition and lot more things can not be done with it because VCV
>> matrix is singular. Then how can that process be a VAR process?
>>
>
> --
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> http://www.nabble.com/VAR-process-tp21530701p21678648.html
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>
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