Mark

This should be reasonably straightforward. In the simplest case you wih to

draw a random complex number in the unit circle. This is best done in polar

coordinates.

If r is a random mumber on (0,1) and theta a random number on (0, 2 Pi)

then if x=r cos(theta) and y= r sin(theta), x + i y is inside the unit

circle. As such roots come in conjugate pairs a second is x-iy. If you

then need an odd number of roots the final can simply be a random number on

(0,1). You do not need to use a uniform distribution but can use any

distribution on the required intervals or restrain more or the eigenvalues

to be real.

John

On Sunday, 15 January 2012, Mark Leeds wrote:

hi john. I think I follow you. but , in your algorithm, it is

straightforward to

generate a set of eigenvalues with modulus less than 1 ? thanks.

On Sat, Jan 14, 2012 at 5:31 PM, John C Frain wrote:

Mark, statquant2

As I understand the question it is not to test if a VAR is stable but how

to construct a VAR that is stable and automatically satisfies the condition

Mark has taken from Lutkohl. The algorithm that I have set out will

automatically satisfy that condition.The matrix that should be "estimated

by the algorithm is A on the last line of page 15 of Lutkepohl.

Incidentally the corresponding matrix for the example on page 15 is

singular. The algorithm that I have set out will only lead to systems with

a non-singular matrix.

I still don't see how a matrix generated in this way corresponds to a

real economic system. Of course you may have some other constraints in

mind that would make the generated system correspond to something more real.

John

On Saturday, 14 January 2012, Mark Leeds wrote:

Hi statquant2 and john: In the first chapter of Lutkepohl, it is shown

that stability f

a VAR(p) is the same as

det(I_k - A1z - .... Ap Z^p ) does not equal zero for z < 1.

where I_k - A1z - ... Ap z^p is referred to as the reverse

characteristic polynomial.

So, statquant2, given your A's, one way to do it but be would be to

check the roots of the

polynomial implied by taking the determinant of the your polynomial.

There's an example on pg 17 of lutkepohl if you have it. If you don't, I

can fax it to you

over the weekend if you want it.

On Fri, Jan 13, 2012 at 8:34 PM, John C Frain wrote:

I think that you must approach this in a different way.

1 Draw a set of random eigenvalues with modulus < 1

2 Draw a set of random eigenvalues vectors.

3 From these you can, with some matrix manipulations, derive the

corresponding Var coefficients.

If your original coefficients were drawn at random I suspect that the

VAR

would not be stable. I am curious about what you are trying to do.

John

On Friday, 13 January 2012, statquant2 wrote:

Hello Paul

Thanks for the answer but my point is not how to simulate a VAR(p)

process

and check that it is stable.

My question is more how can I generate a VAR(p) such that I already

know

that it is stable.

We know a condition that assure that it is stable (see first message)

but

this is not a condition on coefficients etc...

What I want is

generate say a 1000 random VAR(3) processes over say 500 time periods

that

will be STABLE (meaning If I run stability() all will pass the test)

When I try to do that it seems that none of the VAR I am generating

pass

this test, so I assume that the class of stable VAR(p) is very small

compared to the whole VAR(p) process.

--

View this message in context:

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John C Frain

Economics Department

Trinity College Dublin

Dublin 2

Ireland

--

John C Frain

Economics Department

Trinity College Dublin

Dublin 2

Ireland

www.tcd.ie/Economics/staff/frainj/home.html

mailto:

frainj@tcd.iemailto:

frainj@gmail.com