A VaR question

Thu, Nov 26, 2009 12:52 PM

Hi RON70

Not sure about what you want to do... So you want to simulate a VAR with 
those parameters? There is the function TVAR.simin package tsDyn that 
allow you to simulate a VAR, see:

co<-matrix(c(0.985,  0.283, -1.714,  0.125,  1.100, -1.491,0.071, 
-0.089,  1.388), ncol=3, byrow=TRUE)
library(tsDyn)

TVAR.sim(B=cbind(co,co),nthresh=0, inc="none", lag=2)

Hope this is what you wanted?

MAT2009

RON70 a ?crit :

I think I should be more clear on what I would like to do. From that
estimated model I would like to get a "Real" VAR DGP, perhaps by tweaking
some of the coeficients. Which I would use to simulate artrificial data, for
some study. If somebody shows me some lihgt on how I can achive that, I
would be truly grateful.

Best,



RON70 wrote:

Hi all,

My problem seems to be bizzare, however I want to do like that. Here I
have estimated a VECM model from my dataset (seems not stationary) and
once I converted those into a VAR representation I have following
estimates :

A1; A2; A3; A4

     V1     V2     V3
1 0.985  0.283 -1.714
2 0.125  1.100 -1.491
3 0.071 -0.089  1.388
      V1     V2     V3
1  0.258 -0.493  1.459
2  0.252 -0.387  1.165
3 -0.057  0.076 -0.536
     V1     V2    V3
1 0.332 -0.459 0.251
2 0.482 -0.686 0.313
3 0.112 -0.104 0.218
      V1    V2     V3
1 -0.532 0.624 -0.006
2 -0.619 0.714 -0.044
3 -0.129 0.121 -0.069

Now I took them as an original DGP process and checked the solution of
it's ch. equation. I got following :

 library(PolynomF)
 z = polynom()


p11 <- 1 - A1[1,1]*z - A2[1,1]*z^2 - A3[1,1]*z^3 - A4[1,1]*z^4
p12 <- 0 - A1[1,2]*z - A2[1,2]*z^2 - A3[1,2]*z^3 - A4[1,2]*z^4
p13 <- 0 - A1[1,3]*z - A2[1,3]*z^2 - A3[1,3]*z^3 - A4[1,3]*z^4


p21 <- 0 - A1[2,1]*z - A2[2,1]*z^2 - A3[2,1]*z^3 - A4[2,1]*z^4
p22 <- 1 - A1[2,2]*z - A2[2,2]*z^2 - A3[2,2]*z^3 - A4[2,2]*z^4
p23 <- 0 - A1[2,3]*z - A2[2,3]*z^2 - A3[2,3]*z^3 - A4[2,3]*z^4

p31 <- 0 - A1[3,1]*z - A2[3,1]*z^2 - A3[3,1]*z^3 - A4[3,1]*z^4
p32 <- 0 - A1[3,2]*z - A2[3,2]*z^2 - A3[3,2]*z^3 - A4[3,2]*z^4
p33 <- 1 - A1[3,3]*z - A2[3,3]*z^2 - A3[3,3]*z^3 - A4[3,3]*z^4

p <- p11*(p22*p33 - p23*p32) - p12*(p21*p33 - p23*p31) + p13*(p21*p32 -
p22*p31)


abs(solve(p))

 [1] 1.521516 2.102119 2.102119 4.912478 4.912478 1.000233 1.000233
1.502034 1.502034 1.228100 2.536582 5.342635
    
Now if I assume (upto a few significant digits) "1.000233 1.000233 " both
equal to "1" then, I am actually getting two unit roots here. Therefore I
am wondering how to tackle it as VAR is defined on max one unit root
process. 

Am I missing anything? Can anyone please help me?

Best

A VaR question

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