[R-SIG-Finance] [R-sig-finance] A VaR question

RON70 ron_michael70 at yahoo.com
Wed Nov 25 10:29:15 CET 2009

```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.
>